Rajasthan Board RBSE Class 12 Maths Chapter 1 Composite ...

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Rajasthan Board RBSE Class 12 Maths Chapter 1 Composite Functions Ex 1.1 Question 1. If f : R → R and g : R → R be two functions defined as below, then find (fog) (x) and (gof)(x): (i) f(x) = 2x + 3, g(x) = x2 + 5 (ii) f(x)= x2 +8, g(x) = 3x3 +1 (iii) f(x) = x, g(x) = |x| (iv) f(x) = x2 + 2x + 3, g(x) = 3x – 4 Solution: (i) Given, f(x) = 2x + 3 and g(x) = x2 + 5 (fog)(x) = f(g(x)) = f(x2 + 5) = 2(x2 + 5) + 3 = 2x2 + 10 + 3 = 2x2 + 13 (gof)(x) = g(x)) = g(2x + 3) = (2x + 3)2 + 5 = 4x2 + 9 + 12x + 5 = 4x2 + 12x + 14 (ii) Given, f(x)= x2 + 8 and g(x) = 3x3 + 1 (fog)(x) = f(g(x)) = f(3x3 + 1) = (3x3 + 1)2 +8 = 9x6 + 6x3 + 1 +8 = 9x6 + 6x3 +9 (gof)(x) = g(f(x)) = g(x2 + 8) = 3(x2 + 8)3 + 1 (iii) Given, f(x)=x and g(x) = |x| (fog)(x) = f(g(x)) = f(|x|) = |x| (gof)(x)=g(f(x)) = g(x) = |x| (iv) Given, f(x)= x2 + 2x + 3 and g(x) = 3x – 4 (fog)(x) = f(g(x)) = f(3x – 4) = (3x – 4)2 + 2(3x – 4) + 3 = 9x2 – 24x + 16 + 6x – 8 + 3 = 9x2 – 18x + 11 (gof)(x) = g(f(x)) = g(x2 + 2x + 3) = 3(x2 + 2x + 3) – 4 = 3x2 + 6x + 9 – 4 = 3x2 + 6x + 5. Question 2. If A = {a,b,c}, B = {u, v, w}. If f: A → B and g : B → A, defined as f = {(a, v), (b, u), (c, w)}



g = {(u, b), (v, a), (w, c)} then find (fog) and (gof). Solution: Given, f= {(a, v), (b, u), (c, w)} g= {(u, b), (v, a), (w, c)} f(a)= v and g(u) = b f(b)= u and g(v) = a f(c)= w and g(w) = C So, from fog(x) = f(g(x)] fog(u) = f(g(u)] = f(b) = u fog(v) = f(g(v)] = f(a) = v fog(w)= f[g(w)] = f(c) = w So, fog = {(u, u), (v, v), (w, w)} gof(a) = g[f(a)] = g(v) = a gof(b) = g[fb)] = g(u) = b gof(c) = g[(c)] = g(w) = C gof = {(a, a), (b, b), (c, c)} Question 3. If f: R+ → R+ and g: R+ → R+, defined as f(x) = x2 and g(x)= √x, then find gof and fog. Are they identity functions ? Solution: Given, f : R+ → R+, 4(x) = x2 g : R+ → R+, g(x) = √x (gof)(x) = g[f(x)] = g(x2) = √x2 = x (fog)(x) = f(g(x)] = f(√x) = (√x)2 = x So, (fog)(x) = (gof)(x) = x, ∀ x ∈ R+ Hence, (fog) and (gof) are identity function. Question 4. If f : R → Rand g : R → R be such two functions that defined as f(x) = 3x +4 and g(x)

= (x – 4), then find (fog)(x) and (gof)(x), also find (gog)(1). Solution: Given, f : R → R, f(x) = 3x + 4



Question 5. If f, g, h be three functions from R to R, defined as f(x) = x2, g(x) = cos x and h(x) = 2x + 3, then find {ho(gof)} (√2π). Solution: Given function, f(x) = x2, g(x) = cos x, h(x) = 2x + 3 ∴ {ho(gof)}(x) = hog{f(x)} = h[g{f(x)}] = h[g(x2)] = h(cos x2) = 2 cos x2 + 3 {ho(gof)}√2π = 2 cos (√2π)2 + 3 = 2 cos 2π + 3 = 2 x 1 + 3 = 5 Question 6. If functions f and g be defined as below, then find (gof)(x) : f : R → R, f(x) = 2x + x-2 g : R → R, g(x) = x4 + 2x + 4 Solution: Given, f: R → R f(x) = 2x + x-2 g: R → M R, g(x)= x4 + 2x + 4. ∴ (gof)(x) = g(f(x)} = g{2x + x-2} = (2x + x-2)4 + 2(2x + x-2) + 4 Question 7. If A = {1, 2, 3, 4}, f : R → R, f(x) = x2 + 3x + 1 g : R → R, 8(x) = 2x – 3, then find (i) (fog)(x) (ii) (gof)(x) (iii) (fof)(x) (iv) (gog)(x) Solution: Given,



f : R→ R, f(x) = x2 + 3x + 1 g : R → R, g(x) = 2x – 3 (i) (fog)(x) = f{g(x)} = f{2x – 3} = (2x – 3)2 + 3(2x – 3) + 1 = 4x2 – 12x + 9 + 6x – 9 + 1 = 4x2 – 6x + 1 (ii) (gof)(x) = g{f(x)} = g(x2 + 3x + 1) = 2(x2 + 3x + 1) – 3 = 2x2 + 6x + 2 – 3 = 2x2 + 6x – 1 (iii) (fof)(x) = f{f(x)} = f(x2 + 3x + 1) = (x2 + 3x + 1)2 + 3(x2 + 3x + 1)+1 = x4 + 9x2 + 1 + 6x3 + 6x + 2x2 + 3x2 + 9x + 3 + 1 = x4 +6x3 + 14x2 + 15x + 5 (iv) (gog)(x) = g{g(x)} = g(2x – 3) = 2(2x – 3) – 3 = 4x – 6 – 3 = 4x – 9

Rajasthan Board RBSE Class 12 Maths Chapter 1 Composite Functions Ex 1.2 Question 1. If A = {1, 2, 3,4}, B = {a,b,c,d}, then define four bijection from A to B and also find their inverse functions. Solution: Given A = {1, 2, 3, 4), B = {a, b, c, d} (a) f1 = {(1, a),(2, b), (3, c), (4, d)} f1

-1 = {(a, 1), (1, 2), (c, 3), (d, 4)} (b) f2 = {(1, a), (2, c), (3, b), (4, d)} f2

-1 = {(a, 1), (C, 2), (6, 3), (d, 4)} (c) f3 = {(1, b), (2, a), (3, d), (4, b)} f3

-1 = {(b, 1), (a, 2), (d, 3), (6,4)} (d) f4 = {(1, c), (2, d), (3, a),(4, b)} f4

-1 = {(c, 1), (d, 2), (a, 3), (b, 4)} Question 2. If f : R → R, such that f(x) = x3 – 3, then prove that f-1 exists and find its formula. Thus, find f-1(24) and f-1(5). Solution: Given, f : R → R, f(x) = x3 – 3 One-one/many-one: Let a, b ∈ R ∴ f(a) = f(b) ⇒ a3 – 3 = b3 – 3 ⇒ a3 = b3

⇒ a = b



Hence, . f(a) = f(b) ⇒ a = b ∴ f is a one-one function. Onto/into: Let y ∈ R (Co-domain) f(x) = y ⇒ x3 – 3 = y ⇒ x= (y + 3)1/3 ∈ R, ∀ y ∈ R Here, for each value of y, x exists in domain R. Thus, Range of F= co-domain of f. So, ‘f’ is onto function. It is clear that ‘f is one-one onto function. Hence, f-1: R → R exists. f-1(y) = x ⇒ f(x) = y But f(x)= x3 – 3 ∴ x3 = 3 = y

⇒ x3 = y + 3 ⇒ x = (y + 3)1/3 ⇒ f-1(y) = (x + 3)1/3 ⇒ f-1(x) = (x + 3)1/3, ∀ X ∈ R For x = 24 ∴ f-1(24)= (24 + 3)1/3 = (27)1/3 = 33 x 1/3 = 3 For x = 5 f-1(5) = (5 + 3)1/3 = (8)1/3 = 23 x 1/3 = 2 Question 3. If f : R → R, defined as (i) f(x) = 2x – 3 (ii) f (x) = x3 + 5 Then, prove that f is bijection in both conditions, Also find f-1. Solution: (i) Given function, f : R → R, f(x) = 2x – 3 One-one/many-one: Let a, b ∈ R f(a) = f(b) ⇒ 2a – 3 = 2b – 3

⇒ 2a = 2b ⇒ a = b

So, f(a) = f(b) ⇒ a = b, ∀ a, b ∈ R ∴ f is one-one function. onto/into: Let y ∈ R (co-domain) f(x)= y ⇒ 2x – 3 = y

⇒ x = ∈ R , ∀ y ∈ R So, pre-image for each value of y exists in domain R. Thus, function ‘f’ is onto



function. It is clear that ‘f’ is one-one into function. Hence f-1 : R → R exists. Let x ∈ R (domain of f) and y ∈ R (co-domain of f) Let f(x) = y, then f-1(y) = x ⇒ f(x) = y

⇒ 2x – 3 = y

(ii) According to question, f : R → R, f(x) = x3 + 5 One-one/onto : Let a, b ∈ R f(a) = f(b) ⇒ a3 + 5 = b3 + 5

⇒ a3 = b3 a = b So, f(a) = f(b) ⇒ a = b, ∀ a, b ∈ R ∴ f is one-one function Onto/into: Let y ∈ R (co-domain) f(x) = y ⇒ x3 + 5 = y ⇒ x3 = y – 5 ⇒ x = (y – 5)1/3 ∈ R, ∀ x ∈ R So, pre-image for each value of y exists in domain R. Thus, range of f = co-domain of f. So, function ‘f’ is onto function. Hence, we can say that f is one-one onto function. So, f-1: R → R defined as f-1(y)= x ⇔ f(x) = y ………(i) ⇒ f(x) = y

⇒ x3 + 5 = y x3 = y – 5 ⇒ x = ( y – 5)1/3 ⇒ f-1(y)= (y – 5)1/3 [from (1)] ⇒ f-1(x) = (x – 5)1/3 Question 4. If A = {1, 2, 3, 4), B = {3, 5, 7, 9), C = {7, 23, 47, 79} and f: A → B, g : B → C such that f(x) = 2x + 1 and g(x) = x2 – 2, then find (gof)-1 and f-1og-1 in ordered form. Solution: A = {1, 2, 3, 4), B = {3, 5, 7,9} and C = {7, 23, 47, 79}



f : A → B, f (x) = 2x + 1 g : B → C, g(x) = x2 – 2 Now, gof (x)=g{f(x)} = g(2x + 1) = (2x + 1)2 – 2 = 4x2 + 4x – 1 ∴ gof (x) = 4x2 + 4x – 1 On putting x = 1, 2, 3, 4 gof = {(1,7), (2, 23), (3,47),(4, 79)} ∵ gof is bijection function. ∴ Its inverse is possible

⇒ (gof)-1 = {(7, 1), (23, 2), (47,3), (79,4)} ⇒ f-1og-1 = {(7,1), (23, 2) (47, 3), (79,4)} [∵ (gof)-1 = f-1og-1 by theorem. Question 5. If f : R → R, such that f(x) = ax + b, a ≠ 0, then prove that f is a bijection function. Also, find f-1. Solution: Given function f : R→ R and ax + b, a ≠ 0 f-1 exists if f : R → R be a bijection function, before this we have to prove f be a bijection function. One-one/many-one: Let p, q ∈ R f(p) = f(q) ⇒ ap + b = aq + b ⇒ ap = aq ⇒ p = q So, f(p) = f(q), ∀ p, q ∈ R ∴ f is a bijection function. Onto/into: Let f(x) = y, y ∈ R ax + b = y

⇒ x = ∈ R So, pre-image of every value of y exist in domain R, ∴ ‘f’ is onto function. Thus, range of f = co-domain of f. So, f is a bijection function So, f-1 exists. Let, y ∈ R and f-1(y)= x, then f(x) = y ⇒ ax +b = y



Question 6. If f : R → R,f(x) = cos (x + 2), is f-1 exists. Solution: Given function f : R → R, f(x) = cos (x + 2). Putting x = 2π f(2π) = cos (2π+ 2) = cos (2) Putting x = 0 f(0) = cos (0 + 2) = cos 2 Here, only one image is obtained for 0 and 2π. So, ‘f’ is not one-one. Thus, ‘f’ is not one-one onto. Hence, f-1 : R → R does not exist. Question 7. Find f-1 (if exists), where f: A → B, such that (i) A = {0,-1,- 3, 2}, B = {-9, – 3,0, 6}, f(x) = 3x (ii) A = {1, 3, 5, 7,9}, B = {0, 1, 9, 25, 49, 81), f(x) = x2 (iii) A = B = R, f(x) = x3 Solution: (i) Give function f : A → B, f(x) = 3x where, A = {0,-1,-3, 2} B = {-9, – 3,0, 6} Now, f(0)= 3 x 0 = 0 f(-1) = 3 x – 1 = -3 f(-3) = 3 x (- 3) = -9 f(2)= 3 x 2 = 6 So, f= {(0,0), (- 1, -3), (- 3,-9), (2, 6)} Since, different element of A have different image in B under f. So ‘f is one-one function. Here, range of f = {-9, -3,0, 6} = B (co-domain). So, f is onto function. It is clear that ‘f’ is one-one onto function. So, f-1 : B → A exist. f-1 = {(0,0), (-3,- 1), (-9, – 3), (6, 2)} (ii) Given function f : A → B, f(x) = x2 where, A = {1, 3, 5, 7, 9} B = {0, 1, 9, 25, 49, 81} Now, f(1) = 12 = 1 f(3) = 32 = 9 f(5)= 52 = 25 f (7) = 72 = 49 f(9) = 92 = 81 So, f= {(1, 1), (3,9), (5, 25), (7,49), (9,81)} Since, different element of A have different image in B under f. So ‘f’ one-one function. Here, range of R = {1,9, 25, 49, 81} ≠ B (co-domain)



∴ ‘f’ is not onto function. Hence, f-1 does not exist. (iii) Given, A = B = R, f(x) = x3 f(x)= x3 Let a, b ∈ R and f(a) = f(b) ⇒ a3 = b3

a = b, ∀ a, b ∈ R Let f(x)= y and y ∈ B or y ∈ R ∴ x3 = y ⇒ x = y1/3 Since, different element of A have different image of B under f. So ‘f’ is one-one function. and range of f= co-domain of f. So, ‘f’ is onto function. Hence, it is clear that f is one-one function f-1 exists. If y is image of x under ‘f’, then f-1 : B → A is defined as f-1(y) = x ⇒ f(x) = y ⇒ x3 = y ⇒ x = y1/3 ⇒ f-1(y) = y1/3 ⇒ f-1(x) = x1/3

So, f-1: B → A, f-1(x) = x1/3, ∀ x ∈ B.

Rajasthan Board RBSE Class 12 Maths Chapter 1 Composite Functions Ex 1.3 Question 1. Determine whether each of the following operation define a binary operation on the given set or not. Also, Justify your answer. (i) a*b = a, on N (ii) a*b = a + b – 3, on N (iii) a*b = a + 3b, on N (iv) a*b = a/b, on Q (v) a*b = a – b, on R Solution: (i) a*b = a, on N Here, * is a binary operation because a, b ∈ N

= a*b = a ∈ N Here, a*b ∈ N Hence, * is a binary operation. (ii) a*b = a + b – 3 ∈ N Here, a*b is not a binary operation because, 1 ∈ N, 2 ∈ N then 1 + 2 – 3 = 0 ∈ N (iii) a*b= a + 3b ∈ N Here, * is a binary operation because 1 ∈ N, 2 ∈ N

then 1 + 3 × 2 = 1 + 6 = 7 ∈N



(iv) a*b = ∈ Q Here, a*b is not a binary operation because, Let a = 22 ∈ Q and b= 7 ∈ Q But

(v) a*b= a – b ∈ R Here, * is a binary operation because, a ∈ R, b ∈ R ⇒ a – b ∈ R, ∀ a, b ∈ R Question 2. Determine which of the following binary operation is commutative and which is associative : (i) * on N defined as a*b = 2ab (ii) * on N defined as a*b = a + b + aab (iii) * on Z defined as a*b = a – b (iv) * on Q defined as a*b = ab + 1 (v) * on R defined as a*b = a + b – 7 Solution: (i) Given a*b = 2ab Commutativity: Let a, b ∈ N a*b = 2ab = 2b.a = b*a So, a*b = b*a ∴ * is a commutative operation.

Associativity: Let a, b, c ∈ N (a*b)*c = 2(ab)*2c = 2ab + c = 2c*2(ab) = 2c +ab a*(b*c) = 2a*2(bc) = 2a + bc 2ab+c ≠ 2a+bc It is clear that (a*b)*c ≠ a*(b*c) So, (a*b)*c is not an associative operation. Hence, a*b = 2ab is commutative but not associative. (ii) Given a*b= a + b + a2b Commutativity: Let a, b ∈ N a*b = a + b + a2b b*a = b + a + b2a a*b ≠ b*a . So, * is not a commutative operation. Associativity: Let a, b, c ∈ N (a*b)*c = (a + b + a2b)*c a*(b*c) = a*(b + c + b2c) t is clear that (a*b)*c ≠ a*(b*c) So * is not an associative operation. Hence, a*b = a + b + a2b is neither commutative nor associative.



(iii) Given, a*b = a – b Commutativity: a*b = a – b, (a, b ∈ Z) b*a = b – a, (a, b ∈ Z) a*b* ≠ b*a So * is not a commutative operation. Associativity : (a*b)*c = (a – b)*c = a – b-c a*(b*c) = a*(b-c) = a – b + c ∵ (a*b)*c ≠ a*(b*c) So, it is not associative operation. It is clear that a*b = a – b is neither commutative nor associative. (iv) Given, a*b = ab + 1 Commutativity: Let a, b ∈ Q a*b = ab + 1 and : b*a= ba + 1 ⇒ a*b= b*a ∴ It is commutative. ∴ Addition and multiplication of rational number is commutative.

Associativity: Let a, b, c ∈ Q (a*b)*c = (ab + 1)*c = ab + 1 + c (b*c)*a = (bc + 1) +a = (a*b)*c ≠ (b*c)*a So, * is not associative. It is clear from above that a*b = ab + 1 is commutative but not associative. (v) Given, a*b = a + b – 7 Commutativity: In R, a*b = a + b – 7 = b + a – 7 = b*a’ Associativity : (a*b)*c = (a + b – 7)*c = (a + b – 7) + c – 7 = a + b + c – 14 a*(b*c) = a*(b + c – 7) = a + (b + c – 7) – 7 = a + b + c – 14 So, (a*b)*c = a*(b*c) Hence, it is clear that a*b = a + b – 7 are commutative and associative. Question 3. If * be an operation on Z, defined as a*b = a + b + 1, ∀ a, b ∈ Z then prove that * is commutative and associative, find its identity element. Also, find inverse element of any integer in Z. Solution: Given a*b = a + b + 1, ∀ a, b ∈ z Commutativity : a*b = a + b + 1



a*b = b + a + 1 = b*a ∴ a*b = b*a ∴ * is commutative operation. Associativity : (a*b) * c = (a + b + 1)*c = a + b + 1 + c +1 a + b + c + 2 Again a*(b*c) = a*(b + c + 1) = a + b + c + 1 + 1 = a + b + c + 2 a*(b*c)= (a*b)*c ∴ * is associative operation

Identity : If e is identity element, then a*e = a ⇒ a + e + 1 = a ⇒ e = -1 So, – 1 ∈ Z is identity element. Inverse : Let inverse of a is x, then by definition a*x = -1 [∵ – 1 is identity]

⇒ a + x + 1 = -1 ⇒ x = -(a + 2) ∈ Z Inverse element a-1 = -(a + 2). Question 4. If * be a binary operation defined on R – {1}. as a*b = a + b – ab, ∀ a, b ∈ R – {1} prove that * is commutative and associative. Find identity element and also find inverse of a. Solution: If a, b ∈ R – {1} by definition a*b = a + b – ab = b + a – ba = b*a ∴ * is a binary operation. Again, (a*b)*c = (a + b – ab)*c = (a + b – ab) + c – (a + b – ab)c = a + b – ab + c – ac – bc + abc = a + b +c – ab – bc – ac + abc …(i) and a*(b*c) = a*(b + c – bc) = a + (b + c – bc) – a (b + c – bc) = a + b + c – bc – ab – ac + abc = a + b + c – ab – bc – ac + abc … (ii) From eqs. (i) and (ii), (a*b)*c= a*(b*c) ∴ * is associative operation. Let e is identity of *, then for a ∈ R, a*e = a (from definition of identity) ⇒ a + e – ae = a ⇒ e(1 – a)= 0 ⇒ e = 0 ∈ R – {1}

∵ 1 – a ≠ 0



∴ 0 is identity of *. Let b is inverse of a. a*b = e a + b – ab = 0.e b – ab = -a b(1 – a) = -a

Question 5. Four functions are defined on set Ro, Such that, f1(x) = x, f2(x) = -x, f3(x) = 1/x, f4(x) = – 1/x Construct the composition table for the composition of functions f1, f2, f3, f4. Also, find identity element and inverse of every element. Solution: Given, f1(x) = x, f2(x) = – x,



∴ Table for above operations

Hence, it is clear that f1 is identity function of f1, f2, f3, f4. Hence, inverse element is itself too.

Rajasthan Board RBSE Class 12 Maths Chapter 1 Composite Functions Miscellaneous Exercise Question 1. If f : R → R, f(x) = 2x – 3; g : R → R, g(x) = x3 + 5, then the value of (fog)-1 (x):



Solution: Given, f : R → R, f(x) = 2x – 3 g : R → R, g(x)= x3 + 5 ∴ (fog)(x) = f(g(x)] = f(x3 +5) = 2(x3 + 5) – 3 = 2x3 + 10 – 3 = 2x3 + 7 Let y = (fog)(x) = 2x3 + 7 ∴ (fog)-1(y) = x

So, option (d) is correct. Question 2. If

, then the value of f(y): (a) x (b) x – 1 (c) x + 1 (d) 1 – x Solution:

So, no any option is correct.



Question 3. If

, then f{f{f(x)}] is equal to : (a) x (b) 1/x (c) – x (d) – 1/x Solution:

So, option (a) is correct. Question 4. If f(x) = cos (log x) then,

is equal to: (a) -1 (b) 0 (c) 1/2 (d) – 2



Solution: Given, f(x) = cos (log x)

So, option (b) is correct. Question 5. If f : R → R, f(x) = 2x + 1 and g : R → R, g(x) = x3, then (gof)-1(27) is equal to : (a) 2 (b) 1 (c) -1 (d) 0 Solution: Let (gof)-1(27) = x ∴ (gof)(x) = 27 g{f(x)} = 27 g{2x + 1} = 27 (2x + 1)3 = 27 2x + 1 = 271/3 2x + 1 = 33 x 1/3 2x + 1 = 3 ∴ 2x = 2 x = 1 So, option (b) is correct. Question 6. If f: R → R and g: R+ R, where f(x) = 2x + 3 and g(x) = x2 + 1, then (gof)(2): (a) 38 (b) 42 (c) 46 (d) 50 Solution: Let (gof)(2) = y ∴ y = g{{{2}} = g{2 x 2 + 3} = g(7) = (7)2 + 1



= 49 + 1 = 50 So, option (d) is correct. Question 7. If * is an operation on Q0, defined by a*b = ab/2, ∀ a, b ∈ Q0, then identity element of this operation is : (a) 1 (b) 0 (c) 2 (d) 3 Solution: If e is identity element For a ∈ Q0 a*e = a

⇒ = a e = 2 So, option (c) is correct. Question 8. A binary operation on R is defined by a*b = 1 + ab, ∀ a, b ∈ R, then identity element of this operation is : (a) commutative but not associative (b) associative but not commutative (c) neither commutative nor associative (d) commutative and associative both Solution: Given, a*b = 1 + ab, ∀ a, b ∈ R Commutativity a*b = 1 + ab = 1 + b.a = b*a ∵ Set of real numbers is commutative. So, a*b = b*a ∴ is commutative Associativity (a*b)*c = (1 + ab)*c = (1 + ab)c = 2 + abc a*(b*c) = a*(1 + bc) = 1 + a*(1 + bc) = 1 + a + abc It is clear that (a*b)*c ≠ a*(b*c) So, * operation is not associative. So, option (a) is correct. Question 9. Subtraction is an operation on Z, which is (a) commutative and associative (b) associative but not commutative (c) neither commutative and nor associative (d) commutative but not associative Solution: Let a, b ∈ z Commutativity : As we know, a-b ≠ b – a



Associativity : (a – b) – c ≠ a – (b – c) So, option (c) is correct. Question 10. * be an operation on Z definied by a*b = a + b – ab, ∀ a, b ∈ Z, for any a ∈ Z, a(≠ 1), inverse of a with respect to * is :

Solution: Given, a*b = a + b – ab, ∀ a, b ∈ z

Let a ∈ Z is possible and x is inverse of a, then According to question a*x = 0 (∵ 0 is identity)

⇒ a + x – ax = 0 ⇒ x(1 – a) = – a

Hence, option (a) is correct. Question 11. Which of the following operations is commutative in R: (a) a*b = a2b (b) a*b = ab (c) a*b = a – b + ab (d) a * b = a + b + a2b Solution: (a) ∵ a*b=ab and b*a = b2a ∴ a*b ≠ b*a So, operation is not commutative. (b) ∵ a*b = ab and b*a = ba ∴ a*b ≠ b*a So, operation is not commutative. (c) ∵ a*b = a – a + ab and b*a = b – a + ba ∴ a*b* ≠ b*a So, operation is not commutative. (d) ∵ a*b= a + b + a2b and : b*a = b + a + b2a ∴ a*b* ≠ b*a So, operation is not commutative. Hence, any of these operation is not commutative. Question 12. Verify the associative law for the composite function of following three functions : f : N → Z0, f(x) = 2x g : Z0 → Q, g(x) = 1/x h : Q → R, h(x) = ex Solution:



Given, f : N→ Z0 g : Z0 → Q h : Q → R Then, ho(gof) : N → R and (hog)of : N → R Hence, domain and co-domain of ho(gof) and go(hof) are same because both functions are defined from N to R. Hence, we have to prove [ho(gof)](x) = [(hog)of)(x), ∀ X ∈ N Now, [ho(gof)](x)= h[(gof)(x)] = h[g{f(x)] = h[g(2x)]

From eqs. (i) and (ii), (hog)of = ho(gof) Hence, associativity of f, g, h is proved. Proved. Question 13. If f: R+ → R+ and g : R+ → R+, defined as f(x) = x2, g(x) = √x, then find gof and fog wheather are they equivalent ? Solution: Given, f: R+ → R+, f(x) = x2 g : R+ → R+, g(x)= √x Then, (fog): R+ → R+ and (gof): R+ → R+ are defined ∴ (gof)(x) = g[f(x)] = g(x2) = √x2 = x (fog)(x) = f(g(x)] = f(√x)= (√5)2 = x Based on above, (gof) and (fog) are of same domain and co-domain. (fog)(x) = (gof)(x) ∀ x ∈ R+ So, (fog) and (gof) are equivalent functions. Question 14. If f : R → R, f(x) = cos (x + 2), justify whether f is inverse function or not. Solution: Given function, f : R → R, f(x) = cos (x + 2) Putting x = 2π f(2π) = cos (2π + 2) = cos (2)



Putting x = 0 f(0) = cos (0 + 2) = cos 2 Since, only one image is obtained for 0 and 2π, so function is not one-one. Hence, function is not one-one onto. Hence, f-1 : R → R does not exist. Question 15. If A = {-1,1} fand g are two functions defined on A, where f(x) = x2, g(x) = sin (πx/2), prove that g-1 exists but f-1 does not exist, also find g-1. Solution: Given, A = {-1, 1} f(x)= x2, g(x) = sin πx/2 One-one/many-one : f : A → A, f(x) = x2 f(-1) = f(1) = 1 ⇒ Image of – 1 and 1 is same. So, f is not one-one. Onto/into : Elements of co-domain unpaired with any element of domain. So ‘f’ is not onto. So, f is neither one-one nor onto. Hence, ‘f-1‘ does not exist. Thus, for g(x) = sin πx/2 One-one/many-one : Let x1, x2 ∈ A, such that

Hence, function in one-one. Let y ∈ R (co-domain), if possible then pre image is x in R, then

Since, pre-image of each value of y exist in domain R, then g(x) is onto function. Hence, g(x) is one-one onto function, so g-1 exist.

Question 16. If f : R → R and g : R → R, such that f(x) = 3x + 4 and g(x) = (x – 4)/3 then (fog)(x) and (gof)(x). Also, find the value of (gog) (1). Solution: Given, f : R → R, f (x) = 3x + 4



Rajasthan Board RBSE Class 12 Maths Chapter 2 Inverse Circular Functions Ex 2.1 Question 1. Find the principal value of the following angles:

Solution: (i) sin-1(1)



Prove the following : (Q. 2 to 8) Question 2.



Solution:

Question 3.



Solution:

Question 4.

Solution:



Question 5. sec2 (tan-1 2) + cosec2 (cot-1 3) = 15 Solution: Let tan-12 = θ ⇒ tan θ = 2 sec2 θ = 1 + tan2 θ = 1 + (2)2 = 1 + 4 = 5 ∴ sec2 (tan-12) = 5 Let cot-13 = Φ = cot Φ = 3 ∴ cosec2 Φ = 1 + cot2 Φ = 1 + (3)2 = 1 + 9 = 10 ∴ cosec2(cot-1 3) = 10 …..(ii) Adding eq. (i) and (ii), we get sec2 (tan-12) + cosec2 (cot-13) = 5 + 10 ⇒ sec2 (tan-12) + cosec2(cot-13) = 15 Hence proved. Question 6.

Solution:



Question 7.



Solution:

Question 8.

Solution:



Question 9. If cos-1x + cos-1y + cos-1z = π, then prove that x2 + y2 + x2 + 2xyz = 1. Solution: cos-1x + cos-1y + cos-1z = π (According to question) cos-1x + cos-1 y = π – cos-1z

Question 10. If sin-1x + sin-1 y + sin-1z = π, then prove that:

Solution: Let sin-1 x = A ⇒ x = sin A sin-1 y = B ⇒ y = sin B sin-1 z = C ⇒ z = sin C ∵ sin-1 x + sin-1 y + sin-1 z = π

∴ A + B + C = π ∴ To prove



Question 11. If tan-1x + tan-1y + tan-1z = π/2, then prove that xy + yz + zx = 1. Solution:



According to question,

Question 12. If

then prove that x + y + z = xyz. Solution: Let x = tan A, y = tan B, z = tan C



Question 13. If

then prove that x + y + z = xyz. Solution:



According to question

Question 14. Prove that : tan-1x + cot-1(x + 1) = tan-1(x2 + x + 1). Solution: L.H.S. = tan-1x + cot-1(x + 1)

Question 15. If tan-1x, tan-1y, tan-1z are in arithmetic progression, then prove that y2(x +z) + 2y(1 – xz) – x – z = 0.



Solution: tan-1x, tan-1y, tan-1z are in A.P.

Question 16. If α, β, γ be the roots of equation x3 + px2 + qx + p = 0, then prove that, except a special condition, tan-1α + tan-1 β + tan-1γ = nπ also find the special condition, when it does so on. Solution: Given, α, β, γ are root of equation x3+ px2 + qx + p = 0

Special Situation: When sum and product of roots is not equal, then tan-1 α + tan-1 β + tan-1γ ≠ nπ Solve the following euqations : (Q. 17 to 25)



Question 17.

Solution:

Question 18.



Solution:

Question 19.

Solution:



Question 20.

Solution:



Question 21.

Solution:



Question 22.

Solution:



Question 23. sin 2 (cos-1 {cot (2 tan-1x)} = 0 Solution: Given equation



Question 24.

Solution:



Question 25.

Solution: According to question,



Rajasthan Board RBSE Class 12 Maths Chapter 2 Inverse Circular Functions Miscellaneous Exercise Question 1. Principal value of tan-1(- 1) is : (a) 45° (b) 135° (c) – 45° (d) – 60° Solution: ∵ tan-1(- x)= – tan-1 x tan-1 (-1) = – tan-1(1) Let tan-11 = θ ∴ tan θ = 1 ⇒ tan θ = tan 45°



∴ θ = 45° ∴ tan-1(-1) = – 45° So, option (c) correct. Question 2. 2 tan-1(1/2) is equal to :

Solution:

So, option (a) is correct. Question 3. If tan-1 (3/4) = θ, then sin is :


Question 4. cot (tan-1 α + cot-1 α) is equal to: (a) 1 (b) ∞



(c) 0 (d) none of those Solution: cos (tan-1 a + cot-1 a)

So, option (c) correct. Question 5.

If sin-1 ( ) = x, then general value of x is :

Solution: Given,

Hence, option (d) is correct. Question 6. 2 tan (tan-1x + tan-1 x3) is :

Solution:

Hence, option (a) is correct. Question 7.

If tan-1(3x) + tan-1 (2x) = , then x is:



Solution:

Question 8.

Value of sin-1( ) + 2 cos-1 ( )is :

Solution:

Question 9.

If tan-1(1) + cos-1 ( ) = sin-1 x, then value of x is:

Solution:



Question 10.

If cot-1 x + tan-1 = then x is : (a) 1 (b) 3

(c) (d) none of these Solution:

Hence, option (c) is correct. Question 11. If 4 sin-1 x + cos-1x = π, then find x. Solution: 4 sin-1x + cos-1x = π

Question 12.



Solution:

Question 13.

If sin-1 ( ) + sec-1 ( ) = x, then find x. Solution:

Question 14.

Find: sin-1 ( ) + 2 tan-1 ( ) Solution:



Question 15.

If sin-1( ) + sin-1 ( ) = 90°, then find x. Solution:

Question 16. Prove that:

Solution:

Question 17. tan-1x + tan-1y + tan-1z = π, then prove x + y + z = xyz



Solution: tan-1x + tan-1y + tan-1z = π

Question 18.

Prove that tan-1( tan 2A) + tan-1 (cot A) + tan-1 (cot2 A) = 0. Solution:



Question 19. Prove that tan-1x = 2 tan-1(cosec (tan-1 x) – tan (cot-1 x)]. Solution: Let tan-1x = θ

Question 20.

then prove that value of Φ – θ is 30°. Solution:




Solution:



Rajasthan Board RBSE Class 12 Maths Chapter 3 Matrix Ex 3.1 Question 1. If A = [aij]2 x 4, then write the number of elements in A. Solution: 8 Question 2. Write identity matrix of order 4 x 4. Solution: Unit matrix of 4 x 4

Question 3.

If , then find the value of a. Solution: Given,

∴ On comparing ∴ k +4 = a …….(i) k – 6 = -4 …….(ii) From (i), k = -4 + 6 = 2 From (ii),



a = k + 4 = 6 ∴ a = 6 Question 4. What are the possible order of matrix, having 6 elements ? Solution: 6 x 1, 1 x 6, 2 x 3 and 3 x 2. Question 5. Construct a 2 x 2 matrix A = [aij], where elements are given by :

Solution:



Question 6.

Construct a matrix A = [aij] of order 2 x 3, whose elements are a[aij] = | 2i – 3i| . Solution: In 2 x 3, there are 2 rows and 3 columns So, i = 1, 2 and j = 1, 2, 3



Question 7.

then find ‘a’ and ‘b’. Solution:



On comparing, a + b = 6 ……(i) ab = 8 From equation (i) and (ii), a(6 – a)= 8 ⇒ 6a – a2 – 8=0 ⇒ a2 – 2a – 4a + 8 = 0

⇒ a2 – 2a – 4a + 8 = 0 2 (a – 2)(a – 4) = 0 So, a = 2, 4 From ab = 8 we get b = 4, 2 Question 8.

, then find the value of x, y, z and p. Solution :

On comparing 2x = 4 ⇒ x = 2 ⇒ 3x + y = 5 y = 5 – 3 x 2 = 5 – 6 = -1 -x + z = -4 z = -4 + x = -4 + 2 = – 2 ⇒ 3y – 2p = -3 ⇒ 2p = 3y + 3 = 3 x -1 + 3 = 0 So, p = 0 x = 2, y = -1, z = -2, p = 0. Question 9. Find the values of a, b and c if matrices A and B are equal, where

Solution : Given, A = B

On comparing a – 2 = b ⇒ a – b = 2 …..(i) 3 = c 12c = 6b

⇒ b = = 6



⇒ b = 6 From b + 2 = a, a – b = 2 ∴ a = 2 + b = 2 + 6 = 8 So, a = 8, b = 6, c = 3.

Rajasthan Board RBSE Class 12 Maths Chapter 3 Matrix Ex 3.2 Question 1.

then find A+B and A – B. Solution: Given,

Question 2.

