Maths CBSE 2021 Board Prelim Paper Solution - Vidyalankar ...

XII/TSCPT/21/Maths/FT/CBSE/Test.1_Soln/Pg.1

Std. XII − Mathematics : Solution to Prelim Test − 1 (CBSE)

Section − A

Q.1 (A) [1 M]

By definition

Q.2 (A) [1 M]

A = adj (A) A adj A=

Also, n 1

adj A A−

=

n 1 n 2

A A A 1 A 1− −

= = =

Q.3 (D) [1 M]

Since a b a b+ = − 2 2

a b a b+ = −

( ) ( )2 2

a b a b+ = −

2 2 2 2a b 2a.b a b 2a.b+ + = + −

4a.b 0= a.b 0 a b= ⊥

Angle between a and b is 2

.

Q.4 (D) [1 M]

7

x 1

P(X x) 1=

= =

K + 2K + 2K + 3K + K2 + 2K2 + 7K2 + K = 1

10K2 + 9K − 1 = 0 (10K − 1)(K + 1) = 0

1

K10

= or K = −1

But Probability cannot be negative 1

K10

=

Q.5 (A) [1 M]

Given points are P(1, 5, 4) and Q(4, 1, −2).

Direction ratios of PQ = 4 − 1, 1 − 5, −2 −4, i.e. 3, −4, −6

Q.6 (D) [1 M]

cos−1 1

2

+ 2 sin−1 1

2

= cos−11cos 2 sin sin

3 6

− +

= 2

23 6 3

+ =

Q.7 (B) [1 M]

cos 2 + cos 2 + cos 2

= 2 cos2 − 1 + 2 cos2 − 1 + 2 cos2 − 1

= 2 (cos2 + cos2 + cos2 ) − 3

= 2(1) − 3

= 2 − 3 = −1

(2) Vidyalankar : Std. XII (CBSE) − Mathematics


Q.8 (A) [1 M]

Let f(x) = cos x n1 x

1 x

+

− . Then,

f(−x) = cos (−x) n1 x

1 x

−

+ = − cos x n

1 xf (x)

1 x

+ = −

−

Since f (x) is an odd function,

1/2

1/2

1 xcosx n dx 0

1 x−

+ =

−

Q.9 (A) [1 M]

Let the equation of the plane be x y z

1a b c

+ + = … (1)

Then on using the given information, we have

1 1 1

k,a b c

+ + =

where k is a constant and a, b, c are the intercepts made by the plane with the axes.

1 1 1

1ak bk ck

+ + =

1 1 1 1 1 1

. . 1a k b k c k

+ + =

This shows that plane (1) passes through the fixed point 1 1 1

, , .k k k

Q.10 (B) [1 M]

Let the plane intersects the coordinate axes at (a, 0, 0), (0, b, 0), (0, 0, c) respectively

a 0 0

3

+ + = 1,

0 b 0

3

+ + = 2,

0 0 c

3

+ + = −1 i.e. a = 3, b = 6, c = −3

equation of the plane is

x

3 +

y

6 +

z

3− = 1, i.e. 2x + y − 2z = 6

Q.11 Yes [1 M]

Given that, A = {1, 2, 3} and B = {4, 5, 6, 7}

Now, f : A → B is defined as

f = {(1, 4), (2, 5), (3, 6)}

Therefore, f(1) = 4, f(2) = 5 and f(3) = 6

It is seen that the images of distinct elements of A under f are distinct.

f is one−one as for x1, x2 A, x1 ≠ x2 f(x1) ≠ f(x2)

Q.12 1 [1 M]

Given differential equation is

4 2

2

dy d y3x 0

dx dx

+ =

Order = 2 Degree = 1

Q.13 1 [1 M]

Two matrices are equal, if its corresponding elements are equal

a − b = − 1 …….. (i)

2a − b = 0 …….. (ii)

On solving equations (i) and (ii), we get,

− a = −1

a = 1

Mathematics (CBSE) : Solution to Prelim Test (3)


Q.14 3

2− [1 M]

We have y2 + 3x − 7 = 0 dy dy 3

2y 3dx dx 2y

−= − =

( )h,k

dy 3

dx 2k

− =

and as per condition,

3 3

1 k2k 2

− −= =

OR

Q.14 −72 units/sec. [1 M]

We have y = 5x − 2x3

2dy5 6x

dx= −

( )2d dy d dx5 6x .

