XI-Mathematics SOLUTION - Faizan Ahmed
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Transcript of XI-Mathematics SOLUTION - Faizan Ahmed
BOARD OF INTERMEDIATE EDUCATION, KARACHI
INTERMEDIATE EXAMINATION, 2018 (ANNUAL) (Science pre-Engineering and Science General Groups)
XI-Mathematics
SOLUTION
FROM THE DESK OF: FAIZAN AHMED
Code: MT-09 Time: 20 Minutes
SECTION โAโ (MCQs-Multiple Choice Questions) (20 Marks)
Q1. Choose the correct answer for each from the given options.
i. If 1,x-1, 3 are in A.P., then ๐ฅ =:
a. 0 b. 1 c. 2 d. 3 ii. The H.M.between 3 and 6 is:
a. 1
4 b.
9
2 c. ยฑ 18 d. 4
iii. If ๐โ๐
๐โ๐=
๐
๐, then a,b,c are in:
a. A.P. b. ๐บ. ๐ . c. H.P. d. A.G.P iv. The number of permutations of the letters of the word COMMITTEE
a. 9
2,2,2 b.
62,2,2
c. 9
2,2,1 d.
2,2,29
v. The middle term in the expansion of 2๐ฅ โ1
๐ฅ2 20
is:
a. Ninth term b. tenth term c. ๐๐๐๐ฃ๐๐๐ก๐ ๐ก๐๐๐ d. twelvth term
vi. If ๐ = 0, then ๐+1 !
๐ !=:
a. 0 b. 1 c. n d. โ vii. ๐ ๐๐600๐๐๐ 300 โ ๐๐๐ 600๐ ๐๐300 =:
a. 1
2 b. โ
3
2 c.
3
2 d. โ
1
2
viii. If arc length s is equal to the radius r , then the central angle ๐ is:
a. 0 radian b. 1
2 radian c. 2 radian d. 1 ๐๐๐๐๐๐
ix. In a triangle ABC, if ๐พ = 900, then the law of cosine reduces to:
a. ๐2 = ๐2 + ๐2 b. ๐2 = ๐2 โ ๐2 c. ๐2 = ๐2 + ๐2 d. ๐2 = ๐2 โ ๐2
x. In an escribed triangle ABC, โ
๐3=:
a. s b. (๐ โ ๐) c. (๐ โ ๐) d. (๐ โ ๐)
xi. If ๐๐๐๐ ๐ = 4 and ๐๐ ๐๐๐ = 3, then ๐ =:
a. 3 b. 5 c. 6 d. 2 xii. 10.5 0 =:
a. ๐
18radians b.
7๐
120 radians c.
10.5
๐ radians d. 5๐ radians
xiii. If ๐ด = {2,3} and ๐ต = {3,4}, then ๐ด โ ๐ต โฉ ๐ต =:
a. โ b. {โ } c. {2} d. {3}
xiv. ๐ด๐๐ดโฒ โฒ =:
a. ๐ด b. ๐ดโฒ c. โ d. ๐
xv. The imaginary part of ๐ 3 + 5๐2 is:
a. โ2๐ b. 3๐ c. โ2 d. โ5 xvi. If ๐ง is a complex number, then ๐ง๐ง =:
a. ๐ง2 b. ๐ง 2 c. |๐ง| d. ๐ง 2
xvii. The product of the roots of the equation ๐ฆ2 + 1 = 7๐ฆ โ 7
a. 4 b. 8 c. 7 d. 1 xviii. If ๐ is a complex cube root of unity, then 2 โ ๐ โ ๐2 2 =:
a. โ1 b. 1 c. 3 d. 9 xix. If A,B,C are non-singular matrices, then ๐ถ๐ต๐ด โ1
a. ๐ดโ1๐ตโ1๐ถโ1 b. ๐ถโ1๐ตโ1๐ดโ1 c. ๐ด๐ต๐ถ โ1 d. ABC xx. ๐ด ๐ดโ1 =:
a. ๐ด๐ดโ1 b. ๐ด ๐ผ3 c. ๐๐๐๐ด d. ๐ด2
SECTION B
COMPLEX NUMBER, ALGEBRA, MATRICES
Q.2(i) QUESTION: Solve the complex equation for x and y: ๐ฅ + 2๐ฆ๐ 2 = ๐ฅ๐. SOLUTION: (๐ฅ + 2๐ฆ๐)2 = ๐ฅ๐
(๐ฅ)2 + 2 ๐ฅ 2๐ฆ๐ + (2๐ฆ๐)2 = ๐ฅ๐
๐ฅ2 + 4๐ฅ๐ฆ๐ + 4๐ฆ2๐2 = ๐ฅ๐