, then find matrix A and B. Solution:



Question 3.

, then find C, where A + 2B + C = 0. Solution: Given,



A + 2B + C = O

Question 4.

then find 3A2 – 2B. Solution: According to question,



Question 5.

, then show that AB ≠ BA. Solution:



Hence, AB ≠ BA. Hence Proved. Question 6.

, then show that: f(A)(B) = f(A + B). Solution:



Question 7.

, then prove : (AB)T = BTAT Solution:



From (i) and (iv),(AB)T = BTAT Hence Proved. Question 8. Prove that

Solution:

= (ax + hy + gz)x + (hx + by + fz)v + (gr + fy + cz)z = ax2 + hxy + gzx + hxy + by2 + fyz + gzx + fyz + cz2 = ax2 + by2 + cz2 + 2hxy + 2fyz + 2gzx = R.H.S. Hence Proved.



Question 9.

and ‘I’ is unit matrix of order 3 x 3, then prove

Solution: Given,



Question 10.

, where O is zero matrix, then find a. Solution:

Question 11.

, (A + B)2 = A2 + B2, then find a, b. Solution: Given.



Question 12.

and I is unit matrix of order 2 x 2, then prove that

Solution:



Question 13.

, then find K, where A2 = 8A + KI. Solution: Let A2 = 8A + KI



Question 14.

, then find A. Solution:



Question 15.



Solution:

Rajasthan Board RBSE Class 12 Maths Chapter 3 Matrix Miscellaneous Exercise Question 1.

, then find A. Solution:



Question 2.

, then find (A – 2I) (A – 3I). Solution:



Question 3.

, then find AB. Solution:

Question 4.

, then find BA. Solution:

Question 5.

, then find matrices A and B. Solution:



Question 6.

, then find the value of x and y. Solution: On comparing, x + 2 = -2 ∴ x = -4 – y – 2 = 5 ⇒ y = -7 Hence, x = -4, y = -7 Question 7. Order of matrix A is 3 x 4 and B is a matrix, such that ATB and ABT defined, then write the order of B. Solution: ∴ Order of A = 3 x 4

∴ Order of AT = 4 x 3 But ATB and ABT is defined. So, order of B is 3 x 4. Question 8.

, is a symmetric matrix, then determine x. Solution: Given,

On comparing aij = aji a32 = a23 ⇒ -x = 4

∴ x = -4 Question 9. Construct a matrix of order 3 x 3, B = [bij], whose elements are bij= (i) (j). Solution: B11 = 1 x 1 = 1 B12 = 1 x 2 = 2 B13 = 1 x 3 = 3 B21 = 2 x 1 = 2 B22 = 2 x 2 = 4 B23 = 2 x 3 = 6 B31 = 3 x 1 = 3 B32 = 3 x 2 = 6 B33 = 3 x 3 = 9



Question 10.

Solution:

Question 11. Express matrix A as the sum of symmetric and skew-symmetric matrices, where

. Solution: Given,



Question 12.

then prove that : (i) (AT)T = A (ii) A + AT is a symmetric matrix. (iii) A – AT is a skew-symmetric matrix. (iv) AAT and ATA are symmetric matrix. Solution:



(i) (AT)T = A

(ii) A + AT is symmetric

So, A + AT is symmetric matrix. Hence Proved.



So, A – AT is skew symmetric matrix. Proved.



Here, a21 = a12 = 0 a31 = a13 = 0 a23 = a32 = 6 So, AAT is symmetric matrix.

Here, a12 = a21 = 0 a13 = a31 = 0



a32 = a23 = 4 So, ATA is symmetric matrix. Question 13.

, and 3A – 2B + C is a null matrix, then determine matrix ‘C’. Solution:

Question 14. Construct a matrix B = [bij] of the order 2 x 3, whose elements are bij = (i +2j)2/2 Solution:



Given, B = [bij] whose elements are

Question 15.

, then find the element of 1st row of ABC. Solution:



So, element of 1st row is 8. Question 16.

, then find AAT. Solution: Given,



Question 17.

, then find x. Solution:

Question 18.

, then prove

Solution: Given,



= (bc – ad)I2 = R.H.S. Hence Proved. Question 19.

, then find the matrix form of the following (aA + bB) (aA – bB). Solution:



Givn,

Question 20.

, then prove that (A – B)2 ≠ A2 – 2AB + B2. Solution: Given,



From (i) and (ii),



(A – B)2 + A2 – 2AB + B2 Hence Proved. Question 21.

, then find k, where A2 = kA – 2IA2 . Solution: Given,

On comparing corresponding element From 3k – 2 = 1

3k = 3 ⇒ k = ⇒ k = 1. Question 22. i = √-1 then prove that : (1) A2 = B2 = C2 = -I2 (ii) AB = – BA = -C Solution:



Question 23.

and f(A) = A2 – 5A + 7I then find f(A).



Solution:

Question 24. Prove that

Solution:



Rajasthan Board RBSE Class 12 Maths Chapter 4 Determinants Ex 4.1 Question 1. For which value of k, det

will be zero ? Solution:



Question 2.

, find x : y. Solution:

Question 3.

, then find the value of x and y. Solution :

Question 4. Find x, if

Solution :



⇒ (x – 1) (x – 3) – x(x – 2) = 0 ⇒ x2 – 3x – x + 3 – x2 + 2x = 0

⇒ -2x + 3 = 0 ⇒ – 2x = -3 So, x = 3/2 Question 5. Write minors and co-factors of following determinants corresponding to first column, also find the value of determinants:

Solution:

= – 2 – 10 = -12 Cofactor of a F11 = (- 1)2 M11 = 1 x (- 12) = -12 Minor of a21

= -6 – 10 = -16 Cofactor of a21 F21 = (- 1)3 M21 = -1 x (-16) = 16 Minor of a31

= – 6 – (- 2) = – 6 + 2 = -4 Cofactor of a31 F31 = (- 1))4M31 = 1 x (-4) = -4 So, Determinant M = a11F11 + a21 F21 + a31 F31 = 1 .(- 12) +4·(16) + 3.(-4) = – 12 + 64 – 12 = 40 So, M11 = -12, M21 = – 16, M31 =-4 F11 = – 12, F21 = 16, F31 = -4 |A| = 40



(ii) Solution: Minor of a11

= a . (bc – f2) + h. (fg – hc) + g. (hf – bg) = abc – af2 +fgh – h2c + fgh – bg2 = abc + 2fgh – af2 – bg2 – ch2 So M11 = bc – f2, M21= hc – fg, M31 = hf – bg F11 = bc – f2, F21 = fg – hc, F31 = hf – bg |A| = abc + 2fgh – af2 – bg2 – hc2 Question 6. Find the value of determinant

Solution:

= -5 (0 – 3) = -5 x (-3) = 15



Question 7. Prove that :

Solution:

= 1 + c2 + a2 – abc + abc + b2 = 1 + a2 + b2 + c2 = R.H.S. Proved.

Rajasthan Board RBSE Class 12 Maths Chapter 4 Determinants Ex 4.2 Question 1.

, then find l : m. Solution:

⇒ l x 3 – 2 x m = 0

⇒ 3l – 2m= 0 ⇒ 3l = 2m

⇒ = So, l : m = 2 : 3 Question 2. Find the minors of elements of second row of determinant

. Solution:

Question 3. Find the value of determinant



Solution:

Question 4. Write the effect on value of determinant, if first and third columns of any determinant are interchanged. Solution: Sign of determinant will be changed. Question 5. Prove that



Solution:

Question 6. Find the value of determinant

.



Solution:

Question 7. Solve the equation :

Solution:

Now, according to question, changed form of given eq. i.e., eq. (i) is equal to zero.



img ⇒ 1(- 145 + 143) – 1(- 58 + 13x + 104) + 2( – 22 – 5x +40) = 0 – 2 + 13x – 46 – 10x + 36 = 0 ⇒ 3x – 12 = 0

⇒ = 4. Question 8. Prove without expansion that,

Solution: Let



Question 9. Prove that,



Solution:

= (a + b + c)[0 – 0 + 1{(a – c)(b – c) – (a – b) (b – a)}] = (a + b + c){(ab – ca – bc + c2) – (ab – a2 – b2 + ab)} = (a + b + c)(ab – ca – bc + c2 – ab + a2 + b2 – ab) = (a + b + c)(a2 + b2 + c2 – ab – bc – ca) = a3 + b3 + c3 – 3abc – R.H.S. Proved Question 10. Find

Solution:



Question 11. If ω is a cube root, then find

Solution:

= 1(1 – ω2) – 1(1 – ω3) + ω2(ω – ω2) = 1 – ω2 – 1 + ω3 + ω3 – ω4 = 1 – ω2 – 1 + 1 + 1 – ω3.ω (∵ ω3 = 1) = 3 – ω2 – 1 – ω (∵ ω3 = 1) = 3 – (1 + ω + ω2) = 3 – 0 = 3 (∵ 1 + ω + ω2 = 0) Question 12. Prove that

Solution:



Question 13. If in determinant img, A1, B1, C1, ……….. etc are co-factors of elements a1, b1, C1, … respectively, then prove that :



Solution:

Rajasthan Board RBSE Class 12 Maths Chapter 4 Determinants Miscellaneous Exercise Question 1. Value of determinant

is: (a) 0 (b) 1



(c) -1 (d) none of these Solution:

= cos 80° sin 10° – (- cos 109) sin 80° = cos 80° sin 10° + cos 10° sin 80° = sin (10° + 80°) = sin 90° = 1 So, option (b) is correct. Question 2. Co-factors of first column of determinant

(a) -1, 3 (b) -1,- 3 (c)- 1, 20 (d) – 1,- 20 Solution:

Co-factor of a11 F11 = (- 1)2 M11 = 1 x (- 1) = – 1 Co-factor of a12 F21 = (-1)11 M21 = (- 1) × 20 = – 20 So, option (d) is correct. Question 3.

If , then the value of will be : (a) – 2∆ (b) 8∆ (c) -8∆ (d) -6∆ Solution:



So, option (c) is correct. Question 4. Which of the following determinant is identical to determinant

:

Solution:

So, option (c) is correct. Question 5. Value of

(a) 0



(b) 1 (c) 1/2 (d) – 1/2 Solution:

= cos 50° cos 10° – sin 50° sin 10° = cos (50° + 10°)

= cos 60° = So, option (c) is correct. Question 6. Value of

(a) ab + bc + ca (b) 0 (c) 1 (d) abc Solution:

Question 7. If ω is a cube root of unity, then value of

(a) ω2 (b) ω (c) 1 (d) o 11 04 081



Solution:

Question 8.

If , then x will be: (a) 6 (b) 7 (c) 8 (d) 0 Solution:

⇒ (4 – 2)2 = (3x – 2) – (x + 6) ⇒ (2)2 = 3x – 2 – x – 6 ⇒ 4 = 2x – 8

⇒ 4 + 8 = 2x x = 6 So, option (a) is correct. Question 9.

If and F11, F12, F13, …, are the corresponding cofactors of a11, a12, a13, …, then which of the following is true : (a) a12F12+ a22F22 + a32F32 = 0 (b) a12F12 + a22F22 + a32F32 ≠ ∆



(c) a12F12 + a22F22 + a32F32 = ∆ (d) a12F12 + a22F22 + a32F32 = – ∆. Solution: a12F12 + a22F22 + a32F32 = ∆ So, option (c) is correct. Question 10. Value of determinant

(a) x + y + z (b) 2(x + y + z) (c) 1 (d) 0 Solution:

So, option (d) is correct. Question 11. Solve the following equation :

Solution:

⇒ 1(9x – 48) – 2(36 – 42) + 3(32 – 7x) = 0

⇒ 9x – 48 + 12 + 96 – 21x =0 ⇒ – 12x + 60 = 0 ⇒ – 12x = -60

⇒ x = = -5 So, x = 5.



Question 12. Find the value of determinant :

Solution:

= 1(27 – 1) – 3(9 – 9) + 9(3 – 81) = 26 – 0 – 702 = -676. Question 13. Find the value of determinant

Solution:


Solution:



= a2b2c2 [- 1(1 – 1) – 1(-1 – 1) + 1(1 + 1)] = a2b2c2 (0 + 2 + 2) = 4a2b2c2 = R.H.S. Hence Proved. Question 15. Prove that x = 2 is a root of following equation. Also find its remaining roots :

Solution:

∵ R1 = R2

∴ Determinant will be zero.



∴ Root of equation is 2.




Solution:


. Solution:



= (a + b + c)[(- b – c – a) (- c – a – b) – 0] = (a + b + c)(b + c + a)(c + a + b) = (a + b + c)3 = R.H.S. Proved. Question 18. Prove that :



Solution:


Solution:



= (a – b) (b – c) [(b2 + c2 + bc) – (a2 + b2 + ab)] = (a – b) (b – c) (b2 + c2 + bc – a2 – b2 – ab) = (a – b)(b – c) [bc + c2 – a2 – ab] = (a – b)(b – c) [bc – ab + c2 – a2 ] = (a – b) (b – c) [b(c – a) + (c2 – a)] = (a – b)(b – c)(c – a)(b + c + a) = R.H.S. Proved. Question 20. Prove that:

Solution:



Question 21. If a + b + c = 0, then solve:

Solution:



⇒ (-x) [(c – a) (a – c + x) – (b – c + x) (b – x – a)] = 0 ⇒ (-x) [(ac – c2 + cx – a2 + ac – ax) ⇒ (b2 – bx – ab – bc + cx + ac – xb – x2 – ax)] = 0 = (-x) [x2 – (a2 + b2 – ab – bc – bc – ca)]= 0 If – x = 0, then x = 0 Now, if x2 = (a2 + b2 + c2 + bc – ab – ca) = 0 .




Solution:

Question 23. If p + q + r = 0, then prove that



Solution:




Solution:

Rajasthan Board RBSE Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations Ex 5.1 Question 1. For which value of x, matrix

is singular ? Solution:

⇒ 1 (- 6 – 2) + 2 (-3 – x) + 3(2 – 2x) = 0 ⇒ -8 – 6 – 2x + 6 – 6x = 0 ⇒ -8 = 8

⇒ x = = -1 Hence, x = -1



Question 2.

If matrix , then find adj.A, Also prove that, A(adj.A) = |A| I3 = (adj.A)A. Solution:

Matrix made from adjoint of matrix A,



Question 3. Find the inverse matrix of the following matrix:

Solution: (i) Let



= 1(1 + 3) – 2(-1 + 2) + 5(3 + 2) = 4 – 2 + 25 |A| = 27 ≠ 0 So. A-1 exists. Cofactors of matrix A,




Cofactors of matrix A,




Cofactors of matrix A.




Question 4. If matrix

then find A-1 and prove that: (i) A-1A= I3 (ii) A-1 = F(-α) (iii) A(adj.A) = |A|I = (adj A).A Solution:

Then, |A| = cos α (cos α – 0) + sin α (sin α – 0) + 0(0 – 0) = cos2 α + sin2 α |A|= 1 ≠ 0 So, A-1 exists. Cofactors of matrix A,



Matrix made from adjoint of matrix A



So, A(adj.A) = |A|I = (adj.A) A Hence Proved. Question 5.

, then prove that : A-1 = AT. Solution: Let



Question 6.

If matrix , then prove that A-1 = A3. Solution: Given,

So, A-1 exists. On finding adjoint of matrix A, a11 = – 1, a12 = – 2, a21 = 1, a22 = 1 Matrix formed by adjoint of A,



From (i) and (ii), A-1 = A3 Hence proved. Question 7.

then find (AB)-1. Solution: Given

Then, |A| = 5(3 – 4) – 0(2 – 2) + 4(4 – 3) = – 5 – 0 + 4



|A| = -1 ≠ 0 So, A-1 exists. On finding adjoint of matrix A,

Matrix formed by adjoint of A



Question 8.

If Solution: Given



Question 9.

Show prove that matrix satisfies equation A2 – 6A + 17I = O. Thus find A-1. Solution: Given



Question 10.

If matrix , then show that A2 + 4A – 42I = O. Hence A2. Solution: Given



So, given matrix satisfies A2 + 4A – 42I = O Now A2 + 4A – 42I = O ⇒ A2 + 4A = 42I ⇒ A-1 (A2 + 4A) = 42A-1.I

⇒ A-1.A2 + 4A-1.A = 42A-1.I ⇒ A + 4I = 42A-1



(∵ A-1.A = I and A-1.I = A-1)

Rajasthan Board RBSE Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations Ex 5.2 Question 1. Find area of triangle, whose vertices are: (i) (2, 5), (- 2, – 3) and (6,0) (ii) (3, 8), (2, 7) and (5, – 1) (iii) (0, 0), (5, 0) and (3, 4) Solution: (i) Area of triangle



(ii) Area of triangle

(iii) Area of triangle

Question 2. Use determinant to find the area of triangle whose vertices are (1, 4), (2, 3) and (-5, – 3), Are these points collinear ? Solution:



Area of triangle

These points are not collinear because area of triangle is not equal to zero. Question 3. Find the value of k, if area of triangle is 35 sq. unit and vertices of triangle are (k, 4), (2, -6) and (5, 4). Solution: Given points (k, 4), (2, -6) and (5, 4) and area of triangle = 35 sq. unit

taking +ve sign ⇒ – 10k + 50 = 70 ⇒ – 10k = 70 – 50 ⇒ – 10k = 20

⇒ k = -2 taking -ve sign – 10k + 50 = – 70 ⇒ 10k = – 70 – 50 ⇒ – 10k = – 120 ⇒ k = 12

⇒ k= -2, 12. Question 4. Use determinant to find k, if points (k, 2 – 2k), (-k + 1, 2k) and (-4 – k, 6 – 2k) and collinear. Solution:



Given points (k, 2 – 2k), (-k + 1, 2k) and (-4 – k, 6 – 2k) are collinear.

⇒ k [2k – ( 6 – 2k)] – (2 – 2k) [(-k + 1) – (-4 – k)] + 1 [(-k + 1) (6 – 2k) – (-4 – k) (2k)] = 0 ⇒ k [2k – 6 + 2k] – (2 – 2k) [-k + 1 + 4 + k ] + 1[- 6k + 2k2 + 6 – 2k + 8k + 2k2] = 0 ⇒ K(4k – 6) – (2 – 2k) (5) + 1(4k2 + 6) = 0 ⇒ 4k2 – 6k – 10 + 10k + 4k2 + 6 = 0 ⇒ 8k2 + 4k – 4 = 0

⇒ 2k2 + k – 1 = 0 ⇒ 2k 2 + (2 – 1)k – 1 = 0

⇒ 2k(k + 1) – 1(k + 1) = 0 ⇒ (k + 1) (2k -1)=0 Hence, k = -1, 1/2. Question 5. If point (3,-2), (x, 2) and (8, 8) are collinear, then find the value of x, using determinant. Solution: Given (3,-2), (x, 2) and (8, 8) are collinear.

⇒ 3(2 – 8) + 2(x – 8) + 1(8x -16) = 0 ⇒ – 18 + 2x – 16 + 8x – 16=0

⇒ 10x – 50 = 0 ⇒ 10x = 50 ⇒ x = 50/10 Hence, x = 5 Question 6. Find the equation of line passing through the two points (3, 1) and (9, 3), using determinant, also find the area of traingle if third point is (– 2, – 4). Solution: Equation of line passing through (3, 1) and (93),

⇒ x(1 – 3) – (3 – 9) + 1(9 – 9) = 0 ⇒ – 2x + 6 + 0 = 0 ⇒ – 2(x – 3y) = 0 ⇒ x – 3y = 0 Area of triangle



Area of triangle ∆ = 10 sq. unit (leaving -ive sign) Question 7. Solve the following system of equations by Cramer’s rule : (i) 2x + 3y = 9, 3x – 2y = 7 (ii) 2x – 7y – 13 = 0, 5x + 6y – 9 = 0 Solution: Given equations 2x + 3y = 9 3x – 2y = 7 Here,



Solution of equation be x = 3, y = -1. Question 8. Prove that following system of equation are inconsistent : (i) 3x + y + 2z = 3 2x + y + 3z = 5 x – 2y – z = 1 (ii) x + 6y + 11 = 0 3x + 20y – 67 + 3 = 0 6y – 187 + 1 = 0 Solution: Given equations 3x + y + 2z = 3 2x + y + 3z = 5 x – 2y – z= 1

= 3(- 1 + 6) – 1(-2 – 3) + 2 (- 4 – 1) = 15 + 5 – 10 = 10 ≠ 0 ∵ ∆ ≠ 0 ∴ Solution is not possible. Hence, system of equation are in consistent. Hence proved. (ii) Given equation x + 6y + 11 = 0 or x + 6y = – 11 3x + 20y – 6z + 3 = 0 or 3x + 20y – 6z = -3 6y – 18z = -1 Here



∵ ∆ = 0, ∆1 ≠ 0, ∆2 ≠ 0 and ∆3 ≠ 0 ∴ Solution of equations is not possible. Hence, system of equation are inconsistent. Hence proved. Question 9. Solve the following system of equations by Cramer’s rule : (i) x + 2y + 4z = 16 4x + 3y – 2z = 5 3x – 5y + z = 4 (ii) 2x + y – z = 0 x – y + z = 6 x + 2y + z = 3 Solution: (i) Given equations x + 2y + 4z = 16 4x + 3y – 2z= 5 3x – 5y + z = 4



(ii) Given equations 2x + y – z = 0 x – y + z = 6 x + 2y + z = 3

= 2(-1 – 2) – 1(1 – 1) – 1(2 + 1) = -6 – 0 – 3 = -9



= 0(- 1 – 2) – 1(16 – 3) – 1(12 + 3) = 0 – 3 – 15 = -18

= 2(6 – 3) – 0(1 – 1) – 1(3 – 6) = 6 – 0 + 3 = 9

= 2(-3 – 12) – 1(3 – 6) + 0(2 + 1) = -30 + 3 + 0 = -27 Using Cramer’s rule,

Hence, x = 2, y = -1, z = 3. Question 10. Solve the following system of equation using determinants : (i) 6x + y – 3z = 5 x + 3y – 2z = 5 2x + y + 4z = 8

Solution: (i) Given equations 6x + y – 3z = 5 x + 3y – 2z = 5 2x + y + 4z = 8



= 6(12 + 2) – 1(4 + 4) – 3(1 – 6) = 84 – 8 + 15 = 91

= 5(12 + 2) – 1(20 + 16) – 3(5 – 24) = 70 – 36 + 57 = 91

= 6(20 + 16) – 5(4 + 4) – 3(8 – 10) = 216 – 40 + 6 = 182

= 6(24 – 5) – 1(8 – 10) + 5(1 – 6) = 114 + 2 – 25 = 91



Using Cramer’s rule



Question 11. Use matrix method to solve following system of equations : (i) 2x – y = -2 3x + 4y = 3 (ii) 5x + 7y + 2 = 0 4x + 6y + 3 = 0 (iii) x + y – 7 = 1 3x + y – 2z = 3 x – y – z = -1 (iv) 6x – 12y + 25z = 4 4x + 15y – 20z = 3



2x + 18y + 15z = 10 Solution: (i) Given equations 2x – y = -2 3x + 4y = 3 Let AX = B

So, A– 1 exists. On finding adjoint of matrix A F11 = 4, F12 = -3, F21 = 1, F22 = 2 Matrix formed by adjoint of A,

(ii) Given equations 5x + 7y + 2 = 0 or 5x + 7y = -2 4x + 6y + 3 = 0 or 4x + 6y = – 3 ⇒ AX = B …(i)



So, A– 1 exists. On finding adjoint of matrix A, F11 = 6, F12 = – 4, F21 = – 7, F22 = 5 Matrix formed by adjoint of A,

(iii) Given equations, x + y – z = 1 3x + y – 2z = 3 x – y – z = -1 Let AX = B ……(i)



(iv) Given equations 6x – 12y + 25z = 4 4x + 15y + 15z = 3 2x + 18y + 15z = 10 AX = B ……….(i)

= 6(225 + 360) + 12(60 + 40) + 25(72 – 30) = 3510 + 1200 + 1050 = 5760 = 0 So, A– 1 exists. On finding adjoint of matrix A,



Question 12. If img then, find A– 1 and solve the following system of linear equations : x – 2y = 10, 2x + y + 3x = 8, – 2y + z = 7 Solutin: Given



= 1(1 + 6) + 2(2 – 0) + 0(- 4 – 0) = 7 + 4 + 0 |A| = 11 ≠ 0 So, A– 1 exists. On finding adjoint of A,

Matrix formed by adjoint of A,



Now, given equations x – 2y = 10 2x + y + 3z = 8 – 2y + z = 7



In matrix form

Question 13. Find the product of matrices

with the help of it, solve the following system of linear equations : x – y + z = 4, x – 2y – 2z = 9, 2x + y + 3z = 1 Solution: Let



Matrix form of equations



Question 14. Find the inverse matrix of matrix

and with the help of it, solve the following system of equations :

Solution:

= 1(1 + 3) + 1(2 + 3) + 1(2 – 1) = 4 + 5 + 1 = 10 ≠ 0 So, A– 1 exists. On finding adjoint of A,



Matrix form of equations

Question 15. If (x1, y1), (x2, y2), (x3, y3) are vertices and a is side of an equilateral triangle respectively then prove that

Solution:



Rajasthan Board RBSE Class 12 Maths Chapter 5 Inverse of a Matrix and Linear Equations Miscellaneous Exercise Question 1.

If A = , then find A-1. Solution:

So, A-1 exists. On finding adjoint of A, F11 = 3, F12 = – 2, F21 = 1, F22 = 1



Matrix formed by adjoint A,

Question 2.

If , then find A-1. Solution:

So, A-1exists. On finding adjoint of A,



Question 3.

If matrix is a singular matrix, then find x. Solution:

⇒ (- 6 – 2) + 2(- 3 – x) + 3(2 – 2x) = 0 ⇒ – 8 – 6 – 2x + 6 – 6x = 0 ⇒ – 8 – 8x = 0

⇒ x =

⇒ x = -1 ⇒ Hence, x = -1 Question 4. Use Cramer’s rule to solve the following system of equations : (i) 2x – y = 17 3x + 5y = 6 (ii) 3x + ay = 4 2x + ay= 2, a ≠ 0 (iii) x + 2y + 37 = 6 2x + 4y + z = 7 3x + 2y + 9z = 14 Solution: (i) Given, 2x – y = 17 3x + 5y = 6

∵ ∆ ≠ 0, ∆1 ≠ 0 and ∆2 ≠ 0 ∴ Solution will be unique. By using Cramer’s rule



Hence, solution is x = 7, y = -3 (ii) Given, 3x + ay = 4 2a + ay = 2, a ≠ 0

∵ ∆ ≠ 0, ∆1 ≠ 0 and ∆2 ≠ 0 ∴ Solution will be unique. Using Cramer’s rule

(iii) Given, x + 2y + 2x = 6 2x + 4y + z = 7 3x + 2y + 92 = 14



= 1(56 – 14) – 2(28 – 21) + 6(4 – 12) = 42 – 12 – 48 = -20 ∵ ∆ ≠ 0, ∆1 ≠ 0 , ∆2 ≠ 0 and ∆3 ≠ 0

∴ Solution will be unique. Using Cramer’s rule,

Hence, solution is x = 1, y = 1, z = 1. Question 5. Use Cramer’s rule and prove that following system of equations are inconsistent : (i) 2x – y = 5 4x – 2y = 7 (ii) x + y + z = 1 x + 2y + 37 = 2 3x + 4y + 5y = 3



Solution: (i) Given, 2x – y = 5 4x – 2y = 7

Hence, system of equations is inconsistent. (ii) Given, x + y + z = 1 x + 2y + 3z = 2 3x + 4y + 3z = 3

∵ ∆ ≠ 0, ∆1 ≠ 0 , ∆2 ≠ 0 and ∆3 ≠ 0 Hence, system of equations is inconsistent.



Question 6. Find a matrix A of order 2, where

Solution:

Then, AB = C ABB-1 = CB-1 AI = CB-1 [∵ BB-1 = I] A = CB-1 …….(i) [∵ AI = A]

So, B-1 exists. On finding adjoint of B, F11 = 4, F12 = -1, F21 = 2, F22 = 1 Matrix formed by adjoint of B,

From equation (i), A = C.B-1

Question 7.

If then prove that A2 + 4A – 42I = 0, also find A-1. Solution:



Question 8.

If then prove that A-1 = A. Solution:

So, A-1exists. On finding adjoint of A, F11 = – 2, F12 = – 5, F21 = – 3, F22 = 2 Matrix formed by adjoint of A,

Question 9.

If , then find A-1 and show that A-1.A= I3. Solution:



= 1(16 – 9) – 3(4 – 3) + 3(3 – 4) = 7 – 3 – 3 |A|= 1 ≠ 0 So, A-1 exists. On finding adjoint of A,




Again to prove that

Hence, A-1A = I3. Question 10.

If , then prove that A2 – 4A – 5I = 0, also find A-1. Solution:



Question 11. Use matrix method to solve following system of equations : (i) 5x – 7y = 2 7x – 5y = 3 (ii) 3x + y + z = 3 2x – y – z = 2 – x – y + z = 1 (iii) x + 2y – 2z + 5 = 0 – x + 3y + 4 = 0 – 2y + 7 – 4 = 0 Solution: (i) Given equations 5x – 7y = 2 7x – 5y = 3 In matrix form AX= B ….(i)…(i)



Determinant of matrix A,

= – 25 + 49 = 24 ≠ 0 So, A-1 exists. On finding adjoint of matrix A F11 = – 5, F12 = – 7, F21 = 7, F22 = 5 Matrix formed by adjoint of A

(ii) Given equations 3x + y + z = 3 2x – y – z = 2 – x – y + z = 1 In matrix form AX = B ….(i)



= 3(-1 – 1) – 1(2 – 1) + 1(-2 – 1) = -6 – 1 – 3 = -10 ≠ 0 So, A-1 exists. On finding adjoint of A,




Hence, x = 1, y = -1, z = 1 (iii) Given equations, x + 2y – 2z + 5 = 0 or x + 2y – 2z = – 5 -x + 3y + 4 = 0 or – x + 3y = – 4 – 2y +z – 4 = 0 or – 2y + z = 4 In matrix form, AX= B ….. (i)

= 1(3 – 0) – 2(-1 – 0) – 2(2 – 0)



= 3 + 2 – 4 = 1 ≠ 0 So, A-1exists. On finding adjoint of A,




Hence, x = 1, y = – 1, z = 2. Question 12. Find area of ∆ABC, if vertices are : (i) A(- 3, 5), B(3, – 6), C(7, 2) (ii) A(2, 7), B(2, 2), C(10, 8) Solution: (i) Vertices of ∆ABC A = (-3,5), B = (3, -6), C = (7, 2) Area of ∆ABC



(ii) Vertices of ∆ABC, A = (2, 7), B = (2, 2), C = (10,8) Area of ∆ABC

Question 13. If points (2, – 3), (λ, – 2) and (0, 5) are collinear, then find λ. Solution: Given points (2, – 3), (λ, – 2) and (0, 5) are collinear, then

⇒ 2(-2 – 5) + 3(λ – 0).+ 1(5λ – 0) = 0 ⇒ – 14 + 3λ + 5λ = 0

⇒ 8λ = 14

⇒ λ = = Question 14.

Find matrix A, if Solution:

Given, equation BAC = I ……(i)



On finding adjoint of C, F11 = 5, F12 = – 3, F21 = – 7, F22 = 4 Matrix formed by adjoint of C,

From equation (i) BAC = I ⇒ B-1(BAC)C-1 = B-1C-1 ⇒ (B-1B)A(CC-1) = B-1C-1 ⇒ IAI = B-1C-1 ⇒ (IA)I = B-1C-1

⇒ AI = B-1C-1 ⇒ A = B-1C-1



Question 15.

If , then find A-1 and solve the following system of equations, using it : x + y + z = 2, x + 2y – 37 = 13, 2x – y + 37 = -7. Solution: Given

= 1(6 – 3) – 1(3+6) + 1(- 1 – 4) = 3 – 9 – 5 = -11 ≠ 0 so, A-1 exists. On finding adjoint of A,




Given equations,



x + y + z = 2 x + 2y – 3z = 13 2x – y + 3z = -7 In matrix form,

Question 16.

If , then find A-1 and show that aA-1 = (a2 + bc + 1)I – aA. Solution: Given,

So, A-1 exists.



On finding adjoint of A,

Question 17. Using determinant solve the following system of equations: x + y + z = 1 ax + by + cz = k



a2x + b2y + c2z = k2 Solution: Given equations, x + y + z = 1 ax + by + cz = k a2x + b2y + c2z = k2



Question 18.

If , then find A-1 and solve the following system of equations using it : x + 2y – 3z = – 4, 2x + 3y + 2z = 2, 3x – 3y – 4z = 11



Solution:

= 1(- 12 + 6) – 26 – 8 – 6) – 3(- 6 – 9) = -6 + 28 + 45 = 67 ≠ 0 So, A-1 exists. On finding adjoint of A,



Given equations, x + 2y – 32 = -4 2x + 3y + 2z = 2 3x – 3y – 4z = 11



In matrix form,

Question 19.

, then find the value of X. Solution:

Then, AX = B X = A-1B …..(i) Determinants of matrix A,

So, A-1 exists. F11 = – 2, F12 = – 3, F21 = 4, F22 = 1




Question 20. Find the value of a and b, if following system of equations has infinite solutions : 2x + y + az = 4 5x – 2y + z = -2 5x – 5y + z = -2 Solution: Given equations, 2x + y + az = 4 : bx – 2y + z = -2 5x – 5y + z = -2 ∵ For infinite solution ∆ = 0, ∆1 = 0, ∆2 = 0 and ∆z = 0.

⇒ 2(- 2 + 5) – 1(b – 5) + d(-5b + 10) = 0 ⇒ 6 – b + 5 – 5ab + 10a = 0 ⇒ 10a – b – 5ab = -11 …(1)



4(-2 + 5) – 1(-2+2) + a(10 – 4) = 0 ⇒ 12 + 6a=0

⇒ 16a = -12

a = = -2 From (i), 10a – b – 5ab = -11 ⇒ 10(-2) – b – 5(-2)b = -11 ⇒ -20 – b + 10b = -11

⇒ 9a = 9

⇒ b = Hence, a = -2, b = 1.

Rajasthan Board RBSE Class 12 Maths Chapter 6 Continuity and Differentiability Ex 6.1 Question 1. Examine the following functions for continuity :



Solution: (a) Given function.

Left hand limit

Right hand limit



Left hand limit

Right hand limit

Hence, given function is not continuous at x = 0.



and at x = 3 f (3) = 1 + x ⇒ 1 + 3 = 4 ∴ f (3 – 0) = f(3 + 0) = f(3) = 4 Hence, given function is continuous at x = 3. (d) Given function,

Left hand limit f(0 – 0) = limh→0 f(0 – h) = limh→0 sin (0 – h) = limh→0 sin (- h) = limh→0 (-sin h) [∵ sin (-θ) = -sin θ] = 0 Right hand limit

Hence, given function is continuous at x = 0. (e) Given function,



Left hand limit

Right hand limit

Hence, given function is not continuous at x = 0. (f) Given function,

Left hand limit



Right hand limit

Hence, given function is not continuous at x = a.



(g) Given function,

Left hand limit

Right hand limit

Hence, given function is continuous at x = a. Question 2. Test the continuity of f(x) at x = 3 if f(x) = x – [x]. Solution: Given function f(x) = x – [x], at x = 3 Left hand limit f (3 – 0) = limh→0 f(3 – h) = limh→0 (3 – h) – [3 – h] = 3 – 2



= 1 [∵ 2 is just before greatest integer 3] Right hand limit f (3 + 0) = limh→0 f(3 + h) = limh→0 (3 + h) – [3 + h] = 3 – 3 = 0 ∴ f (3 – 0) ≠ f (3 + 0) Hence, given function at not continuous at x = 3. Question 3.

continuous at x = 2, then find k. Solution : Left hand limit



Right hand limit

Question 4.

the continuity of f (x) is closed interval [- 1, 2]. Solution: Here, we will test the continuity of function at x = 0 and 0 ∈ [-1,2]. Left hand limit f(0 – 0) = limh→0 f(0 – h) = limh→0 (0 – h)2 = limh→0 h2 = 0 Right hand limit f (0 + 0) = limh→0 f(0 + h) = limh→0 4(0 + h) – 3 = limh→0 Ah – 3 = 0 – 3 = – 3 ∴ f (0 – 0) ≠ f(0 + 0) LHL ≠ RHL So, function is not continuous at x = 0 and x ∈ [- 1, 2] At x = 1 Left hand limit



f(1 – h) = limh→0 4(1 – h) – 3 = limh→0 4 – 3 – Ah =4 – 3 – 0 = 1 Right hand limit f(1 + h) = limh→0 5(1 + h)2 – 4 (1 + h) = limh→0 5(1 + h2 + 2h) – (4 + Ah) = limh→0 5h2 +10h + 5 – 4 – 4h = 5 × 0 + 10 × 0 + 1 – 4(0) = 1 Value of function at x = 1 f(1) = 4 × 1 – 3 = 1 ∵ limh→0 f(1 – A) = f(1) = limh→0 f(1 + h) ∴Function is continuous at x = 1. Hence, function is not continuous in interval [-1, 2],

Rajasthan Board RBSE Class 12 Maths Chapter 6 Continuity and Differentiability Ex 6.2 Question 1. Show that following functions are differentiable for every value of x : (i) Identity function, f (x) = x (ii) Constant function,f (x) = c, where c is a constant (iii) f(x) = ex (iv) f(x) = sin x. Solution: (i) Given, f (x) = x, (identity function) where, x ∈ R Let a be arbitrary constant, then At x = a, Left hand derivative of f (x)



So, for every x, identity function f(x) is differentiable. (ii) Given, constant function f(x) = c, where c is constant. Domain of function f(x) is set of real numbers (R). Let a be any arbitrary real number, then At x = a, Left hand derivative of f (x)

So, for every x, identity function f(x) is differentiable. (iii) Given function f (x) = ex, where x ∈ R Let a be an arbitrary constant then at x = a, Left hand derivative of f (x)



Again, at x = a, Right hand derivative of f (x)

Hence,f(x) = ex is differentiable for every x. (iv) Given function f(x) = sin x, where x ∈ R Let a be any arbitrary real number. At x = a, Left hand derivative of (x)



Again, at x = a, Right hand derivative of f (x)

Hence, for every x, function will be differentiable. Question 2. Show that the function f (x) = | x | is not differentiable at x = 0.