dt dx dt dt

= −

= ( ) ( )( )( )dx

12x 12 3 2 72dt

− = − = −

Q.15 i 5

k26 26

+ [1 M]

Let ( ) ( )ˆ ˆ ˆ ˆ ˆ ˆc a b 2i j 2k i j 3k= + = − + + − + +

= ˆ ˆ ˆi 0j 5k+ +

Now, 2 2c a b 1 5 26= + = + =

( )a b

ca b

+=

+

aa

a

=

= i 5

k26 26

+

OR

Q.15 1

a

[1 M]

We have, a = 1 a = 1 = 1

a

Q.16

102 18 36 17 3 6

A 1 3 4 6 1 3 4

17 3 6 17 3 6

= = (By taking 6 common from R1)

= 0 [ R1 and R3 are identical] [1 M]

Q.17 Given integral is 2

2 3sin xdx

cos x

− [1 M]

= 2 2

2 3sin xdx

cos x cos x

−

= ( )22sec x 3sec x tan x dx−



= 22 sec x dx 3 secx tan xdx−

= 2 tan x − 3 sec x + C [1 M]

Q.18 Let x + sin x = t (1 + cos x) dx = dt

( )dt

I 3 3log t c 3log x sinx ct

= = + = + + [1 M]

OR

Q.18 −2 2cos 2x sin 2x dx

cos 4x dx = sin4x

c4

+ [1 M]

Q.19 Given family of curves is y = mx …….. (i) where, m is a parameter.

Now, differentiating equation (i) w.r.t. x, we get,

dy

mdx

=

Again, differentiating w.r.t. x, we get,

2

2

d y0

dx= , which is the required differential equation. [1 M]

Q.20 We have, f (x) = x

f (x) = ( )f ' x dx = x dx = ( )

3/2xC

3 / 2+ = 3/22

x C3

+

Also, f (1) = 2 2 = ( )3/22

1 C3

+ C = 2

23

− = 4

3

f (x) = 3/22 4x

3 3+ [1 M]

Section − B

Q.21 The given function is

f(x) = 2x3 − 9x2 + 12x − 15

Differentiating w.r.t. x, we get,

f (x) = 6x2 − 18x + 12

Putting f (x) = 0, we write,

6x2 − 18x + 12 = 0 6 (x2 − 3x + 2) = 0 6 (x− 1) (x − 2) = 0

x − 1 = 0 or x − 2 = 0 x = 1, 2 [1 M]

Now, we find intervals in which f(x) is increasing or decreasing.

Interval f (x) = 6 (x − 1) (x − 2) Sign of f (x)

x < 1 (+) (−) (−) + ve

1 < x < 2 (+) (+) (−) − ve

x > 2 (+) (+) (+) + ve

Now, we know that a function f (x) is increasing when f (x) 0 and

f (x) is decreasing when f (x) ≤ 0

So, the given function is increasing on intervals (−, 1) and (2, ) [1 M]

OR

Q.21 The given function is

f(x) = 2x3 − 9x2 + 12x − 15



Differentiating w.r.t. x, we get,

f (x) = 6x2 − 18x + 12

Putting f (x) = 0, we write,

6x2 − 18x + 12 = 0 6 (x2 − 3x + 2) = 0 6 (x− 1) (x − 2) = 0

x − 1 = 0 or x − 2 = 0 x = 1, 2 [1 M]

Now, we find intervals in which f(x) is increasing or decreasing.

Interval f (x) = 6 (x − 1) (x − 2) Sign of f (x)

x < 1 (+) (−) (−) + ve

1 < x < 2 (+) (+) (−) − ve

x > 2 (+) (+) (+) + ve

Now, we know that a function f (x) is increasing when f (x) 0 and

f (x) is decreasing when f (x) ≤ 0

So, the given function is decreasing on (1, 2). [1 M]

Q.22 Given that, t

y a cos t log tan ;2

= +

x = a sin t

Differentiating both equations w.r.t. t, we get

( )dy d d t

a cos t log tandt dt dt 2

= +

dy 1 d t

a sin t tantdt dt 2

tan2

= − +

2dy 1 t 1a sin t sec

tdt 2 2tan

2

= − +

2

tcos

1 12a sin t

t t2sin cos

2 2

= − +

1

a sin tt t

2sin .cos2 2

= − +

1

a sin tsin t

= − +

[1 M]