๐ฅ2 + 4๐ฅ๐ฆ๐ + 4๐ฆ2(โ1) = ๐ฅ๐ โต ๐ = โ1 โ ๐2 = โ1
๐ฅ2 + 4๐ฅ๐ฆ๐ โ 4๐ฆ2 = ๐ฅ๐
๐ฅ2 โ 4๐ฆ2 + 4๐ฅ๐ฆ๐ = 0 + ๐ฅ๐
By comparing
4๐ฅ๐ฆ = ๐ฅ โ 4๐ฆ = 1
๐ฆ =1
4
Now, ๐ฅ2 โ 4๐ฆ2 = 0
๐ฅ2 โ 4 1
4
2
= 0
๐ฅ2 โ1
4= 0
๐ฅ2 =1
4
๐ฅ = ยฑ1
2
๐. ๐ = ยฑ1
2,
1
4 OR ๐. ๐ =
1
2,
1
4 , โ
1
2,
1
4
OR Q.2(i)
QUESTION: Solve the complex equation for x and y: ๐ฅ 1 + 2๐ + ๐ฆ 3 + 5๐ = โ3๐ SOLUTION: ๐ฅ 1 + 2๐ + ๐ฆ 3 + 5๐ = โ3๐ ๐ฅ + 2๐ฅ๐ + 3๐ฆ + 5๐ฆ๐ = โ3๐ ๐ฅ + 3๐ฆ + 5๐ฆ๐ + 2๐ฅ๐ = โ3๐ ๐ฅ + 3๐ฆ + ๐ 5๐ฆ + 2๐ฅ = 0 โ 3๐ By Comparing ๐ฅ + 3๐ฆ = 0 ๐ฅ = โ3๐ฆ ---- 1 And 5๐ฆ + 2๐ฅ = โ3 5๐ฆ + 2(โ3๐ฆ) = โ3 5๐ฆ โ 6๐ฆ = โ3 โ๐ฆ = โ3
๐ฆ = 3
1 โ ๐ฅ = โ3๐ฆ ๐ฅ = โ3 3 = โ9 ๐. ๐ = โ9,3
2(ii) QUESTION:
Solve: ๐ฅ +1
๐ฅ
2
= 4 ๐ฅ โ1
๐ฅ
SOLUTION:
๐ฅ +1
๐ฅ
2
= 4 ๐ฅ โ1
๐ฅ
๐ฅ 2 + 2 ๐ฅ 1
๐ฅ +
1
๐ฅ
2
= 4 ๐ฅ โ1
๐ฅ
๐ฅ2 + 2 +1
๐ฅ2= 4 ๐ฅ โ
1
๐ฅ
๐ฅ2 โ 2 +1
๐ฅ2+ 4 = 4 ๐ฅ โ
1
๐ฅ
๐ฅ โ1
๐ฅ
2
+ 4 = 4 ๐ฅ โ1
๐ฅ ----(1)
Let ๐ฆ = ๐ฅ โ1
๐ฅ
1 โ ๐ฆ2 + 4 = 4๐ฆ ๐ฆ2 โ 4๐ฆ + 4 = 0 ๐ฆ โ 2 2 = 0 ๐ฆ โ 2 = 0
๐ฆ = 2
But ๐ฅ โ1
๐ฅ= ๐ฆ
๐ฅ โ1
๐ฅ= 2
๐ฅ2 โ 1
๐ฅ= 2
๐ฅ2 โ 1 = 2๐ฅ ๐ฅ2 โ 2๐ฅ โ 1 = 0
๐ฅ =โ๐ยฑ ๐2โ4๐๐
2๐
๐ฅ =โ(โ2)ยฑ (โ2)2โ4 1 (โ1)
2(1)
๐ฅ =2ยฑ 4+4
2=
2ยฑ 8
2=
2ยฑ 4ร2
2=
2ยฑ2 2
2=
2 1ยฑ 2
2
๐ฅ = 1 ยฑ 2
๐. ๐ = 1 ยฑ 2
2(iii) QUESTION: For what values of โmโ will the equation have equal roots? ๐ + 1 ๐ฅ2 + 2 ๐ + 3 ๐ฅ + 2๐ + 3 = 0 SOLUTION: ๐ + 1 ๐ฅ2 + 2 ๐ + 3 ๐ฅ + 2๐ + 3 = 0 ----(1)
๐ด = ๐ + 1, ๐ต = 2 ๐ + 3 , ๐ถ = 2๐ + 3
As roots of (1) are equal
i.e. ๐ต2 โ 4๐ด๐ถ = 0
2 ๐ + 3 2 โ 4 ๐ + 1 2๐ + 3 = 0
4 ๐2 + 6๐ + 9 โ 4 2๐2 + 3๐ + 2๐ + 3 = 0
รท ๐๐๐ ๐๐ฆ 4, ๐ค๐ ๐๐๐ก
๐2 + 6๐ + 9 โ 2๐2 + 5๐ + 3 = 0
๐2 + 6๐ + 9 โ 2๐2 โ 5๐ โ 3 = 0
โ๐2 + ๐ + 6 = 0 โน ๐2 โ ๐ โ 6 = 0
๐2 โ 3๐ + 2๐ โ 6 = 0
๐ ๐ โ 3 + 2 ๐ โ 3 = 0 ๐ โ 3 ๐ + 2 = 0
๐ธ๐๐ก๐๐๐: ๐ โ 3 = 0 โ ๐ = 3
๐๐: ๐ + 2 = 0 โ ๐ = โ2
2(iv) QUESTION:
If ๐จ = ๐๐๐๐ฝ โ๐๐๐๐ฝ๐๐๐๐ฝ ๐๐๐๐ฝ
and ๐ฉ = ๐๐๐๐ฝ ๐๐๐๐ฝ
โ๐๐๐๐ฝ ๐๐๐๐ฝ then prove that ๐จ๐ฉ = ๐ฉ๐จ = ๐ฐ๐
SOLUTION:
๐ด = ๐ ๐๐๐ โ๐๐๐ ๐๐๐๐ ๐ ๐ ๐๐๐
and ๐ต = ๐ ๐๐๐ ๐๐๐ ๐
โ๐๐๐ ๐ ๐ ๐๐๐
??๐ด๐ต = ๐ต๐ด = ๐ผ2
AB = ๐ ๐๐๐ โ๐๐๐ ๐๐๐๐ ๐ ๐ ๐๐๐
๐ ๐๐๐ ๐๐๐ ๐
โ๐๐๐ ๐ ๐ ๐๐๐
= ๐ ๐๐2๐ + ๐๐๐ 2๐ ๐ ๐๐๐๐๐๐ ๐ โ ๐ ๐๐๐๐๐๐ ๐๐ ๐๐๐๐๐๐ ๐ โ ๐ ๐๐๐๐๐๐ ๐ ๐๐๐ 2๐ + ๐ ๐๐2๐
= 1 00 1
= ๐ผ2
BA= ๐ ๐๐๐ ๐๐๐ ๐
โ๐๐๐ ๐ ๐ ๐๐๐
๐ ๐๐๐ โ๐๐๐ ๐๐๐๐ ๐ ๐ ๐๐๐
= ๐ ๐๐2๐ + ๐๐๐ 2๐ โ๐ ๐๐๐๐๐๐ ๐ + ๐ ๐๐๐๐๐๐ ๐โ๐ ๐๐๐๐๐๐ ๐ + ๐ ๐๐๐๐๐๐ ๐ ๐๐๐ 2๐ + ๐ ๐๐2๐
= 1 00 1
= ๐ผ2
๐ด๐ต = ๐ต๐ด = ๐ผ2
2(v) QUESTION: Using the properties of determinants, evaluate the determinant:
1 ๐ฅ ๐ฆ๐ง1 ๐ฆ ๐ง๐ฅ1 ๐ง ๐ฅ๐ฆ
=
1 1 1๐ฅ ๐ฆ ๐ง
๐ฅ2 ๐ฆ2 ๐ง2
SOLUTION:
??