Solution: For differentiability at x = 0. Left hand derivative

Right hand derivative

Hence, f(x) is not differentiable at x = 0. Question 3. Examine for differentiability of function, f (x) = | x – 1 | + | x | at x = 0, 1. Solution: Given function can be written as

For differentiability at x = 0,



Left hand derivative


So, function f(x) is not differentiable at x = 0 For differentiability at x = 1 Left hand derivative




So, function f(x) is not differentiable at x = 1. Hence, function f(x) is not differentiable at x = 0 and x = 1. Question 4. Examine the function for differentiability in interval [0, 2], if f(x) = | x – 1 | + | x – 2 | Solution: Given function can be written as

Here, we will test the differentiability at x = 1. Since 1 ∈ [0, 2] For differentiability at x = 1 Left hand derivative




Hence, f (x) is not differentiable at x = 1. So, it is not differentiable in interval [0,2]. Question 5. Examine the function for differentiability at

Solution: For differentiability at x = 0, Left hand derivative


Hence, function is differentiable at x = 0. Question 6. Examine the function f(x) for differentiability at x = 0, if



Solution: For differentiability at x = 0, Left hand derivative




Hence, function is not differentiable at x = 0. Question 7. Show that the following function

(a) Continuous at x = 0 if m > 0 (b) Differentiable at x = 0 if m > 1. Solution: (a) Continuity at x = 0, (i) At x = 0,f (0) = 0 (ii) At x = 0, Left hand limit

(iii) At x = 0, Right hand limit



At x = 0, function f(x) will be continuous if (i) and (ii) will be zero.

Since, -1 < cos ( ) < 1 Therefore, both limits will be zero if m > 0 Hence, function f(x) will be continue at x = 0 if m > 0. (b) Differentiability at x = 0 Left hand derivative


Given, f(x) is differentiable at x = 0, then f'(0 – 0) = f'(0 + 0)



Which is possible only when m – 1 > 0 or m > 1 Hence, given function f(x) is differentiable at x = 0 if m >1. Question 8. Examine the function f(x) for differentiability at x = 0 if

Solution: At x = 0, Left hand derivative


Hence, function is not differentiable at x = 0. Question 9. Examine the function f(x) for differentiability



Solution: At x = 0, Left hand derivative


Hence, given function is not differentiable at x = 0. Question 10.

Examine the function f(x) for differentiability at x = , if



Solution:

At x = , Left hand derivative




Question 11. Find the value of m and n, if function

is differentiable at every point. Solution: Given that at x = 1, f(x) is differentiable. We know that every differentiable function is continuous. So, at x = 1, function is continuous also. Left hand limit



Right hand limit

Left hand derivative




∴ Function is differentiable at x = 1 then f'(1 – 0) = f'(1 + 0) 5 = n ⇒ n = 5 From equation (i), m – 5 = – 2 ⇒ m = – 2 + 5 ⇒ m = 3 Hence, m = 3 and n = 5

Rajasthan Board RBSE Class 12 Maths Chapter 6 Continuity and Differentiability Miscellaneous Exercise Question 1.

If function f (x) = is continuous at x = 3, then value of (3) will be : (a) 6 (b) 3 (c) 1 (d) 0 Solution:



Function is continuous at x = 3, so f(3) = f(3 + 0) f(3) = 6 Hence, option (a) is correct. Question 2. If function f(x)

is continuous at x = 2 then value of m will be : (a) 3 (b) 1/3 (c) 1 (d) 0 Solution:

Question 3.

is continuous at x = 0, then the value of k will be : (a) 0 (b) m + n (c) m – n (d) m.n Solution:



Question 4.

is continuous at x = 3 then the value of λ will be : (a) 4 (b) 3 (c) 2 (d) 1 Solution: Left hand limit f (3 – 0) = limh→0 f (3 – h) = limh→0 (3 – h) + λ = (3 – 0) + λ = 3 + λ ∵ At x = 3, function is continuous then f (3 – 0) = f (3) 3 + λ = 4 ⇒ λ = 4 – 3 λ = 1 Hence, option (d) is correct.



Question 5.

If f(x) = cot x is discontinuous at x = , then : (a) n ∈ Z

(b) n ∈ N

(c) ∈Z (d) Only n = 0 Solution:

∵ Function is not continuous at x =

Question 6. Function f(x) = x | x | is differentiable in interval: (a) (0, ∞) (b) (-∞, 0) (c) (-∞, 0) (d) (-∞, 0) ∩ (0, ∞) Solution: Given function can be written as



Hence, function is differentiable at x = 0, so function is differentiable in (-∞, ∞). Hence, option (b) correct. Question 7. Which of the following function is not differentiable at x = 0 : (a) x | x | (b) tan x (c) e-x (d) x + | x | Solution:



Question 8. Function f(x)

then value of left hand derivative of f(x) at x = 2 is : (a) -1 (b) 1 (c) -2 (d) 2



Solution:

Question 9. Function f(x) = [x] is not differentiable : (a) at every integer (b) at every rational number (c) at x = 0 (d) everywhere Solution: [x] is not continuous at all integers. We know that at every intiger discontinuous function is not differentiable. Hence, correct option is (a). Question 10. Function

is differentiable at x = 0, then right derivative of f(x) at x = 0 is : (a) -1 (b) 1 (c) 0 (d) infinite Solution:



Question 11. Examine the function f (x) = | sin x | + | cos x | + | x |, ∀ X ∈ R for continuity. Solution:



Let x ∈ c be any arbitrary constant, then at x = c for continuity of function f(x).

Question 12. If function,

is continuous at x = 0, then find m. Solution:



At x = 0,

Question 13. Find m and n if following function is continuous

Solution: For continuity at x = 2, Left hand limit f (2 – 0) = limh→0 f(2 – A) = limh→0 [(2 – h)2 + m(2 – h) + n] = (2 – 0)2 + m(2 – 0) + n = 4 + 2m + n Right hand limit f(2 + 0) = limh→0 f(2 + h) = limh→0 [4(2 + h) – 1 ] = 4(2 + 0) – 1 = 8 – 1 = 7 At x = 2, value of f (x), f(2) = 4 × 2 – 1 = 8 – 1 = 7 Given function is continuous at x = 2, then f(2 – 0) = f(2 + 0) = f(2)



4 + 2m + n = 7 = 7 ⇒ 4 + 2m + n = 7 ⇒ 2m + n = 7 – 4 or 2m + n = 3 For continuity at x = 4, Left hand limit f (4 – 0) = limh→0 f (4-h) = limh→0 [4(4 – h) – 1] = 4 (4 – 0) – 1 = 16 – 1 = 15 Right hand limit f (4 + 0) = limh→0 f (4 – h) = limh→0 [m(4 + h)2 + 17n] = m (4 + 0)2 + 17n = 16m + 17n At x = 4, value of function f(4) = 4 × 4 – 1 = 16 – 1 = 15 Given function is continuous at x = 4, f (4 – 0) = f (4 + 0) = f (4) 15 = 16m + 17n = 15 ⇒ 16m + 17n = 15 ……(ii) Putting n = 3 – 2m from (i) in 16m + 17(3 – 2m) = 15 16m + 51 – 34m = 15 – 18m = 15 – 51 – 18m = – 36 m = 2 From equation (i), 2 × 2 + n = 3 n = 3 – 4 Hence, m = 2,n = – 1. Question 14. Examine the function

for continuity at x = 0. Solution :



Continuity at x = 0,

Question 15. Examine the function

for continuity at x = 1 and 3. Solution: For continuity at x = 1



= | 1 + 0 – 3 | = | – 2 | = 2 Value of function f(x) at x = 1, f(1) = | 1 – 3 | = | -2 | = 2 ∴ f (1 – 0) = f(1 + 0) = f (1) = 2 So, f(x) is continue at f (x), x = 1 For continuity at x = 3 Left hand limit f (3 – 0) = limh→0 f (3 – h) = limh→0 – {(3-A)-3} = limh→0 (3 – 3 + h) = 0 Right hand limit f(3 + 0) = limh→0 f(3 + h) = limh→0 (3 + h – 3) = 0 f(3) = 3 – 3 = 0 ∴ f (3 – 0) = f(3 + 0) = f(0) = o Hence, function is continue at x = 3. Question 16. Find a, b, c if the function



is continuous at x = 0. Solution: At x = 0, given function is continuous then,



Question 17. Test the continuity of the function

Solution:

At x = , for continuity



Question 18. Test the continuity of the function f(x) = | x | + | x – 1 | in interval [- 1, 2]. Solution: Given, f(x) = | x | + | x – 1 | It can be written as



Here, we will test of continuity at x = 0 and x = 1 only. Test of continuity at x = 0 Here, f(0) = 1 Left hand limit f(0 – 0) = limh→0 f(0 – h) = limh→0 1 – 2(0 – h) = 1 Right hand limit f(0 + 0) = limh→0 f(0 + h) = 1 ∴ f(0) = f (0 – c) = f (0 + 0) So, at x = 0,f(x) is continuous Now, at x = 1 test of continuity Here, f(1) = 2 × 1 – 1 = 1 Left hand limit f (1 – 0) = limh→0 f(1 – h) = limh→0 1 Right hand limit f (1 + 0) = limh→0 f(1 + h) = limh→0 2(1 + h) – 1 = 2(1 + 0) – 1 = 2 – 1 = 1 ∴ f (1) = f(1 – 0) = f(1 + 0) So, at x = 1, f(x) is continuous. ∴ Function is continuous at x = 0 and x = 1 Hence, function is continuous in interval [-1,2]. Question 19. Find the value of f(0) if function

is continuous at x = 0. Solution:



Function is continuous at x = 0

Question 20. Examine the function fix) for continuity at

Solution:



Question 21. For which value of x, f(x) = sin x is differentiable. Solution: Given function f(x) = sin x Domain (f) = R Let a ∈ R be any arbitrary real number. Then at x = a,



Question 22. Examine the function f(x) for differentiablity

Solution:



∵ f’ (0 – 0) = f'(0 + 0) Hence, At x = 0 function is differentiable, ∵ x ∈ R, then for every x ∈ R function will be differentiable and f'(0) = 0. Question 23. Examine the function for differentiability at x = a if

Solution:



Question 24. Prove that the function

is not differentiable at x = 1. Solution:



Question 25. Examine the function for differentiability at x = 0, if



Solution:

Question 26. Show that the function

is differentiable at x = 0 Solution:



Question 27. Check the differentiablity of function f(JC) = |x – 2| + 2 | x – 3 | in the interval [1, 3]. Solution: Given function can be written as :



Hence, at x = 2, function is not differentiable. So, we can say that this function is not differentiable in interval [1, 3], Question 28. If flnction f(x) = x3 is differentiable at x = 2, then find f’ (2). Solution: At, x = 2,



Hence, at x = 2, function is differentiable, and f’ (2) = 12. Question 29. Show that the greatest integer function f(x) = [x] is not differentiable at x = 2 Solution:



At x = 2,

Question 30.



Solution:

Rajasthan Board RBSE Class 12 Maths Chapter 7 Differentiation Ex 7.1 Find the derivative of following functions w.r.t. x : Question 1. sin (x2) Solution: Let y = sin (x2) DifF. w.r.t.x,

Question 2. tan (2x + 3) Solution: Let y = tan (2x + 3) DifF. w.r.t.x,

= (tan(2x + 3))

= sec2 (2x + 3) (2x + 3) = sec2 (2x + 3).(2 × 1 + 0) = 2 sec2 (2x + 3). Question 3. sin {cos (x2)} Solution: Let y = sin {cos (x2)} DifF. w.r.t. x,

= sin (cos(x2))



cos (cos x2). (- sin x2) x2 = -cos (cos x2) sin (x2). 2x = -2x sin (x2) cos (cos x2) Question 4.

Solution:

Question 5.

Solution:



Question 6. sin x° Solution:



Question 7.

Solution:

Question 8. sec x° Solution:



Question 9.



Solution:

Question 10.

Solution:



Question 11.

Solution:



Question 12.



Solution:

Question 13. a tan3x Solution:

Question 14. loge (sec x + tan x) Solution :



Question 15. sin3 x. sin 3x. Solution:

= 12 sin3 x . cos x – 24 sin5 x cos x = 12 sin3 x cos x (1 – 2 sin2 x) = 12 sin3 x cos x . cos 2x = 6 sin2 x . 2 sin x . cos x . cos 2x = 6 sin2 x . sin 2x . cos 2x = 3 sin2 x . 2 sin 2x . cos 2x = 3 sin2 x sin 2(2x) = 3 sin2 x . sin 4x.

Rajasthan Board RBSE Class 12 Maths Chapter 7 Differentiation Ex 7.2 Find derivative of following functions w.r.t. x :



Question 1.

Solution:



Question 2.



Solution:



Question 3.

Solution:



Question 4.

Solution:



Question 5.

Solution:



Question 6.

Solution:



Question 7.

Solution:



Rajasthan Board RBSE Class 12 Maths Chapter 7 Differentiation Ex 7.3 Find of following functions : Question 1. (i) 2x + 3y = sin y (ii) x2 + xy + y2 = 200 Solution: (i) 2x + 3y- sin y Diff. w.r.t. x on both sides,



(ii) x2 + xy + y2 = 200 Diff. w.r.t. x on both sides,

Question 2.

(i) + = (ii) tan (x + y) + tan (x – y) = 4 Solution:

(i) + = DifF. w.r.t. x on both sides,

(ii) tan (x + y) + tan (x – y) = 4



DifF. w.r.t. x on both sides,

Question 3. (i) sin x + 2 cos2 y + xy = 0

(ii)x + y = 1 Solution: (i) sin x + 2 cos2 y + xy = 0 DifF. w.r.t. x on both sides,



(ii) x + y = 1



DifF. w.r.t. x on both sides,

Question 4. (i) (x2 + y2)2 = xy

(ii) sin (xy) + = x2 – y Solution: (i) (x2 + y2) = xy Diff. w.r.t. x on both sides,



(ii) sin (xy) + = x2 – y



Diff. w.r.t. x on both sides,

Question 5. (i) x3 + y3 = 3axy (ii) xy + yx = ab Solution: (i) x3 + y3 = 3axy DifF. w.r.t. x on both sides,



(ii) xy + yx = ab Solution : Let u = xy and v = yx Then, u + v = ab Again, u = xy and v = yx Taking log on both sides, log u = y log x and log v = x log y DifF. w.r.t. x,



Question 6. (i) y = xy (ii) xa . yb = (x – y)a+b Solution: (i) y = xy Taking log on both sides, log y = log xy = y log x Diff. w.r.t. x on both sides,



Question 7.

Solution:



Question 8.

Solution:



Question 9.

Solution:



Question 10.

Solution:



Rajasthan Board RBSE Class 12 Maths Chapter 7 Differentiation Ex 7.4 Find , if Question 1. (i) x = a sec t, y = b tan t (ii) x = log t + sin t, y = et + cos t Solution: (i) x = a sec t, y = b tan t x = a sec t Diff. w.r.t. t on both sides,



= (a sec t) = a sec t tan t and y = b tan t Diff. w.r.t. t on both sides,

(ii) x = log t + sin t, y = et + cos t x = log t + sin t Diff. w.r.t. t on both sides,



Question 2. (i) x = log t, y = et + cos t (ii) x = a cos θ, y = b sin θ Solution: (i) x= log t, y = et + cos t ∵ x = log t



Diff. w.r.t. t on both sides,

(ii) x = a cos θ, y = b sin θ ∵ x = a cos θ Diff. w.r.t. θ on both sides,

Question 3. (i) x = cos θ – cos 2θ, y = sin θ – sin 2θ (ii) x = a (θ – sin θ), y = a (1 + cos θ)



Solution: (i) x = cos θ – cos 2θ y = sin θ – sin 2θ x = cos θ – cos 2θ ∵ Diff. w.r.t. θ on both sides,

(ii) x = a(θ – sin θ), y = a(1 + cos θ) ∵ x = (θ – sin θ) Diff. w.r.t. θ on both sides,



Question 4.

Solution:



Question 5.

Solution:



Question 6.

If x3 + y3 = t – and x6 + y6 = t2 + then prove that x4y2 = 1 Solution:

Rajasthan Board RBSE Class 12 Maths Chapter 7 Differentiation Ex 7.5 Question 1.

(i) y = x3 + tan x



(ii) y = x2 + 3x + 2 (iii) y = x cos x (iv) y = 2 sin x + 3 cos x (v) y = e-x cos x (vi) y = a sin x – b cos x Solution: (i) y = x3 + tan x Diff. w.r.t. x on both

Again diff. w.r.t. x on both sides

(ii) y = x2 + 3x + 2 Diff. w.r.t. x on both sides,



(iii) y = x cos x Diff. w.r.t. x on both sides,

(iv) y = 2 sin x + 3 cos x Diff. w.r.t. x on both sides,



(v) y = e-ex cos x Diff. w.r.t. x on both sides,

(vi) y = a sin x – b cos x Diff. w.r.t. x on both sides,

Question 2. If y = a sin x + b cos x, then prove that:

Solution: Given y = a sin x + b cos x …..(i)




Question 3. If y = sec x + tan x, then prove that:

Solution: Given y = sec x + tan x Diff. w.r.t. x on both sides,



Question 4. If.y = a cos nx + b sin nx, then prove that:



Solution: Given y = A cos nx + b sin nx Diff. w.r.t. x on both sides,

Question 5.

If x = a cos3 θ, y = a sin3 θ, then find at θ = Solution: Given y = a sin3 θ and x = a cos3 θ Diff. w.r.t. θ on both sides,



Question 6. If x3 + y3 – 3axy = 0, then prove that

Solution: Given,



x3 + y3 – 3axy = 0 Diff. w.r.t x on both sides,

Question 7. If y = sin-1 x, then prove that:

Solution: Given y = sin-1 x



Diff. w.r.t. x on both sides

Question 8. If y = (sin-1 x)2, then prove that:

Solution: Given y = (sin-1 x)2




Rajasthan Board RBSE Class 12 Maths Chapter 7 Differentiation Ex 7.6 Question 1. Verify the Rolle’s theorem, for the following functions :



Solution: (i) Given function

∵ f(x) is polynomial in x.

∴ It is differentiable and continuous every-where.

Hence, f(x) satisfies Rolle’s theorem in interval [π/4, 5π/4] From(1), f'(x) = ex (cos x + sin x) + (sin x – cos x).ex f'(x) = ex (cos x + sin x + sin x – cos x) Similarly, ec 2 sin c = 0 ⇒ 2sinc = 0 ⇒ sin c = 0 ⇒ c = π c = π ∈ (π/4, 5π/4),f(c) = 0 ∴ Hence, Rolle’s theorem satisfied. (ii) f(x) = (x – a)m (x – b)n, x ∈ [a, b], m, n ∈ N Here, (x – a)m and (x – b)n both are polynomial. On its multiplication a polynomial of power (m + n) is obtained. A polynomial function is continuous everywhere always. So, f(x) is continuous in [a, b]. Polynomial function is also differentiable. ∴ f'(x) = m(x- a)m-1 .(x – b)n + n(x – d)m (x – b)n-1 = (x – a)m-1 (x – b)n-1 × [m(x – b) + n(x – a)]



= (x – a)m-1 (x – b)n-1 × [(m + n) x – mb – na] Here,f'(x) exists. ∴ f(x) is differentiable in interval (a, b). Again f(a) = (a – a)m (a – b)n = 0 f (b) = (b – a)m (b – b)m-n = 0 ∴ f(a) = f(b) = 0 Hence, Rolle’s theorem satisfies,then in interval (a, b) at least any one point is such that f'(c) = 0.



Function is not diffenentiable at x = 0. So, Rolle’s theorem does not satisfied. (iv) Given function f(x) = x2 + 2x – 8, x ∈ [- 4, 2] Here, it is clear that function f(x) is continuous in interval [- 4, 2] and f’ (x) is finite and exist at every point of interval, (- 4, 2), hence given function is differentiable in interval (-4, 2). ∵ f (- 4) = 0 = f (2) ⇒ f (- 4) = f(2) So, from above f'(x) satisfies all the three conditions of Rolle’s theorem in given interval.



(vi) Given function f(x) = [x], x ∈ [- 2, 2] ∵ f(x) is not continuous in [- 2, 2] as greatest integer function is neither continuous nor differentiable. Hence, Rolle’s theorem does not satisfied. Question 2. Prove Rolle’s theorem for following functions :



Solution: (i) Given function f(x) = x2 + 5x + 6, x ∈ [- 3, – 2] ∵ f(x) = x2 + 5x + 6 is a polynomial function. Hence, it is continuous in [- 3, – 2], Now f'(x) = 2x + 5, exists ∀ x ∈ (-3,-2) ∴ f (x) is differentiable in (- 3, – 2)

∵ f (- 3) = 0 = f (- 2) ⇒ f (- 3) = f (- 2) All the condition of Rolle’s theorem is satisfied, then a point c ∈ (-3,-2) exists in such a way that f'(c) = 0.

All the conditions of Rolle’s theorem satisfied, then a point c in (0,π) exists in such a way that f'(c) = 0.



(iii) Given function

is finite and exist at every point in interval (0, 1). Hence,f(x) is differentiable in (0, 1). ∵ f(0) = 0 = f(1)

⇒ f (0) = f (1) So, f (x) satisfies all the conditions of Rolle’s theorem. Hence, f'(c) = 0

(iv) Given function f (x) = cos 2x, x ∈ [0, π] ⇒ f (x) = cos 2x is defined in [0,π]. ∵ cosine is continuous in its domain. So, it is continuous in [0, π]. Then, f'(x) = – 2 sin 2x exists,



where x ∈ (0, π) ∴ f (x) is differentiable in (0, π). Now f(0) = cos 0 = 1 and f(π) = cos 2π = 1 ∴ f(0) = f(π) = 1 Thus, all the conditions of Rolle’s theorem, are satisfied then at least one point c such that c ∈ (0, π) and f’ (c) = 0. f'(c) = 0 ∴ f'(c) = – 2 sin 2c = 0 ⇒ sin 2c = 0 ⇒ 2c = π c = π/2 is an element of (0,π)

∴ c = ∈ (0,π) such that f'(c) = 0

Thus for c = , Rolle’s theorem satisfied. Question 3. Verify the Lagrange’s mean value theorem for the following functions :

Solution:



Clearly, f(x) is continuous in interval [0, 2] and f'(x) is finite and exists. So,f(x) is differentiable in (0, 2). Hence f(x) satisfies both conditions of Langrange’s mean value theorem.



∵ c is an imaginary number. Hence, Langrange’s mean value theorem does not satisfied. (iii) Given function f(x) = x2 – 3x + 2, x ∈ [- 2, 3] Clearly,f(x) is continuous in interval [-2, 3] and f'(x) is finite and exists in (-2, 3). So, f(x) is differentiable in (- 2, 3). Hence f(x) satisfies both conditions of Langrange’s mean value theorem.



Clearly,f(x) is continuous in [ 1,4] and f'(x) is finite and exist in interval (1,4). Hence,f(x) is differentiable in (1, 4). f(x) satisfies both conditions of Langrange’s mean value theorem.

Thus, Langrange’s mean value theorem satisfied.

Rajasthan Board RBSE Class 12 Maths Chapter 7 Differentiation Miscellaneous Exercise Differentiate given functions from question 1 to 10 w.r.t. x : Question 1.

sin-1 (x ), 0 ≤ x ≤ 1. Solution:

Let y = sin-1 (x )




Question 2.

Solution:



Question 3.

Solution:

Question 4. x3,ex,sin x.



Solution: Let y = x3.ex.sin x Diff. w.r.t. x on both sides,

Question 5.

log ( ) Solution:

Question 6. (x log x)log x, Solution: Let y = (x log x)log x Taking log of both sides, log y = log (x log x)log x ⇒ log y = log x.log (x log x)




Question 7.

Solution:



Question 8. xx2 – 3 + (x – 3)x2, x > 3 Solution:



Question 9. y – 12(1 – cos t),x = 10(t – sin t).



Solution: Diff. w.r.t. x on both sides,

Question 10.

sin-1 x + sin-1 Solution:

Question 11. If cos-1

a, then prove that =



Solution:

Question 12. If, y = x sin (a + y) then prove

Solution:



Question 13.

If y = (sin x – cos x)(sin x- cos x) then find Solution: Let y – (sin x – cos x)(sin x – cos) Taking log on both sides, log y = (sin x – cos x) log (sin x – cos x)




Question 14. If y = sin (sin x), then show that:

Solution:



Question 15. (i) If y = eax sin bx, then show that

Solution:



Question 16. Prove the Rolle’s theorem for the following :

Solution:



which is defined and exist at every point in interval (0, 2). So, function is differentiable in interval (0, 2). ∵ f(0) = 0 = f(2) ⇒ f(0) = f(2) Here,f(x) satisfies Rolle’s theorem in given interval.

(b) Given f(x) = (x – 1) (x – 3), x ∈ [1, 3] Here, f (x) is continuous in interval [1, 3] and f'(x) = 2x – 4, which is defined and exist at every point in interval (1, 3). So,f(x) is differentiable in interval (1, 3). ∵ f(1) = 0 = f(3) ⇒ f(1) = f(3) Here,f(x) satisfies Rolle’s theorem in given interval. Hence f'(c) = 0 f'(c) = 2c – 4 = 0 ⇒ 2c = 4 ⇒ c = 2 ∴ c = 2 ∈ (1,3) such that f'(c) = 0 Hence, Rolle’s theorem verified for c = 2. Question 17. Verfiy the Langrange’s theorem for the following:

Solution: (a) Given function f(x) = (x – 1) (x – 2) (x – 3), x ∈ [0, 4] ⇒ f(x) = x3 – 6x2 + 11x – 6, x ∈ [0, 4] Clearly,f(x) is continuous in intemval [0,4] and f'(x) is defined and exists in interval (0,4). Hence, function satisfies Lagrange’s mean value theorem.



Hence,f(x) is continuous and differentiable in interval (1, 3) excepting x = 2.



So, function f(x) is not differentiable at x = 2. Here condition for Lagrange’s theorem does not satisfied. Hence, Lagrange’s does not verified.

Rajasthan Board RBSE Class 12 Maths Chapter 8 Application of Derivatives Ex 8.1 Question 1. Find the rate of change of the area of a circle with respect to its radius r, when r = 3 cm and r = 4 cm. Solution:



Let, area of circle = A

Question 2.

A particle moves along y = x3 + 1. Find the points on the curve at which y – coordinate is changing twice as fast as the x – coordinate. Solution: Given equation

y = x3 + 1 …..(i) Let position of particle at time ’t’ is p(x, y) and hence p(x, y) is situated on curve (i).

y = x3 + 1

Question 3. A ladder 13 m long leans against a wall. The foot of the ladder is pulled along the ground away from the wall, at the rate of 1.5 m/sec. How fast is the angle θ between



the ladder and the ground changing when the foot of the ladder is 12 m away from the wall. Solution: Let AB is a ladder, its end ‘A’ is x distance far from wall and upper end is at y from the ground. Let slope between ladder and ground is θ.



Hence, angle between ground and ladder is decreasing at the rate of rad/s Question 4. An edge of a variable cube is increasing at the rate of 3 cm/sec. How is the volume of the cube increasing when the edge is 10 cm long ? Solution: Let at any time ‘t’ length of edge of cube is x and its volume is V, then V = x3 …..(i)



and this edge is increasing at the rate of 3 cm/s.

∴ = 3 cm/s. We have to find, rate of change of volume V When x = 10cm

Question 5. A Balloon which is always spherical, is being inflated by pumping in 900 cm3 of gas per second. Find the rate at which the radius of the balloon is increasing when the radius is 15 cm. Solution: Let at any time t, radius of balloon is r and its volume is V.

Question 6.

A balloon which is always spherical, has diameter (2x +1) Find rate of change of its volume w.r.t. x. Solution: Let volume of balloon is V. According to question,



Question 7. The total cost C(x) associated with the production of x units of a product is given by C(x) = 0.005x3 – 0.02x2 + 30x + 5000 Find the marginal cost when 3 units are produced. Solution: Let cost of production of x units is C(x), where C(x) = 0.005x3 – 0.02x2 + 30x + 5000 Marginal cost = MC



= 0.005 × 3x2 – 0.02 × 2x + 30 × 1 = 0.005 × 3x2 – 0.02 × 2x + 30 For x = 3, MC = 0.005 × 3 × (3)2 – o.02 × 2 × (3) + 30 = 0.005 × 27 – 0.02 × 6 + 30 = 0.135 – 0.12 + 30 = 30.015 or 30.02 (Approx.) Hence, marginal cost is ₹ 30.02 (Approx). Question 8. The radius of a soap bubble is increasing at the rate of 0.2 cm/sec. At what rate is the surface area of a bubble increasing when the radius is 7 cm and at what rate is the volume of that bubble increasing when the radius is 5 cm. Solution: Let radius of soap bubble is r and surface area is S. According to question,

Hence, rate of increasing surface area of soap bubble is 11.2π cm2/s. Again, let at any time t, radius of soap bubble is r and volume is V.




Question 9. Sand is pouring from a pipe at the rate of 12 cm3/sec. The falling sand forms a cone of the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is height of the sand cone increasing when the height is 4 cm ? Solution: Let at any time “t” volume of cone of sand is V, height h and radius r. According to question,



Question 10. The total revenue received from the sale of x units of a product is given by R(x) = 13x2 + 26x + 15 Find marginal cost when x = 15 Solution: Given, R(x) = 13x2 + 26x + 15

Marginal cost MR(x) = {R(x)}



= (13x2 + 26X + 15) = 26x + 26 Putting x = 15 Then MR(x) = 26 × 15 + 26 = 390 + 26 MR (7) =416 Hence, Marginal cost = ₹ 416

Rajasthan Board RBSE Class 12 Maths Chapter 8 Application of Derivatives Ex 8.2 Question 1. Prove that f(x) = x2 is increasing in interval (0, ∞) and decreasing in interval (-∞,0). Solution: Let x1, x2 ∈ [0,∞] is such that x1 < x2 ∴ x1 < x2 ⇒ x1

2< x1x2 …..(i)

and x1 < x2 ⇒ x1x2 < x22……(i)

From (i) and (ii), x1 < x2 ⇒ x1

2 < X22

⇒ f(x1) < f(x2) ∴ X1 < X2 ⇒ f(x1) < f(x2) where x1, x2 ∈ [0, ∞] Hence, f(x) is continuously increasing in [0, ∞) Again, let x1, x2 ∈ (-∞, 0) is such that x1 < x2 Then x1 < x2 ⇒ x1

2> x1x2 [∵ – 3 < – 2, (- 3) (- 3) = 9 (- 3) × (- 2) = 6 ∴ 9 > 6 x1

2 > x1 x2] Again x1 < x2 ⇒ x1x2 > x2

2 Again – 3 < – 2 (- 3) × (- 2) = 6 (- 2) × (- 2) = 4 6 > 4 ∴ x1x2 > x2

2] From (i) and (ii), x1 < x2 ⇒ X1

2 > x22

⇒ f(x1) > f(x2) ∴ x1 < x2 ⇒ f(x1) > f(x2) Hence, f(x) is continuously decreasing in (-∞, 0). Question 2. Prove that f(x) = ax, 0 < a < 1, is decreasing in R. Solution: Let x1,x2 ∈R is such that x1 < x2



Then, x1 < x2 ⇒ ax1 > ax2 [∵ 0 < a < 1 and x < x2 ⇒ ax1 > ax2] ⇒ f(x1) > f(x2)

∴ x1 < x2 ⇒ f(x1) > f(x2) ∀ x1, x2 ∈ R Hence, function f(x) = ax, 0 < a < 1, R is decreasing in R. Question 3.

f(x) = log sin x, x ∈ (0, ) Solution:

Hence, in interval (0, ) function is continuously increasing Question 4.

f(x) = x100 + sin x + 1- x ∈ (0, ) Solution: f(x) = x100 + sin x + 1 Diff. w.r.t. x, f'(x) = 100x99 + cos x

In interval (0, ) f'(x) = 100x99 + cos x > 0 [∵ cos x > 0 and 100x99 > 0]

⇒ f'(x) > 0

Hence, function in interval (0, ) is increasing. Question 5. f(x) = (x – 1)ex + 1, x > 0. Solution: f(x) = (x – 1)ex+1 Diff. w.r.t. x,



Hence, at x = 0, function is increasing. Question 6. f(x) = x3 – 6x2 + 12x – 1,x ∈ R. Solution: f(x) = x3 – 6x6

2 + 12x – 1 DifF. w.r.t x, f'(x) = 3x2 – 12x + 12 ⇒ f'(x) = 3(x2 – 4x + 4) ⇒ f'(x) = 3(x – 2)2

⇒ f'(x) = 3(x – 2)2 ≥ 0 ⇒ f'(x) ≥ 0 [∵(x – 2)2 > 0] Hence, function f(x) is increasing in R. Question 7. f(x) = tan-1 x – x, x ∈ R. Solution: f(x) = tan-1 x – x Diff. w.r.t. x,

Hence, function fix) decreasing in R. Question 8.

f(x) = sin4 x + cos4 x, x ∈(0, ) Solution: f(x) = sin4 x + cos4 x Diff. w.r.t x, f'(x) = 4 sin3 x cos x + 4 cos3 x(-sin x) ⇒ f'(x) = 4 sin x cos x (sin2 x – cos2 x) ⇒ f'(x) = -2.2 sin x cos x (cos2 x – sin2 x) ⇒ f'(x) = – 2sin 2x cos 2x



⇒ f'(x) = – sin 4x

Question 9.

f(x) = + 5, x ∈ R, x ≠ 0. Solution:

Hence, function f(x) is decreasing in x ∈ R when x ≠ 0. Question 10. f(x) = x2 – 2x + 3, x < 1. Solution: f(x) = x2 – 2x + 3 Diff. w.r.t. x, f'(x) = 2x – 2 ⇒ f'(x) = 2(x – 1) ⇒ f'(x) = 2(x – 1) < 0 [When x < 1]

⇒ f'(x) < 0 Hence function f(x) is decreasing in interval x < 1. Question 11. f(x) = 2x3 – 3x2 – 36x + 7 Solution: Given f(x) = 2x3 – 3x2 – 36x + 7 ⇒ f'(x) = 6x2 – 6x – 36 f'(x) = o ⇒ 6x2 – 6x – 36 = 0 ⇒ 6(x2 – x – 6) = 0 ⇒ 6(x – 3) (x + 2) = 0

⇒ x – 3 = 0 or x + 2 = 0 ⇒ x = 3 or x = -2 Hence, point x = – 2, x = 3 divides the real number line into three intervals (-∞,-2), (-2, 3) and (3,∞).