21 sin t

asin t

−=

=2cos t

asin t

and ( )dx d

asin t a cos tdt dx

= =

2a cos t

sin tdy dy dtcot t

dx dx dt a cos t

= = = [1 M]

Q.23 Let A (−2, 4, 7) and B(3, −5, 8) and c divides seg AB at yz place is ratio K : 1 then

c(o, y, z) = 2 1 2 1 2 1kx 1x ky y kz z, ,

k 1 k 1 k 1

+ + +

+ + +

o = 3k 2

k 1

−

+ k =

2

3 [1 M]

y = 2( 5) 3 (4)

2 5

− +

+ =

2

7



z = 2(8) 3(7) 37

2 3 5

+ =

+

Ratio is 2 : 3 and point of division is 2 37

0, ,5 5

[1 M]

Q.24 ( ) ( )a b c . a b c 0+ + + + =

a.a a.b a.c b.a b.b b.c c.a c.b c.c 0+ + + + + + + + =

( )22 2

a b c 2 a.b b.c c.a 0+ + + + + = [1 M]

( )2 2 23 5 7 2 a.b b.c c.a 0+ + + + + = ( ) ( )2 a.b b.c c.d 9 25 49+ + = − + +

( )83

a. b b.c c.d2

+ + = − [1 M]

Q.25 line lies in the plane point on line satisfies equation of plane

pt on the line is (4, 3, k) [1 M]

put x = 4, y = 2, z = k is 2x − 4y + z = 7

8 − 8 + k = 7

k = 7 [1 M]

Q.26 ( )80 4

P A ,100 5

= = ( )90 9

P B100 10

= =

P (Agree) = P (Both speaking truth or both telling lie)

= ( )P AB or AB [1 M]

= ( ) ( )P A P B or ( ) ( )P A P B

= 4 9 1 1

5 10 5 10

+

=

36 1 37

50 50

+= =

7474%

100= [1 M]

SECTION C

Q.27 ( )x 0

lim f x−→

= ( )h 0

lim f 0 h−→

− = 1/h

1/hh 0

e 1lim

e 1−

−

−→

−

+ [1 M]

= 1/h

h 0

1/h

11

elim

11

e

−→

−

+

= 0 1

0 1

−

+= −1 [1 M]

and ( )x 0

lim f x+→

= ( )h 0

lim f 0 h+→

+ = 1/h

1/hh 0

e 1lim

e 1+→

−

+ [1 M]

= 1/h

h 01/h

11

elim1

1e

+→

−

+

= 1 0

1 0

−

+= 1

( ) ( )x 0 x 0

lim f x lim f x− +→ →

( )x 0lim f x→

does not exist.

f (x) is discontinuous at x = 0 [1 M]

Q.28 Given relation is

S = {(a, b) : a, b R and a ≤ b3}

(i) Reflexive



As

31 1

3 3

is not true

1 1, S

3 3

S is not reflexive. [1 M]

(ii) Symmetric

As −2 ≤ (3)3 is true but 3 ≤ (−2)3 is not true. i.e., (−2, 3) S but (3, −2) S

S is not symmetric. [1 M]

(iii) Transitive

As

33

32

and

33 4

,2 3

but

34

33

is not true. [1 M]

3

3, S2

and

3 4, S,

2 3

but

43, S

3

S is not transitive. [1 M]

Hence, S is none of these i.e., reflexive, symmetric and transitive.

OR

Q.28 The given relation is

R = {(a, b) : f(a) = f(b)}, f : X → Y

To show that R is an equivalence relation, we show that it is reflexive, symmetric and transitive.

(i) Reflexive : Since, for every x X, we have

f(x) = f(x)

By using definition of R i.e., f(a) = f(b); a, b X

(x, x) R, X

Thus R is reflexive. [1 M]

(ii) Symmetric : Let (x, y) R, x, y X

f(x) = f(y)

f(y) = f(x)

(y, x) R or (x, y) R

Thus R is symmetric. [1 M]

(iii) Transitive : Let x, y, z X

Let (x, y) R and (y, z) R

f(x ) = f(y); x, y X …….. (i)

and f(y) = f(z); y, z X …….. (ii)

From equations (i) and (ii), we get,

f(x) = f(z); x, z X

(x, z) R; x, z X [1 M]

We have shown that

(x, y) R and (y, z) R

(x, z) R; x, y, z X

R is transitive relation.