1 ๐ฅ ๐ฆ๐ง1 ๐ฆ ๐ง๐ฅ1 ๐ง ๐ฅ๐ฆ
=
1 1 1๐ฅ ๐ฆ ๐ง
๐ฅ2 ๐ฆ2 ๐ง2
L.H.S=
1 ๐ฅ ๐ฆ๐ง1 ๐ฆ ๐ง๐ฅ1 ๐ง ๐ฅ๐ฆ
ร ๐๐๐ ๐ 1 ๐๐ฆ ๐ฅ, ๐ 2 ๐๐ฆ ๐ฆ ๐๐๐ ๐ 3 ๐๐ฆ ๐ง
=1
๐ฅ๐ฆ๐ง
๐ฅ ๐ฅ2 ๐ฅ๐ฆ๐ง
๐ฆ ๐ฆ2 ๐ฅ๐ฆ๐ง
๐ง ๐ง2 ๐ฅ๐ฆ๐ง
Taking common ๐ฅ๐ฆ๐ง from ๐ถ3
=๐ฅ๐ฆ๐ง
๐ฅ๐ฆ๐ง ๐ฅ ๐ฅ2 1๐ฆ ๐ฆ2 1
๐ง ๐ง2 1
= ๐ฅ ๐ฅ2 1๐ฆ ๐ฆ2 1
๐ง ๐ง2 1
Taking transpose
=
๐ฅ ๐ฆ ๐ง
๐ฅ2 ๐ฆ2 ๐ง2
1 1 1
Interchanging ๐ 2 and ๐ 3
= (โ1)
๐ฅ ๐ฆ ๐ง1 1 1๐ฅ2 ๐ฆ2 ๐ง2
Interchanging ๐ 1 and ๐ 2
= (โ1)(โ1)
1 1 1๐ฅ ๐ฆ ๐ง
๐ฅ2 ๐ฆ2 ๐ง2
=
1 1 1๐ฅ ๐ฆ ๐ง
๐ฅ2 ๐ฆ2 ๐ง2 = ๐ . ๐ป. ๐
SECTION B
GROUPS, SEQUENCES & SERIES, COUNTING PROBLEMS
Q.3(i) QUESTION: Let ๐บ = {1, ๐, ๐2}, where ๐ is a complex cube root of unity. Show that ๐บ, . is an abelian group, where โ.โ Is an ordinary multiplication. Note: For a group:
a) Associative law holds b) There exist an identity element w.r.t multiplication c) Every element has an inverse in G
For Abelian: multiplication is Commutative SOLUTION: ๐บ = {1, ๐, ๐2}
a) Associative law holds 1 ร ๐ ร ๐2 = 1 ร ๐ ร ๐2 1 ร ๐3 = ๐ ร ๐2 ๐3 = ๐3
1 = 1 โ ๐บ b) 1 is the identity element with respect to multiplication
As, 1 ร 1 = 1 โ ๐บ 1 ร ๐ = ๐ โ ๐บ 1 ร ๐2 = ๐2 โ ๐บ
c) Each element has an inverse in G Inverse of 1 is 1โ ๐บ Inverse of ๐ is ๐2 โ ๐บ Inverse of ๐2 is ๐ โ ๐บ
๐บ, . is a group For Abelian:
1 ร ๐ = ๐ ร 1 = ๐ โ ๐บ 1 ร ๐2 = ๐2 ร 1 = ๐2 โ ๐บ
๐ ร ๐2 = ๐2 ร ๐ = ๐3 = 1 โ ๐บ โ.โ Is multiplication ๐บ, . is a Abelian group
Q.3(ii) QUESTION: If three books are picked at random from a shelf containing 3 novels, 4 book of poems and a dictionary. What is the probability that: (i) dictionary is selected (ii) one novel and 2 book of poems are selected SOLUTION: Total novels = 3 Total book of poems = 4 Total dictionary = 1 Total books = 3 + 4 + 1 = 8 Three books are selected at random ๐ ๐ = ๐ถ3.
8
(i) dictionary is selected Let A be the event that a dictionary is selected
๐ ๐ด =๐ ๐ด
๐ ๐
=๐ถ2.
7 ๐ถ1.1
๐ถ3.8
=21
56
=3
8
(ii) one novel and 2 book of poems are selected Let B be the event that one novel and 2 book of poems are selected
๐ ๐ต =๐ ๐ต
๐ ๐
=๐ถ1.
3 ๐ถ2.4
๐ถ3.8
=3 ร 6
56
=9
28
OR Q.3(ii)
QUESTION: In how many ways can a party of 5 students and 2 teachers be formed out of 15 students and 5 teachers. SOLUTION: Total students = 15 Total teachers = 5 A party of 5 students and 2 teachers be formed ๐๐๐ก๐๐ ๐ค๐๐ฆ๐ = ๐ถ5.
15 ๐ถ2.5
= 3003 ร 10 = 30030
Q.3(iii) QUESTION: Prove by the principle of Mathematical Induction. 1
1.2+
1
2.3+
1
3.4+. . . +
1
๐(๐+1)=
๐
๐+1, โ๐ โ โ
SOLUTION: PROOF I:
Verifying p(n) for ๐ = 1 1
1.2=
1
1+1
1
2=
1
2 Verified
๐ ๐ ๐๐ ๐ก๐๐ข๐ ๐๐๐ ๐ = 1
PROOF II:
Assuming that p(๐) is true for ๐ = ๐
So we have, 1
1.2+
1
2.3+
1
3.4+. . . +
1
๐(๐+1)=
๐
๐+1 ----(1)
Kth term =1
๐(๐ + 1)
k + 1 th term =1
(๐ + 1)(๐ + 2)
Adding 1
(๐+1)(๐+2) both sides in (1)
1
1.2+
1
2.3+
1
3.4+. . . +
1
๐ ๐+1 +
1
๐+1 ๐+2 =
๐
๐+1 +
1
๐+1 ๐+2
=๐ ๐+2 +1
๐+1 ๐+2
=๐2+2๐+1
๐+1 ๐+2
= ๐+1 2
๐+1 ๐+2
=๐+1
๐+1+1
๐๐๐๐๐ ๐๐ ๐ก๐๐ข๐ ๐๐๐ ๐ = ๐ + 1
๐ป๐๐๐๐, ๐ ๐ ๐๐ ๐ก๐ข๐๐ ๐๐๐ ๐๐๐ ๐๐๐ก๐ข๐๐๐ ๐๐ข๐๐๐๐๐ ๐.