(a) For (-∞, -2) interval. f'(x) = 6x2 – 6x – 36 > 0 ∵ At x = – 3 f'(x) = 6(- 3)2 – 6(- 3) – 36 = 6 × 9 + 6 × 3 – 36 = 54 + 18 – 36 = 36 >0 So,f'(x) > 0 can be shown like this by taking other points. Hence, function is continuously increasing in interval (-∞, – 2). (b) For (-2, 3) f'(x) = 6x2 – 6x – 36 < 0 ∴ At x = 1 f'(x) = 6(1)2 – 6 × 1 – 36 = 6 – 6 – 36 = -36 < 0 At x = 0, f'(x) = 6(0)2 – 6 × 0 – 36 = – 36 < 0 So,f'(x) < 0 can be shown like this by taking other points. Hence, function is continuously decreasing for x ∈ (- 2, 3). (c) For (3, ∞) f'(x) = 6x2 – 6x – 36 > 0 ∵ At x = 4, f'(x) = 6 × (4)2 – 6 × 4 – 36 = 96 – 24 – 36 = 96 – 60 = 36 > 0 At x = 5, f'(x)= 6(5)2 – 6 × 5 – 36 = 6 × 25 – 30 – 36 = 150 – 30 – 36 = 84 > 0 So,f'(x) > 0 can be shown like this for other points. Hence, function f(x) is continuous increasing in interval (-∞, -2) ∪ (3,∞). {∵ f'(x) > 0} Function is continuously decreasing in interval (- 2, 3). Question 12. f(x) = x4 – 2x2. Solution: Given f(x) = x4 – 2x2 f'(x) = 4x6

3 – 4x (∵f(x) = 0) ⇒ 4x3 – 4x = 0 ⇒ 4x(x2 – 1) = 0

⇒ 4x = 0 or x2 – 1 = 0 ⇒ x = 0 or x = ± 1 Here, point x = 0, 1, – 1 divides real number line into four interval (-∞, -1), (-1, 0), (0, 1) and (1,∞). (a) For interval (-∞, -1) f'(x) = 4 x3 – 4x At x = – 2, f'(x) = 4 x (- 2)3 – 4(- 2) = -32 + 8 = -24 < 0 Like this it can be shown for other points (f'(x) <0). (b) For interval (-1,0) f'(x) = 4x3 – 4x



For x = -0.5 f'(x) = 4 × (- 0.5)3 – 4 × (- 0.5) = -0.5 + 2.0 = 1.5 >0 Like this it can be shown for other points {f'(x) > 0} (c) For interval (0, 1) f'(x) = 4x3 – 4x For x = 0.5 f'(x) = 4 × (0.5)3 – 4 × (0.5) = 4 × 0.125 – 2.0 = 0.5 – 2.0 = – 1.5 < 0 Like this it can be shown for other points {f'(x)< 0} (d) For interval (1, ∞) f'(x) = 4x3 – 4x For x = 2 f'(x) = 4 × (2)3 – 4 × 2 = 32 – 8 = 24 > 0 Like this it can be shown for other points {f'(x) > 0} Hence, function is decreasing in interval (-∞, – 1) ∪ (0, 1) and function is increasing in interval (-1, 0) ∪ (1, ∞). Question 13. f(x) = 2x3 – 9x2 + 12x + 5. Solution: Given f(x) = 2x3 – 9x2 + 12x + 15 f'(x) = 6x2 – 18x + 12 = 6(x2 – 3x + 2) x = 1 and x = 2 divides the real number line into three intervals (-∞, -1), (1, 2) and (2,∞). when x ∈ (-∞, 1) then f'(x) = + ve when x ∈ (1,2) then f'(x) = – ve

when x ∈ (2,∞) then f'(x) = + ve So, function is increasing in interval (-∞, 1) ∪ (2,∞) and decreasing in interval (4, 2). Question 14. f(x) = – 2x3 + 3×2 + 12x + 5. Solution: Given, f(x) = -2×3 + 3×2 + 12x + 5 f'(x) = – 6x2 + 6x + 12 = – 6(x2 – x – 2) = – 6(x – 2)(x + 1) x = – 1 and x = 2 divides the real number line into three intervals (-∞, -1), (- 1, 2) and (2,∞). when x ∈ (-∞,-1), then f'(x) = -ve when x ∈ (-1,2), then f'(x) = +ve

when x ∈ (2,∞), then f'(x) = -ve So, function is increasing in interval (-1, 2) and decreasing in interval (-∞, -1) ∪ (2,∞). Question 15. Find the minimum value of a, such that function f(x) = x2 + ax + 5, is increasing in interval [1, 2] Solution: f(x) = x2 + ax + 5 ⇒ f'(x) = 2x + a



x ∈ [1, 2] ⇒ 1 ≤ x ≤ 2 ⇒ 2 ≤ 2x ≤ 4

⇒ 2 + a ≤ 2x + a ≤ 4 + a ⇒ 2 + a ≥ 0 ⇒ a ≥ -2 Hence, minimum value of a is – 2. Question 16.

Prove that, function f(x) = tan-1 (sin x + cos x), is increasing function in interval (0, ) Solution:

Hence, function is continuously increasing in interval (0, π/4).

Rajasthan Board RBSE Class 12 Maths Chapter 8 Application of Derivatives Ex 8.3 Question 1. For curve y = x3 – x find slope of tangent at x = 2.



Solution:

Question 2.

For curve y = , x ≠ 2 find slope of tangent at x = 10. Solution:

Question 3. Find all points where slope of tangent to the curve

Solution:



Question 4.

Find all equations of lines which are tangent to the curve y + = 0 and slope of those is 2. Solution:



Putting x = 4 in eq. (i)

y = = -2 Then, point = (4, – 2) Again putting, x = 2 in eq. (i)

y = = 2 Then, point = (2, 2) Now, tangent at ponit (4, – 2), y – (- 2) = 2 (x – 4) ⇒ y + 2 = 2x – 8

⇒ 2x – y – 10 = 0 Again, tangent at point (2, 2), y – 2 = 2 (x – 2) ⇒ y – 2 = 2x – 4 ⇒ 2x – y – 2 = 0 Hence, required equations are 2x – y – 10 = 0 and 2x – y – 2 = 0. Question 5. For curve

find those points where tangent is (i) Parallel to x-axis (ii) Parallel to y-axis. Solution:



(i) When tangent is parallel to x-axis then,

Hence, tangents are parallel to x-axis at points (0, ± 5). (ii) When tangent is parallel to y-axis then,

Hence, tangents are parallel to 7-axis at points (± 2, 0) Question 6.

For curve x – a sin3 t,y = b cos3 t find equation n of tangent at t = Solution:



So, at t = or at (a, 0), equation of tangent of given curve y – 0 = 0 (x – a) or y = 0 Question 7. For curves = sin2 x, find equation of normal at

Solution:



Question 8. Find equation of tangent and normal following curves, at given points.



Solution: (a) Given curve, y = x2 + 4x + 1 …..(i) Putting x = 3 in equation (i), y = (3)2 + 4 (3) + 1 y = 22 So, point of contact = (3, 22) DifF. (i) w.r.t. x,

⇒ 10y – 220 = – x + 3

⇒ x + 10y = 223 So, equation of normal x + 10y = 223 (b) Given curve, y2 = 4ax …..(i) Putting at x = a in equation (i) y2 = 4a (a)



⇒ y2 = 4a2 ⇒ y = ± 2a Hence, points of contact are (a, 2a) and (a, – 2a). DifF. (i) w.r.t. x,

Hence, at (a, 2a) equation of tangent is x – y + a = 0 and equation of normal is x + y – 3a = 0



⇒ m2 (my – 2d) = – (m2x – a) ⇒m3y – 2am2 = – m2x + a

⇒ m2x + m3 y = a (2m2 + 1) Hence, equation of normal m2x + m3y = a (2m2 + 1)



⇒ ay tan θ – ab tan2 θ = bx sec θ – ab sec2 θ ⇒ bx sec θ – ay tan θ = ab(sec2 θ – tan2 θ)



Equation of tangent y – 2at = 1 (x – at2) At t = 1 y – 2a = x – a ⇒ x – y + a = 0 Hence, equation of tangent x – y + a = 0 Again, equation of normal y – 2at = – 1 (x – a t2) At t = 1 y – 2a = – (x – a) ⇒ x + y – 3a = 0 Hence, equation of normal x + y – 3a = 0



(g) Given, x = θ + sin θ, y – 1 – cos θ Diff. w.r.t. θ,



Rajasthan Board RBSE Class 12 Maths Chapter 8 Application of Derivatives Ex 8.4



Use differential to approximate the following : [Q. 1 to Q. 11] Question 1. (0.009)1/3 Solution: Let y = x1/3 and x = 0.008 ∆x = 0.009 – 0.008 = 0.001 ∵ y = x1/3 Diff. w.r.t. x,

Question 2. (0.999)1/10 Solution: Let y = x1/10, x = 1, y = 1 ∆x = 0.999 – 1 = -0.001 ∵ y = x1/10



Diff. w.r.t. x

Question 3.

Solution:

Let y = , x = 0.0036, y = 0.06 ∆x = 0.0037 – 0.0036 = -0.001

∵ y = Diff. w.r.t. x



Question 4.

Solution:

Question 5. (15)1/4 Solution: Let y = x1/4 x = 16, y = (16)1/4 = 2 ∆x= 15 – 16 = – 1 ∵ y = x1/14 Diff. w.r.t. x



Question 6.

Solution: Let y = x1/2, x = 400, y = 20, ∆x = 401 – 400 = 1 ∵ y = x1/2 Diff. w.r.t. x



Question 7. (3.968)3/2 Solution: Let y = x3/2, x = 4 y = (4)3/2 = (22)3/2 = 23 = 8 ∆x = 3.968 – 4 = -0.032 ∵ y = x3/2



Diff. w.r.t. x

Question 8. (32.15)1/5 Solution: Let y = x1/5, x = 32, y = (32)1/5 = 2, ∆x = 32.15 – 32 = 0.15 ∵ y = x1/5



Diff. w.r.t. x

Question 9.

Solution:

Let y = , x = 0.64, y = 0.8 ∆x = 0.6 – 0.64 = -0.04

∵ y =



Diff. w.r.t. x

Question 10. log10 (10.1), when log10 e = 0.4343 Solution: Let y = log10 x where x = 10, ∆x = 0.1 ⇒ x + ∆x = 10.1 y = log10 x = log10e. loge x Diff. w.r.t. x,

Now, from equation (i) y + ∆y = log10 (x + ∆x)



⇒ log10 x + ∆y = log10 (x + ∆x) ⇒ log10 10 + 0.004343 = log10 (10.1) ⇒ log10 (10.1)= 1 + 0.004343 log10 (10.1)= 1.004343 Hence, approximate value of log10 (10.1) is 1.004343. Question 11. loge (10.02), when loge10 = 2.3026 Solution: Let y = loge x Where x = 10, ∆x = 0.02 and x + ∆x = 10.02 ∵ y = loge x Diff. w.r.t. x,

From equation (i), y + ∆y = loge (x + ∆x) loge x + ∆y = loge (x + ∆x) ⇒ loge 10 + 0.002 = loge (10.02)

⇒ loge (10.02) = 2.3026 + 0.002 ⇒ loge (10.02)= 2.3046 Hence, approximate value of loge (10.02) is 2.3046. Question 12. If y = x2 + 4 and x changes from 3 to 3.1, then use differential to approximate change in y. Solution: Given, y = x2 + 4 At x = 3 y = (3)2 + 4 = 9 + 4 = 13 ∆x = 3.1 – 3 = 0.1 ∵ y = x2 + 4 Diff. w.r.t. x,



Hence, approximate value of change in y is 0.6. Question 13. Show that the relative error in computing the volume of a cubical box, due to an error in measuring the edge, is approximately equal to the three times the relative error in the edge. Solution: Let length of edge of cubical box is x and volume is V, then V = x3 Diff. w.r.t. x,

Question 14. The radius of a sphere shrinks from 10 cm to 9.8 cm. Find approximate decrease in its volume. Solution: Given Radius of sphere = 10 cm ∆r = radius shrinks = 9.8 – 10 = – 0.2 cm

Volume of sphere, V = πr3 Diff. w.r.t. r,

= π 3r2 = 4 πr2 ∵ Approximation error in calculation of volume of sphere,



dV = × (∆r) dV= 4πr2 × (∆r) ⇒ dV= 4π × (10)2 × (- 0.2) ⇒ dV = – 400 × 0.2 π cm3 = -80 π cm3 Hence approximation error in volume is 80 π cm3

Rajasthan Board RBSE Class 12 Maths Chapter 8 Application of Derivatives Ex 8.5 Question 1. Find maximum and minimum values of following function : (a) 2x3 – 15x2 + 36x + 10 (b) (x – 1)(x – 2)(x – 3) (c) sin x + cos 2x (d) x4 – 5x4 + 5x3 – 1 Solution: (a) Let y = 2x3 – 15x2 + 36x + 10

⇒ = 6x2 – 30x + 36 For maxima or minima,

= 0 ⇒ 6x2 – 30x + 36 = 0

⇒ 6(x2 – 5x + 6) = 0 ⇒ (x – 3) (x – 2) = 0 so, x = 2,3

At x = 2, maximum value of function = 2(2)3 – 15(2)2 + 36 × 2 +10 = 16 – 60 + 72 + 10 = 38

(b) (x – 1) (x – 2) (x – 3) Let y = (x – 1) (x – 2) (x – 3) Then, = x3 – 6x2 + 11x – 6



(c) Let y = sin x + cos 2x ……(i) Diff. w.r.t. x,

= cos x – 2 sin 2x …..(ii)

For maximum or minimum value = 0 or cos x – 2 sin 2x = 0 ⇒ cos x – 4 sin x cos x = 0 ⇒ cosx (1 -4 sinx) = 0



(d) Let y = x5 – 5x4 + 5x3 – 1

= 5x4 – 20x3 + 15x2

⇒ 5x2 (x2 – 4x + 3) 5x2(x2 – 3x – x + 3) = 5x2 [x(x – 3) -1 (x – 3)] = 5x2 (x – 3)(x – 1)



For maxima or minima

Question 2. Find maximum and minimum values of following function, if exists : (a) – | x + 1 | + 3 (b) | x + 2 | – 1 (c) | sin 4x + 3 | (d) sin 2x + 5 Solution: (a) Let g(x) = – | x + 1 | + 3 – |x + 1 | < 0, So function has no minimum value Again minimum value of – | x + 1 | is 0 ∴ -| x + 1 | = 0 ⇒ x = -1 then maximum value of g(x) is 3 [∵ g(-1) = – | – 1 + 1 | + 3 = 3] (b) Let f(x) = | x + 2 | – 1 |x + 2 | > 0, So, f(x) does not have any maximum value. Again, minimum value of | x + 2 | = 0 ∴ | x + 2 | = 0 ⇒ x = -2, then minimum value of f(x) is – 1 (∴ f (- 2) = | -2 + 2 | – 1 = 0 – 1 = – 1) (c) Let f(x) = | sin 4x + 3 | sin 4x has maximum value as 1. ∴ Maximum value of f(x) is | -1 + 3 | = 4 Again, minimum value of sin 4x Minimum value of f(x) | -1 + 3 | = | 2 | = 2



(d) Let h(x) = sin 2x + 5 sin (2x) has maximum value as 1 So, h(x) has maximum value as 1 + 5 = 6 Again, sin (2x) has minimum value as – 1 So, minimum value of h (x) will be -1 + 5 = 4 Question 3. Find maximum and minimum values of following functions in given interval: (a) 2x3 – 24x + 107, x ∈ [1,3] . (b) 3x4 – 2x3 – 6x2 + 6x + 1, x ∈ [0, 2] (c) x + sin 2x, x ∈ [0, 2π]

(d) x3 – 18x2 + 96x, x ∈ [0, 9] Solution: (a) Let y = 2x3 – 24x + 107, x ∈ [1,3]

= 6x2 – 24 For maxima or minima

= 0 ⇒ 6x2 – 24 = 0 ⇒ x = ± 2. ∵ x ∈ [1, 3] ∴ x = 2 Now, y1 = 2(1)3 – 24(1) + 107 = 2 – 24 + 107 = 85 y2 = 2(2)3 – 24(2) +107 = 16 – 48 + 107 = 75 y3 = 2(3)3 – 24(3) + 107 = 54 – 72 + 107 = 89 Hence, maximum value is 89 at x = 3 where as minimum value is 75 at x = 2. (b) Let y = 3x4 – 2x3 – 6x2 + 6x + 1, x ∈ (0, 2)

= 12x3 – 6x2 – 12x + 6 = 6(2x3 – x2 – 2x + 1) For maxima or minima

Now y1 = 3(1)4 – 2(1)3 – 6(1)2 + 6(1) + 1 ⇒ y1 = 3 – 2 – 6 + 6 + 1 = 2 ⇒ y-1 = 3 (- 1)4 – 2 (-1)3 – 6(- 1)2 + 6(- 1) + 1 ⇒ y-1 = 3 + 2 – 6 – 6 + 1 = -6 ⇒ y0 = 3(0)4 – 2(0)3 – 6(0)2 + 6(0) + 1 ⇒ y0 = 1



⇒ y2 = 3(2)4 – 2(2)3 – 6(2)2 + 6(2) + 1 = 48 – 16 – 24 + 12 + 1 = 21



∴ Hence, maximum value of function is 2π at x = 2π and minimum value of function is 0 at x = 0.

y4 = (0)3 – 18(0)2 + 96(0) = 64 – 288 + 384 = 160 y8 = (8)3 – 18(8)2 + 96 × 8 = 512 – 1152 + 768 = 128



y9 = (9)3 – 18(9)2 + 96(9) = 729 – 1458 + 864= 135 So, at x = 0 minimum value = 0 and at x = 4 maximum value = 160 Question 4. Find extreme value of following functions : (a) sin x cos 2x (b) a sec x + b cosec x, 0 < a < b (c) (x)1/x, x > 0

(d) log x, x ∈ (0,α) Solution: (a) Let y = sin x.cos 2x …..(i)

= sin x.(- sin 2x) + cos 2x.cos x = – 4 sin2 x.cos x + (1 – 2 sin2 x).cos x = – 4 sin2 x cos x + cos x – 2 sin2 x cos x = cos x (1 – 6 sin2 x) …..(ii)



So, at x = function will be maxima and maximum value of function



Hence, on putting x = e in given function we get maximum value of function which is



(e)1/e.

Question 5.

Prove that value of is maximam at x = cos x. Solution:



Question 6.

Prove that value of sin2 x (1 + cosx) is maximum x = Solution: Let y = sin2 x.(1 + cosx) …..(1) Diff.wrt.x,

⇒ = sin2 x(- sin x) + (1 + cos x)

⇒ = sinx (-sin2 x + 2 cos x + 2 cos2 x) …..(2)

For maxima of minima = 0



Hence, at cos x = function has maximum value. Question 7.

Prove that ,y = sinP θ cosq θ is maximum at tan θ = Solution:



+ (p cot θ – q tan θ) × 0



= -y {(q + p) + (p + q)} + 0 = -2y (p + q) = -2 (sinp θ cosq θ) (p + q) = -ve So, y will be maximum Hence, y will be of maximum value if

Rajasthan Board RBSE Class 12 Maths Chapter 8 Application of Derivatives Ex 8.6 Question 1. Show that the triangle of maximum area that can be inscribed in a circle is an equilateral triangle. Solution: Let ∆ABC is inscribed in a circle of radius r. A perpendicular E is drawn of AC from the vertex B of triangle which meets AC at D and circle at E. Let base of triangle AC = 2x and height of triangle BD = y



Hence, ∆ABC is an equilateral triangle. Question 2. Given the sum of the perimeters of a square and a circle, then prove that the sum of their area is least when one side of the square is equal to diameter of the circle. Solution: Let side of square is x and radius of circle is r. Perimeter of circle (circumference) = 2πr Sum of both perimeters = 4x + 2πr = k …..(i) Area of circle, A2 = πr2 Area of square, A2 = x2 ∴ Sum of areas, A = πr2 + x2



Hence, side of square is equal to diameter of circle, when sum of area of both is minimum. Question 3. Show that the cone of the greatest volume which can be inscribed in a given sphere

has an altitude equal to of the diameter of the sphere. Solution: Let radius of sphere = r height of cone = x radius of base of cone = y

Diameter of sphere and axis of cone will coincide with each other that by maximum height can be taken.



= 2 : 3 or height of cone is of diameter of the sphere.



Question 4. The expenses per hour for plying steamer in a river is proportionate to cube of its speed. If the speed of the water current is x kin per hour, then prove that the

maximum normal (prudent) speed of the steamer will be km per hour if it flied against the current of the water. Solution: Let velocity of steamer is u km/h and distance covered is S km. According to question Speed of water current = x km/h and relative velocity of steamer = (u – x) k/h time take to cover the distance

T = hour Given expense per hour = Ku3 {K is constant}

Total expenses E = Ku3 .

Question 5. If sum of hypotenuse and one side of right angled triangle is given, then area of

triangle is maximum if angle between these sides is Solution: Let, in right angle triangle, hypotenuse is h and base is x, angle between them is θ.



Hence, Area is maximum when θ = π/3 Question 6. If a circle of radius ‘a’ is inside an equilateral triangle, then prove that minimum

perimeter of triangle will be 6 a. Solution: Let ∆ABC is an equilateral triangle. In which AB = AC Then, ∠ABC = ∠ACB = 2θ (Let)



In ∆OBD, BD = a cot θ Similarly CD = BE = CF = a cot θ In ∆AOE, AE = a cot (90° – 2θ) = a tan 2θ = AF AB = AE +BE = a tan 2θ + a cot θ BC = BD +CD = a cot θ + a cot θ = 2a cot θ CA = AF + CF = a tan 2θ + a cot θ Let perimeter of ∆ABC is P, then P = AB + BC + CA = (a tan 2θ + a cot θ) + 2a cot θ + a tan 2θ + a cot θ) = 2a tan 2θ + 4a cot θ = 2a(tan 2θ + 2 cot θ) …..(i) Diff w.r.t. 0,

= 2a (2 sec2 2θ – 2 cosec2 θ)

For maxima and minima = 0 ⇒ 2a(2 sec2 2θ – 2 cosec2 θ) = 0 ⇒ sec2 2θ – cosec2 θ = 0

⇒ sec 2θ = cosec θ ⇒ cot 2θ = sin θ



Hence, required minimum perimeter of isosceles triangle = 6 a Question 7. If a normal is drawn a point ‘P’ of ellipse

then Prove that the maximum distance from centre of ellipse will be a – b. Solution: We have to show that maximum value of OM is a – b. Now equation of normal at point P (a cos θ, b sin θ), ax sec θ – by cosec θ = a2 – b2

Perpendicular distance between normal and centre of ellipse,



will be minimum. Here (a2 – b2) is constant. Now, Let y = a2 sec2 θ + b2 cosec2 θ Note : Here, we will prove that (a + b)2 is minimum value of a2 sec2 θ + b2 cosec2 θ

= 2a2 sec2 θ tan θ – 2b2 cosec2 θ cot θ using = 0 ⇒ 2[a2 sec2 θ tan θ – b2 cosec2 θ cot θ] = 0

Putting tan2 θ = in y = a2 sec2 θ + b2 cosec2 θ y = a2 sec2 θ + b2 cosec2 θ = a2 (1 + tan2 θ) + b2 (1 + cot2 θ)

= a2 [1 + ] + b2 [1 + ] = a2 + b2 + 2ab = (a + b)2 So, y = a2 sec2 θ + b2 cosec2 θ has minimum value (a + b)2 From equation (i), Maximum value of p

Hence, maximum distance between normal and centre of ellipse is (a - b)

Rajasthan Board RBSE Class 12 Maths Chapter 8 Application of Derivatives Miscellaneous Exercise Question 1. If radius and height of a cylinder is r and A, then find the rate of change of surface area with respect to its radius. Solution: Radius of cylinder = r and Height of cylinder = h



Rate of change of surface area w.r.t. r, = Surface area of cylinder S = 2πr2 + 2πrh

Question 2. Find the value of x and y for function, y = x3 + 21 where, the rate of change ofy is three times as the rate of change of x. Solution: Given, function y = x3 + 21 Diff w.r.t. t,

= 3x2 …..(i) According to question

= 3 …..(ii) From equation (i) and (ii) 3x2 = 3 ⇒ x = ± 1 In equation y = x3 + 21 Putting x = 1 y = 13 + 21 = 1 + 21 =22 Putting x = -1 y = (- 1)3 + 21 = -1 + 21 = 20 So, x = ± 1 and y = 22, 20 Question 3. Prove that exponential function (ex) is an increasing function, Solution: y = ex

Then, = ex = + ve ∀ x ∈ R So, for x ∈ R, ex is an increasing function. Question 4.

Prove that f(x) = log (sinx) is increasing in (0, ) and decreasing in ( ,π) Solution: f(x) = log (sin x) Diff w.r.t. t,



Question 5.

If tangent OX and OY of curve – = at some point cut the axis at point P and Q, then show that OP + OQ – a, where O is origin. Solution: Given,

Equation – = …..(i) Let, tangent cut the axis OX and OY at point P and Q, point on curve contact with tangent is (h, k). Since, (h, k) situated on curve.

So, – = …..(ii) DifF. (i) w.r.t. x,

So, from y – y1 = m(x – x1) equation of tangent at the point (h, k).



Question 6. Find equation of tangent of curve y = cos (x + y), x ∈ [- 2π, 2π] which is parallel to the line x + 2y = 0 Solution:



Hence, equation (i) and (ii) are the required equations of tangents. Question 7. Find the percentage error in calculating the volume of a cubical box if an error of 5%



is made in measuring the length of edge of the cube. Solution: Let side of cube is x and volume is V.

Percentage error in volume = 3 × 5% = 15% Hence, required percentage error is 15% Question 8. A circular metal plate expands under heating so that its radius increased by 2%. Find the approximate increase in the area of the plate, if the radius of the plate before heating is 10 cm. Solution: Let radius of circular plate is r and surface area S.

Hence, approximate increase in the area of plate is 4π cm2. Question 9.

Prove that the volume of the largest cone inside a sphere is of the volume of sphere.



Solution: Let r and h be radius and height of the cone respectively inscribed in a sphere of radius R.



Question 10. Prove that semivertical angle of a cylinderical cone of given surface area and

maximum volume is sin-1 ( ). Solution: Let, Slant height of cone = l Height = h



radius = r T.S.A. = S Volume = V and Semi vertical angle = α

It is clear that, for S when V is maximum or minimum, then V2 be also maximum or minimum



Hence, when semi-vertical angle of cone is sin-1 ( ) then its volume will be maximum.

Rajasthan Board RBSE Class 12 Maths Chapter 9 Integration Ex 9.1 Question 1. Integrate the following with respect to x :

(a) 3 (b) e3x

(c) ( )x (d) a2 loga x Solution:



Find the value of the integrates given below : Question 2.

Solution:

Question 3.



Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6. ∫ax da Solution:

Question 7.



Solution:

Question 8.

Solution:

Question 9. ∫sec x (sec x + tan x) dx Solution: ∫sec x (sec x + tan x) dx = ∫sec2 x dx + ∫sec x tan x dx = tan x + sec x + C Question 10. ∫(sin-1 x + cos-1 x) dx Solution :



Question 11.

Solution :

Question 12. ∫ tan2 x dx Solution : ∫ tan2 x dx = ∫ (sec2 x – 1) dx = ∫ sec2 x – ∫ dx = tan x – x + C Question 13. ∫ cot2 x dx Solution : ∫ (cosec2 x – 1) dx = ∫ cosec2 x dx – ∫dx = – cot x – x + C Question 14.

Solution :

Question 15. ∫ (tan2 x – cot2 x) dx Solution :



∫ (tan2 x – cot2 x) dx = ∫ (sec2 x -1 – cosec2 x +1) dx = ∫ sec2 x dx – ∫ cosec2 x dx = tan x + cot x + C Question 16.

Solution :

Question 17.

Solution :



Question 18.

Solution :

Question 19. ∫ cot x (tan x – cosec x) dx Solution : ∫ cot x (tan x – cosec x) dx = ∫ cot x tan x dx – ∫ cot x cosec x dx = ∫ 1 dx – ∫ cosec x cot x dx = x + cosec x + C Question 20.

Solution :

Question 21. ∫ logx x dx Solution :



Question 22.

Solution :

Question 23.

Solution :

Question 24.

Solution :

Rajasthan Board RBSE Class 12 Maths Chapter 9 Integration Ex 9.2 Question 1. Integrate the following w.r.t.x : (a) x sin x2

(b) x Solution :



Question 2. Find the following :



Solution:


Solution:




Solution:



Question 7. Find the following : (a) ∫ sin 3x sin 2x dx

(b) ∫ dx



Solution:

Question 8. Find the following : (a) ∫ cos4 x dx (b) ∫ sin3 x dx Solution:




Solution:




Solution:



Question 16.

Solution:



Rajasthan Board RBSE Class 12 Maths Chapter 9 Integration Ex 9.3 Integrate the following functions w.r.t. x: Question 1.



Solution:

Question 2.

Solution:



Question 3.



Solution:

Question 4.

Solution:



Question 5.

Solution:



Question 6.



Solution:

Question 7.

Solution:



Question 8.

Solution:



Question 9.

Solution:



Question 10.

Solution:



Question 11.

Solution:



Question 12.

Solution:



Question 13.

Solution:

Rajasthan Board RBSE Class 12 Maths Chapter 9 Integration Ex 9.4 Integrate the following functions w.r.t. x. Question 1.



Solution:

Question 2.

Solution:

Question 3.

Solution:



Question 4.



Solution:



Question 5.



Solution:

Question 6.



Solution:

Question 7.

Solution:



Question 8.



Solution:

Question 9.



Solution:

Question 10.

Solution:



Question 11.



Solution:

Question 12.



Solution:

Question 13.

Solution:



Question 14.



Solution:

Question 15.

Solution:



Question 16.



Solution:

Question 17.

Solution:



Question 18.

Solution:



Question 19.



Solution:

Question 20.

Solution:



Question 21.

Solution:



Question 22.

Solution:



Question 23.



Solution:

Question 24.



Solution:

Rajasthan Board RBSE Class 12 Maths Chapter 9 Integration Ex 9.5 Integrate the following functions w.r.t.x. Question 1.

Solution:

Question 2.

Solution:



Question 3.



Solution:

Question 4.

Solution:

Question 5.

Solution:



Question 6.



Solution:

Question 7.

Solution:



Question 8.

Solution:



Question 9.



Solution:

Question 10.



Solution:

Question 11.

Solution:



Question 12.



Solution:

Question 13.

Solution:



Question 14.

Solution:



Question 15.

Solution:

Question 16.

Solution:



Question 17.

Solution:



Question 18.



Solution:

Question 19.

dx Solution:



Question 20.

Solution:



Question 21.

Solution:

Question 22.

Solution:



Rajasthan Board RBSE Class 12 Maths Chapter 9 Integration Ex 9.6 Exalnate the following : Question 1. (a) ∫ x cos x dx (b) ∫ x sec2 x dx Solution :



Question 2. (a) ∫ x3 e-x dx (b) ∫ x3 sin x dx Solution:



Question 3. (a) ∫ x3 (log x)2 dx (b) ∫ x3 ex2 dx Solution:



Question 4. (a) ∫ e2x eex dx (b) ∫ (log x)2 dx Solution:



Question 5.

Solution:



Question 6.

Solution: (a) Let I = ∫ sin-1 (3x – 4x3) dx Now, let x = sin t, then dx = cos t dt ∴ I = ∫ sin-1 (3 sin t – 4 sin3 t).cos t dt = ∫ sin-1 (sin 31) cos t dt = ∫ 3t cos t dt = ∫ 3t cos t dt I II



Question 7.

Solution:



Question 8.

Solution:



Question 9.

Solution:



Question 12.

Solution:

Question 13.



Solution:

Question 14.



Solution:

Question 15. ∫ex [log (sec x + tan x) + sec x] dx Solution: ∫ex [log (sec x + tan x) + sec x] dx = ∫ex log (sec x + tan x) dx + ∫ex sec x dx



Question 16. ∫ex (sin x + cos x) sec2 x dx Solution: ∫ex (sin x + cos x) sec2 x dx

Question 17.



Solution:

Question 18.

Solution:



Question 19.

Solution:

Question 20.

Solution:



Question 21.



Solution:

Question 22. ∫(sin-1 x)2 dx Solution: Let I = ∫ (sin-1 x)2 dx Let sin-1 x = θ ⇒ x = sin θ ∴ dx – cos θ dθ∴

∴ I= ∫θ2 . cos θ dθ = θ2 sin θ – ∫20 sin θ dθ + C

Rajasthan Board RBSE Class 12 Maths Chapter 9 Integration Ex 9.7 Evalute the following : Question 1. ∫ e2x cos x dx



Solution:

Question 2. ∫ sin (log x) dx Solution: Let I = ∫ sin (log x) dx



Again let log x = t ⇒ x = et ⇒ dx = et dt

Question 3.

Solution:



Question 4.

Solution:



Question 5. ∫ ex sin2 x dx Solution:



Question 6. ∫ ea sin-1 x dx Solution:



Question 7.



Solution:

Question 8. ∫ e4x cos 4x cos 2x dx Solution:



Question 9.



Solution:

Question 10.

Solution:

Question 11.

Solution:



Question 12.

Solution:



Question 13.



Solution:

Question 14.

Solution:



Question 15.

Solution:



Question 16.

Solution:



Rajasthan Board RBSE Class 12 Maths Integration Miscellaneous Exercise Integrate the following functions with respect to x : Question 1. [1+2 tan x (tan x + sec x)] Solution: ∫ [1 + 2 tan x (tan x + sec x)] dx = ∫ [1 + 2 tan2 x + 2 tan x sec x] dx = ∫ [2(1 + tan2 x) + 2 sec x tan x -1] dx = 2 ∫ (sec2 x + sec x tan x) dx – ∫ dx = 2(tan x + sec x) – x + C Question 2. ex sin3 x Solution:

Question 3. x2 log (1 – x2) Solution:



Question 4.

Solution:



Question 5.

Solution:



Question 6.

Solution:



Question 7.

Solution:

Question 8.



Solution:

Question 9.

Solution:

Question 10.



Solution:

Question 11.



Solution:

Question 12.

Solution:

Question 13.



Solution:

Question 14.

Solution:

Question 15.

Solution:



Question 16.

Solution:



Question 17.



Solution:

Question 18.

Solution:



Question 19.



Solution:

Question 20.



Solution:

Question 21.



Solution:

Question 22.

(a) tan x + x + C (b)cot x + x + C (c) tan x – x + C (d) cot x – x + C Solution:

Hence correct option is (c). Question 23.



Solution:

Question 24.

Solution:

Question 25.



Solution:

Rajasthan Board RBSE Class 12 Maths Chapter 10 Definite Integral Ex 10.1 Find the value of definite integerals, as limit of sum (by first principle) Question 1.

Solution:



Question 2.

Solution:



Question 3.

Solution:



Question 4.



Solution:

Question 5.

Solution:



Question 6.

Solution:



Rajasthan Board RBSE Class 12 Maths Chapter 10 Definite Integral Ex 10.2 Find the value of the following integrals : Question 1.

Solution:

Question 2.



Solution:

Question 3.

Solution:



Question 4.

Solution:

Question 5.

Solution:



Question 6.



Solution:

Question 7.



Solution:

Question 8.



Solution:

Question 9.



Solution:

Similar Question



Solution:

Question 10.

Solution:



Question 11.

Solution:



Question 12.



Solution:

Question 13.

Solution:



Question 14.



Solution:

Question 15.

Solution:



Question 16.

Solution:



Question 17.



Solution:

Question 18.

Solution:



Question 19.



Solution:

Question 20.

Solution:



Question 21.



Solution:

Question 22.

Solution:



Question 23.

Solution:



Question 24.



Solution:

Question 25.

Solution:



Question 26.

Solution:

= [log (2 + 1) – log (1 + 1)] – [log (2+ 2)-log (1 +2)] = log 3 – log 2 – log 4 + log 3 = 2 log 3 – (log 2 + log 4) = 2 log 3 – log 8

= log = log

Rajasthan Board RBSE Class 12 Maths Chapter 10 Definite Integral Ex 10.3 Find the value of the following integrals : Question 1.



Solution:

Question 2.



Solution:

Question 3.

Solution:

Question 4.

[x] dx, where [.] is greatest integer function.



Solution:

Question 5.

Solution:

Here f (x) = x5 cos2 x Now f (-x) = (- x)5 cos2 (- x) = – x5 cos2 x = -f(x) Thus, this is an odd function.

Question 6.

Solution:

[∵ sin x is odd function and cos x is even function ∴ sin (- x) = – sin x cos (- x) = cos x] = -f (x)



Thus this is a odd function

Question 7.

Solution:



Question 8.

Solution:

Question 9.

Solution:



Question 10.

Solution:

Question 11.



Solution:

Question 12.



Solution:

Question 13.

Solution:



Question 14.



Solution:

Question 15.

Solution:



Where t = tan x and dt = sec2 x dx When x = 0, then t = tan 0 = 0

When x = ,then t = tan = 1

Question 16.

Solution:



Question 17.

Solution:



Question 18.

Solution:



Question 19.

Solution:



Question 20.

Solution:



Question 21.

Solution:

Question 22.



Solution:

Rajasthan Board RBSE Class 12 Maths Chapter 10 Definite Integral Miscellaneous Exercise Question 1. The value

(a) 2 sin 3x.x dx (b) 0 (c) a2 (d) 1 Solution:



Question 2.

(a) 3 (b) 2

(c)

(d)



Solution:

Question 3.

Solution: Option (b) is correct Question 4.

If A(x) = θ2 dθ, then value of A(3) will be : (a) 9 (b) 27 (c) 3 (d) 81 Solution:



Thus Option (a) is correct Question 5.

Solution:



Question 6.

Solution:

Question 7.