Since, R is reflexive, symmetric and transitive, it is an equivalence relation. [1 M]

Q.29 f(x) = 2x3 − 15x2 + 36 x + 1

f (x) = 6x2 − 30x + 36

f (x) = 12x − 30 [1 M]

for extreme values

f (x) = 0

6x2 − 30x + 36 = 0

6(x2 − 5x + 6) = 0

(x − 2) (x − 3) = 0

x = 2, x = 3 [1 M]



Now, f(2) = 2(2)3 − 15(22) + 36(2) + 1 = 29 [1 M]

f(3) = 2(3)3 − 15(32) + 36(3) + 1 = 28

f(1) = 2(1)3 − 15(12) + 36(1) + 1 = 24

f(5) = 2(5)3 − 15(52) + 36(5) + 1 = 56

Absolute maximum value of f(x) is 56 at x = 5 and minimum value is 24 at x = 1. [1 M]

Q.30 Given, ( )T

70P P ,

100= ( )T

70 30P P 1

100 100= − =

( )T

80P Q ,

100= ( )T

80 20P Q 1

100 100= − = [1 M]

P (A and B are agree to each other)

= ( ) ( )T T T TP P Q P P Q + = ( ) ( ) ( ) ( )T T T TP P .P Q P P .P Q+ [1 M]

=70 80 30 20

. .100 100 100 100

+ =5600 600

10000 10000+ =

6200 62

10000 100=

Thus the percentage of A and B agree with each other = 62

100 62%100

= [1 M]

But they agree does not mean they are speaking truth only. Either both are speaking truth or both

are speaking lies.

Q.31 tan−1 1 32sin 2 cos

2

−

= tan−1 12sin 2 cos cos6

−

[1 M]

= tan−1 2sin 24

= tan−1 2 sin3

[1 M]

= tan−1 3

22

= tan−1 3 = tan−1 tan3

=

3

[1 M]

Q.32 Let the number of pedestal lamps sold = x

and the number of wooden shades sold = y

The given data can be put into tabular form as

Pedestal Lamp Wooden Shade Total hours available

Grinding / Cutting 2 1 12

Sprayer 3 2 20

Profit Rs. 5 Rs. 3

The required LPP is given as

Maximize profit z = 5x + 3y

Subject to constraints

2x + y ≤ 12 , 3x + 2y ≤ 20, x, y 0 [1 M]

Now, table for line 2x + y = 12 is

x 6 0

y 0 12

So, it passes through points (6, 0) and (0, 12).

Table for line 3x + 2y = 20 is

x 20/3 0

y 0 10



So, it passes through points 20

,03

and (0, 10).

Now, we solve the equations of lines to find their point of intersection.

3 (2x + y = 12) 6x + 3y = 36

2 (3x + 2y = 20) 6x + 4y = 40 − −

− y = − 4 y = 4

Putting y = 4 in 2x + y = 12, we get 2x = 8 or x = 4

Lines intersect at point (4, 4).

Now, graph of above LPP is as follows : [1 M]

From the figure, we see that the region OABC

is the feasible region.

The corner points of the feasible region are

O(0, 0), A(0, 10), B(4, 4), C(6, 0).

Now we find the value of Z at the corner points.

The maximum profit is Rs. 32 at the point B (4, 4) when 4 lamps and 4 shades are sold. [1 M]

Section − D

Q.33 Given that,

Number of members for honesty award = x

Number of members for helping others award = y

Number of members for supervising the workers to keep the colony neat and clean award = z

Now by first condition,

x + y + z = 12 …….. (i)

Second condition, 2x + 3(y + z) = 33

2x + 3y + 3z = 33 …….. (ii)

and third condition, (x + z) = 2y

x − 2y + z = 0 …….. (iii)

Hence, system of equation becomes

x + y + z = 12,

2x + 3y + 3z = 33,

and x − 2y + z = 0 [1 M]

Here, coefficient matrix,

1 1 1

A 2 3 3

1 2 1

= −

Let,

x

X y

z

=

and

12

B 33

0

=

Now by matrix method,

AX = B

Thus, X = A−1B …….. (iv)

Now, ( ) ( ) ( )A 1 3 6 1 2 3 1 4 3= + − − + − − = 9 + 1 − 7 = 10 − 7 = 3 [1 M]

Now, cofactors of A :