OR Q.3(iii)
QUESTION: Find the sum of the following series: 212 + 222 + 232 + . . . + 502 SOLUTION: 212 + 222 + 232 + . . . + 502
= 502 โ 212
โต ๐2 =๐ ๐ + 1 (2๐ + 1)
6
=50 50 + 1 (2 ร 50 + 1)
6โ
21 21 + 1 (2 ร 21 + 1)
6
=50 51 (101)
6โ
21 22 (41)
6
= 42925 โ 3157 = 39768
Q.3(iv) QUESTION: Find the sum of an A.P., of nineteen terms whose middle term in 10. SOLUTION: Number of terms =n=19
๐๐๐๐๐๐ ๐ก๐๐๐ =๐+1
2๐ก๐ ๐ก๐๐๐
๐. ๐ =20
2= 10๐ก๐ ๐ก๐๐๐
In an A.P. nth term = Tn=a+(n-1)d ๐10 = ๐ + 10 โ 1 ๐ 10 = ๐ + 9๐ ๐ + 9๐ = 10 ---(1) In an A.P.
Sum of n terms = ๐๐ =๐
2{2๐ + ๐ โ 1 ๐}
Sum of 19 terms = ๐19 =19
2{2๐ + 19 โ 1 ๐}
๐19 =19
2{2๐ + 18๐}
๐19 =19
2ร 2 ๐ + 9๐
๐19 = 19 10
๐19 = 190 [๐๐ ๐๐๐ (1)]
Q.3(v) QUESTION:
Find the value of n so that ๐๐+1+๐๐+1
๐๐ +๐๐ may become the H.M. between a and b.
SOLUTION:
H.M. between a and b is 2๐๐
๐+๐
According to the question: ๐๐+1 + ๐๐+1
๐๐ + ๐๐=
2๐๐
๐ + ๐
๐ + ๐ ๐๐+1 + ๐๐+1 = 2๐๐ ๐๐ + ๐๐ ๐๐+2 + ๐๐๐+1 + ๐๐+1๐ + ๐๐+2 = 2๐๐+1๐ + 2๐๐๐+1 ๐๐+2 + ๐๐๐+1 โ 2๐๐๐+1 + ๐๐+1๐ โ 2๐๐+1๐ + ๐๐+2 = 0 ๐๐+2 โ ๐๐๐+1 โ ๐๐+1๐ + ๐๐+2 = 0 ๐๐๐+1 โ ๐๐+1๐ โ ๐๐ + ๐๐๐+1 = 0 ๐๐+1 ๐ โ ๐ โ๐๐+1 ๐ โ ๐ = 0 ๐ โ ๐ ๐๐+1 โ ๐๐+1 = 0 ๐๐+1 โ ๐๐+1 = 0 ๐๐+1 = ๐๐+1 ๐๐+1
๐๐+1= 1
๐
๐ ๐+1
= ๐
๐
0
By comparing ๐ + 1 = 0
๐ = โ1
OR Q.3(v)
QUESTION: Find the first term of a G.P., whose second term is 3 and sum to infinity is 12. SOLUTION: In a G.P.: ๐๐ = ๐๐๐โ1 ๐2 = ๐๐2โ1 = 3 ๐๐ = 3 --- 1
๐ =๐
1 โ ๐ = 12
๐
1 โ ๐ = 12
ร ๐๐๐ ๐๐ฆ ๐, ๐๐๐ก๐ ๐ ๐๐๐๐ ๐๐
1 โ ๐ = 12๐
3
1 โ ๐ = 12๐ [๐ข๐ ๐๐๐ (1)]
1
1 โ ๐ = 4๐
1 = 4๐ 1 โ ๐ 1 = 4๐ โ 4๐2 4๐2 โ 4๐ + 1 = 0 2๐ โ 1 2 = 0 2๐ โ 1 = 0
๐ =1
2
1 โน ๐๐ = 3
๐ 1
2 = 3
๐ = 6
SECTION B
TRIGONOMETRY
Q.4(i) QUESTION: A belt 24.75 meters long passes around a 3.5 cm diameter pulley. As the belt makes 3 complete revolutions in a minute, how many radians does the wheel turn in two second? SOLUTION: Length of belt = 24.75๐
Diameter of the wheel of Pulley = 3.5๐๐
โต ๐ =๐ท๐๐๐๐๐ก๐๐
2
๐ =3.5
2๐๐ = 1.75๐๐
๐ =1.75
100= 0.0175๐
As the belt makes three complete revolutions in a minute
๐ = ๐๐๐ ๐ก๐๐๐๐ ๐๐๐ฃ๐๐๐๐ ๐๐ฆ ๐ก๐๐ ๐๐๐๐ก ๐๐ ๐ ๐๐๐๐ข๐ก๐ = 3(24.75)
= 74.25๐
๐ = ๐๐๐ ๐ก๐๐๐๐ ๐๐๐ฃ๐๐๐๐ ๐๐ฆ ๐ก๐๐ ๐๐๐๐ก ๐๐ ๐ ๐ ๐๐๐๐๐ =74.25
60
๐ = 1.2375๐
โต ๐ = ๐๐
๐ =๐
๐=
1.2375
0.0175= 70.714
For 2 seconds ๐ = 70.714 ร 2
๐ = 141.428๐๐๐๐๐๐
๐ ๐๐๐๐๐
Q.4(ii) QUESTION: Find the period of ๐ก๐๐๐ฅ. SOLUTION: ๐ฟ๐๐ก ๐ ๐ฅ = ๐ก๐๐๐ฅ ๐ ๐ฅ + ๐ = ๐ก๐๐(๐ฅ + ๐)
=๐ก๐๐๐ฅ + ๐ก๐๐๐
1 โ ๐ก๐๐๐ฅ๐ก๐๐๐
Replacing p with ๐, we get
๐ ๐ฅ + ๐ =๐ก๐๐๐ฅ + ๐ก๐๐๐
1 โ ๐ก๐๐๐ฅ๐ก๐๐๐
= ๐ก๐๐๐ฅ + 0
1 โ ๐ก๐๐๐ฅ(0)
= ๐ก๐๐๐ฅ
1 โ 0
= ๐ก๐๐๐ฅ ๐ ๐ฅ + ๐ = ๐(๐ฅ)
๐ป๐๐๐๐, ๐ ๐ฅ ๐๐ ๐ก๐๐ ๐๐๐๐๐๐๐๐ ๐๐ข๐๐๐ก๐๐๐ ๐๐ ๐๐๐๐๐๐ ๐.