Solution:

Question 8.



Solution:

Question 9.

Solution:



Question 10.

Solution:



Question 11.



Solution:

Question 12.

Solution:



Question 13.

Solution:



Question 14.

Solution:



Question 15.

Solution:

and x = cot θ ∴ dx = – cosec2 θ dθ

when x = 0, then θ = when x = ∞, then θ = 0



Question 16.



Solution:


Solution:



Rajasthan Board RBSE Class 12 Maths Chapter 11 Application of Integral:Quadrature Ex 11.1 Question 1. Find the area enclosed by parabola y2 = 4ax and its latus rectum. Solution:



Area bounded by parabola y2 = 4ax and its latus rectum is symmetric about x-axis.

Question 2. Sketch the circle x2 + y2 = 4, find area enclosed by y – axis and x = 1. Solution: Circle x2 + y2 = 4, whose centre is (0,0) and radius 2.



Question 3. Find the area enclosed by curve y = sin x and x – axis, whereas 0 < x < 2π. Solution : The area enclosed by curve y = sin x and x = π and x = 2π has shown in given following figure by shaded part. Table of y = sinx for various volues of x is given



below:

Question 4.

Find the area enclosed by curve y = 2 and x = 0, x = 1. Solution :

y = 2 ⇒ y2 = 4x, which is a parabola. The area enclosed by curve y2 – 4x, x = 0, x = 1 has shown by shaded part in the following figure.



If we take area between only positive coordinates, then we will solve like this

y = 2 , x = 0, x = 1 Required Area = Area APOA,

Question 5. Find the area enclosed by y = | x |, x = – 3, x = 1 and x-axis.



Solution : y = | x |, x = – 3, x = 1 Thus y = + x and y = -x Graph of these lines is given below

Question 6. Find the area enclosed by curve x2 = 4ay, x-axis and line x = 2. Solution:



Curve x2 = 4ay is a parabola. Its graph is given below.

Question 7.

Find the area enclosed by ellipse + = 1 and lie above the x – axis Solution:

The area enclosed by ellipse + = 1 and lie above the x axis is shown in the



following figure :

Question 8.

Find the total area of ellipse + = 1. Solution:

Total area of ellipse + = 1 is shown in the following figure.



Question 9.

Find the area enclosed by line – = 2 and co-ordinate axis. Solution:



Graph of line – = 2 is as follows:

Question 10. Find the area enclosed by line x + 2y = 8, x = 2, x = 4 and x-axis. Solution:



Question 11. Find the area enclosed by the curve y – x2, ordinates x = 1, x = 2 and x-axis. Solution: Parabola x2 = y is symmetric about x axis. Its vertex is origin (0, 0) Area enclosed by lines x = 1, x = 2, x-axis and curve x2 = y is shown in figure by shaded part.



Question 12. Find the area of the region in the first quadrant enclosed by y = 4x2, x = 0, y = 1 and y = 4. Solution: ∵ y = 4 x2

⇒ x2 = y This is a equation of parabola Thus x = 0, y = 1, y = 4 area enclosed by and curve y = 4x2 is shown in figure by



shaded part.

Rajasthan Board RBSE Class 12 Maths Chapter 11 Application of Integral:Quadrature Ex 11.2 Question 1. Find the area between the region of parabola. y2 = 2x and x2 + y2 = 8. Solution: Required area is shaded in following figure.



Question 2. Find the area of the region enclosed by parabola 4y = 3x2 and line 3x – 2y + 12 = 0. Solution: Parabola 4y = 3x2 and line 3x – 2y + 12 = 0 intersect each other at point, A(-2, 3) and



B(4, 12). Required area is shaded in the follows figure.

Question 3.

Find the area of the region enclosed by curve y = , x = y and x-axis. Solution:

Curve y = is circle whose vertex is origin and radius is 2.



Question 4. Find the area of the region in the first quadrant enclosed by circle x2 + y2 = 16 and line y = x. Solution: Centre of circle x2 + y2 = 16 is origin and radius is 4 line y = x passes through origin



and cuts the circle at A.

Question 5. Find the area of common region between parabola y2 = 4x and x2 = 4y. Solution: Equation of given parabolas are y2 = 4x …..(i) x2 = 4y ……(ii)



Solving these, we get (0, 0) and (4, 4) their intersecting points as

Question 6. Find the area of the region in the first quadrant enclosed by curve x2 + y2 – 1 and x + y = 1 equation. Solution: Given circle x2 + y2 = 1 its centre passes through origin and radius is 1. x + y = 1 is equation of line which passes through points (1,0) and (0, 1).



Question 7. Find the area of the region enclosed by curve y2 = 4ax, and line y = 2a and y – axis. Solution: Shaded part of following figure shows area enclosed by curve y2 = 4ax, line y = 2a



and y axis.

Question 8. Find the area of that portion of circle x2 + y2 = which lies outside the parabola y2 = 6x. Solution: Radius of given circle x2 + y2 = 16 is 4 unit and it passes through origin. Let parabola y2 = 6x is interested by circle at points P and Q then solving two equations. x2 + 6x = 16 (∵y2 – 6x) ⇒ x2 + 6x – 16 = 0

⇒ x + 8x – 2x – 16 = 0 (x + 8) (x – 2) = 0

Thus x = – 8, + 2 Here we take positive value of x. Thus we will take limits 0 and 2, 2 and 4



Area POQSP = 2 × area PORSP = 2 [area PORP + area PRSP]

Question 9. By using integrating method, find the area of ∆ABC whose coordinates of vertices are A(2, 0), 5(4, 5), C(6, 3). Solution:



In figure, ∆ABC is shaded.



Question 10. By using intersection method, find the area of triangular prism equations of whose sides are 3x – 2y + 3 = 0, x + 2y – 7 = 0 and x – 2y + 1 = 0. Solution: Given lines 3x – 27 + 3 = 0 …..(i) x + 27 – 7 = 0 …..(ii)



and x – 27 + 1 = 0 …..(3) Solving equation (1) and (2) x = 1, y = 3 Solving equation (2) and (3) x = 3, y = 2 Solving equation (3) and (1) x = -1, y = 0 Now draw the graph of threee lines



Rajasthan Board RBSE Class 12 Maths Chapter 11 Application of Integral:Quadrature Miscellaneous Exercise Question 1.

Area of the region bounded by the curve y = and y = x is : (a) 1 sq. unit

(d) sq. unit

(c) sq. unit

(d) sq. unit Solution:

Curve y = is a parabola whose equation is y2 = x and centre is at origin y = x is a line which passes through origin which is shown in following figure.

Question 2. Area of region enclosed by y2 = x and x2 = y is :

(a) sq. unit (d) 1 sq. unit

(c) sq. unit



(d) 2 sq. unit Solution: Sllving given equations y2 = x …..(i) x2 = y …..(ii) We get integration (0,0) and (4,4)

Question 3. Area of region enclosed by and its latus rectum parabola x2 = 4y is :

(a) sq. unit

(d) sq. unit

(c) sq. unit

(d) sq. unit Solution: Graph of parabola x2 = 4y is given below :



Here, a=1 Thus, latus rectum intersects y axis at (0, 1). Solving equation of latus rectum y = 1 and equation of parabola x2 = 4y is x = ± 2 Required area = Area AOCBA = 2[Area OQCBO – Area OQCO)

Question 4.

Area of region enclosed by y = sin x, < x < and x-axis is : (a) 1 sq. unit (d) 2 sq. unit

(c) sq. unit (d) 4 sq. unit Solution: The area enclosed by curve y = sin x and x = π/2 and x = 3rc/2 is shown by shaded part in figure for various values of x, table for values of y = sin x is given below :



Question 5. Area of region enclosed by y2 = 2x and circle x2 + y2 = 8 is :

Solution: y2 = 2x is equation of parabola whose centre (0, 0) and circle whose centre is (0, 0)

and radius is 2 units. Their graph of enclosed area is shown by shaded part.



Question 6. Find the area of the region bounded by parabola y2 = x and line x + y = 2. Solution: Area of region bounded by parabola y2 = x and x + y = 2 is shown by shaded part in the following graph.



Question 7. Find the area of the region bounded by curve y2 = 2ax – x2 and y2 = ax



Solution: Area enclosed by curve y2 = 2ax – x2 and y2 = ax is shown by shaded part in the following graph.

Question 8. Find the area of region bounded by parabola y = x2 and y = | x |. Solution: Curve y = x2 is a parabola whose vertex is (0, 0) and is symmetric about y-axis. Equation y = | x | represents two lines When x > 0, then y = x When x < 0, then y = -x Intersection points of y = x and parabola y = x2 are O(0, 0) and A (1, 1).



Intersection points of y = – x and parabola y = x2 and O(0, 0) and B (- 1, 1). The region bounded by lines y = x and y = – x and parabola y = x2 is shown in the

following figure. Question 9. Find the area of the common region bounded by x2 + y2 = 16 and parabola y2 = 6x. Solution: Centre of circle x2 + y2 = 16 is origin and radius is 4 unit. Vertex of parabola y2 = 6x is origin. Common part of these curves is shown in figure.



Two curves intersect at point P and Q by solving equations, coordinates of these can be obtained. Equation of curves x2 + y2 = 16 …..(i) y2 = 6x …..(ii)



Question 10. Find the area of the region bounded by curve x2 + y2 = 1 and x + y > 1. Solution: Area enclosed by circle x2 + y2 = 1 and line x + y > 1 is shown by shaded part in



figure.

Question 11. By using intersection method, find the area of the triangle whose vertices are (- 1, 0), (1, 2) and (3, 2). Solution: Graph of triangle is given below. Required Area is shaded in the figure given below:



Now, putting values of area ∆ABP, area of trapezium BPAQ and area of ∆AQC in equation (i). Required Area of ∆ABC = 3 + 5 – 4 = 4 sq. unit ∴ Area of ∆ABC = 4 sq. unit



Question 12. Find the area of region bounded by line y = 3 x + 2, x axis and ordinates x = – 1 and x = 1. Solution: Area enclosed by line y = 3x + 2, x axis and ordinates x = – 1 and x = 1 is shown below by shaded part.

Question 13. Find the area of the region bounded by y2 = 2x, y = 4x – 1 and y > 0. Solution:



Question 14. Find the area of the region bounded by curve y2 = 4x, y-axis and line y = 3. Solution: Curve y2 = 4x is a parabola whose vertex is origin and symmetric about x-axis. Area



enclosed by curved y2 = 4x and line y = 3 is shown by shaded part in figure given below.

Question 15. Find the area of the region bounded by two circles x2 + y2 = 4 and (x – 2)2 + y2 = 4. Solution: Equations of given circles x2 + y2 = 4 ….(i) and (x – 2)2 + y2 = 4 ….(ii) Centre of circle from equation – (i) is at origin (0, 0) and radius is 2 unit. Centre of circle of equation (ii) is (2, 0) and at x axis and radius is 2 unit. Required enclosed area is shown by shaded part in figure.



Solving equation (i) and (ii)

Obtained intersecting points of circles are p(1, ) and Q (1 – ) Out of two circles one symmetric about x axis ∴ Required area = 2(Area OPACO) = 2 [Area OPCO + Area CPAC] = 2[Area OPCO (part of circle (x – 2)2 + y2 = 4) + Area CPAC (part of circle x2 + y2 = 4) ∴ Required Area = 2 ∫ y dx (for circle (x – 2)2 + y2 = 4) + ∫ y dx (for circle x2 + y2 = 4)



Rajasthan Board RBSE Class 12 Maths Chapter 12 Differential Equation Ex 12.1 Find the order and degree of the following differential equations :



Question 1.

= sin 2x + cos 2x Solution:

= sin 2x + cos 2x

The highest derivative of y = = ( )1 Thus, order = 1, degree = 1. Question 2.

= sin x + cos x Solution:

= sin x + cos x

The highest derivative of y = = ( )1 Thus, order = 2, degree = 1. Question 3.

Solution:

The highest derivative of y =

Thus, order = 2, degree = 2. Question 4.

Solution:



Question 5.

Solution:

Question 6. xdx + ydy = 0 Solution: x dx + y dy = 0

Thus, order = 1, degree = 1. Question 7.

Solution:

Question 8.



Solution:

Thus, order = 1 degree = 2.

Rajasthan Board RBSE Class 12 Maths Chapter 12 Differential Equation Ex 12.3 Question 1. Prove that y2 = 4a (x + a) is the solution of:

Solution:



Question 2. Prove that y = ae-2x + bex is the solution:



Solution:

Question 3.

Prove that y = is the solution of (1 + x2) + (1 + y2) = 0 Solution:



Question 4. Prove that y = a cos (log x) + b sin (log x) is the solution of

Solution: y = a cos (log x) + b sin (logx)



Differentiating w.r.t. x,

Question 5. Prove that xy = log y + c is the solution of

Solution:

Rajasthan Board RBSE Class 12 Maths Chapter 12 Differential Equation Ex 12.4 Solve the following differential equations :



Question 1. (ey + 1) cos x dx + ey sin x dy = 0. Solution: Given equation is

Question 2. (1 + x2) dy = (1 + y2) dx. Solution: Given

Question 3.

(x + 1) = 2xy



Solution:

Question 4.

Solution:

Question 5. (sin x + cos x) dy + (cos x – sin x) dx = 0. Solution: Given (sin x + cos x) dy + (cos x – sin x) dx



Question 6.

Solution:

Question 7. sec2 x tan y dy + sec2 y. tan x dx = 0. Solution:



Given,

Question 8.



Solution:

Question 9. (1 + cos x) dy = (1 – cosx) dx Solution:

Question 10.



Solution:

Rajasthan Board RBSE Class 12 Maths Chapter 12 Differential Equation Ex 12.5 Solve the following differential equations : Question 1.

(x + y)2 = a2 Solution: Given equation



Question 2.

= Solution:



Question 3. cos (x + y) dy = dx Solution:



Question 4.

ex+y = 1 +



Solution:

Question 5. (x + y) (dx – dy) = dx + dy Solution:



Question 6.

Solution:



Question 7.

x+ +y = sin-1 ( )



Solution:

Question 8.

= + 1 Solution:



Question 9.

= sec (x + y) Solution:



Question 10.



Solution:

Rajasthan Board RBSE Class 12 Maths Chapter 12 Differential Equation Ex 12.6 Solve the following differential equations : Question 1. x2 ydx – (x3 + y3) dy = 0 Solution:



Question 2.

Solution:



Question 3.

Solution:



Question 4.

Solution:



Question 5.

Solution:



So f(x,y) is homogeneous function. ∴ Given equation is homogeneous Now y = vx …..(2)



Question 6. (x2 +y2) dx = 2xydy Solution:



Question 7.

Solution:



Thus f (x, y) is homogeonous function. ∴ Given differential equation is homogeneous Putting x = vy …..(2)



Question 8. (3xy + y2) dx + (x2 + xy) dy = 0 Solution:



Question 9.

Solution:



Question 10. x (x-y) dy = y(x+y) dx



Solution:


Solution:



h and k are such that 3h + 2k – 5 = 0 and 2h + 3k – 5 = 0 Putting these values in equation h = 1 ,k = 1 This is homogeneous equation



Question 2.

Solution:

Question 3. (2x + y + 1) dx + (4x + 2y – 1) dy = 0 Solution: (2x + y + 1) dx + (4x + 2y – 1) dy = 0



Question 4.

Solution:



∴ Integrating x + C = – 2v – 2 log (1 – v) or x + C + 2(x + y) + 2 log (1 – x – y) = 0 [∵ v = x + y] or 3x + 2y + 2 log (1 – x – y) + C = 0, This is required solution. Question 5.

Solution:



Rajasthan Board RBSE Class 12 Maths Chapter 12 Differential Equation Ex 12.8 Solve the following differential equations :



Question 1.

Solution:

Question 2.

Solution:



Question 3.

Solution:



Question 4.



Solution:

Question 5.

Solution:



Question 6.

Solution:



Question 7.

Solution:



Question 8.

Solution:



Question 9. dx + xdy = e-y sec2 y dy Solution: dx + x dy = e-x sec2 y dy



Question 10.



Solution:


+ xy = x3y3.



Solution:

Question 2.

= ex-y (ex – ey) Solution:



Question 3.

– y tan x = – y2 sec x



Solution:

Question 4.

tan x cos y + sin y + esin x = 0 Solution:



Question 5.

+ x sin 2y = x3 cos2 y



Solution:

Question 6.



Solution:

Question 7.

Solution:



Rajasthan Board RBSE Class 12 Maths Chapter 12 Differential Equation Miscellaneous Exercise Question 1. Solution of differential equation



Solution:

Hence, option (b) is correct. Question 2.

Solution:

Thus Option (a) is correct. Question 3.

Solution of + cos x tan y = 0 is : (a) log sin y + sin x + C (b) log sin x sin y = C



(c) sin y + log sin x + C (d) sin x sin y + C Solution:

Hence, option (a) is correct. Question 4.

Solution:

Thus Option (b) is correct. Question 5.

Solution:



Thus Option (a) is correct. Question 6.

Solution:


Solution = cos2 y is : (a) x + tan y = C (b) tan y = x + C (c) sin y + x = C (d) sin y – x = C Solution:




Solution:

Thus Option (d) is correct. Question 9. By which displacement the differenting equation

will be correct into linear equation : (a) y = 1 (b) y2 = t



(c) = t

(d) = t Solution: Hence, option (c) is correct. Question 10. By which displacement the differential equation

will be correct into linear equation :

(a) = v (b) y-2 = v (c) y-3 = v (d) y3 = v Solution: Hence, option (b) is correct. Question 11. Find the general solution of differential equation

Solution:

This is required solution Question 12.

Find the integrating factor of differential equation + y tan x = sin x. Solution:



Question 13. Find the integrating factor of differential equation

Solution:

Question 14.

Write the form of differential equation dy cos (x + y) = 1. Solution: Given equation is of the form of converting variables separately. Question 15. Write the form of differential function



Solution: Linear equation Question 16. Write the general solution of the following differential equation

Solution:



Similar Question



If we take equation

Question 17.

Solution:



Question 18.



Solution:

Question 19.

Solution:



Question 20.



Solution:

Rajasthan Board RBSE Class 12 Maths Chapter 13 Vector Ex 13.1 Question 1. Compute the magnitude of the following vectors :

Solution:



Question 2. Write two different vectors having same magnitude.



Solution:

Question 3. Write two different vectors having same direction. Solution:



Question 4.

Find the values of x and y so that the vectors 2 + 3 and x + y are equal. Solution: Two vectors are equal if their corresponding coefficients are equal



⇒ 2 = x and 3 = y x = 2 and y = 3. Question 5. Find the scalar and vector components of the vector with initial point (2,1) and terminal point (- 5,7). Solution: Coordinates of initial point A are (2, 1). and coordinates of terminal point B are (- 5, 7) Now from formula

Question 6.

Solution:

Question 7.

Find the unit vector in the direction of the vector = + + 2



Solution:

Question 8.

Find the unit vector in the direction of vector where P and Q are the points (1,2,3) and (4,5,6). Solution:



Question 9. For the given vectors,

and

Find the unit vector in the direction of the vector



Solution:

Question 10. Find a vector in the direction of vector

which has magnitude 8 units.



Solution:

Question 11. Show that the vectors

and are collinear.

Solution:



Question 12. Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are

and

respectively in the ratio 2:1. (i) internally (ii) externally. Solution:



Question 13. Find the position vector of the midpoint of the vector joining the points P(2,3,4) and Q(4, 1, -2).



Solution:

Question 14. Show that the points A, B and C with position vectors,

and

respectively from the vertices of a right angled triangle. Solution: Let O is the origin, then according to question,



Hence ∆ABC is a right angled triangle, or point A, B and C are the vertices of a right angled triangle.

Rajasthan Board RBSE Class 12 Maths Chapter 13 Vector Ex 13.2 Question 1. If magnitude of two vectors be 4 and 5 units, then find scalar product of them, where angle between them be : (i) 60° (ii) 90° (iii) 30° Solution:



Question 2.

Find . , if and are as follow :



Solution:




Solution:

Question 4. If coordinates of P and Q are (3, 4) and (12, 4) respectively, then find ∠POQ where O is origin.



Solution:

Question 5.

For which value of λ, vectors and are mutually perpendicular :



Solution:

Question 6. Find the projection of the vector

on the vector

Solution:



Question 7. If

and

then find a vector , so that , , represents the sides of a right angled triangle. Solution: Given that



Question 8.

If | + | = | – |, then prove that and are mutually perpendicular vectors. Solution:




Question 9. If coordinates of points A, B, C and D are (3, 2, 4), (4, 5, -1), (6, 3, 2) and (2, 1, 0)

respectively, then prove that lines and are mutually erpendicular. Solution: Given that coordinates of points A, B, C and D are (3,2,4), (4,5,-1), (6,3,2) and (2,1,0) respectively. Then position vectors of A, B, C and D with respect to origin are



Question 10.

For any vector , prove that

Solution:



Question 11. Use vectors to prove that sum of square of diagonal of a parallelogram is equal to the sum of square of their side. Solution: Let OACB is a parallelogram. Taking O as origin, the position vectors of A and B

are and respectively.

∴ Sum of the squares of diagonals = sum of the squares of sides. Hence the sum of square of diagonals of a parallelogram is equal to the sum of square of their sides.



Rajasthan Board RBSE Class 12 Maths Chapter 13 Vector Ex 13.3 Question 1. Find vector product of vectors

and

Solution:

Question 2. Find perpendicular unit vector of vectors

and

Solution:



Question 3.

For vectors and , prove that



Solution:





Question 5.

If , , are unit vectors, such that

. = 0 = . and angle between and is , then prove that = ± 2 (

× ) Solution: Given that



Question 6. Find the value of

Solution:

We know that if and are two vectors and θ is the angle between them, then



Question 7. Find vector perpendicular to the vectors

and



whose magnitude is 9 unit. Solution:

Question 8. Show that:

also, explain geometrically. Solution:



= 2(vector area of parallelogram ABCD). Thus we conclude that area of parallelogram whose adjacent sides are diagonals of given parallelogram is twice the area of given parallellogram. Question 9.

For any vector , prove that



Solution:

Question 10. If two adjacent sides of a triangle are represented by vectors

and

then find the area of triangle.



Solution:

Rajasthan Board RBSE Class 12 Maths Chapter 13 Vector Ex 13.4 Question 1. Prove that:

Solution:



Question 2.



Solution:

Question 3. Prove that vectors

and

are coplanar Solution:



Question 4. For which value of λ, following vectors are coplanar :

Solution:



Question 5. Prove that the following four points are coplanar: (i) A(- 1,4,-3), B(3,2,-5), C(- 3,8,-5), D(- 3,2,1) (ii) A(0,-1,0), B(2,1,-1), C(1,1,1), D(3,30) Solution: (i) Given points are A(-1,4,-3), B(3,2,-5), C(- 3,8,-5) and D(-3,2,1)




and

are vector sides of right angled triangle. Solution:



Given sides are

Question 7. Find the volume of parallelopiped, whose coterminous edges are given by following vectors :

Solution: (i) The coterminous edges are



(ii) The coterminous edges are

Rajasthan Board RBSE Class 12 Maths Chapter 13 Vector Ex 13.5 Question 1. Find the value of

Solution:




Solution:



Question 3. Evaluate the formula

Solution:



Question 4.

For any vector , prove that:

Solution:



We know that


Solution:



Question 6.

Prove that , , are coplanar if × , × , × [/latex], are coplanar Solution:




Solution:

Question 8.

If magnitude of two and are √3 and 2 respectively and. = √6, then find

the angle between and Solution:



Question 9. Find the angle between the vectors

and

Solution:

Question 10. Find the projection of vector



Solution:

Question 11. Find projection vector on vector

on vector



Solution:

Question 12. Find the value of

Solution:

Question 13.

Find the magnitude of two vectors and , if their magnitude are equal and angle

between them is 60° and their scalar product is Solution:




Question 14.

If for a vector , ( – ).( + ) = 12, then | |. Solution:




Question 15.

such that + λ is perpendicular on , then find the value of λ. Solution:



According to question + λ is a perpendicular to

Question 16.

If vertices , , are such that

+ + =

then, find the value of . + . + . Solution:



Question 17. If vectors/1, B, C of triangle ABC are (1, 2,3), (-1,0, 0,), (0, 1, 2) respectively, then find ∠ABC. Solution: Let O be the origin.

Rajasthan Board RBSE Class 12 Maths Chapter 14 Three Dimensional Geometry Ex 14.1 Question 1. Find the direction cosines of a line which makes equal angles with the coordinate axis. Solution: Let the line makes equal angle 0 with the coordinate axis. ∴ Direction cosines are



l = cos θ, m = cos θ, n = cos θ but l2 + m2 + n2 = 1

Question 2. Find the direction of the line joining two points (4, 2, 3) and (4, 5, 7). Solution: Let dc’s of line joining two points P(x1,y1,z1) and Q(x2,y2,z2) are

Question 3. If a line has direction ratios 2, -1, -2 determine its direction cosines. Solution: Given : a = 2, b = -1, c = -2



Let dC’s of line are l,m and n.

Question 4.

A vector makes angles of 45°, 60°, 120° with X, Y and Z-axis respectively. If

magnitude of is 2 units then find Solution:

Rajasthan Board RBSE Class 12 Maths Chapter 14 Three Dimensional Geometry Ex 14.2 Question 1. Find the equation of the line which passes through the points (5, 7,9) and is parallel to the following axis : (i) A-axis



(ii) F-axis (iii) Z-axis Solution: Let position vector of point (5, 7, 9)

(i) The line parallel to X-axis passes through point (1,0,0) whose position vector

∴ Vector equation of required line.

Cartesian equation of line :

(ii) Let the line parallel to F-axis passes through point (0, 1,0). ∴ Position vector of point (0, 1, 0).

∴ required vector equation of line



(iii) The line parallel to Z-axis passes through (0, 0, 1).

∴ Required vector equation of line

Question 2. Find the equation of the line in vector and in cartesian form that passes through the point with vector

and

is parallel to the vector



Solution: Position vector of the given point is

Question 3. Find the equation of the line parallel to the vector

and passes through the point (5,-2,4). Solution : Line passes through (5, -2, 4). ∴ Position vector of (5, – 2, 4)



Question 4. Find the equation of the line passes through the point (2,-1,1) and is parallel to the

line = = Solution:

The equation of line passing through point (2,-1, 1) and parallel to the given line

= = because dc’s of parallel lines are same.

For vector equation of line passing through (2, – 1, 1) and parallel to , The position vector of (2, -1, 1).

Question 5.

If cartesian equation of line is = = , then find vector equation of line. Solution:

Question 6. Find the equation of a line in cartesian form which passes through (1, 2, 3) and is

parallel to = = Solution: Let a, b, c are dc’s of line passing through (x1,y1,z1) then equation of the line



Here line passes through point (1, 2, 3) and parallel to line

Question 7. Coordinates of three vertices of parallelogram ABCD are A(4, 5, 10), B(2, 3, 4) and C(1, 2, – 1). Find vector and cartesian equation of AB and AC. Also Find coordinates of D. Solution: Let O is origin. ∴ Position vector of points A, B and C are respectively.



(iii) For coordinates of point D. Let (x1, y1, z1,) are the coordinates of point D. ∵ ABCD is a parallelogram whose diagonals AC and BD bisect each other at point D. Therefore mid points of AC and BD will coincide. So the coordinate of mid point of



AC.

Question 8. Cartesian equation of a line is 3x + 1 = 6y – 2 = 1 -z. Find the point from where it passes and also find its direction ratios and vector equation. Solution: The equation of given line



Question 9. Find an equation passes through (1,2,3) and is parallel to the vector

Solution:



Question 10. Find the equation of line in vector and cartesian form that passes through the point with position vector

and is in the direction

Solution:

Question 11. Find the cartesian equation of a line passes through the point (- 2, 4, -5) and is parallel to



Solution: Let line passes through point (x1,y1,z1) whose dc’s are a,b,c then equation of line is

Question 12. The cartesian equation of a line is

Write its vector form. Solution: The cartesian equation of line is

Question 13. Find the equation of line in vector and cartesian form that passes through origin and (5, -2, 3). Solution:

(i) Position vector of origin O(0,0,0) is = and position vector equation of (5, – 2, 3) is



(ii) Line passing through origin O(0, 0, 0) whose dc’s are 5,-2, 3.

Question 14. Find the equation of line in vector and cartesian form that passes through the points (3, -2,-5) and (3,-2,6). Solution: Let line passes through point (3,-2,-5) and (3,-2,6).

(ii) Line passes through A(3, -2,-5) and 5(3, – 2, 6). Cartesian equation of line AB



Rajasthan Board RBSE Class 12 Maths Chapter 14 Three Dimensional Geometry Ex 14.3 Question 1. Find the angle between the lines

Solution:

Question 2. Find the angle between the lines



Solution:

Question 3. Show that the line passes through the points (1,-1,2) and (3,4,-2) is perpendicular to the line passes through the points (0, 3, 2) and (3, 5, 6). Solution: The equation of line passing through points (1,-1,2) and (3, 4,-2) is



Question 4.

Find k, if lines = = and = = are perpendicular to each other.



Solution:

Question 5. Find the vector equation of a line passing through point (1,2,- 4) and perpendicular to the both lines

Solution:



Question 6. Find the cartesian equation of a line that passes through point (- 2, 4, -5) and is parallel to

Solution:



Let Sine passing through (x1, y1, z1) whose dc’s are a, b and c is

Rajasthan Board RBSE Class 12 Maths Chapter 14 Three Dimensional Geometry Ex 14.4 Question 1.

Show that the lines = = and = = z, intersects. Find their point of intersection. Solution:

It is clear from above that both lines intersect each other and the coordinates of point of intersection are (-1, -1, -1). Question 2. Check, whether following lines intersect or not



Solution:

Question 3.

Find the foot of perpendicular drawn from point (2, 3, 4) to the line = = Also find the perpendicular distance from point to line. Solution: The equation of given line is



Question 4. Find the vector equation of line that passes throguh point (2, 3, 2) and is parallel to



the line

Also find the distance between the line. Solution: Line Passes through point (2,3,2) ∴ Position vector of the point is



Rajasthan Board RBSE Class 12 Maths Chapter 14 Three Dimensional Geometry Ex 14.5 Question 1. Find the shortest distance between the lines

Solution:



Question 2. Find the shortest distance between the lines



Solution: Minimum distance between the lines



Question 3. Find the shortest distance between the lines, whose vector equations are

Solution:



Question 4. Find the shortest distance between the lines, whose vector equations are



Solution: Comparing the line



Question 5. Find the shortest distance between the lines,

Also, find the equation of line with shortest distance.



Solution: Given lines are



Rajasthan Board RBSE Class 12 Maths Chapter 14 Three Dimensional Geometry Ex 14.6 Question 1. Find the equation of the plane which is perpendicular to x-axis and that passes through the point (2,-1,3). Solution: Equation of plane passing through point (2,-1,3) a(x – 2) + b(y + 1) + c(z – 3) = 0 ∵ Plane is perpendicular to T-axis. ∴ b = 0, c = 0 Hence required equation of plane a(x – 2) + 0(y + 1) + 0(z – 3) = 0 ⇒ a(x – 2) = 0

⇒ x – 2 = 0 (∵a ≠ 0) Question 2. Find the equation of the plane that passes through X-axis and point (3,2, 4). Solution: Equation of plane passing through (3, 2, 4) a (x – 3) + b(y – 2) + c(z – 4) = 0 …..(1) ∵ Plane passes through X-axis ∴ a = 0,d = 0 ⇒ by + cz = 0 …..(2) From (1), a = 0 so b (y – 2) + c(z – 4) = 0 …..(3) ⇒ by – 2b + cz – 4c = 0

⇒ by + cz – 2b – 4c = 0 ⇒ – 2b = c [∵ by + cz = 0 from (2)] ⇒ b = -2c

∴ Equation of plane passing through (3,2,4) and X-axis b(y- 2) + c(z – 4) = 0 ⇒ – 2c(y – 2) + c(z – 4) = 0 ⇒ -2y + 4 + z – 4 = 0 ⇒ 2y – z = 0 Question 3. A variable plane passes through the point (p, q, r) and meets the coordinate axis at point A, B and C. Show that the locus of a common point of plane passing through A, B and C and parallel to the coordinate planes, will be

Solution: Let equation of plane is

∴ Plane passes through point (p,q,r)



Again plane meets the coordinate points. Coordinates of points are (α,0,0) Coordinates of point B are (0, β, 0) and coordinates of point C are (0, 0, γ) ∴ Equation of planes, parallel to coordinate axis and passing through Point A is x = α …..(3) Point B is y = β …..(4) Point C is z = γ …..(5) ∴ Locus of the point of intersection is

+ + = 1 Question 4. Find the vector equation of a plane which is at a distance of 7 unit from the origin

and has as the unit vector normal to it. Solution: Given unit vector along normal

= i and distance from origin (0, 0, 0) = 7 units ∴ from vector equation of plane

Question 5. Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector 6i + 3j – 2k. Solution: Unit vector along 6i + 3j – 2k

Question 6.

Reduce the equation 3x – 4y + 12z = 5 or .(3i – 4j + 12k) = 5 to normal form and hence find the length of perpendicular from the origin to the plane. Also find direction cosines of the normal to the plane.



Solution: First method : Given

II method : On dividing 3x – 4y + 12z – 5 by its absolute value 5.



Question 7. Find the vector equation of a plane which is at a distance of 4 units from the origin and direction cosines of the normal to the plane are 2, – 1,2. Solution: Given Dc’s of the normal to the plane are 2,-1, 2.

Question 8. Find normal form of the plane 2x – 3y + 6z + 14 = 0. Solution: Equation of given plane is 2x – 3y + 6z + 14 = 0 Dc’s of normal plane are (2, 3, 6). ∴ Dc’s of normal are

Question 9. Find the equation of plane perpendicular of the origin from the plane is 13 and direction ratios of this perpendicular are 4,-3,12. Solution: Given DR’s of normal on plane are 4, -3, 12. ∴ DC’s of normal are



Question 10. Find a unit normal vector to the plane x + y + z – 3 = 0. Solution: Let given x + y + z – 3 = 0

Rajasthan Board RBSE Class 12 Maths Chapter 14 Three Dimensional Geometry Ex 14.7 Question 1. Find the angle between the planes :



Solution:



Question 2. Find the angle between the planes : (i) x + y + 2z = 9 and 2x – y + z = 15 (ii) 2 x – y + z = 4 and x + y + 2z = 3 (iii) x + y – 2z = 3 and 2x – 2y + z = 5 Solution:



If plane are a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 then angle between

them is



Question 3. Show that following planes are at right angles :

Solution: x – 2y + 4z = 10 and 18x + 17y + 4z = 49 a1 = 1,b1 = -2, c1 = 4 and a2 = 18 b2 = 17, c2 = 4 (i) Planes will be perpendicular if a1 a2 + b1b2 + c1c2 = 0 L.H.S. = 1 × 18 + (-2) × 17 + 4 × 4 = 18 – 34+16



= – 34 + 34 = 0 ∴ L.H.S. = R.H.S.

Question 4. Find λ, if following planes are perpendicular to each other:



Solution:

Question 5.

Find the angle between the line = = and the plane 2x + y – 3z + 4 = 0 Solution:



Normal vector of the plane 2x + y – 3z + 4 = 0

Question 6.

Find the ngle between the line = = and the plane 3x + 4y + z + 5 = 0.



Solution:

Question 7. Find the angle between

Solution:



We know that the angle between the line and the plane is

Question 8. Find the angle between

Solution: We know that angle between the line



Question 9. Determine the value of m, if line



Solution:

Question 10. Find m, if line

is parallel to the plane

Solution: If given line is parallel to given plane

Rajasthan Board RBSE Class 12 Maths Chapter 14 Three Dimensional Geometry Miscellaneous Exercise Question 1. Which of the following group is not direction cosines of a line : (a) 1,1,1 (b) 0,0, -1 (c)-1,0,0 (d)0,-1,0 Solution: Direction cosines of a line are proportional to direction ratio’s.



Let a, b and c are direction ratio’s, then according to question

Question 2. Consider a point P such that OP = 6 and makes angle 45° and 60° with OX and OY – axis respectively, then position vector of P will be :

Solution:



Question 3. Angle between two diagonals of a cube is :

Solution: Let the adjacent cores of cube of side ‘a’ are OA, OB, OR to be taken as coordinate axis.



Then the coordinates of the vertices of cube are following :

Question 4. Direction cosines of 3i be (a) 3,0,0 (b) 1,0,0 (c)-1, 0,0 (d)-3,0,0



Solution: Given vector

whose direction ratio’s are 3, 0, 0.