Corner points Z = 5x + 3y

O (0, 0) Z = 5(0) + 3(0) = 0

A (0, 10) Z = 5(0) + 3(10) = 30

B (4, 4) Z = 5 (4) + 3 (4) = 20 + 12 = 32

(maximum)

C (6, 0) Z = 5(6) + 3(0) = 30

[1 M for graph]



( )1 1

11

3 3A 1 3 6 9

2 1

+= − = + =

−

( ) ( )1 2

12

2 3A 1 2 3 1

1 1

+= − = − − =

( )1 3

13

2 3A 1 4 3 7

1 2

+= − = − − = −

−

( ) ( )2 1

21

1 1A 1 1 2 3

2 1

+= − = − + = −

−

( )2 2

22

1 1A 1 1 1 0

1 1

+= − = − = [1 M]

( ) ( )2 3

23

1 1A 1 2 1 3

1 2

+= − = − − − =

−

( )3 1

31

1 1A 1 3 3 0

3 3

+= − = − =

( ) ( )3 2

32

1 1A 1 3 2 1

2 3

+= − = − − = −

( )3 3

33

1 1A 1 3 2 1

2 3

+= − = − =

11 12 13

21 22 23

31 32 33

A A A

C A A A

A A A

=

9 1 7

3 0 3

0 1 1

−

= − −

[1 M]

T

9 3 0

adjA C 1 0 1

7 3 1

−

= = − −

1

9 3 0adjA 1

A 1 0 1A 3

7 3 1

−

−

= = − −

[1 M]

Now from equation (iv),

9 3 0 121

X 1 0 1 333

7 3 1 0

−

= − −

108 991

123

84 99

−

= − +

9 31

12 43

15 5

= =

x 3

X y 4

z 5

= =

On comparing, we get

x = 3, y = 4 and z = 5

The management of the colony must include the bravery award for some of its members. Because in

this category we appreciate the brave members of the colony for their bravery and make aware the

other members (men, women and children) of the colony.

(Students may write any other value) [1 M]

OR



Q.33 Given,

1 2 1

A 1 1 1

1 3 1

= −

−

( ) ( ) ( )1 2 11 3 3 1A 1 1= − ++ −− − = 4 + 4 + 2 = 10

A 0, hence A−1 exists.

1 1

A .adj AA

− = [1 M]

Now to find adj A.

( )

T11 12 13

21 22 23

31 32 33

A A A

adj A A A A

A A A

=

( )2

11

1 1A 1 1 3 4

3 1= − = + =

−

( ) ( )3

12

1 1A 1 1 1 2

1 1

−= − = − − − =

( )4

13

1 1A 1 3 1 2

1 3

−= − = − =

−

( ) ( )3

21

2 1A 1 2 3 5

3 1= − = − + = −

− [1 M]

( )4

22

1 1A 1 1 1 0

1 1= − = − =

( ) ( )5

23

1 2A 1 3 2 5

1 3= − = − − − =

−

( )4

31

2 1A 1 2 1 1

1 1= − = − =

( ) ( )5

32

1 1A 1 1 1 2

1 1= − = − + = −

−

( )6

33

1 2A 1 1 2 3

1 1= − = + =

− [1 M]

4 2 2 4 5 1

adj A 5 0 5 2 0 2

1 2 3 2 5 3

−

= − = − −

1

4 5 11

A 2 0 210

2 5 3

−

−

= −

[1 M]

From the given system of equations, we have,

1 2 1

A 1 1 1 ,

1 3 1

= − −

4

B 0

4

=

and

x

X y

z

=

[1 M]

1

4 5 1 41

X A B 2 0 2 010

2 5 3 4

−

−

= = −



( )

16 0 4 20 21 1

8 0 8 0 010 10

8 0 12 20 2

+ +

+ + − = = + +

x = 2, y = 0 and z = 2 [1 M]

Q.34 Let 3

3

tan x tan xI dx

1 tan x

+=

+

( )2

3

tan x 1 tan xdx

1 tan x

+=

+

2

3

tan x sec xdx

1 tan x

=

+ [ 1 + tan2x = sec2x] [1 M]

Put tan x = t sec2x dx = dt

3

t dtI

1 t=

+

= ( )( )2

t dt

1 t 1 t t+ + − [ (a3 + b3) = (a + b) (a2 − ab + b2)]