Q.4(iii) QUESTION: If a=b=c then prove that r: R : r1 = 1 : 2 : 3, where r, R, r1 have their usual meanings. SOLUTION: ? ? r: R โถ r1 = 1 โถ 2 โถ 3
๐ =โ
๐ , ๐ =
๐๐๐
4โ, ๐1 =
โ
๐ โ ๐
Given a=b=c
โ= ๐ ๐ โ ๐ ๐ โ ๐ ๐ โ ๐ ----(1)
Where, ๐ =๐+๐+๐
2
Here, ๐ =๐+๐+๐
2=
3๐
2
๐ โ ๐ = ๐ โ ๐ = ๐ โ ๐ =3๐
2โ
๐
1=
๐
2
๐ โ ๐ =๐
2, ๐ โ ๐ =
๐
2, ๐ โ ๐ =
๐
2
1 โ โ= 3๐
2ร
๐
2ร
๐
2ร
๐
2
โ= 3๐4
16
โ= 3 ๐2
4
Now, ๐ =โ
๐ =
3 ๐2
4
3๐
2
= 3 ๐2
4ร
2
3๐=
3๐
2ร3 =
3๐
6
๐ =๐๐๐
4โ=
๐. ๐. ๐
4 ร 3 ๐2
4
=๐
3=
๐
3ร
3
3=
3๐
3
๐1 =โ
๐ โ ๐=
3 ๐2
4
๐2
= 3 ๐2
4ร
2
๐=
3๐
2
Now, ๐: ๐ : ๐1 = 3๐
6โถ
3๐
3 โถ
3๐
2
๐๐๐๐ ๐ . ๐ป. ๐ ๐๐ฆ 6
3๐, ๐ค๐ ๐๐๐ก
๐: ๐ : ๐1 = 3๐
6 ร
6
3๐:
3๐
3ร
6
3๐ โถ
3๐
2ร
6
3๐
๐: ๐ : ๐1 = 1 โถ 2 โถ 3
Q5(iv) QUESTION: ๐ก๐๐2๐๐๐๐ก๐ = 3 SOLUTION: ๐ก๐๐2๐๐๐๐ก๐ = 3
2๐ก๐๐๐
1 โ tan2 ๐ร
1
๐ก๐๐๐= 3
2
1 โ tan2 ๐= 3
2 = 3 1 โ tan2 ๐ 2 = 3 โ 3 tan2 ๐ 3 tan2 ๐ = 3 โ 2 3 tan2 ๐ = 1
tan2 ๐ =1
3
๐ก๐๐๐ = ยฑ1
3
Either:
๐ก๐๐๐ =1
3
Or:
๐ก๐๐๐ = โ1
3
๐ = tanโ1 1
3
๐ =๐
6
๐ = tanโ1 โ1
3
๐ = โ๐
6
๐บ. ๐ = 2๐๐ +๐
6 โช 2๐๐ โ
๐
6
๐ โ โค
Q.4(v) QUESTION: Prove that: ๐๐๐โ1 1
5+ ๐๐๐โ1 1
4= ๐๐๐โ1 9
19
SOLUTION:
? ? ๐๐๐โ11
5+ ๐๐๐โ1
1
4= ๐๐๐โ1
9
19
๐ฟ๐๐ก ๐ฆ = ๐๐๐โ1 1
5+ ๐๐๐โ1 1
4= ๐ด + ๐ต ------(1)
Where, ๐ด = ๐๐๐โ1 1
5 and ๐ต = ๐๐๐โ1 1
4
So, ๐๐๐๐ด =1
5 and ๐๐๐๐ต =
1
4
1 โ ๐ฆ = ๐ด + ๐ต Taking Tan of both sides ๐๐๐๐ฆ = ๐๐๐(๐ด + ๐ต)
๐๐๐๐ฆ =๐๐๐๐ด + ๐๐๐๐ต
1 โ ๐๐๐๐ด ๐๐๐๐ต
๐๐๐๐ฆ =
15
+14
1 โ15
ร14
=
4 + 520
1 โ1
20 =
9201920
๐๐๐๐ฆ =9
19
๐ฆ = ๐๐๐โ19
19
๐๐๐โ1 1
5+ ๐๐๐โ1 1
4= ๐๐๐โ1 9
19 ๐น๐๐๐ 1
OR Q.4(v)
QUESTION:
Prove that: ๐๐๐โ1๐ด + ๐๐๐โ1๐ต = sinโ1 ๐ด 1 โ ๐ต2 + ๐ต 1 โ ๐ด2
SOLUTION: ๐ฟ๐๐ก ๐ก = ๐๐๐โ1๐ด + ๐๐๐โ1๐ต = ๐ฅ + ๐ฆ ------(1) Where, ๐ฅ = ๐๐๐โ1๐ด and ๐ฆ = ๐๐๐โ1๐ต
So, ๐๐๐๐ฅ = ๐ด ๐๐๐2๐ฅ + cos2 ๐ฅ = 1 ๐ด2 + cos2 ๐ฅ = 1 cos2 ๐ฅ = 1 โ ๐ด2
๐ถ๐๐ ๐ฅ = 1 โ ๐ด2
and ๐ ๐๐๐ฆ = ๐ต
๐๐๐2๐ฆ + cos2 ๐ฆ = 1 ๐ต2 + cos2 ๐ฆ = 1 cos2 ๐ฆ = 1 โ ๐ต2
๐ถ๐๐ ๐ฆ = 1 โ ๐ต2
1 โ ๐ก = ๐ฅ + ๐ฆ Taking Sin of both sides ๐ ๐๐๐ก = sin(๐ฅ + ๐ฆ) ๐ ๐๐๐ก = sinxcosy + cosxsiny
๐ ๐๐๐ก = A 1 โ ๐ต2 + B 1 โ ๐ด2
๐ก = sinโ1 ๐ด 1 โ ๐ต2 + ๐ต 1 โ ๐ด2 Proved.