Question 5. vector form of line

(a) (3i + 4j – 7k) + ?(-2i – 5j + 13k) (b) (- 2j – 5j + 13k) + ?(3i + 4j – 7k) (c) (- 3i – 4j + 7k) + ?(- 2i – 5j + 13k) (d) None of these Solution:

∴ Position vector of point A

∴ Direction ratio of line are -2,-5, 13

∴ Vector equation of line

Hence, (a) is the correct option. Question 6. If lines

are perpendicular to each other than value of ? is : (a) 0 (b) 1 (c) -1 (d) 2



Solution:

Question 7. Shortest distance between lines

(a) 10 unit (b) 12 unit (c) 14 unit (d) None of these Solution:



Question 8. Angle between line



Solution: We know that angle between two lines

Question 9. If equation lx + my + nz = p is normal form of a plane, then which of the following is not true : (a) l, m, n are direction cosines of normal to the plane (b) p is perpendicular distance from origin to plane (c) for every value of p, plane passes through origin (d) l2 + m2 + n2 = 1 Solution: ∵ P is distance of the plane from origin. So, plane can pass through origin only if p = 0 otherwise not for other values. Hence, (c) is correct option. Question 10. A plane meets axis in A, B and C such that centroid of ? ABC is (1, 2, 3) then equation of plane is :



Solution:

Let equation of plane + + = 1 which meets the coordinate axis on points A

(a,0,0), B(0,b,0) and C (0,0,c), then centroid of ∆ABC will be ( , , )

Question 11. Position vectors of two points are

Equation of plane passing through Q and perependicular of PQ is

Solution: Let position vector of point P.

and position vector of point Q.

then = position vector of Q- position of vector of P



∴ Equation of plane passing through point Q ( ) perpendicular to PQ is

Question 12. Relation between direction cosines of two lines are l – 5m + 3n = 0 and 7l2 + 5m2 – 3n2 = 0 Find these lines. Solution: Given



Question 13. Projection of a line on axis are – 3, 4, – 12. Find length of line segment and direction cosines. Solution: Projection of a line coordinate axis are the direction ratios of a line. If direction cosines are l, m, n then

Question 14. Prove that the line joining the points (a, b, c) and (a’ b’, c’) passes through origin, if aa’+ bb’+ cc’ = pp’ where p and p’ are distance of points from origin. Solution: According to question, distance of points (a, b, c) and (a’, b’, c’) from origin.



Question 15. Find the equation of plane, passes through P (-2,1,2) and is parallel to the two vectors

Solution: ∵ Plane passes through point P(- 2, 1, 2).

∴ Equation of plane is a(x + 2) + b(y – 1) + c(z – 2) = 0



But plane is travelled to the vector

Rajasthan Board RBSE Class 12 Maths Chapter 15 Linear Programming Ex 15.1 Solve the following linear programming problem by graphical method : Question 1. Minimize Z = -3x + Ay subject to the constraints x + 2y ≤ 8 3x + 2y ≤ 12 and x ≥ 0, y ≥ 0 Solution: Converting the given inequations into equations x + 2y = 8 3x + 2y = 12 Region represented by x + 2y ≤ 8 : The line x + 2y = 8 meets the coordinate axis at



A(8, 0) and B(0, 4). x + 2y = 8

X 8 0

y 0 4

A(8,0),B(0,10) Join the points A and B to obtain the line. Clearly (0, 0) satisfies the inequation x + 2y = 8. So the region containing the origin represents the solution set of the inequation. Region represented by 3x + 2y ≤ 12 : The line 3x + 2y = 12 meets the coordinate axis at C(4, 0) and D(0, 6). 3x + 2y = 12

X 4 0

y 0 6

C(4, 0); D(0,6) Join the points C and D to obtain the line. Clearly (0, 0) satisfies the inequation 3x + 2y = 12. So the region containing. The origin represents the solution set of the inequations. Region represented by x ≥ 0, y ≥ 0 : Since every point in the first quadrant satisfies these inequations. So the first quadrant is the region represented by the inequations x ≥ 0, y ≥ 0. The shaded region OCEB represents the common region of the above inequations. This region is the feasible region of the given linear programming problem.

The coordinates of the comer points of the feasible region are O(0, 0), C(4, 0), E(2, 3) and 5(0, 4). The point E(2, 3) has been obtained by solving the equations of the corresponding intersecting lines simultaneously. The values of the objective function on these points are given in the following table :

Point x-coordinate y-coordinate Objective function Z = 3x + 4y

O 0 0 Z0 = -3(0) + 4(0) = 0

C 4 0 Zc = -3(4) + 4(0) = -12

E 2 3 ZE = -3(2) + 4(3) = 6



B 0 4 ZB = -3(0) + 4(4) = 16

It is clear from the above table that the objective function has minimum value at comer point C(4, 0). So the minimum value of given linear programming problem at x = 4 and y = 0 is – 12. Question 2. Maximize Z = 3x + 4y subject to the x + y ≤ 4 constraints x ≥ 0, y ≥ 0 Solution: Converting the given inequations into equation x + y = 4 Region represented by x + y ≤ 4 : The line x + y = 4 meets the coordinate axis at A(4, 0) and B(0, 4).

X 4 0

y 0 4

A(4, 0); B(0,4) Join the points A and B to obtain a line. Clearly (0, 0) satisfies the inequation x + 2y ≤ 4. So the region containing the origin represents the solution set of the inequation. Region represented by x ≥ 0,y ≥ 0 : Since every point in the first quadrant satisfies these inequations. So the first quadrant is the region represented by the inequations x ≥ 0 and y ≥ 0. The shaded region OAB represents the common region of the above inequations. This region is the feasible region of the given Linear Programming Problem.

The coordinates of the corner points of the shaded feasible region are O(0, 0), A(4.0) and B(0, 4). The values of the objective function of these points are given in the following table :


O 0 0 Z0 = 3(0) + 4(0) = 0

A 4 0 ZA = 3(4) + 4(0) = 12

B 0 4 ZB = 3(0) + 4(4) = 16

It is clear from the table, the objective function has maximum value at point B(0, 4). So, the required solution of the given LPP, is x = 0, y = 4 and the maximum value is Z = 16.



Question 3. Minimize Z = – 50x + 20y Subject to the constraints 2x – y ≥- -5 3x + y ≥ 3 2x- 3y ≤ 12 and x ≥ 0, y ≥ 0 Solution: Converting the given inequations into equations 2x – y = – 5 …..(1) 3x + y = 3 …..(2) 2x – 3y = 12 …..(3) Region represented by 2x – y ≥ – 5 : The line 2x – y = -5 meets the coordinate axis

at A( ,0) and B(0,5) 2x – y = -5

X -5/2 0

y 0 5

A( ,0) ;B (0,5) Join the points A and B to obtain the line. Clearly (0, 0) satisfies the inequation 2x – y ≥ – 5. So the region containing the origin, represents the solution set of the inequation. Region represented by 3x + y ≥ 3 : The line 3x + y = 3 meets the coordinate axis at A(1, 0) and B(0, 3). 3x + y = 3

X 1 0

y 0 3

C(1,0);D(0,3) Join the points C to D to obtain the line. Clearly the point (0, 0) does not satisfy the inequation. So the region opposite to the origin represents the solution set of the inequation. Region represented by 2x – 3y ≤ 12 : The line 2x – 3y = 12 meets the coordinate axis at E(6, 0) and F(0, -4). 2x – 3y = 12

X 6 0

y 0 -4

E(6,0);F(0,-4) Join the points E and F to obtain the line. Clearly (0, 0) satisfies the inequation 2x – 3y ≤ 12. So the region containing the origin represents the solution set of the inequation. Region represented by x ≥ 0, y ≥ 0 : Since every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, y ≥ 0.



The point of intersection of lines 2x – y = -5 and 3x + y = 3 is G ( , ).

The shaded area is an open and common region which satisfies the given constraints. This is the proper solution of the given LPP. The value of the objective function on These points are given in following table:

Point x Coordenate y Coordinal z = -50x + 2y

B 0 5 zB = -50 x 0 + 20 x 5 = 100

D 0 3 zD = -50 x 0+ 20 x 3 = 60

C 1 0 zC = -50 x 1 + 20 x 0 = -50

E 6 0 zE = -50 x 6 + 20 x 0 = -300

Clearly Z is minimum at x = 6 and y = 0 ∵ Minimum value of Z = – 300 Question 4. Minimize Z = 3x + 5y Subject to the constraints x + 3y ≥ 3 x + y ≥ 2 and x ≥ 0,y ≥ 0 Solution: Converting the given inequations into equations x + 3y = 3 x + y = 2 Region represented by x + 3y ≥ 3 : The line x + 3y = 3 meets the coordinate axis at



A(3, 0) and B(0, 1). x + 3y = 3

X 3 0

y 0 l

A(3, 0); B(0, 1) Join the points A to 5 to obtain a line. Clearly (0,0) does not satisfy the inequation x + 3y ≥ 3. So the region opposite to the origin, represents the solution set of the inequation. Region represented by x + y ≥ 2 : The line x + y = 2 meets the coordinate axis at points C(2, 0) and D(0, 2). x + y = 2

X 2 0

y 0 2

C(2, 0); D(0, 2) Join the points C to D to obtain the line. Clearly (0, 0) does not satisfy the inequation x + y ≥ 2. So the region opposite to origin, represents the solution set of the inequation. Region represented by x ≥ 0, y ≥ 0: Since every point in the first quadrant satisfies these inequations. So the first quadrant is the region represented by the equations x ≥ 0, y ≥ 0.

The point of intersection of lines x + 3y = 3 and x + y = 2 is E ( , ).

The shaded region A ED is an open and common region of given inequations. This is the proper solution of the given linear programming problem.

The coordinates of the shaded region are A(3, 0), E ( , ) and D(0, 2). The values of the objective function of these points are given in following table :



Clearly Z is minimum at x = and y =

Therefore x = and y = is the required solution of the L.P. problem and the minimum value of Z is 7. Question 5. Evaluate maximum and minimum value, where Z = 3x + 9y Subject to the constraints x + 3y ≤ 60 x + y ≥ 10 x – y ≥ 0 and x ≥ 0, y ≥ 0 Solution: Converting the given inequations into equations x + 3y = 60 …..(1) x + y = 10 …..(2) x = 0, y = 0 ….(3) Region represented by x + 3y ≤ 60: The line x + 3y = 60 meets the coordinate axis at A(60,0) and B(0, 20). x + 3y = 60

X 60 0

y 0 20

A(60, 0); B(0, 20) Join the points A to B to obtain the line. Clearly (0,0) satisfies the inequation x + 3y ≤ 60. So the region containing the origin, represents the solution set of the inequation. Region represented by x + y ≥ 10 : The line x + y = 10 meets the coordinate axis at point C(10,0) and D(0, 10). x + y = 10

X 10 0

y 0 10

C(10, 0);D(0, 10) Join point C to D to obtain the line. Clearly (0, 0) does not satisfy the inequation x + y ≥ 10. So the region opposite to origin, represents the solution set of the inequation. Region represented by x – y ≥ 0 : The line x – y = 0 meets the coordinate axis at O(0, 0), E(5, 5).

X 0 5

y 0 5

O(0, 0); E(5, 5) Join point O to E to obtain the line. Clearly (0, 0) satisfies the x – y ≥ 0. So the region cointaining origin represents the solution of inequation.



Region represented by x ≥ 0, y ≥ 0 : Since every point in the first quadrant satisfies these inequations. So the first qudrant is the region represented by the inequation x ≥ 0 and y ≥ 0.

The shaded region BDEF represents the solution region of the above inequations. This region is the feasible region of the given L.P.P. The coordinates of the comer points of the shaded feasible region are (0, 10), (5, 5), (15, 15) and (0, 20). The point C(15, 15) is the point of intersection of lines x =y and x + 3y=60 and point E is the intersection point of lines x + y = 10 and x = y. The values of the objective function on these points are given in the following table:


D 0 10 ZP = 3(0) + 9(10) = 90

E 5 5 ZE = 3(5) + 9(5) = 60

F 15 15 ZF = 3(15) + 9(15) = 180

B 0 20 ZH = 3(0) + 9(20) = 180

Clearly value of objective function Z is minimum at point E(5,5) where x = 5 and y = 5 and Z= 60 and the Z is maximum at two points C( 15, 15) and D(0, 20) where Z = 180. Hence the Minimum value of Z = 60 and Maximum value of Z = 180. Question 6. Minimize Z = x + 2y Subject to the constraints 2x + y ≥ 3 x + 2y ≥ 6 and x ≥ 0, y ≥ 0 Solution: Converting the given inequations into equations 2x + y = 3 ….(1) x + 2y = 6 ….(2) Region represented by 2x + y ≥ 3 : The line 2x + y – 3 meets the coordinate axis at

points A ( ,0) and B(0. 3). 2x + y = 3

X 3/2 0

y 0 3

A( ,0);B(0,3) Join the points ,A and B to get the line. Clearly (0,0) does not satisfy the inequation 2x + y ≥ 3. So the region opposite to the origin represents the solution set of the inequation.



Region represented by x + 2y ≥ 6 : The line x + 2y = 6 meets the coordinate axis at point C(6, 0) and D(0,3). x + 2y = 6

X 6 0

y 0 3

C(6, 0); D(0, 3) Join point C to D to obtain the line. Clearly (0,0) does not satisfy the inequation x + 2y ≥ 6. So the region opposite to the origin represents the solution set of the inequation. Region represented by x ≥ 0 and y ≥ 0 : Since every point in the first quadrant satisfies the inequations. So the first quadrant is the region represented by the inequations x ≥ 0 and y ≥ 0.

In the shaded region on CB, each and every point on line CB is satisfying the given inequations. So the values of the objective function on these points are given in the following table :

Point x-coordinate v-coordinate Objective function Z = x + 2y

5 0 3 ZB= 0 + 2(3) = 6

C 6 0 Zc = 6 + 2(0) = 6

It is clear from the table that required solution of given L.P.P. every point is on line BC and the minimum value of Z = 6. Question 7. Find maximum and minimum value where Z = 5x + 10y Subject to the constraints x + 2y ≤ 120 x + y ≥ 60 x – 2y ≥ 0 and x ≥ 0, y ≥ 0 Solution: Converting the given inequations into equations x + 2y = 120



x + y = 60 x – 2y = 0 Region represented by x + 2y ≤ 120 : The line x + 2y = 120 meets the coordinate axis at points ,4(120, 0) and B(0, 60). x + 2y = 120

X 120 0

y 0 60

A(120,);B(0,60) Join points A and B to obtain the line, clearly (0, 0) satisfies the given inequation. So the region containing the origin represents the solution set of the inequation x + 2y ≤ 120. Region represented by x + y ≥ 60 : The line x + y = 60 meets the coordiante axis at point C(60,0) and B(0, 60). x + y = 60

X 60 0

y 0 60

C(60,0) and D(0, 60). Join the points C to 5 to obtain a line. Clearly (0,0) does not satisfy the given inequation x + y ≥ 60. So the region opposite to the origin represents the solution set of the inequation. Region represented by x – 2y ≥ 0 : The line x = 2y meets the coordinate axis on origin (0,0) and any other points (60, 30). x = 2y

X 0 60

y 0 30

O(0,0);E(60,30) Join the points (0,0) and E(10,5) to obtain a line. Clearly (0, 0) satisfies the given inequation. So the region containing the origin represents the solution set of the inequation x – 2y ≥ 0. Region represented by x ≥ 0, y ≥ 0 : Since every point in the first quadrant satisfies these inequation. So, the first quadrant is the region represented by the inequation x ≥ 0 and y ≥ 0. The shaded region ACEF represents the common region of the above inequations. This region is the feasible region of the given LPP.



The point of intersection of lines x + 2y = 120 and x + y = 60 is (0, 60). The point of intersection of lines x + 2y = 120 and x – 2y = 0 is (60,30) and the point of intersection of lines x + y = 60 and x – 2y = 0 is (20, 40). The coordinates of the vertices (comer points) of the shaded feasible region are 4(120,0), C(60,0), £(40, 20) and F(60, 30). The values of the objective function at these points are given in the following table.


A 120 0 ZA = 5(120) + 10(0) = 600

C 60 0 ZC = 5(60) + 10(0) = 300

E 40 20 ZE = 5(40) + 10(20) = 400

F 60 30 ZF = 5(60) + 10(30) = 600

Clearly the value of Z is minimum at C(60, 0) = 300 and the maximum value of Z= 600 at every point of line AF. Question 8. Maximize Z = x + y Subject to the constraints x – y ≤ -1 -x + y ≤ 0 and x ≤ 0,y ≤ 0 Solution: Converting the given inequations into equations x – y = – 1 -x + y = 0 Region represented by x – y ≤ -1 : Following are the points obtained by the line x – y = -1

X -1 0

y 0 1

A(-1,0);B(0,1) Join the points A and B to obtain the line. Clearly (0, 0) does not satisfy the inequation x – y ≤ -1. So the region opposite to the origin represent the solution set of the inequation.



Region represented by -x + y ≤ 0 : The line -x + y = 0 meets the coordinates axis at B(0,0) and C( 1,1). -x + y = 0

X 0 1

y 0 1

O(0,0);C(1,1) Join the points O and C to obtain this line. Clearly (0,0) satisfies the inequation -x + y ≤ 0. So, the region containing the origin represents the solution set of this inequation. Region represented by x ≥ 0, y ≥ 0 : Since every point in the quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0 and y ≥ 0.

It is clear from the figure that line joining point A(-1, 0) and B(0, 1) is parallel to the line joining point C(1,1) to D(2, 2). Hence there is no finite solution of the given LPP. Minimum and Maximum value do not exist. Question 9. Minimize Z = 3x + 2y Subject to the constraints x + y ≥ 8 3x + 5y ≤ 15 and x ≥ 0,y ≥ 0 Solution: Converting given inequations into equations x + y = 8 ……(1) 3x + 5y = 15 …..(2) Region represented by x + y ≥ 8 : The line x + y = 8, meets the coordinate axis at A(8, 0) and B(0, 8) x + y = 8

X 8 0

y 0 8



A(8,0);B(0,8) Join the points A and B to obtain the line. Clearly, (0, 0) does not satisfy the inequation x + y ≥ 8. So the region opposite of the origin represents the solution set of the inequation x + y ≥ 8. Region represented by 3x + 5y ≤ 15 : The line 3x + 5y = 15 meets the coordinate axis at A(5, 0) and B(0, 3). 3x + 5y = 15

X 5 1

y 0 3

A(5, 0);B(0, 3) Join points A and B to obtain the line. Clearly (0,0) satisfies the inequation 3x + 5y ≤ 15. So, the region containing the origin represent the solution of the inequation 3x + 5y ≤ 15. Region represented by x ≥ 0, y ≥ 0 : Since every point in the first quadrant satisfies the inequations. So, the first quadrant is the region represented by the inequations x ≥ 0 and y ≥ 0.

It is clear from the graph that there is no common feasible solution region. Therefore no value exists for given objective function. Question 10. Maximize Z = -x + 2y Subject to the constraints x ≥ 3 x + y ≥ 5 x + 2y ≥ 6 and x ≥ 0,y ≥ 0 Solution: Converting the given inequations into equations x = 3 …..(1) x +y = 5 …..(2) x + 2y = 6 …..(3) y = 0 …..(4) Region represented by x + y ≥ 5 : The line x + y = 5



meets the coordinate axis on points A(5,0) and B(0, 5). x + y = 5

X 5 0

y 0 5

A(5,0);B(0,5) Join the points A to 5 to obtain the line. Clearly (0, 0) does not satisfy the inequation 0 + 0 = 0 ≥ 5. So the opposite region to the origin represents the feasible solution region. Region represented by x + 2y ≥ 6 : The line x + 2y = 6 meets the coordinate axis at point C(6,0) and D(0, 3). x + 2y = 6

X 6 0

y 0 3

C(6,0);D(0,3) Join point C to D to obtain the line. Clearly (0,0) does not satisfy the inequation 0 + 2(0) = 0 ≥ 6. So the region opposite to the origin represents the solution region of the inequation. Region represented by x ≥ 3 and y ≥ 0 : Since every point in the first quadrant satisfies these inequations. So the first quadrant is the region represented by the inequations x ≥ 3 and y ≥ 0.

It is clear from the graph that there is no any common region of the given inequations. Hence there does not exist any maximum value for the given inequations.

Rajasthan Board RBSE Class 12 Maths Chapter 15 Linear Programming Ex 15.2 Question 1. A dietician wishes to mix two types of foods in such a way that vitamin contents of the mixture contain atleast 8 units of vitamin A and 10 units of vitamin C. Food I contains 2 units/kg of vitamin A and 1 unit/kg of vitamin C. Food II contains 1 unit/kg of vitamin A and 2 units/kg of vitamin C. It costs ₹5/kg to purchase food ‘I’ and ₹7/kg



to purchase food ‘II’. Formulate this problem as a LPP to minimize the cost of such a mixture and solve it graphically. Solution: Let x kg of food I andy kg of food II in mixture. According to question, Cost of x kg food @ ₹5 per kg = ₹5x and cost of y kg food @ ₹ 7 per kg = ₹ 7y ∴ Total cost of mixture = ₹(5x + ly) Total units of vitmain A in x kg of food I in mixutre = 2x and total units of vitamin A in y kg of food II in mixture = y ∴ according to question, 2x + y ≥ 8 …..(1) Total units of vitamin C in x kg of food I in mixture = x and total units of vitamin C iny kg of food II in mixture = 2y ∴ According to question, x + 2y ≥ 10 …..(2) and x ≥ 0, y ≥ 0 ∴ Mathematically formulation of LPP is Minimum value of cost function Z = 5x + 7y Subject to the constraints 2x + y ≥ 8 x + 2y ≥ 10 x ≥ 0, y ≥ 0 Converting the inequation into the equation 2x + y = 8 x + 2y = 10 Region represented by 2x + y ≥ 8 : The line 2x + y = 8 meets the coordinate axis at A(4, 0) and B(0, 8). 2x + y = 8 Join the points A to B to obtain the line. Clearly (0, 0) does not satisfy the given inequation 2(0) + 0 = 0 ≥ 8. So the region opposite to the origin represents the solution set of the inequation. Region represented by x + 2y ≥ 10 : The line x + 2y = 10 meets the coordinate axis at the points C(10, 0) and D(0, 5). x + 2y = 10

X 10 0

y 0 5

C(10,0);D(0,5) Join the points C and D to obtain the line. Clearly (0, 0) does not satisfy the given inequation 0 + 2(0) = 0 ≥ 10. So the region opposite to the origin represents the solution set of the inequation. Region represented by x ≥ 0,y ≥ 0 : Since every point in first quadrant satisfies these equations. So the first quadrant is the region, represented by the inequation x ≥ 0 and y ≥ 0. The coordinate of the point of intersection of lines 2x + y = 8 and x + 2y = 10 are x = 2 and y = 4.



Shaded region CEB represents the common region of the inequations. This region is the feasible solution region of the inequations. The coordinates of the comer points are C(10, 0), E(2,4) and B(0,8). The value of the objective function on these points are given in the following table :


C 10 0 Zc= 5 x 10+7 x 0 = 5

E 2 4 ZE = 5 x 2+7 x 4 = 38

B 0 8 ZB = 5 x 0 + 7 x 8 = 56

In the above table value of objective function Z is minimum and Z = ₹38 at x = 2 and y = 4. Because feasible region is open. Therefore open region of inequation 5x + 7y ≤ 38 does not have any common feasible point. So the mixture has 2 kg of food I and 4 kg of food II in mixture and its minimum cost = ₹38. Question 2. A housewife wishes to mix together two kinds of food, X and F in such a way that the mixtures contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin contents of one kg of food are given below : Vitamin A Vitamin B Vitamin C

Food X 1 2 3

Food Y 2 2 1

One kg of food X costs ₹6 and one kg of food F cost ₹10. Find the least cost of the mixture which will produce the diet. Solution : Let x kg of food X and Y kg of food F are mixed together to make the mixture. Since one kg of food X contains one unit of Vitamin A and one kg of food Y contains 2 units of Vitamin A. Therefore x kg of food X and y kg of food Y will contain x + 2y units of Vitamin A. But the mixture should contain at least 10 units of Vitamins A. Therefore, x + 2y ≥ 10



Similarly, x kg of food Xand y kg of food F will produce 2x + 2y units of Vitamin B and 3x + y units of Vitamin C. But minimum requirements of Vitamin B and Vitamin C are respectively of 12 and 8 units. Therefore, 2x + 2y ≥ 12 and 3x + y ≥ 8 Since, the quantity of food X and food y cannot be negative. ∴ x ≥ 0, y ≥ 0 It is given that one kg of food X costs ₹6 and one kg of food Y costs ₹10. So, x kg of food X and y kg of food Fwill cost ₹(6x + 10y). Thus, the given linear programming problem is Minimize Z = 6x +10y Subject to the constraints, x + 2y ≥ 10 2x + 2y ≥ 12 3x + y ≥ 8 and x ≥ 0, y ≥ 0 Converting the inequation into equations x + 2y – 10 2x + 2y = 12 3x + y = 8 Region represented by x + 2y ≥ 10 : The line x + 2y = 10, meets the coordinate axis at ,A(10, 0) and B(0, 5). x + 2y = 10

X 10 0

y 0 5

A (10, 0); 5(0,5) Join the points A and B to obtain the line. We find that, the point (0, 0) does not satisfy the inequation x + 2y ≥ 10. So, the region opposite to the origin represents the solution set of the inquation. Region represented by 2x + 2y ≥ 12 : The line 2x + 2y = 12, meets the coordinate axis at C(6, 0) and D(0, 6). 2x + 2y = 12

X 6 0

y 0 6

C(6, 0); D(0, 6) Joint the points C and D to obtain the line. We find that, the point (0, 0) does not satisfy the inequation 2x + 2y ≥ 10. So, the region opposite to the origin, represents the solution set of the inequation. Region represented by 3x + y ≥ 8 : The line 3x + y = 8 meets the coordinate axis at E

( ,0);F(0,8) 3x + y = 8

X

0

y 0 8

E ( ,0); F(0,8) Join the points E and F to obtain the line. We find that, the point (0,0) does not satisfy the inequation 3x + y ≥ 8. So, the region opposite to the origin repesents the solution set of the inequation.



Region represented by x ≥ 0, y ≥ 0: Since every point in the first quadrant satisfies the inequations. So, the first quadrant is the region represented by the inequation x ≥ 0 and y ≥ 0. The shaded region AGHB represents the region of the above inequations. This region the feasible region of the given LLP.

The coordinates of the comer points of the shaded feasible region are .4(10, 0), G(2, 4), H(1, 5) and B(0, 8). Where the points G and H have been obtained by solving the equations of the corresponding intersecting lines, simu-ltaneously. The value of the objective function at these points are given in the following table :

Point x-coordinates y-coordinates Objective function Z = 6x + 10y

A 10 0 ZA = 6(10) + 10(0) = 60

G 2 4 ZG = 6(2) + 10(4) = 52

H 1 5 ZE = 6(1) + 10(5) = 56

B 0 8 ZB = 6(0)+ 10(8) = 80

Clearly, Z is minimum at x = 2 and y = 4. The minimum value of Z is 52. We observe that the open half plane represented by 6x + 10y < 52 does not have points in common with the feasible region. So, Z has minimum value equal to 52. Hence, the least cost of the mixture is ₹52. Question 3. One kind of cake requires 300 kg of flour and 15 g of fat and another kind of cake requires 150 g of flour and 30 g of fat, find the maximum number of cakes which can be made from 7.5 kg of flour and 600 g of fat assuming that there is no shortage of the other ingredients used in making the cakes. Formulate the LPP solve the problem by graphical method. Solution: Let x cakes are made of one kind and y cakes are made of another kind. Therefore objective function of maximum limit = x + y. ⇒ Z = x + y



Firstly according to question, there is 300x gm flour in one kind and 150y gm in another kind. ∴ 300x + 150y ≤ 7500 gm Secondly according to question, there is 15x gm fats in one kind of cake and 30y gm fats is in another kind. ∴ 15x + 30y ≤ 600 gm ∵ The number of cakes can never be negative so x ≥ 0 and y ≥ 0. Therefore mathematically formulation of Linear Programming Problem is the following : Maximum Z = x + y Subject to the constraints 300x + 150y ≤ 7500 15x + 30y ≤ 600 x ≥ 0, y ≥ 0 Converting the given inequations into equations 300x + 150y ≤ 7500 ⇒ 2x + y = 50 …..(1) and 15x + 30y = 600 ⇒ x + 2y = 40 …..(2) Region represented by 2x + y ≤ 50 : The line 2x + y = 50, meets the coordinate axis at A(25, 0) and B(0, 50). 2x + y = 50

X 25 0

y 0 50

A(25, 0); B(0, 50) Join the points A and B to obtain the line. Clearly, (0,0) satisfies the inequation 2 × 0 + 0 ≤ 50. So, the region containing the origin represents the solution set of the inequation. Region represented by x + 2y ≤ 40 : The line x + 2y = 40 meets the coordinate axis at C(40,0) and D(0, 20). x + 2y = 40

X 40 0

y 0 20

C(40, 0); D(0, 20) Join the points C and D to obtain the line. Clearly (0, 0) satisfies the inequation (0 + 2 × 0) < 40. So, the region containing the origin represents the solution set of the inequation. Region represented by x ≥ 0, y > ≥ : Since every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0 and y ≥ 0. The shaded region OAED represents the common region of the above inequations. This region is the feasible region of the given LPP.



The coordinates of the comer-points of the shaded feasible region are O(0,0), C(25,0), E(20,10) and B(0, 20). The point E has been obtained by solving the equations of the corresponding intersecting lines, simultaneously. The value of the objective function of these points are given in the following table :

Point x-coordinate y-coordinate Objective function Z = x + y

O 0 0 Z = x + y

A 25 0 ZA = 25 + 0 = 25

E 20 10 ZE = 20 + 10 = 30

D 0 20 ZD= 0 + 20 = 20

It is clear from the table that objective function has maximum value at point E(20, 10). Hence there are 20 cakes of one kind and 10 cakes of another kind. Maximum number of cakes = 20 + 10 = 30. Question 4. A manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hour on machine B to produce a package of nuts, it takes 3 hour of work on machine A and 1 hour on machine B to produce a package of bolts. He earns a profit of ₹ 2.50 per pacakge on nuts and ₹ 1 per package on bolts. How many pacakges of each should be produced each day so as to maximise his profit, if he operates his machines for at the most 12 hours a day ? Translate this problem mathematically and then solve it Solution: Let he should make x packets of nuts and y packets of bolts to get maximum profit. He gets profit of ₹ 2-50 per package of nuts and ₹1 per package of bolts. ∴ Objective function, Z = 2.50 x + y He works maximum of 1 hours on machine A and 3 hours on machine B.

∴ + ≤ 12 ⇒ 3x + y ≤ 36 …..(1)



and he works maximum of 3x hour on machine A and 1 hour on machine B.

∴ + ≤ 12 ⇒ x + 3y ≤ 36 ……(2) Because x and y are the number of nut and bolt so it can not be negative. ∴ x ≥ 0 and y ≥ 0 Mathematically formulation of Linear Programming Problem is as following : Maximum Z = 2.50x + y Subject to the constraints x + 3y ≤ 36 3x + y ≤ 36 x ≥ 0, y ≥ 0 Converting the inequations into equations x + 3y = 36 3x + y = 36 Region represented by x + 3y ≤ 36 : Line x + 3y =12 meets the coordinate axis at the points A( 12,0) and B(0, 4). x + 3y = 36

X 36 0

y 0 12

A(36,0);B(0, 12) Join the points A to 5 to obtain the line. Clearly (0, 0) satisfies the given inequation 0 + 3(0) = 0 ≤ 36. So the origin containing the region represents the solution set of the inequation x + 3y ≤ 36. Region represented by 3x + y ≤ 36 : The line 3x + y = 36, meets the coordinate axis at C(12,0) and D(0, 12). 3x + y = 36

X 12 0

y 0 36

C(12, 0); D(0, 36) Join the points C and D to obtain the line. Clearly, (0, 0) satisfies the inequation 3x + y ≤ 36. So, the region containing the origin represents the solution set of the inequation 3x + y ≤ 36. Region represented by x ≥ 0, y ≥ 0 : Since every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0 and y ≥ 0.



E (9, 9) is the point of intersection of lines 3x + y = 36 and x + 3y = 36 shaded region OCEB represents the common region of the inequations. This region is the feasible solution set the given L.P.P. The comer points of this feasible solution region are O(0, 0), C(12,0), E{9, 9) and B(0, 12). The value of objective function on these points is as following :

Point x-coordinate y-coordinate Objective function Z = 2.50 x + y

O 0 0 ZO = 2.50(0) + 0 = 0

C 12 0 Zc = 2.50(12) + 0 = 30

E 9 9 ZE = 2.50(9) + 9 = 31.50

B 0 12 ZB= 2.50(0) + 12 = 12

It is clear from the table that objective function Z = 2.50x + y has maximum value at point E(9,9). Hence manufacture should make nine packets of each nut and bolts daily to get maximum profit ₹ 31.50. Question 5. A dealer wishes to purchase a number of fans and sewing machines. He has only ₹5760 to invest and has space for at most 20 items. A fan costs him ₹360 and a sewing machine ₹240. His expectation is that he can sell a fan at a profit of ₹22 and a sewing machine at a profit of ₹18. Assuming that he can sell all the items that he can buy. How should he invest his money in order to maximize his profit ? Formulate this problem mathematically and then solve it. Solution: Let dealer should purchase x fans andy sewing machines. ∴ the cost of x fans = ₹ 360x and the cost of y sewing machines = ₹ 240y ∴ According to question, 360x + 240 ≤ 5760 According to place available to have x fans andy sewing machines x + y ≤ 20. The profit earned by dealer on x fans = ₹22x and profit earned on y sewing machines = ₹18y ∴ To earn maximum profits objective function Z = 22 x + 18y Mathematically formulation of LPP is as following : Maximum Z = 22x + 18y



Subject to the constraints 360x + 240y ≤ 5760 x + y ≤ 20 x ≥ 0, y ≥ 0 Converting the given inequations into equations 360x + 240y = 5760 ⇒ 3x + 2y = 48 …..(1) and x + y = 20 ……(2) Region represented by 3x + 2y ≤ 48 : The line 3x + 2y = 48, meets the coordinate axis at A(16, 0) and B(0, 24). 3x + 2y = 48

X 16 0

y 0 24

A(16,0);B(0,24) Join the points A and B to obtain the line. Clearly (0, 0) stisfies the inequation 3 × 0 + 2 × o = 0 ≤ 48. So, the region containing the origin represents the solution set of the inequation. Region represented by x + y ≤, 20 : The line x + y = 20 meets at the point E(20, 0) and F(0, 20). x + y = 20

X 20 0

y 0 20

C(20, 0); D(0, 20) Join the points E and F to obtain the line. Clearly (0,0), satisfies the inequation 0 + 0 ≤ 20. So, the region containing the origin represents the solution set of the inequation. Region represented by x ≥ 0, y ≥ 0 : Since every point in the first quadrant satisfies these inequation. So, the first quadrant is the region represented by the inequations x ≥ 0 and y ≥ 0. The shaded region OAED represents the common region of the above inequations. This region is the feasible region of the given LPP.

The coordinates of the comer points of this feasible region are O(0,0), A(16,0), E(8,



12) and D( 0,20) where E( 8,12) is the point of intersection of lines 3x + 2y = 48 and x + y = 20. The value of objective function on these points are given in the following table:


O 0 0 ZO = 22 x 0 + 18 x 0 = 0

A 16 0 ZA = 22 x 16 + 18 x 0 = 352

E 8 12 ZE = 22 x 8 + 18 x 12 = 392

D 0 20 ZD = 22 x 0 + 18 x 20 = 360

From the table it is clear that the objective function has maximum value ₹392 at point E(8,12) where x = 8 and y = 12. Hence dealer should purchase 8 fans and 12 sewing machines to get maximum profit ₹392. Question 6. A factory manufactures two types of screws, A and B. Each type of screw requires the use of two machines an automatic and a hand operated. It takes 4 minutes on the automatic and 6 minutes on hand opereated machines to manufacture a package of screw A, while it takes 6 minutes on automatic and 3 minutes on hand operated machines to manufacture a package of screw B. Each, machine is available for almost 4 hours on any day. The manufacturer can sell a package of screws A at a profit of 70 paise and screws B at ₹1. Assuming that he can sell all the screws he manufactures, how many packages of each type should the factory owner produce in a day in order to maximise his profit ? Solution: Let manufacturer should produce x packets of screw of type A and y packets of screw of type B. ∴ Profit on x packets = ₹0-70x and profit on y packets = ₹y ∴ Objective function to earn maximum profit Z = 0.70 x + y Time to produce x screws of type A = 4x minute and time to produce y screws of type B = 6y minute But automatic machine is available for 4 hours only. ∴ According to question. 4x + 6y ≤ 4 hour ⇒ 4x + 6y ≤ 240 minute Similarly time taken to prepare screws of type A on handmade machine = 6x minutes and time taken to prepare screws of type B on handmade machine = 3y minutes But handmade machine is available for 4 hours only a day. ∴ 6x + 3y ≤ 4 hour ⇒ 6x + 3y ≤ 240 minute ∵ x and y are the number of screws. ∴ x ≥ 0 and y ≥ 0 Mathematically formulation of this Linear Programming Problem is as the following : Maximize Z = 0.70x + y Subject to the constraints 4x + 6y ≤ 240 6x + 3y ≤ 240 x ≥ 0, y ≥ 0 Converting the given inequations into equations



4x + 6y = 240 …..(1) 6x + 3y = 240 …..(2) Region represented by 4x + 6y ≤ 240 : The line 4x + 6y = 240 meets the coordinate axis at A(60, 0) and B(0, 40). 4x + 6y = 240

X 60 0

y 0 40

A(60, 0); B(0, 40) Join the points A and 5 to obtain the line. Clearly (0, 0) satisfies the inequations 4 × 0 + 6 × 0 = 0 ≤ 240. So, the region containing the origin represents the solution set of the inequation. Region represented by 6x + 3y ≤ 240 : The line 6x + 3y = 240 meets the coordinate axis at points C(40, 0) and D(0, 80). 6x + 3y = 240

X 40 0

y 0 80

C(40,0);D(0,80) Join C and D to obtain the line. Clearly (0,0) satisfies the given inequation 6 × 0 + 3 × 0 = 0 ≤ 240. So the region containing the origin represents the solution set of the inequation. Region represented x ≥ 0 and y ≥ 0 : Since every point in the first quadrant satisfies these inequations. So the first quadrant in the region represented by the inequation x ≥ 0 and y ≥ 0.