Let ( )( ) 22

t A Bt C

1 t t t 11 t 1 t t

+= +

+ − ++ + − [1 M]

t = A (t2 − t + 1) + (Bt + C) (1 + t)

t = (A + B)t2 + (−A + B + C) t + (A + C)

Putting t = −1

−1 = A + B + A − B − C + A + C = 3A

1

A3

−=

Equating the coefficient of t2 and constant terms, we get, A + B = 0 and A + C = 0

When A + B = 0 When A + C = 0

1

B 03

−+ =

1B

3=

1C 0

3

−+ =

1C

3= [1 M]

2

1 1t

1 3 3 3I dt dt1 t t t 1

+−

= ++ − +

( )

2 2

2 t 11 dt 1 t 1 1 dt 1 1dt dt

3 1 t 3 3 t 1 3 2t t 1 t t 1

++ −= − + = +

+ +− + − +

2

1 dt 1 2t 1 3dt

3 1 t 6 t t 1

− += − +

+ − +

2 2

1 dt 1 2t 1 3 dtdt

3 1 t 6 6t t 1 t t 1

−= − + +

+ − + − + [1 M]

Let z = t2 − t + 1 dz = (2t − 1) dt

( )22

2t 1 dzdt log z log t t 1

zt t 1

−= = = − +

− +

2

22

1 1 1 dtI log 1 t log t t 1

3 6 2 1 3t2 2

= − + + − + +

− +

[1 M]

2 1

1t1 1 1 1

log 1 tan x log tan x tan x 1 . tan C23 6 2 3

3 22

−

− = − + + − + + +

[ t = tanx]



2 11 1 1 2tan x 1log 1 tan x log tan x tan x 1 tan C

3 6 3 3

− − = − + + − + + +

[1 M]

Q.35. Area

4 4

2 2

y dx 3 x dx = = 3. ( )4

3/22

2x

3 [1 M]

= 2 ( )3/2 3/24 2 − [1 M]

= 2(8 − 2 2 ) [1 M]

= 16 − 4 2 sq. units [1 M]

Q.36 We know that equation of line passing through the points (x1, y1, z1) and (x2, y2, z2) is given by

1 1 1

2 1 2 1 2 1

x x y y z z

x x y y z z

− − −= =

− − − …….. (i) [1 M]

Here, (x1, y1, z1) = (0, 6, −9) and (x2, y2, z2) = (−3, −6, 3)

Equation of line AB is given by

x 0 y 6 z 9

3 0 6 6 3 9

− − += =

− − − − +

x y 6 z 9

3 12 12

− += =

− −

x y 6 z 9

1 4 4

− += =

− − [1 M]

Next, we have to find coordinates of foot of perpendicular D.

Now let x y 6 z 9

1 4 4

− += = =

− − [Say]

Hence x = − , y − 6 = −4 and z + 9 = 4

x = − , y = −4 + 6 and z = 4 − 9 [1 M]

Let coordinates of

D = (−, −4 + 6, 4 − 9) …….. (ii)

Now, dr’s of line CD are

(− − 7, −4 + 6 − 4, 4 − 9 + 1) = (− − 7, −4 + 2, 4 − 8)

Now, CD ⊥ AB

a1a2 + b1b2 + c1c2 = 0 [1 M]

where, a1 = − − 7, b1 = − 4 + 2, c1 = 4 − 8 [dr’s of line CD]

a2 = −1. b2 = −4, c2 = 4 [dr’s of line AB]

−1(− − 7) − 4 (−4 + 2) + 4 (4 − 8) = 0

+ 7 +16 − 8 + 16 − 32 = 0 33 − 33 = 0 33 = 33 = 1 [1 M]

Putting = 1 in equation (ii), we get required foot of perpendicular,

D = (−1, 2, −5)

Also, we have to find equation of line CD, where, C (7, 4, −1) and D (−1, 2, −5).

Required equation is

x 7 y 4 z 1

1 7 2 4 5 1

− − += =

− − − − + [Using equation (i)]

x 7 y 4 z 1

8 2 4

− − += =

− − −

x 7 y 4 z 1

4 1 2

− − += =

− − −

7 x 1 z

4 y4 2

− − −= − = [1 M]

❑ ❑ ❑ ❑ ❑ ❑

C (7, 4, −1)

B (−3, −6, 3) A (0, 6, −9) D

Maths CBSE 2021 Board Prelim Paper Solution - Vidyalankar ...

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