SECTION C Q.5(a) QUESTION:
SOLUTION: Let five shares in A.P. are: ๐ โ 4๐, ๐ โ 2๐, ๐, ๐ + 2๐, ๐ + 4๐ A/c to the Ist condition: ๐ โ 4๐ + ๐ โ 2๐ + ๐ + ๐ + 2๐ + ๐ + 4๐ = 600 5๐ = 600
๐ = 120 A/c to the 2nd condition:
๐ โ 4๐ + ๐ โ 2๐ =1
7 ๐ + ๐ + 2๐ + ๐ + 4๐
2๐ โ 6๐ =1
7 3๐ + 6๐
14๐ โ 42๐ = 3๐ + 6๐ 14๐ โ 3๐ = 42๐ + 6๐ 11๐ = 48๐ โ 48๐ = 11๐
๐ =11 ร 120
48=
55
2
Five shares are when a=120 and d=55
2:
๐ โ 4๐, ๐ โ 2๐, ๐, ๐ + 2๐, ๐ + 4๐
120 โ 4 55
2 , 120 โ 2
55
2 , 120, 120 + 2
55
2 , 120 + 4
55
2
10, 65, 120, 175, 230
OR Q.5(a)
QUESTION: Note: BIEK has made mistake here, by 35th term, it is impossible question. It is 34th term.
๐พ๐๐๐๐ ๐ธ๐๐๐๐๐๐๐: ๐ผ๐ ๐๐ ๐ป. ๐. , ๐ก๐๐ 10๐ก๐ ๐ก๐๐๐ ๐๐ 35, 35๐ก๐ ๐ก๐๐๐ ๐๐ 25.๐ผ๐ ๐ก๐๐ ๐๐๐ ๐ก ๐ก๐๐๐ ๐๐ 2, ๐๐๐๐ ๐ก๐๐ ๐๐ข๐๐๐๐ ๐๐ ๐ก๐๐๐๐ .
SOLUTION: In an H.P:
pth term = 10th term = ๐ฅ = 35, ๐ = 10
qth term = 35th term = ๐ฆ = 25, ๐ = 35
rth term = last term = ๐ง = 2, ๐ == ๐ =?
We have,
1
๐ฅ
1
๐ฆ
1
๐ง
๐ ๐ ๐1 1 1
= 0
1
35
1
25
1
210 35 ๐1 1 1
= 0
รing ๐ 1 by 350, we get
10 14 17510 35 ๐1 1 1
= 0 ร 350
Expanding by ๐ถ3, we get
175 10 351 1
โ ๐ 10 141 1
+ 1 10 1410 35
= 0
175 10 โ 35 โ ๐ 10 โ 14 + 250 โ 140 = 0 175 โ25 โ ๐ โ4 + 350 โ 140 = 0 โ4375 + 4๐ + 210 = 0 4๐ โ 4165 = 0 4๐ = 4165
๐ง =4165
4 Not a positive integer
Hence, wrong Question
Q.5(b) QUESTION:
Prove law of Cosine: a2 =b
2 + c
2 โ 2bccos๐ผ
PROOF
Law of Cosine: c2 =a
2 + b
2 โ 2abcos๐ธ
Proof: We place a โABC in x-, y- coordinate system such that C(0,0) is at the origin and B(a,0) on
positive x-axis as shown in the figure.
As, cos 1800 โ ๐พ =๐๐๐ ๐
๐๐๐=
๐ถ๐ฟ
๐ด๐ถ
cos 1800 ๐๐๐ ๐พ + sin 1800 ๐ ๐๐๐พ =๐ถ๐ฟ
๐
โ1 ๐๐๐ ๐พ + (0)๐ ๐๐๐พ =๐ถ๐ฟ
๐
๐๐๐ ๐พ =๐ถ๐ฟ
๐
๐ถ๐ฟ = โ๐๐๐๐ ๐พ
As, sin 1800 โ ๐พ =๐๐๐
๐๐๐=
๐ด๐ฟ
๐ด๐ถ
sin 1800 ๐๐๐ ๐พ โ cos 1800 ๐ ๐๐๐พ =๐ถ๐ฟ
๐
0 ๐๐๐ ๐พ โ (โ1)๐ ๐๐๐พ =๐ถ๐ฟ
๐
๐ ๐๐๐พ =๐ถ๐ฟ
๐
๐ถ๐ฟ = ๐๐ ๐๐๐พ
So, coordinates of A are (b.cos๐พ, b.sin๐พ)
We have distance formula as:
d = (๐ฅ2 โ ๐ฅ1)2 + (๐ฆ2 โ ๐ฆ1)2
Here ๐ด๐ต = c = (๐๐๐๐ ๐พ โ ๐)2 + (๐๐ ๐๐๐พ โ 0)2
c = ๐2๐๐๐ 2๐พ โ 2๐๐๐๐๐ ๐พ + ๐2 + ๐2๐ ๐๐2๐พ
c = ๐2๐๐๐ 2๐พ + ๐2๐ ๐๐2๐พ โ 2๐๐๐๐๐ ๐พ + ๐2
c = ๐2(๐๐๐ 2๐พ + ๐ ๐๐2๐พ) โ 2๐๐๐๐๐ ๐พ + ๐2
c = ๐2(1) โ 2๐๐๐๐๐ ๐พ + ๐2 Squaring both sides
c2 = ๐2 โ 2๐๐๐๐๐ ๐พ + ๐2
And hence, c2 = a
2 + b
2 โ 2abcos๐พ
Similarly: a
2 = b
2 + c
2 โ 2bccos๐ผ
b2 = a
2 + c
2 โ 2accos๐ฝ
OR Q.5(b)
QUESTION: Prove fundamental law: cos(๐ผ โ ๐ฝ) = cos๐ผ. cos๐ฝ + sin๐ผ. sin๐ฝ PROOF:
Fundamental Law: Consider a unit circle with centre at O(0,0) as shown in figure.