The coordinate of point of intersection of lines 4x + 6y = 240 and 6x + 3y = 240 are x = 30 and y = 20. The shaded region OAED represent the common region of the inequations. This region is the feasible solution region of the inequations. The comer points of this solution region are O(0,0), A(40, 0), E(30, 20) and D(0, 40). The value of objective function on these points is as following table:

Point x-coordinate y-coordinate Objective function Z = 0.70 x + y

O 0 0 Z0 = 0.70 x 0 + 0 = 0



A 40 0 ZA = 0.70 x 40 + 0 = 28

E 30 20 ZE = 0.70 x 30 + 20 = 41

D 0 40 ZD = 0.70 x 0 + 40 = 40

From the table the value of objective function is maximum at point E(30, 20) = ₹41. Hence the manufacturer should produce 30 packets of screw of type A and 20 packets of screws of type B to get the maximum profit ₹41. Question 7. A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of types A require 5 minute each for cutting and 10 minute each for assembling. Souvenirs of type B require 8 minute each for cutting and 8 minutes each for assembling. There are 3 hour 20 minute available for cutting and 4 hour available for assembling. The profit is ₹ 5 each for type A and ₹ 6 each for type B souvenirs. How many Souvenirs of each type should the company manufacture in order to maximize the profit ? Solution: Let company should make x souvenirs of type A and y souvenirs of type B. Profit earned on x souvenirs = ₹ 5x and profit earned on y souvenirs = ₹ 6y ∴ To earn maximum profit, objective function is Z = 5x + 6y

∵ Time taken for cutting of souvenirs of type A = 5x minute and time taken for cutting of souvenirs of type B = 8y minute ∴ According to question, constraints for total time taken for cutting 5x + 6y ≤ 3 hours 20 minute ⇒ 5x + 6y ≤ 200 minute Similarly time taken for assembling of souvenirs of type A = 10x minute and time taken for assembling of souvenirs of type B = 8y minutes ∴ According to question, constraints for total time taken in assembling 10x + 8y ≤ 4 hour ⇒ 10x + 8y ≤ 240 minute Therefore mathematically formulation of Linear Programming Problem is as following : Maximum Z = 5x + 6y Subject to the constraints 5x + 8y ≤ 200 10x + 8y ≤ 240 x ≥ 0, y ≥ 0 Converting the given inequations into the equations. 5x + 8y = 200 …..(1) 10x + 8y = 240 …..(2) Region represented by 5x + 8y ≤ 200: The line 5x + 8y = 200 meets the coordinate axis on ponits A(40, 0) and B(0, 25). 5x + 8y = 200

X 40 0

y 0 25

A(40, 0); B(0, 25) Join A and D to obtain the line. Clearly (0, 0) satisfies the inequation 5(0) + 8(0) = 0 ≤ 200. So the region containing the origin represents the solution set of the inequation. Region represented by 10x + 8y ≤ 240 : The line 10x + 8 y = 240 meets the coordinate axis at the points A(24,0) and B(0, 30). 10x + 8y = 240



X 24 0

y 0 30

C(24,0); D(0,30) Join the points C and D to obtain the line. Clearly (0, 0) satisfies the given inequation 10(0) + 8(0) = 0 ≤ 240. So the region containing the origin represents the solution set of the inequation. Region represented by x ≥ 0 and y ≥ 0 : Since every point in first quadrant satisfies the given inequation so the first quadrant is the solution region of these inequations x ≥ 0 and y ≥ 0.

The coordinates of the point of intersection of lines 5x + 8y = 200 and 10x + 8y = 240 are x = 8 and y = 20. The shaded region represents the common region of inequations. This region is the feasible solution of the linear programming problem. The corner points of this region are O(0, 0), C(24, 0), E(8, 20) and B(0, 25). The value of objective function on these points is shown in the following table :


O 0 0 Z0 = 5 x 0 + 6 x 0 = 0

C 24 0 Zc = 5 x 24+6 x 0 = 120

E 8 20 ZE = 5 x 8 + 6 x 20 = 160

B 0 25 ZB = 5 x 0 + 6 x 25 = 150

From the table objective function has the maximum value ₹160 at the point E(8, 20) where x = 8 and y = 20. Question 8. A Farmer has two types of fertilizers F1 and F2. F1 consists of 10% nitrogen and 6% phosphoric acid and F2 consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions a farmer find that he needs atleast 14 kg of nitrogen and 14 kg of phosphoric acid for his crop. If F1 costs 60 paise/kg and F2 costs 40 paise/kg, determine how much of each type of fertilizer should be used so that nutrient requirements get a minimum cost. Solution: Let the quantity of fertilizers F1 = x kg and the quantity of fertilizers F2 = y kg. ∵ The cost of F1 is 60 paise per kg and the cost of F2 is 40 paise per kg.



Region represented by ( )x + ( ) y ≤ 14: Line 10x + 5y = 1400 meets the coordinate axis at points ,A(140,0) and B(0, 280). 10x + 5y = 1400

X 140 0

y 0 280

A(140, 0); B(0, 280) Join the points A to B to obtain the line. Clearly (0, 0) satisfies the given inequation

(0) + (0) = 0 ≤ 14 So the region containing origin represents the solution set of the inequation.

Region represented by ( )x + ( ) y ≤ 14: Line 6x + 10y = 1400 meets the

coordinate axis at point C ( ,0)and D(0, 140). 6x + 10y = 1400

X 700/3 0

y 0 140

C ( ,0); and D(0,140) Join C and D to obtain the line. Clearly (0, 0) satisfies the inequation 6(0) + 10(0) = 0 ? 1400. So the region containing origin represents the solution set of the inequation. Region represented by x ≥ 0 and y ≥ 0 : Since every point in first quadrant satisfies the inequations. So the first quadrant is the solution region of the inequations x ≥ 0 and y ≥ 0. Coordinates of the point of intersection of lines 10x + 5y = 1400 and 6x + 10y = 1400 are x = 100 and y = 80.



The shaded region CEB is the common region of the given inequations. This region is the feasible solution region of the inequations.

Corner points of this region are C ( ,0) E (100 80) and B(0,280). The value of objective function on these points are given in the following table :

From the table, it is clear that objective function is minimum at E(100, 80). Hence the quantity of nutrients in F1 must be used 100 kg and in F2 it must be 80 kg and minimum value = ₹ 92. Question 9. A merchant plans to sell two types of personal computers a desktop model and a portable model that will cost ₹ 25000 and ₹ 40000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant should stock to get maximum profit if he does not want to invest more than ₹70 lakhs and if his profit per unit on the desktop model is ₹4500 and on protable model is ₹5000. Solution: Let the number of desktop model = x and the number of portable model = y. ∴ Objective function to earn maximum profit Z = 4500x + 5000y Constraints for the total number of computers is



x + y ≤ 250 ∵ monthly demand of computers is not more than 250. Total cost of computers is 25000x + 40000y ? 70,000,00 ∴ x and y are number of computers. ∴ x ≥ 0 and y ≥ 0 Mathematically formulation of Linear Programming Problem is as following : Maximum Z = 4500x + 5000y Subject to the constraints x + y ≤ 250 25000x + 40,000y = 70,000,00 x ≥ 0 and y ≥ 0 Converting the given inequations, into the equations x + y = 250 …..(1) 25000x + 40000y = 7000000 ⇒ 25x + 40y = 7000 …..(2) Region represented by x + y ≤ 250 : The line x + y = 250 meets the coordinate axis at the points A(250, 0) and B(0, 250). x + y = 250

X 250 0

y 0 250

A(250, 0); B(0, 250) Join point A to B to obtain the line. Clearly (0,0) satisfies the inequation 0 + 0 = 0 ≤ 250. So the region containing origin represents the solution set of the inequation. Region represented by 25x + 40y ≤ 7000 : The line 25x + 40y = 7000 meets the coordinate axis on the points C(280, 0) and D(0, 175).

X 280 0

y 0 175

C(280, 0);D(0, 175) Join point C to D to obtain the line. Clearly (0,0) satisfies the given inequation 25(0) + 40(0) = 0 ≤ 7000. So the region containing the origin represents the solution set of the inequation. Region represented by x ≥ 0, y ≥ 0 : Since every point in the first quadrant satisfies these inequations. So the first quadrant is the region represented by the inequations x ≥ 0 and y ≥ 0. The coordinates of the point of intersection of lines x + y = 250 and 25x + 40y = 7000 are x = 200 and y = 50. The shaded region OAED is the common region of the above inequations. This region is the feasible solution of the Linear Programming Problem. The coordinates of the comer points of this region are O(0, 0), A(250, 0), E(200, 50) and D(0, 175).



The value of objective function on these points is given in the following table :


O 0 0 ZG = 4500 x 0 + 5000 x 0 = 0

A 250 0 ZA = 4500 x 250 + 5000 x 0 = 11,25,000

E 200 50 ZE = 4500 x 200 + 5000 x 50 = 900000 + 250000

= 1150000

D 0 175 ZD = 4500 x 0 + 5000 x 175 = 8,75000

It is clear from the table that the objective function value is maximum at point E(200, 50) and z = 11,50,000 So the merchant should purchase 200 desktop computer and 50 portable computer to earn maximum profit of ?11,50,000. Question 10. Two godowns A and B have grain storage capacity of 100 quintals and 50 quintals respectively. They supply to 3 ration shops, D, E and F whose requirements are 60, 50 and 40 quintals respectively. The cost of transporation per quintal from the godowns to the shop are given in the following table :

To/From A B

D 6 4

E 3 2

F 2.50 3

How should the supplies be transported in order that the transportation cost is minimum ? Solution: Let store A supplies x quintals to shop D and y quintals to shop E. ∴ remaining ration (100 – x – y) quintal will be supplied to shop F. ∴ Transporation charges from store A to shop D = ₹ 6x and transportation charges from store A to shop E = ₹ 3y and transportation charges from store A to shop E

= ₹ (100 – x – y) ∴ Total transportation charges from store A to shops D,



E and F = 6x + 3y + (100 – x – y). The remaining requirement of shop D = (60 -x) quintal The remaining requirements of shop E = (50 – y) quintal and the remaining requirement of shop F = [40 – (100 – x – y)] quintal which will be supplied from store B. ∴ Transportation charges from store B to shop D = ₹ 4 (60 – x). Transportation charges from store B to shop E = ₹2 (50 – y) and transportation charges from store B to shop F = ₹ 3(x + y – 60) ∴ Total trapsortation charges from store B to shops D, E and F = 4(60 – x) + 2(50 – y) + 3(x + y – 60). ∴ Total transportation charges from both stores A and B to the shops D, E and F.

Z = 6x + 3y + (100 – x – y) + 4(60 – x) + 2(50 – y) + 3(x + y – 60) = 2.5x + 1.5y + 410 Total capacity of stored = 100 quintal. ∴ x + y ≤ 100 Ration availed by shop D is x quintal from store A and remaining from store B. ∴ x ≤ 60 Similarly shop E avails y quintal from store A and remaining from store B. ∴ y ≤ 50 Similarly shop F avails (100 – x – y) quintals from store A and remaining from store B. ∴ x + y ≥ 60 ∵ x and y are quantity of ration in quintal. ∴ x ≥ 0 and y ≥ 0 So mathematical formulation of Linear Programming Probelm is as following : Z = 2.5x + 1.5y + 410 x + y ≤ 100 x ≤ 60 y ≤ 50 x + y ≥ 60 x ≥ 0, y ≥ 0 Converting these inequations into the equations x + y = 100 …..(1) x = 60 …..(2) y = 50 …..(3) x + y = 60 …..(4) x = 0 …..(5) y = 0 …..(6) Region represented by x + y ≤ 100 : Line x + y = 100 meets the coordinate axis on points A(100,0) and B(0,100). x + y = 100

X 100 0

y 0 100

A( 100, 0); B(0, 100) Join points A to B to obtain the line. Clearly (0,0) satisfies the inequation 0 + 0 = 0 ≤ 100. So the region containing the origin represents the solution set of the inequation. Region represented by x ≤ 60 : Line x = 60 is parallal toy-axis and its each point will satisfy the inequation in first quadrent so its solution region will be towards origin.



Region represented by y ≤ 50 : Line y = 50 is parallel to x-axis and its each point will satisfy the inequation in first quadrant. So. its solution region will be towards origin. Region represented by x + y ≥ 60: Line x + y = 60 meets the coordinate axis on the points G(60, 0) and H(0, 60). x + y = 60

X 60 0

y 0 60

G(60, 0); H(0, 60) Join G and H to obtain the line. Clearly the origin (0,0) does not satisfying the inequation so the region of solution set is opposite to the origin. Region represented by x ≥ 0 and y ≥ 0 : Since every point in first quadrant satisfies the inequations x ≥ 0 and y ≥ 0 so the solution set of these inequations is in the first quadrant.

The shaded region GJFM is the common region representing the inequations. This region is the feasible solution region of the inequations. Coordinates of comer points of this region are G(60,0), J(60,40), F(50,50) and M(10, 50) where point J is the point of intersection of lines x + y = 100 and x = 60, point F is the point of intersection of x + y = 100 and y = 50 and M is the point of intersection of x + y = 60 and y = 50. The value of objective function on these points is given in following table :

Point x-coordinate y-coordinate Objective function Z = 2.5x + 1.5y + 410

G 60 0 ZG = 2.5(60) + 1.5(0) + 410 = 560

J 60 40 Zj = 2.5(40) + 1.5(40) + 410 = 620

F 50 50 ZF = 2.5(50) + 1.5(50) + 410 = 610

M 10 50 ZM = 2.5(10) + 1.5(50) + 410 = 510

It is clear from the table the objective function has minimum value Z = ₹ 510 at the point M(10,50). Hence for minimum transportation charges merchant should supply the ration from store A to shops D, E and F as 10,50 and 40 quintals respectively and from store B to shops D, E and F as 50,0,0 quintal respectively.



Rajasthan Board RBSE Class 12 Maths Chapter 15 Linear Programming Miscellaneous Exercise Solve the following linear programming problem by graphical method : Question 1. Maximize Z = 4x + y Subject to the constraints, x + y ≤ 50 3x + y ≤ 90 and x ≥ 0, y ≥ 0 Solution: Converting the given inequations into equations x + y = 50 …..(1) 3x + y = 90 …..(2) Region represented by x + y ≤ 50 : The line x + y = 50 meets the coordinate axis at A(50, 0) and B(0, 50). x + y = 50

X 50 0

y 0 50

A(50, 0); B(0, 50) Join the points A and 5 to obtain the line. Clearly (0, 0) satisfies the inequation x + y ≤ 50. So, the region containing the origin represents the solution set of this inequation. Region represented by 3x + y ≤ 90: The line 3x + y = 90 meets the coordinate axis at the points C(30, 0) and D(0, 90). 3x + y = 90

X 30 0

y 0 90

C(30, 0); D(0, 90) Region represented by x ≥ 0 and y ≥ 0 : Since every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequation x ≥ 0 and y ≥ 0. The shaded region OCEB represents the common region of the above inequations. This region is the feasible region of the given LPP. E(20, 30) is the point of intersection of the lines x + y = 50 and lines 3x + y = 90. The coordinates of this region of solution are O(0, 0), C(30, 0), E(20, 30) and B(0, 50).



The value of objective function on these points are given in this table :

Point x-coordinate y-coordinate Objective function Z = 4x + y

O 0 0 ZO = 4(0) + (0) = 0

C 30 0 ZC = 4(30) + 0 = 120

E 20 30 ZE = 4(20) + 30 = 110

B 0 50 ZB = 4(0) + 50 = 50

It is clear from the table that objective fucntion is maximum on point C(30, 0). Hence maximum value of Z = 120. Question 2. Minimize Z = 3x + 2y Subject to the constraints x + y ≥ 8 3x + 5y ≤ 15 and x ≥ 0,y ≤ 15 Solution: Converting the given inequations into the equations : x + y = 8 …..(1) 3x + 5y = 15 …..(2) y = 15 …..(3) Region represented by x + y ≥ 8 : The line x + y = 8 meets the coordinate axis at C(8,0) and D(0, 8). Table for x + y = 8

X 8 0

y 0 8

A(8, 0); B(0, 8) Join the points C and D to obtain the line. We find that the point (0, 0) does not satisfy the inequation x + y > 8. So, the region opposite to the origin represents the solution set to the inequation.



Region represented by 3x + 5y ≤ 15 : The line 3x + 5y = 15 meets the coordinate axis at C(5,0) and D(0, 3). Table for 3x + 5y = 15

X 5 0

y 0 3

C(5,0);D(0,3) Join the points C and D to obtain the line. Clearly (0,0) satisfies the inequation 3x + 5y ≤ 15. So, the region containing the origin represents the solution set of this inequation. Region represented by y ≤ 15 : Line y = 15 is parallel to x-axis, its each point will satisfy the ineqnation in first quadrant. So, its solution region will be towards origin. Region represented by x ≥ 0 and y ≥ 0 : Since every point in the first quadrant satisfies these inequations. So the first quadrant is the region represented by the inequations x ≥ 0 and y ≥ 0.

There is no any common region for solution. Hence there is no feasible solution of this problem. Question 3. Minimize and maximize Z = x + 2y Subject to the constraints, x + 2y ≥ 100 2x – y ≤ 0 2x + y ≤ 200 and x ≥ 0,y ≥ 0 Solution: Converting the given inequations into equations x + 2y = 100 …..(1) 2x – y = 0 …..(2) 2x + y = 200 …..(3) x = 0 …..(4) y = 0 …..(5) Region represented by x + 2y ≥ 100 : The line x + 2y = 100 meets the coordinate



axis at A(10,0) and 5(0, 5). Table for x + 2y = 100

X 100 0

y 0 50

A(100, 0); B(0, 50) Join the points A and B to obtain the line. We find that the point (0, 0) does not satisfy the inequation x + 2y ≥ 100. So, the region opposite to the origin represents the solution set of the inequation. Region represented by 2x – y ≤ 0: The line 2x – y = 0 meets the coordinate points at C(0, 0) and D( 100, 200). 2x – y = 0

X 0 100

y 0 200

C(0, 0), D(100, 200) Join the points C(0,0) to D(100, 200) to obtain the line. Clearly (0, 0) satisfies. The inequation 2(0) – 0 = 0 ≤ 0. So the region containing origin represents the solution set of the inequation. Region represent by 2x + y ≤ 200 : The line 2x + y = 200 meets the coordinate axis at ,A(100,0) and B(0, 200). 2x + y = 200

X 100 0

y 0 200

E(100, 0); F(0, 200) Join points E and F to obtain the line. Clearly (0, 0) satisfies the inequation 2x + y ≤ 200. So the region containing the origin represents the solution set of this inequation. Region represented by x ≥ 0 and y ≥ 0 : Since eveiy point in the first quadrant satisfies these inequations. So the first quadrant is the region represented by the inequations x ≥ 0 and y ≥ 0. The shaded region BEFD represents the common region of the above inequations. This region is the feasible region of the given LPP.



where point E(20,40) is the point of intersection of lines x + 2y = 100 and 2x – y = 0. Point F(50, 100) is the point of intersection of lines 2x + y = 200 and 2x – y = 0. The value of objective function on these comer points of feasible region are E(20, 40), B(0, 50), D(0, 200) and F(50, 100).

Point x-coordinate y-coordinate Objective function Z = x + 2y

E 20 40 ZE = 20 + 2 x 40 = 100

B 0 50 ZB = 0 + 2 x 50 = 100

F 50 100 ZF = 50 + 2 x 100 = 250

D 0 200 ZD = 0 + 2 x 200 = 400

It is clear from the table that minimum value of objective function is = 100 at points E(20, 40) and B(0, 50) which is minimum on every point on line AEB and maximum value of objective function is on the point D(0, 200) where Z = 400. Hence Minimum value of Z = 100 and Maximum value of Z = 400. Question 4. Maximize Z = 3x + 2y Subject to the constraints x + 2y ≤ 10 3x + y ≤ 15 and x ≥ 0,y ≥ 0 Solution: Converting the given inequation into the equations x + 2y = 10 3 x + y = 15 x = 0 y = o Region represented by x + 2y ≤ 10 : The line x + 2y = 10, meets the coordinate axis



at A(10, 0) and B(0, 5). x + 2y = 10

X 10 0

y 0 5

A(10,0);B(0,5) Join point A(10,0) to B(0,5) to obtain the line. Since the origin (0,0) satisfies the inequation 0 + 2(0) = 0 ≤ 10. So the region containing origin represents the solution set of the equation. Region represented by 3x + y ≤ 15 : The line 3x + y = 15 meets the coordinate axis on points C(5,0) and D(0, 15). 3x + y = 15

X 5 0

y 0 15

C(5, 0); D(0, 15) Join point C(5, 0) and D(0, 15) to obtain the line. Since the origin (0, 0) satisfies the given inequation 3(0) + 0 = 0 ≤ 15. So the region containing the origin, represent the solution set of the inequation. Region represented by x ≥ 0,y ≥ 0 : Since every point in first quadrant, satisfies the inequation. So the first quadrant is the region represented by the inequations x ≥ 0 and y ≥ 0. The point of intersection of the lines x + 2y = 10 and 3x + y = 15 is E where x = 4 and y = 3.

Shaded region OCEB represents the common region of the inequations. This region is the feasible solution region of the Linear Programming Problem. The corner points of this common region are O(0, 0), C(5, 0), E(4, 3) and B(0, 5). The value of objective function on these points is as the following table :


O 0 0 ZO = 3(0) + 2(0) = 0

C 5 0 ZC = 3(5) + 2(0) = 15

E 4 3 ZE = 3(4) + 2(3) = 18

B 0 5 ZB =3(0)+ 2(5) = 10



From the table Z = 3x + 2y is maximum at E(4, 3). Hence, maximum value of Z = 18. Question 5. A diet of a sick person must contain at least 4000 units of vitamins, 50 units of minerals and 1400 of calories. Two foods A and B are available at a costs ₹ 4 and ₹ 3 per unit respectively. If one unit of A contains 200 unit of vitamin, 1 unit of mineral and 40 calories and one unit of B contains 100 unit of vitamin, 2 unit of minerals and 40 calories, find what combination of foods should be used to have the least cost ? Solution: Let x unit of food A and y unit of food B are taken in combination of food. ∴ According to question, Objective function to get minimum cost of combination is Z = 4x + 3y Constraints in problem : For vitamin 200x + 100y ≥ 4000 For mineral x + 2y ≥ 50 and for Calorier 40x + 40y ≥ 1400 ∵ x and y are quantity. ∴ x ≥ 0 andy ≥ 0 Converting the given inequations into the equations 200x + 100y = 4000 ⇒ 2x + y = 40 …..(1) x + 2y = 50 …..(2) 40x + 40y = 1400 ⇒ x + y = 35 …..(3) x= 0 …..(4) y = 0 …..(5) Region represented by 200x + 100y ≥ 4000 : The line 2x + y = 40 meets the coordinate axis at points A(20, 0) and B(0, 40). 2x + y = 40

X 20 0

y 0 40

A(20, 0); B(0, 40) Join points A and B to obtain the line. Since the origin (0, 0) does not satisfy the given inequation 2(0) + 0 = 0 ≥ 40. So the region opposite to the origin represents the solution set of the inequations. Region represented by x + 2y ≥ 50 : The line x + 2y = 50 meets the coordinate axis on the points C(50, 0) and D(0, 25). x + 2y = 50

X 50 0

y 0 25

C(50, 0);D(0, 25) Join points C(50,0) and D(0,25) to obtain the line. Since the origin (0, 0) does not satisfy the given inequation 0 + 2(0) = 0 ≥ 50. So the solution set of the inequation in opposite to the origin. Region represented by x + y ≥ 35 : The line x + y = 35 meets the coordinate axis are C(35, 0) and D(0, 35). x + y = 35

X 35 0

y 0 35



E(35, 0); F(0, 35) Join the points E and F to obtain the line. Clearly (0, 0) does not satisfy the inequation x + y ≥ 35. So, the region opposite to the origin represents the solution set of this inequation. Region represented by x ≥ 0 and y ≥ 0 : Since every point in the first quadrant satisfies the inequations. So the first quadrant represents the solution set of the inequations x ≥ 0 and y ≥ 0. The coordinates of the point of intersection of lines 2x + y = 40and x + 2y = 50 are x = 10 and y = 20. The coordinates of the point of intersection of lines x + 2y = 50 and x + y = 35 are x = 20 and y = 15 and the coordinates of the point of intersection of lines 2x + y = 40 are x + y = 35 and x = 5 and y = 30.

The shaded region CHJB represents the common region of the inequations. This is an open feasible solution region whose coordinates of the corner points are C(50, 0), H(20, 15), J(5, 30) and 5(0, 40). The values of objective function on these points are given in the following table :


C 50 0 ZC = 4(50) + 3(0) = 200

H 20 15 ZH = 4(20)+ 3(15) = 125

J 5 30 ZJ = 4(5) + 3(30) = 110

B 0 40 ZB = 4(0) + 3(40) = 120

From table the value of Z is minimum at point J(5, 30) which is 110. ∵ Feasible region is open, so on plotting the graph of line 4x + 3y = 110 which is an open feasible semi region which have a common point J(5,30). Hence the proper solution of LPP is point J(5, 30) to get the minimum price of combination. The minimum price of combination is ₹ 110 and combination has food A = 5 unit and food B = 30 unit. Question 6. A diet is to contain at least 80 unit of vitamin A and 100 unit of minerals. Two foods



F1 and F2 are available. Food F1 costs ₹ 4 per unit and F2 costs ₹ 6 per unit. One unit of food F1 contains 3 unit of vitamin A and 4 unit of minerals. One unit of food F2 contains 6 unit of vitamin A and 3 unit of minerals. Formulate this as a linear programming problem and find graphically the minimum cost for diet that consists of mixture of these foods and also meets the mineral nutritional requirements. Solution: Let x unit of food F1 and y unit of food F2 are required to get the minimum cost and minimum mineral nutritional requirements. The cost of food F1 at the rate of ₹4 per unit = ₹4x and the cost of food F2 at the rate of ₹6 per unit = ₹6y ∴ Minimum cost of combined food = ₹ (4x + 6y) The vitamin A in food F1 = 3x unit and vitamin A in food F2 = 6y unit. ∴ According to question, 3x + 6y ≥ 80 Similarly mineral in food F1 = 4x unit and mineral in food F2 = 3y unit ∴ According to question 4x + 3y ≥ 100 x and y are quantity of food. ∴ x ≥ 0 and y ≥ 0 Therefore the mathematical formulation of Linear Programming Problem is as following : Minimum Z = 4x + 6y Subject to the constraints 3x + 6 ≥ 80 4x + 3y ≥ 100 x ≥ 0, y ≥ 0 Converting the inequations into the equation 3x + 6y = 80 …..(1) 4x + 3y = 100 ……(2) Region represented by 3x + 6y ≥ 80 : The line 3x + 6y = 80 meets the coordinate axis at the points A(80/3,0) and B(0,40/3). Table for 3x + 6y = 80

X 80/3 0

y 0 40/3

A(80/3,0); B(0,40/3) Join the points A(80/3, 0) and B(0, 40/3) to obtain the line. Since the origin (0, 0) does not satisfy the inequation 3(0) + 6(0) = 0 ≥ 80. So the solution region of the inequation is in opposite of origin. Region represented by 4x + 3y ≥ 100 : The line 4x + 3y = 100 meets the coordinate

axis at the points C(25, 0) and D(0, ) Table for 4x + 3y = 100

X 25 0

y 0 100/3

C(25,0);D(0, ) Join points C and D to obtain the line. Since the origin (0, 0) does not satisfy the inequation 4(0) + 3(0) = 0 ≥ 100. So the solution region of the inequation is in opposite to origin.



Region represented by x ≥ 0 and y ≥ 0 : Since every point in the first quadrant satisfy the inequations. So the first quadrant represents the solution set of the inquations. The point of intersection of lines 3x + 6y = 80 and 4x + 3y = 100 is E(24, 4/3).

The shaded region AED is the common region of the inequations. This is an open feasible solution region. The corner points of this open feasible region are A(80/ 3, 0), E(24. 4/3) and D(0, 100/3). The values of object function on these comer points are given in the table:


A 80/3 0 ZA = 4 x 80/3 + 6 x 0 = 106.66

E 24 4/3 ZE = 4 x 24 + 6 x 4/3 = 104

D 0 100/3 ZD = 4 x 0 + 6 x 100/3 = 200

The minimum value of Z = 104 at point E(24, 4/3). ∵ Feasible region is an open region so the inequation 4x + 6y ≤ 104 is satisfied by point E(24, 4/3). Hence the combination of food consists 24 unit of food F1 and 4/3 unit of food F2 to get the minimum price ₹104. Question 7. A manufacturer of furniture makes two products : chairs and tables. Processing of these product is done on two machines A and B. One chair requires 2 hrs on machine A and 6 hrs on machine B. One table requires 4 hrs on machine A and 3 hrs on machine B. There are 16 hrs of time per day available on machine A and 30 hrs on machine B. Profit gained by the manufacturer from a chair and a table is ₹ 3 and ₹ 5 respectively. Find with the help of graph, what should be the daily production of each of the both products so as to maximize his profit ? Solution: Let the manufacturer should produce x chair and y tables per day. ∴ Total profit of manufacturer = ₹ (3x + 5y) Time taken to prepare x chairs on machine A = 2x hrs and on machine B = 6x hrs Time taken to prepare y tables on machine A = 4y hrs and on machine B = 2y hrs ∴ For time taken on machine A 2x + 4y ≤ 16 hrs



and time taken on machine B 6x + 3y ≤ 30 hrs ∴ x ≥ 0 and y ≥ 0 According to question mathematically formulation of Linear Programming Problem is as following : Maximize Z = 3x + 5y Subject to the constraints 2x + 4y ≤ 16 6x + 3y ≤ 30 x ≥ 0 and y ≥ 0 Converting the given inequations into the equations 2x + 4y = 16 6x + 3y = 30 Region represented by 2x + 4y ≤ 16 : The line 2x + 4y = 16 meets the coordinate axis on the points A(8,0) and B(0, 4). Table for 2x + 4y = 16

X 8 0

y 0 4

A(8, 0); B(0, 4) Join the points A and B to obtain the line. Since (0, 0) satisfies the inequation 2(0) + 4(0) ≤ 16 so the region containing the origin represents the solution set of the inequation. Region represented by 6x + 3y ≤ 30 : The line 6x + 3y = 30 meets the coordinate axis on the points C(5,0) and D(0, 10). Table for 6x + 2y = 30

X 5 0

y 0 10

C(5,0);D(0,10) Join the points C and D to obtain the line. Since the origin (0, 0) satisfies the inequation 6(0) + 2(0) = 0 < 30. So the region containing the origin represents the solution set of the inequation. Region represented by x ≥ 0 and y ≥ 0: Since the every point of first quadrant satisfies these inequations. So the first quadrant is the solution set of the inequations. The point of intersection of lines 2x + 4y = 16 and 6x + 3y = 30 is E(4, 2) where x = 4 and y = 2.



The shaded region OCEB represents the common region of inequations. This region represents the feasible solution. The comer points of this region are O(0, 0), C(5,0), E(4, 2) and B(0, 4). The value of objective function on these points is given in the following table :

Point x-coordinate y-coordinate Objective function

O 0 0 ZO = 3(0) + 5(0) = 0

C 5 0 ZC = 3(5) + 5(0) = 15

E 4 2 ZE = 3(4) + 5(2) = 22

B 0 4 ZB = 3(0) + 4(2) = 8

From the table, the value of objective function is maximum on the point E(4,2) where x = 4 and y = 2. Hence to get the maximum profit of ₹ 22, manufacturer should produce 4 chairs and 2 tables. Question 8. A firm manufactures headache pills in two sizes A and B. Size A contains 2 grams of aspirin 5 grams of bicarbonate and 1 gram of cofeine; size B contains 1 grams of asprin of 8 grms of bicarbonate and 6 grains of cofeine. It has been found by users that it requires atleast 12 grams of aspirin 74 grams of bicarbonate and 24 grams of cofeine for providing immediate effects. Deternfine graphically the least number of pills a patient should have to get immediate relief. Solution: Let the patient should have x pills of size A and y pills of size B. ∴ Maximum number of pills = x + y According to question, A type pills have quantity of Aspirin = 2x gramn and B type pills have quantity of Aspirin = 1 y grams ∴ Constraints for quantity of Aspirin 2x + y ≥ 12 grams Similarly constraints for Bicarbonate 5x + 8 ≥ 74 grams and constraints for quantity of cofeine x + 6.6y ≥ 24 grams



∵ x and y are the number of pills. ∴ x ≥ 0 and y ≥ 0 So mathematical formulation of linear programming problem is the following : Minimize Z = x + y Subject to the constraints 2x + y ≥ 12 5x + 8 ≥ 74 x + 6y ≥ 24 x ≥ 0, y ≥ 0 Converting the constraints into the equations 2x + y = 12 …..(1) 5x + 8y = 74 …..(2) x + 6y = 24 …..(3) Region represented by 2x + y ≥ 12 : The line 2x + y = 12 meets the coordinate axis at the points A(6,0) and B(0, 12). Table for 2x + y = 12

X 6 0

y 0 12

A(6,0); B(0, 12) Join the point A to 5 to obtain the line. Since the origin (0, 0) does not satisfy the inequation 2(0) + (0) = 0 ≥ 12. So the region opposite to origin represents the solution set of the inequation. Region represented by 5x + 8y ≥ 74 : The line 5x + 8y = 74 meets the coordinate axis at the point C(74/5, 0) and D(0, 74/8). Tabel for 5x + 8y = 74

X 74/5 0

y 0 74/8

Join the points C and D to obtain the line. Since the origin (0,0) does not satisfy the inequation 5(0) + 8(0) ≥ 74. So the region opposite to origin represents the solution set of inequation. Region represented by x + 6y ≥ 24 : The line x + 6.6y = 24 meets the coordinate axis at the point E(24,0) and F(0,4). The table for x + 6y = 24

X 24 0

y 0 4

A(24, 0); B(0, 4) Join the points E and F to obtain the line. Since the origin (0, 0) does not satisfy the inequation. So the region opposite to origin represents the solution set of the inequation. Region represented by x ≥ 0, y ≥ 0 : Since every point in first quadrant satisfies the inequations so the first quadrant represents the solution set of these inequations. The point of intersection of lines 2x +y – 12 and 5x + 8y = 74 is G(2, 8).

The point of intersection of lines 5x + 8y = 74 and x + 6y = 24 is H ( , ).



The shaded region BGHE is the common region of the inequations. This is an open feasible region of solution. The corner points of this open feasible solution region

are B(0, 12), G(2, 8), H( , ) and E(24, 0). The values of objective function on these comer points are given in the following table :

From the above table the value of objective function is minimum at the point G(2, 8) where x = 2 and y = 8 and z = 2 + 8 = 10. Since feasible region of solution set is an open region. So the graph of x + y ≤ 10 passes through point G(2,8) which satisfies the inequation z + 8 = 10 so the solution of this LPP is G(2, 8) where x = 2 and y = 8. Hence the patient should have minimum 2 pills of type A and 8 pills of type B to get immediate relax. Question 9. A brick manufacturer has two depots A and B with stocks of 30,000 and 20,000 bricks respectively. He receives order from the builders P, Q and R for 15000, 20,000 and 15,000 bricks respectively. The cost in rupee transporting 1000 bricks to the builders from the depots are given below :



To/From P Q R

A 40 20 30

B 20 60 40

How should the manufacturer fulfill the orders so as to keep the cost of transporation minimum ? Solution: Let Depot A supplies x thousand bricks to builder P and y thousand to builder Q. Then remaining bricks (30 – x – y) will be supplied to builder R. Therefore cost of transportation from builder A to builder P, Q and R are ₹40x, ₹20y and ₹ 30(30 – x – y) respectively. From Depot B the number of bricks supplied to builder P = (15 – x) thousand Number of bricks supplied to builder Q = (20 – y) and the number of bricks supplied to builder R = 20 – (15 – x + 20 – y) = (x + y – 15) Therefore the cost of transportation from depot B to builders P, Q and R are ₹20(15 -x), ₹60(20 -x) and ₹40(x + y – 15). So the total cost of transportation from both Depots A and B to the builders P,Q and R. Z = 40x + 20y + 30(30 – x – y) + 20(15 – x) + 60(20 – y) + 40(x + y – 15) = 30x – 30y + 1800 Therefore mathematical formulation of Linear Programming Problem is the following Minimize Z = 30x – 30y + 1800 Subject to the constraints x + y ≤ 30 x ≤ 15 y ≤ 20 x + y ≥ 15 x ≥ 0 y ≥ 0 Converting the given inequations into the equations : x + y = 30 ……(1) x = 15 …..(2) y = 20 …..(3) x + y = 15 …..(4) x = 0 …..(5) y = 0 …..(6) Region represented by x + y ≤ 30 : The line x + y = 30,meets the coordinate axis at ,A(30, 0) and B(0, 30).

X 30 0

y 0 30

A(30,0);B(0,30) Join the points A and B to obtain the line. Clearly (0,0), satisfies the inequation 0 + 0 ≤ 30. So the region containing the origin represents the solution set of the inequation. Region represented by x ≤ 15: The line x = 15 is parallel toy-axis and its every point will satisfy the inequation in first quadrant. The region containing the origin represents the solution set of the inequation. Region represented by y ≤ 20: The line y = 20 is parallel to x-axis and its every point will satisfy the inequation in first quadrant.



Region represented by x + y ≥ 15 : The line x + y = 15 meets the coordinate axis at the points C(15, 0) and D(0, 15). x + y = 15

X 15 0

y 0 15

C(15, 0); D(0, 15) Join the points C to D to obtain the line. Since the origin (0, 0) does not satisfy the inequation 0 + 0 = 0 ≥ 15. So the region opposite to the origin represents the solution set of inequations. Region represented by x ≥ 0 and y ≥ 0 : Since every point in first quadrant satisfy the inequation. So the first quadrant represents the solution of inequations x ≥ 0 and y ≥ 0. The shaded region CJKED represents the feasible solution region of the inequation.

The corner points of this region are C(15, 0), J(15, 15), K(10,20), E(0,20) and D(0, 15) where point C(15, 0), J(15, 15), K(10,20), E(0,20) and D(0, 15) are point of intersection of respective lines. The values of objective function on these comer points are given in the following table :

Point x-coordinate y-coordinate Objective function Z = 30x – 30y +1800

C 15 0 ZC = 30 x 15 – 30 + 0 + 1800 = 2250

J 15 15 ZJ = 30 x 15 – 30 x 15 + 1800 = 1800

K 10 20 ZK = 30 x 10 – 30 x 20 + 1800 = 1500

E 0 20 ZE = 30 x 0 – 30 x 20 + 1800 = 1200 minimum

D 0 15 ZD = 30 x 0 – 30 x 15 + 1800 = 1350

From the table it is clear that the value of objective function is minimum ₹ 1200 at point E(0, 20). So Depot A should deliver the bricks to buiders P, Q and R as 0, 20, 10 thousands and Depot B should deliver 15,0 and 5 thousands . respectively.



Hence,

Question 10. Region represented by the inequation system x + y ≤ 3 y ≤ 6 and x ≥ 0,y ≥ 0 is : (A) Unbounded in first quadrant (B) Unbounded in first and second quadrant (C) Bounded in first quadrant (D) None of the above Solution: Converting the given inequations into equations x + y = 3 …..(1) y = 6 …..(2) Region represented by x + y ≤ 3 : The line x + y = 3 meets the coordinate axes are A(3,0) and B(0,3) respectively, x + y = 3

X 3 0

y 0 3

A(3, 0); B(0, 3) Join points A and B to obtain the line. Clearly, (0, 0) satisfies the inequation x + y ≤ 3. So, the region containing the origin represent the solution set of the inequation. Region represented by y ≤ 6 : The line y = 6 is parallel to x-axis and its every point will satisfy the inquation in first quadrant, region containing the origin represents the solution set of this inequation. Region represented by x ≥ 0 and y ≥ 0 : Since every point in first quadrant satisfy the inequations, so the first quadrant is the solution set of these inequations.



The shaded region is the common region of inequations. This is feasible region of solution which is bounded and is in first quadrant. Hence the region is bounded and is in first quadrant. ∴ Answer (c) is correct.

Rajasthan Board RBSE Class 12 Maths Chapter 16 Probability and Probability Distribution Ex 16.2 Question 1.

If A and B be two events such that P(A) = ; = P(B) = and P(A ∩ B)

= , then find . (1 – P(A)] = P(B) PĀ) = P(Ā)P(B) Solution: Given:



Question 2. If P(A) = 0.4, P(B) = p and P(A ∪ B) = 0.6 and A and B are independent events, then find the value of p. Solution: Given P(A) = 0.4 P(B) = p P(A ∩ B) = 0.6 ∵ A and B are independent events So, P(A ∩ B) = P(A). P(B) P(A ∪ B) = P(A) + P(B) – P(A ∩B)

⇒ 0.6 = 0.4 x p – P(A)P(B) 0.6 = 0.4 x p – 0.4 x p 0.2 = 0.6 x p

P = =

Hence, p = . Question 3. If A and B are independent events, and P(A) = 0.3, P(B) = 0.4, then find: (i)P(A ∩ B) (ii)P(A ∪ B)

(iii) P( )

(iv) P( ) Solution : (i) Given : P(A) = 0.3 P(B) = 0.4 A and B are independent events P( A ∩ B) = P(A) · P(B) = 0.3 x 0.4 = 0.12 (ii) P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = 0.3 + 0.4 – 0.12 = 0.7 – 0.12 = 0.58

Question 4. If A and B are independent events., where P(A) = 0.3, P(B) = 0.6, then find: (i)P(A ∩ B) (ii) P(A ∪ ) (iii) P(A ∪ B) (iv) P( ∩ ) Solution : Given



P(A) = 0.3 P(B) = 0.6 (i) P(A ∩ B) = P(A) × P(B) = 0.3 × 0.6 = 0.18 (ii) P(A ∪ ) = 0.3 – 0.18 = 0.12 (iii) P(A ∪ B) = p(A) + P(B) – P(A ∩B) = 0.3 + 0.6 – 0.18 = 0.90 – 0.18 (iv) P( ∩ ) = [ 1 – P(A)] [1 – P(B)] = [1 – 0.3] [1 – 0.6] = 0. 7 × 0.4 = 0.28 Question 5. A bag contains 5 white, 7 red and 8 black balls. If four balls are drawn one by one without replacement, find the probability of getting all white balls. Solution: Given White ball = 5 Red ball = 7 Black ball = 8 Total number of balls = 5 + 7 + 8 = 20 ∴ Probability of getting first white ball

Probability of getting second white ball

Probability of getting third white ball

Probability of getting fourth white ball

Probability of getting all white balls

Question 6. A die is tossed thrice. Find the probability of getting an odd number at least once. Solution: Number of possible results on a die = {1, 2, 3, 4, 5, 6}



∴ Number of favourable results getting even number (2, 4, 6) = 3

∴ Probability of getting even number = =

∴ Probability of getting even number once =

∴ Probability of getting even number thrice = ∴ Probability of getting at least one odd number on tossing all dice together

= 1 – = Question 7. Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black. Solution: Number of black cards in a pack= 26 Total number of cards = 52 ∴ Probability of getting one black card =

After drawing one black card, remaining cards are 51. Out of them number of black cards = 25 ∴ Probability of getting other black card =

∴Probability of getting both black cards

Question 8. Two coins are tossed. What is the probability of coming up two heads if it is known that at least one head comes up ? Solution: Possible ways of tossing two coins {HH, HT, TH, TT} = 4 Number of ways where at least one head comes up 4 – 1 = 3 Number of getting both heads = 1

∴Probability of getting both heads =

Hence, requried probability = Question 9. In a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English newspapers. A student is selected at random : (i) find the probability that he reads neither Hindi nor English newspapers. (ii) If he reads Hindi newspaper, find the probability that he reads English newspaper. (iii) If he reads English newspaper. Find the probability that he reads Hindi newspaper. Solution: (i) Let H: Event of reading Hindi newspaper and E: Event of reading English newspaper



Probability of reading at least one newspaper is = P(H ∪ E) P(H ∪ E) = P(H) + P(E) – P(H ∩ E) = 0.6 + O.4 – 0.2 = 0.8 Therefore probability of reading neither Hindi nor English newspaper by students = 1 – P(H ∪ E) = 1 – 0.8 = 0.2 = 20% Clearly 20% students do not read newspaper.

∴ Required probability = = (ii) If student read Hindi newspaper English ∴ Probability of reading English newspaper also

(iii) If student reads English newspaper, then probability of reading Hindi newspaper also

Question 10. A can solve 90% of the problems given in a book and B can solve 70%. What is the probability that at least one of them will solve the problem, selected at random from the book ? Solution : Let

P(A) = ; P(B) = ∴ Probability of at least solving by one = P( B) + P(A ) + P(AB) = P( ) x P(B) + P( )P( ) + P(A)x P(B)



= [1 – P(A)] x P(B) + P(A) [1 – P(B)] + P(A) x P(B)]

Question 11.

A problem in mathematics is given to 3 students whose chances of solving it are ,

and What is the probability that the problem is solved ? Solution: Problem will be solved if at least one student can solve.

∴ Probability of solving by one student =

= 1 – =

Probability of not solving by any student = 1 – =

Similarly, probability of not solving by third student = 1 – =

Probability of not solving by any one of them = x x =

∴ Probability of solving by at least one of them = 1 – = Question 12. A bag contains 5 white and 3 black balls. Four balls are successively drawn out without replacement. What is the probability that they are alternately of different colours ? Solution: Total number of balls = 5 + 3 = 8

∴Probability of getting one white ball = ∴ Number of remaining balls = 8 – 1 = 7 out of them there are 4 white and 3 black balls.

∴ Probability of being black ball = Number of remaining balls 7 – 1 = 6 where are 4 white and 2 black balls

∴ Probability of being third white ball = Number of remaining balls = 6 – 1= 5, where are 3 white and 2 black

∴Probability of being fourth black ball = ∵Events are independent every time

∴ Probability of being different colours ball

Question 13.

Probabilities of solving a specific problem independently by A and B are , and respectively. If both try to solve the problem independently, find the probability that



(i) the problem is solved, (ii) exactly one of them solves the problem. Solution:

(i) Probability of solving the problem by A = P(A) = ⇒ Probability of not solving the problem by A P( ) = 1 – P(A)

= 1 – = Probability of solving the problem by B.

P(B) = ⇒ Probability of not solving the problem by B. P( ) = 1 – P(B)

= 1 – = ∴ Probability that problem not is solved P( . ) = P( ).P( )

= x = ⇒ Probability that problem is solved = 1 – P( )

= 1 – = (ii) A and B are independent problems.

Here P(A) = , P( ) =

P( ) = , P(B) = ∴ Probability that exactly one of them solves the problem = P(A ) + P( B) = P(A).P( ) + P( ).P(B)

Rajasthan Board RBSE Class 12 Maths Chapter 16 Probability and Probability Distribution Ex 16.3 Question 1. Given there are two bags I and II. Bag I contains 3 red and 4 black bails while another bag II contains S red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from bag II. Solution: Let E1 be the event that the ball is drawn from the bag I and II and E2 be the event that it is drawn from the bag II and E be the event that the drawn ball is red.

Since the drawn ball is found to be red, so we have to Find P( ) the probability of drawing red ball from bag II. since both the bags are equally likely to be selected to



draw the ball, we have

Probability of getting red ball from bag I

Probability of getting red ball from bag II

Probability of getting red ball from bag III

Hence, using Baye’s Theorem,

Question 2. A doctor has to visit a patient. From the past experience, it is known that the probabilities that he will come by train, bus, scooter or by other means of transport

are respectively , , and . The probabilities that he will be late are , and , if he comes by train, bus and scooter respectively, but if he comes by other means of transport, than he will not be late. When he arrives, he is late. What is the probability that he comes by train ? Solution: Let E be the event “Doctor comes late to visit a patient” and if Doctor comes by Train, Bus, Scooter and any-other means, then probabilities are T1, T2, T3 and T4 respectively.



Question 3. Bag I contains 3 red and 5 black balls and by II contains 4 red and 5 black balls. One ball transferred from bag I to bag II and then a ball is drawn from bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.



Solution: Bag I contain 3 red and 4 black balls, and Bag II contains 4 red and 5 black balls. Let, E1 = Event of drawing red ball from bag I and E2 = Event of drawing black ball from bag II and A : Event of drawing red ball after transfering one ball from I to II

Question 4. A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls, One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag. Solution: Let E1 is the event of drawing from bag I and E2 of drawing from bag II and A represent the event of drawing red ball.

∴ Probability of chosing one bag =

Or P(E1) = P(E2) = There are 4 red and 4 black balls in bag I. ∴ Probability of drawing red ball from it

In second bag, there are 2 red and 6 black balls.



∴ Probability of drawing red ball from it

Now probability of drawing red ball from first bag

Question 5. There are three coins. One is having head on both faces, another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin ? Solution:

Probability of selecting one coin out of three = E1 : event that coin having head on both faces. E2 : event that coin is biased that head come up 75% of the time. E3 : event that third is a unbiased, and A is the event of getting head



Hence, required probability = Question 6. A laboratory blood test is 99% effective in detecting a certain disease when it is if fact, present. However, the test also yields a false positive result for 0-5% of the healthy person tested. If 0.1 percent of the population actually has the disease. What is the probability that a person has the disease given that his test result is positive ? Solution: Let E1 : Event that patient is effected from desease. E2 = Event that patient is not effected from desease and A be the event that blood is tested.



∴ Probability of the person effected from disease

Probability of the person not affected from disease

Probability of those who are patient and their blood are tested

Probability of those who were tested but not patient

∴ Probability that a person has the disease given that his test is positive

Hence, the required probability = Question 7. Of the sutdents in a college, it is known that 60% reside in hostel and 40% are day scholars. Previous year result report that 30% of all student who reside in hostel attain, A grade and 20% of day schloars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has on A grade, what is the probability that the student is a hostlier ?



Solution: Solution : Let E1 : Event that students reside in hostel and E2 : Event that students do not reside in hostel ∴ Probability of students reside in hostel

Probability that students do not reside in hostel

Probability of students securing grade A who reside in hostel

Probability of students who do not reside in hostel but. secure grade A

∴ Probability that student securing graded is a hostelier = P ( )

Hence, required probability : Question 8. An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probabilities of an accident are 0-01, 0*03 and 0-15 respectively. One of the insured person meets with an accident. What is the probability that he is a scooter driver ? Solution: Let E1 : Event that scooter driver is insured



E2 = Event that car driver is insured E3 = Event that truck driver is insured and A : Even that an accident occurs. A company insured 2000 scotter drivers, 4000 car drivers and 6000 truck drivers. ∴ Total number of drivers insured = 2000 + 4000 + 6000 = 12000 ∴ Probability of scooter drivers insured

Probability of car drivers insured

Probability of truck drivers insured

Probability of scooter drivers accidents

Probability of car drivers accidents

Probability of truck drivers accidents

A person meets an accident who is insured. ∴ Probability that the persons is a scooter driver



Hence, required probability : Question 9. In answering a question on a multiple choice test, a student either knows the answer

or guesses. Let be probability that he knows the answer and be the probability that he guesses. Assuming that a student who guesses at the answer will be correct

with probability . What is the probability that the student knows the answer given that he answered it correctly ? Solution: Let E1 : Event that student knows the answer E2 : Event that student guesses and A : Event that he answers correctly

∴ Probability that he answers correctly



Hence requried Probability = Question 10. Suppose 5% men and 0.25% women has white hairs. A person with white hair is chosen at random. Find the probability that a male is selected. Assume that there are equal number of men and women. Solution: Given that number of men and women are equal Let events E1 = To be a man E2 = To be a woman A = White hair



Question 11. Two groups are competing for the position on the Board of directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducting a new product is 0.7 and the corresponding probability is 0-3 if the second group wins. Find the probability that the new product introduced was by the second group. Solution: Let events E1 : First group wins, introduce new product. E2 : Second group wins, introduce new product. Given : Probability that first group wins = P(E1) = 0.6 Probability that second group wins = P(E2) = 0.4 Probability that first group wins introducting new product,

Probability that second group wins introducting new product



∴ Probability that new product introduced was by the second group

Question 12. Suppose a girl throw a dice. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1,2,3 or 4 then tosses a coin once and notes whether a head or tall is obtained. If she obtained exactly one head, what is the probability that she threw 1,2,3 or 4 with the die ? Solution: Number of possible results after throwing a die = (1,2, 3, 4, 5, 6) = 6 Let E1 : Event she gets 5 or 6 E2 : Event she gets 1, 2, 3, 4 A : Event that she gets head if tosses a coin ∴ Probability that 5 or 6 acheived

Probability that 1, 2, 3,4 are acheived

If she gets 5 or 6 she tosses a coin thrice and notes the number of heads. (HHH, HHT, HTH, THH, HIT, THT, TTH, TIT) ∴ Number of ways to get one head = (HTT, THT, TTH) = 3

∴ Probability to get one head =

If she gets 1, 2, 3, 4, then she tosses a coin once and notes whether a head or tail.

∴ Probability to get one head =

If she obtained exactly one head, than probability that she threw 1,2, 3 or 4



Question 13. A card from a pack of 52 card is lost From remaining cards of the pack, two cards are drawn found to be both diamonds. Find probability of the lost card being a diamond. Solution: Let Events E1 = Lost card is diamond E2 = Lost card is not diamond The number of diamond card = 13 out of 52 cards.

and 39 cards are not diamonds

(i) When a card of diamond lost there remaining cards of diamond are 12 out of 52.

Here A represents the cards lost. (ii) When diamonond card is not Lost then probability of drawing 2 cards of diamond

∴ Probability that lost card is a diamond



Hence, required probability = Question 14. A bag contains 3 red and 7 black. Two balls are drawn one by one at a time at random without replacement. If second drawn half is red, then what is the probability that first drawn ball is also red ? Solution: A = Event that drawn ball is Red first time. and B = Event that drawn ball is also red in second time.



Then P(A ∩ B) = P( 1 red and 1 red)

Rajasthan Board RBSE Class 12 Maths Chapter 16 Probability and Probability Distribution Ex 16.4 Question 1. Find which of the following probability distribution is possible for a random variable :

Solution: (i) Sum of probabilities = 0.4 + 0.4 + 0.2 = 1 Hence, given distribution is a probability distribution. (ii) Sum of probabilities = 0.6 + 0.1 + 0.2 = 0.9 ≠ 1 Hence, the given distribution is not a probability distribution. (iii) Here, one of the probability is – 0.1. which is negative. Hence, this distribution is not probability distribution. Question 2. Find the probability distribution of X, the number of heads in a simultaneous toss of



two coins. Solution: Possible values of X are 0, 1, or 2 Now P(X = 0) = P(No head) = P(Tail first time and tail second time) = P(Tail first time).P(Tail in second time)

= × = = P(TH or HT) = P (TH) + P(HT)

= + =

P(x = 2) = P (2 Head) = P(HH) = ∴ Required distribution is

Question 3. 4 rotten oranges are mixed accidently with 16 good oranges. Find the probability distribution of the number of rotten oranges in a draw of two oranges. Solution: 4 bad oranges are mixed with 16 good oranges The number of total oranges = 4 + 16 = 20 2 bad oranges are to be choosen ∴ Probability of choosing a bad orange

= = ∴ Probability of choosing one good orange

= 1 – = ∵ X is the number of bad oranges

Question 4. An urn contains 4 white and 3 red bails. Find the probability distribution of the



number of red balls if 3 balls are drawn at random. Solution: Three balls are taken out of an urn ∴ Sample space is = S {RRR, RRW, RWR, WRR, RWW WRW, WWR WWW} Where R respresents red ball and W, white ball. Let X = number of red balls. So possible value of X are 3,2,1,2,1,0 or 0, 1,2, 3 (No red ball)

Question 5. From a lot of 10 items containing 3 defectives, a sample of 4 items is drawn at random. Let the random variable X denote the number of defective items in the sample. If the sample is drawn randomly, find (i) The probability distribution of X (ii) P(x ≤ 1) (iii) P(x < 1) (iv) P(0 < x < 2)



Solution: Given, from a lot of 10 items, 3 are defective ∴ Good items = 10 – 3 = 7 Let x represents the number of defective items. Clearly values of X are 0, 1,2, 3. P(x = 0) = P(GGGG) = p (good items)

= × × × = P (x = 1) = P (one good and three defective)



Question 6. A die is manufactured in such a way that an even number is twice likely to occur as an odd number. If the die is tossed twice, find the probability distribution of the random variable X representing the perfect square in the two tosses.



Solution: Let X represents the number of perfect squares. S = sample space = { 1, 2, 3. 4, 5, 6} ∴ n(S) 6 n(X) = Possible number of perfect squares = {1,4} = 2 ∴ Probability of getting perfect square

Question 7. An urn contains 4 white and 6 red balls. Four balls are drawn at random from the urn. Find the probability of the number of white balls. Solution: Let X represents white balls. ∴ Total balls =4 + 6 = 10 Four balls are drawn at random. ∴ Expected values of X will be 0,1,2,3,4. ∴ P(X = 0) = P(all red) = P(RRRR)



Question 8. Find the probability distribution of number doublets in three throws of a pair of dice. Solution:



Let x = number of doublets ∴ Expected value of x will be 0, 1, 2, 3 Set of all doublets available on a pair of dice in a throw = (1,1), (2, 2), (3,3), (4,4), (5, 5), (6, 6) Number of all possible ways on throwing a pair of dice. = 6 × 6 = 36

∴ Probablity of getting a doublets on a pair of dice in a throw = =

∴ Probablity of getting a doublets on a pair of dice throw = 1 – =

Question 9. Let a pair of dice be thrown and the random variable X be the sum of the numbers that appear on the two dice. Find the mean of X. Solution: The number of possible outcomes when two dice are thrown = 6 × 6 = 36



Question 10. Find the variance of the number obtained on a throw of an unbiased die. Solution: Let sample space of observation be s = {1,2, 3, 4, 5, 6} X represents the number appeared on dice. Then x is a random variable takes value 1, 2, 3,4, 5 or 6.

Also P(1) = P(2) = P(3) = P(4) = P(5) = P(6) = ∴ Probablity distribution is Following:



Question 11. In a meeting, 70% of the members favour and 30% oppose a certain proposal. A member is selected at random and we take X = 0 if he opposed, and X = 1 if he is in favour. Find the mean and variance of X. Solution: Probability of members, favouring any proposal on X = 1

70% = = 0.70 Probability of members, opposing any proposal on X= 0

30% = = 0.30 ∴ Required probability distribution is following



Question 12. Two card are drawn simultaneously (or successively without replacement) from a well shuffled pack of 52 cards. Find the mean, variance and standard deviation of the number of kings. Solution: Two cards are drawn ∴ Number of ways that two cards are not king.

Number of ways of drawing two cards out of 52 cards = 52C2



∴ Probability of not drawing any king =



Rajasthan Board RBSE Class 12 Maths Chapter 16 Probability and Probability Distribution Ex 16.5 Question 1. If a fair coin is tossed 10 times, then find the probabilities of the following; (i) exactly 6 heads ? (ii) atleast 6 heads ? (iii) atmost 6 heads ? Solution: (i) A fair coin is tossed 10 times and X = numbers of heads.

∴ In distribution X, n = 10 and p = probability of head =



Question 2. An urn contains 5 white, 7 red and 8 black balls. If 4 balls are drawn one by one with replacement, what is the probability that: (i) all are white (ii) only 3 are white (iii) none is white (iv) at least 3 are white Solution: (i) Total number of balls = 5 + 7 + 8 = 20 Number of white balls = 5

Probability of getting white ball is one chance = = ∵ All events are independent

∴ Required probability = × × × = ( )4 (ii) Probability of drawing white ball first time

= 3C1 × × ×

= 3 ×( )3 (iii) P(no ball is white) ∴ Number of other balls = 7 + 8 = 15

∴ Probability of drawing one other colour ball = = ∴ Probability of other colour balls drawn successively (none is white)

= × × ×

= ( )4 (iv) P(at least 3 white) = P (four white) + P(three white)

Question 3. In a hurdle race, a player has to cross 10 hurdles. The probability that he will clear

each hurdle is . What is the probability that he will knock down fewer than 2 hurdles ? Solution : Number of total hurdles (n) = 10

Probability of clearing the hurdles = P =



∴ Probability of non clearing the hurdles

Question 4. five dices are thrown simultaneously. If the occurrence of an even number in a single dice is considered a success, find the probability of at most 3 success. Solution: Sample space on throwing a dice s = {1,2, 3,4, 5,6} ∴ n(S) = 6 Let A represents even numbers ∴A = {2,4,6} n(A) = 3



Question 5. Ten eggs are drawn successively, with replace¬ment, from a lot containing 10% defective eggs. Find the probability that there is at least one defective egg. Solution:



Probability of defective eggs = 10%

Question 6. A person buys a lottery ticket in 50 lotteries, in each of which his chance of winning a

prize is What is the probability that he will win a prize. (i) at least once (ii) exactly once (iii) at least twice Solution:



Question 7. The probability that a bulb produced by a factory will fuse after 150 days on use is 0*05. Find the probability that out of 5 such bulbs (i) none (ii) not more than one (iii) more than one (iv) at least one will fuse after 150 days of use. Solution: Probability that a bulb will fuse after 150 days p = 0.05 Probability that bulb will not fuse after 150 days q = 1 – p = 1 – 0.05 = 0.95



Question 8. In a multiple choice examination with three possible answers for each of the 5 questions out of which only one is correct. What is the probability that a candidate would get four or more correct answers just by guessing ? Solution: ∵ One answer is correct out of 3

∴ Probability of correct answer = p =

∴ Probability of wrong answer = q = 1 – p

= 1 – = Probability that 4 or more answers are correct



Question 9. In a 20 questions true-false examination are asked. Suppose a student tosses a fair coin to determine his answer to each question. If the coin falls heads, his answer ‘true’ If it falls tails, his answer ‘false’. Find the probability that his answer at least 12 questions correctly. Solution:

P (Getting head on toss) P = P (not getting head on toss) q = 1 – person

= 1 – =

∴ Probability of writing correct answer =

and probability of writing incorrect answer =

Question 10. A bag contains 10 balls each marked with one of the digits 0 to 9. If four balls are drawn successively with replacement from the bag, what is the probability that none is marked with the digit 0 ? Solution: A bag contains 10 balls each marked with one of the digit from 0 to 9. Probability that one ball is in marked 0 drawn

P = = 0.1 = 0.1 Probability that ball is not marked 0 q = 1 – p = 1 – 0.1 = 0.9 Now 4 balls are drawn successively with replacement.



∴ Probability that any of them ball is marked 0

Question 11. Five cards are drawn successively with replacement from a well shuffled pack of 52 cards. What is the probability that: (i) all the five cards are spades ? (ii) only 3 cards are spades ? (iii) none is spade ? Solution: There are 13 cards are spade out of all 52 cards. Probability of drawing one card of spade



Question 12.

Suppose X has a binomial distribution B(6, ). Show that x = 3 is the most likely outcome. Solution:



Question 13. A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability of two success. Solution: When a pair of dice in thrown ∴ Number of all possible outcomes n(S) = 6 × 6 = 36 Number of doublets can be make from a pair of dice = 6 [(1,1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)] ∴ Probability that doublet are find

P = = and probability that doublets are not find q = 1 – p

= 1 – = A pair of dice is thrown four times



∴ n = 4

Rajasthan Board RBSE Class 12 Maths Chapter 16 Probability and Probability Distribution Miscellaneous Exercise Question 1. Two events A and B are said to be mutually independent, if:

Solution: Given A and B two independent events

Hence option (iii) is correct Question 2. Two dice are thrown. The probability of getting a pair of even prime number is :

(i) (ii) 0

(iii)

(iv) Solution:

Probability of getting 2 even prime number on one dice =

Probability of getting 2 an even number on other dice = ∴ Probability that 2 even prime number is found when a pair of dice is tossed

= × = Hence option (iii) is correct. Question 3. If A and B be two events, such that A ⊂ B and P(B) = 0, then which of the following



statement is true ?

Solution :

Question 4. Two cards are drawn randomly from a well- shuffled deck of 52 cards. If X denotes the number of aces, then find mean of X:

Solution: Two cards are drawn randomly from 52 cards, firstly number of total ways that two cards are not an ace.

Number of ways drawing 2 cards out of 52 cards

∴ P(X = 0), Probability that no ace is not drawn = Secondly number of ways that there is an ace and there is no ace out of 4C1 × 48C1



Hence option (iv) is correct Question 5. A random variable X, takes the values 0,1,2, and 3. Mean of X is P(x = 3) = 2P(x = 1) and P(x = 2) = 0-3, then P(X=0) is : (i) 0.2 (ii) 0.4 (iii) 0.3 (iv) 0.1 Solution:



Question 6.

Probability of girl to be a racer is What is the probability that 4 girls out of 5 are racer ?



Solution:

Probability that girl to eb a racer = ∴ Probability that girls in not be racer

= 1 – = ∴ Probability distribution of girls to be racer

= ( + )5 Probability that 4 girls are to be racer

= 5C4 ( )4 × ( ) Hence option (iii) is correct Question 7. In a box containing 100 are defective. The probability that out of a sample of 5 bulbs, none is defective is :

Solution: Number of bulbs in box = 100 Number of defective bulbs = 10

∴ Probability of defective bulbs = =

∴ So probability of non defective bulb = 1 – = ∴ Probability that no bulb is defective out a sample of 5

Hence option (iv) is correct Question 8. A couple has two children. Find the probability that (i) both children are male, if it is known that the elder child is a male. (ii) both children are female, it it is known that the elder child is a female. (iii) both children are male, if it is known that at least one of the children is male. Solution:



Question 9. Two integers are selected at random from the 1 to 11 integers. Find the probability that both integers obtained are odd, if it is known that the sum of two numbers is even. Solution: There are 3 even numbers and 6 odd numbers between the numbers 1 to 11 Let A = Event that two odd numbers are choosen and B = Event that sum of two numbers is even. n(A) = number of ways to choose 2 odd numbers out of 6 odd numbers = 6C2 n(B) = Number of ways to choose two numbers whose sum is even = 5C2 + 5C2 ∴ n(A ∩ B) = Number of ways to choose 2 odd numbers whose sum is even = 6C2 Let sample space be S ∴ n(S) = number of ways to choose two numbers out of 11 numbers = 11C2



Question 10. An electronic assembly consists of two subsystem say, A and B. From previous testing procedures, the following probabilities are assumed to be known : P (A fails) = 0.2 P (B fails alone) = 0.15 P (A and B fail) = 0.15 Evaluate the following probabilities (i) P (A fails/B has failed) (ii) P (A fails alone).



Solution:

Question 11. Let A and B be two independent events. The probability of their simultanous

occurrence is and the probability that neither occurs Find P(A) and P(B). Solution:



A and B are independent events.



Question 12. Anil speaks truth in 60% of cases and Anand in 90% of the cases. In what percentage of cases are the likely to contradict each other in stating the same fact ?



Solution:

Question 13. Three persons A, B, C in order toss a coin. The one to throw a head wins. What are their respective chances of winning assuming that the game may continue indefinitely ? Solution:

Chances to be head on tossing a coin = ∵ A starts to play, then A will play again on fourth, seventh ………. then can win. ∴ Possibilities that A wins



Question 14.

Probability of a man to live for next 25 years is and so, his wife is , Find the probabilities that (i) both live for 25 years (ii) at least one of them live for 25 years (iii) Only wife live for 25 years.



Solution : Let A : Event that man live for next 25 years and B : event that wife live for next 25 years. Clearly both are independent events

Question 15. In three groups of children there are 3 girls and 1 boy, 2 girls, 2 boys, 1 girl and 3 boys. A child is randomly selected from each group. Find the probability that selected children are 1 girl and 2 boys. Solution: GI, GII and GIII are three groups of students. Then 1 girl and 2 boys can be selected in following ways.



Here Girls and Boys are represented by G1, G2, G3 and B1, B2, B3 resp. Hence required probability is = P (1 Girl and 2 boys are relected)

Question 16. Bag I contains 3 black and 4 white balls and bag II contains 4 black and 3 white balls. Now an unbiased die throws, if number 1 or 3 shows on die then a ball is drawn from bag I otherwise from bag II. Find the probability that the ball so drawn is black. Solution: Number of total balls in I bag 3 + 4 = 7, where 3 are black and 4 are white. Number of balls in II bag = 4 + 3 = 7, where 4 are black and 3 are white. Total number of result on throwing a die S = {1,2, 3,4, 5, 6} Let E1 be the event to get 1 and 3

Question 17. A person has undertaken a construction job. The probabilities are 0.65 that there will be strike, 0.80 that the construction job will be completed on time if there is no strike



and 0.32 that the construction job will be completed on time if there is a strike. Determine the probability that the construction job will be completed on time. Solution: Probability that there will be strike P(A) = 0.65 Probability that there will not be strike P( ) = 1 – 0.65 = 0.35 Let E the event that work finished in time. Then probability of completion of work in strike.

P ( ) = 0.32 and probability of completion of work when there is no strike

P ( ) = 0.80 ∴ Probability that the construction job will be completed on time

Question 18. Bag I contains 8 white and 7 black balls and bag II contains 5 white and 4 black balls. One ball is randomly transferred from bag I to bag II. Then a ball is drawn from bag II. Find the probability that the ball so drawn is white. Solution: Give that bag I contains 8 white and 7 black balls and Bag II contain 5 white and 4 black balls. One ball from I bag randomly put in II bag. So the possibility is that ball taken out from bag I is white. Then probability that white ball is choosen from bag I

= Now total number of balls in bag II = 5 + 1 = 6 Probability that white ball is choosen from bag II

= When both events occurs together

∴ Probability = × = Another possibility is that ball is choosen from bag I is black. Then probability that black ball is choosen from bag

= Now number of black ball in bag II = 4 + 1 = 5

∴ Probability that black ball is choosen = Probability that both events happen together

= × = ∵ Both events are mutually exclusive so only one event can happen.

∴ Requried probability = +

=



Question 19. In answering a question on a multiple choice test, a student either company the

answer or guesses or he knows the answer. Let be the probability that he copy the

answer and be the probability that he guesses Assuming that a student who copy

the answer will be correct with probability . What is the probability that the student knows the answer given that he answered it correctly. Solution: Probability that student guesses the answer

P(A) = Probability that student copies the answer

P(B) = Probability that student knows the answer

P(C) = 1 – –

= =

= = Let E is the event that answer is correct

Probability that student knows the answer and answered correctly



Question 20. A letter known to have come either from Tatanagar and Calcutta. On the envelope just two consecutive letters TA are visible, what is the probability the letter has come from. (i) Calcutta (ii) Tatanagar Solution: Let E1 be the event that letter came from Calcutta and E2 letter came from tatanagar A = event that just two letters TA are visible.

Question 21. A manufacturer has three machine operators A, B and C. The first operator A



productes 1% defective items, whereas the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B on the job for 30% of the time and C on the job for 20% fo the time. A defective item is produced. What is the probability that it was produced by A? Solution: Let E1 = Product by machine A E2 = Product by machine B E3 = Product by machine C Then E1,E2 and E3 are mutually exclusive events

Question 22. The random variable X has a probability distribution P(X) of the following from, where k is some number

(i) determine the value of k (ii) find P(x < 2), P(x ≤ 2) and P(x ≥ 2) Solution:



According to question Probability distribution of X is

Question 23. A random variable X can take all non-negative values and the probability that X take the value r is proportional to ar where (0 < α < 1). Find P(X = 0). Solution:



Question 24. Let X be random variable which assumes value x1, x2, x3, x4 such that 2P(X = x1) = 3P(X = x2) = 4P(X = x3) = 5P(X = x4) find the probability distribution of X. Solution:



Given 2P(X = x1) = 3p(x = x2) = P(X = x3) = 5 P(X = x4) = K

Question 25. A fair coin is tossed until a head or five tails occur. If X denotes the number of tosses of the coin, find the mean of X. Solution: Given X denotes the number of coins. Coins is tossed until a head or five tails accur therefere, it is clear if on X = 1, head comes then the process will be stopped and if tail comes then coin will be tossed second time. Clearly it will be repeated again and again till 5 tails come maximum. Then the value of X will be 1,2, 3,4, 5 S = {H, TH, TTH, TTTH or TTTTH, TTTTT} ∴ Probability that head comes in first throw

P(X = 1) = Similarly

Rajasthan Board RBSE Class 12 Maths Chapter 1 Composite ...

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