Let P(cos๐ฝ, sin๐ฝ) and Q(cos๐ผ, sin๐ผ) be any two points in unit circle.
We have distance formula as:
d = (๐ฅ2 โ ๐ฅ1)2 + (๐ฆ2 โ ๐ฆ1)2
Here ๐๐ = (๐๐๐ ๐ผ โ ๐๐๐ ๐ฝ)2 + (๐ ๐๐๐ผ โ ๐ ๐๐๐ฝ)2 ------------------(1)
Now rotate the axes so that the positive direction of X-axis passes through the point P. Then with respect to this coordinate system, the coordinates of P and Q become (1,0) and (cos(๐ผ โ ๐ฝ), sin(๐ผ โ ๐ฝ)) respectively.
So, ๐๐ = [ cos ๐ผ โ ๐ฝ โ 1]2 + [ sin ๐ผ โ ๐ฝ โ 0 ]2 ------------------(2)
Comparing (1) and (2), we get
(๐๐๐ ๐ผ โ ๐๐๐ ๐ฝ)2 + (๐ ๐๐๐ผ โ ๐ ๐๐๐ฝ)2 = [ cos ๐ผ โ ๐ฝ โ 1]2 + [ sin ๐ผ โ ๐ฝ โ 0 ]2
or (๐๐๐ ๐ผ โ ๐๐๐ ๐ฝ)2 + (๐ ๐๐๐ผ โ ๐ ๐๐๐ฝ)2 = [ cos ๐ผ โ ๐ฝ โ 1]2 + [ sin ๐ผ โ ๐ฝ โ 0 ]2
or cos2๐ผ โ 2cos๐ผ.cos๐ฝ + cos2๐ฝ + sin2๐ผ โ 2sin๐ผ.sin๐ฝ + sin2๐ฝ = cos2(๐ผ โ ๐ฝ) โ 2cos(๐ผ โ ๐ฝ) + 1 + sin2(๐ผ โ ๐ฝ)
or sin2๐ผ + cos2๐ผ โ 2cos๐ผ.cos๐ฝ โ 2sin๐ผ.sin๐ฝ + sin2๐ฝ + cos2๐ฝ = sin2(๐ผ โ ๐ฝ)+ cos2(๐ผ โ ๐ฝ) + 1 โ 2cos(๐ผ โ ๐ฝ)
or 1 โ 2cos๐ผ.cos๐ฝ โ 2sin๐ผ.sin๐ฝ + 1 = 1 + 1 โ 2cos(๐ผ โ ๐ฝ)
or โ 2cos๐ผ.cos๐ฝ โ 2sin๐ผ.sin๐ฝ = โ 2cos(๐ผ โ ๐ฝ) Dividing by -2, we get or cos๐ผ.cos๐ฝ + sin๐ผ.sin๐ฝ = cos(๐ผ โ ๐ฝ)
Hence, cos(๐ผ โ ๐ฝ) = cos๐ผ.cos๐ฝ + sin๐ผ.sin๐ฝ
Q.6(a) QUESTION:
Show that: 2 = 1 +1
22 +1.3
2!.24 +1.3.5
3!.26 + . . .
SOLUTION:
2 = 1 +1
22 +1.3
2!.24 +1.3.5
3!.26 + . . .
Comparing R.H.S with R.H.S of
1 + ๐ฅ ๐ = 1 + ๐๐ฅ +๐ ๐ โ 1
2!๐ฅ2 + . . .
๐๐ฅ =1
22=
1
4 โ โ โ (1)
Squaring both sides
๐2๐ฅ2 =1
16 โ โ โ (2)
๐ ๐ โ 1
2!๐ฅ2 =
1.3
2!. 24
๐ ๐ โ 1 ๐ฅ2 =3.2!
2! .16
๐ ๐ โ 1 ๐ฅ2 =3
16
รท ๐๐๐ ๐๐ฆ 2 , ๐ค๐ ๐๐๐ก ๐ ๐ โ 1 ๐ฅ2
๐2๐ฅ2=
3
16รท
1
16
๐ โ 1
๐=
3
16ร 16
๐ โ 1
๐= 3 โน ๐ โ 1 = 3๐
โ1 = 3๐ โ ๐ 2๐ = โ1
๐ = โ1
2
1 โน ๐๐ฅ =1
4
โ1
2๐ฅ =
1
4
๐ฅ = โ1
2
Now,
๐ . ๐ป. ๐ = 1 โ1
2
โ12
= 1
2 โ
12
= 2 12
= 2 = ๐ฟ. ๐ป. ๐
Q.6(b) QUESTION: Solve the following system of equations by Cramerโs Rule:
๐ฅ + ๐ฆ + ๐ง = ๐, ๐ฅ + 1 + ๐ ๐ฆ + ๐ง = 2๐, ๐ฅ + ๐ฆ + (1 + ๐)๐ง = 0 SOLUTION:
๐ฅ + ๐ฆ + ๐ง = ๐ ๐ฅ + 1 + ๐ ๐ฆ + ๐ง = 2๐ ๐ฅ + ๐ฆ + 1 + ๐ ๐ง = 0 ๐ โ 0 The determinant of the coefficient is
๐ด = 1 1 11 1 + ๐ 11 1 1 + ๐
Expanding by ๐ 1
= 1 1 + ๐ 1
1 1 + ๐ โ 1
1 11 1 + ๐
+ 1 1 1 + ๐1 1
= 1 + ๐ 2 โ 1 โ 1 + ๐ โ 1 + 1 โ (1 + ๐) = 1 + 2๐ + ๐2 โ 1 โ ๐ โ ๐ ๐ด = ๐2 โ 0, Non-singular
๐ด๐ฅ = ๐ 1 1
2๐ 1 + ๐ 10 1 1 + ๐
Expanding by ๐ถ1
= ๐ 1 + ๐ 1
1 1 + ๐ โ 2๐
1 11 1 + ๐
+ 0
= ๐ 1 + ๐ 2 โ 1 โ 2๐ 1 + ๐ โ 1 + 0 = ๐ 1 + 2๐ + ๐2 โ 1 โ 2๐(๐) = ๐ 2๐ + ๐2 โ 2๐2 = 2๐2 + ๐3 โ 2๐2
๐ด๐ฅ = ๐3
๐ด๐ฆ = 1 ๐ 11 2๐ 11 0 1 + ๐
Expanding by ๐ 3
= 1 ๐ 1
2๐ 1 โ 0 + (1 + ๐)
1 ๐1 2๐
= ๐ โ 2๐ + 1 + ๐ 2๐ โ ๐ = โ๐ + 1 + ๐ ๐ = โ๐ + ๐ + ๐2
๐ด๐ฆ = ๐2
๐ด = 1 1 ๐1 1 + ๐ 2๐1 1 0
Expanding by ๐ 3
= 1 1 ๐
1 + ๐ 2๐ โ 1
1 ๐1 2๐
+ 0
= 2๐ โ ๐(1 + ๐) โ 2๐ โ ๐ = 2๐ โ ๐ โ ๐2 โ ๐
๐ด๐ง = โ๐2
By Cramerโs rule
๐ฅ =|๐ด๐ฅ |
๐ด =
๐3
๐2= ๐
๐ฆ ==|๐ด๐ฆ |
|๐ด|=
๐2
๐2= 1
๐ง ==|๐ด๐ง |
|๐ด|=
โ๐2
๐2= โ1
The solution is: ๐ฅ = ๐, ๐ฆ = 1, ๐ง = โ1
Q.7(a) QUESTION: Find the remaining trigonometric functions using radian function if ๐ ๐๐๐ = 0.6 and ๐ก๐๐๐ is negative. SOLUTION: As, ๐ ๐๐๐ is positive and ๐ก๐๐๐ is negative so ๐(๐) is in second quadrant
In a unit circle
๐ฅ2 + ๐ฆ2 = 1 ----(1)
Where ๐ฅ = ๐๐๐ ๐ and ๐ฆ = ๐ ๐๐๐
Given, sin๐ = ๐ฆ = 0.6
๐ฆ = 0.6 ---(2)
(1) โ ๐ฅ2 + 0.6 2 = 1
๐ฅ2 + 0.36 = 1
๐ฅ2 = 1 โ 0.36
๐ฅ2 = 0.64
๐ฅ = ยฑ0.8
As ๐(๐) is in second quadrant
๐ฅ = โ0.8
๐ ๐๐๐ = ๐ฆ = 0.6
๐๐๐ ๐๐๐ =1
๐ฆ=
1
0.6=
๐๐๐ ๐ = ๐ฅ = โ0.8
๐ ๐๐๐ =1
๐ฅ=
โ1
0.8=
๐ก๐๐๐ =๐ฆ
๐ฅ=
0.6
โ0.8=
๐๐๐ก๐ =๐ฅ
๐ฆ=
Q.7(a) QUESTION: Prove any two of the following:
(i) 1โ๐ ๐๐๐
1โ๐ ๐๐๐= ๐ ๐๐๐ โ ๐ก๐๐๐
(ii) ๐ ๐๐2๐
๐ ๐๐๐โ
๐๐๐ 2๐
๐๐๐ ๐= ๐ ๐๐๐
(iii) ๐ ๐๐7๐โ๐ ๐๐5๐
๐๐๐ 7๐+๐๐๐ 5๐= ๐ก๐๐๐ OR ๐ก๐๐
๐
2 =
๐ ๐๐๐
1+๐๐๐ ๐
SOLUTION:
(i) 1โ๐ ๐๐๐
1+๐ ๐๐๐= ๐ ๐๐๐ โ ๐ก๐๐๐
๐ฟ. ๐ป. ๐ = 1โ๐ ๐๐๐
1+๐ ๐๐๐
ร ๐๐๐ ๐๐๐ รท ๐๐๐ ๐๐ฆ 1 โ ๐ ๐๐๐ , ๐ค๐ ๐๐๐ก
= 1โ๐ ๐๐๐
1+๐ ๐๐๐ร
1โ๐ ๐๐๐
1โ๐ ๐๐๐
= 1โ๐ ๐๐๐ 2
1 2โ ๐ ๐๐๐ 2
= 1โ๐ ๐๐๐ 2
1โsin 2 ๐
= 1โ๐ ๐๐๐ 2
cos 2 ๐
=1โ๐ ๐๐๐
๐๐๐ ๐=
1
๐๐๐ ๐โ
๐ ๐๐๐
๐๐๐ ๐= ๐ ๐๐๐ โ ๐ก๐๐๐ = ๐ . ๐ป. ๐
(ii) ๐ ๐๐2
๐ ๐๐โ
๐๐๐ 2
๐๐๐ = ๐ ๐๐
๐ฟ. ๐ป. ๐ = ๐ ๐๐2๐
๐ ๐๐๐โ
๐๐๐ 2๐
๐๐๐ ๐
=๐ ๐๐2๐๐๐๐ ๐ โ๐๐๐ 2๐๐ ๐๐๐
๐ ๐๐๐๐๐๐ ๐
=๐ ๐๐(2๐ โ ๐)
๐ ๐๐๐๐๐๐ ๐
=๐ ๐๐๐
๐ ๐๐๐๐๐๐ ๐
=1
๐๐๐ ๐
= ๐ ๐๐๐ = ๐ . ๐ป. ๐
(iii) ๐ ๐๐7ฮธ โsin 5ฮธ
๐๐๐ 7ฮธ +cos 5ฮธ= ๐ก๐๐ฮธ
๐ฟ. ๐ป. ๐ =๐ ๐๐7๐ โ ๐ ๐๐5๐
๐๐๐ 7๐ + ๐๐๐ 5๐
= 2 cos
7๐ + 5๐2 sin
7๐ โ 5๐2
2 cos7๐ + 5๐
2cos
7๐ โ 5๐2
โต ๐ ๐๐๐ โ ๐ ๐๐๐ = 2 cos
๐ + ๐
2๐ ๐๐
๐ โ ๐
2
๐๐๐ ๐ + ๐๐๐ ๐ = 2 cos๐ + ๐
2๐๐๐
๐ โ ๐
2
= cos
12๐2 sin
2๐2
cos12๐
2 cos2๐2
= cos 6๐๐ ๐๐๐
cos 6๐๐๐๐ ๐
=๐ ๐๐๐
๐๐๐ ๐
= ๐ก๐๐๐ = ๐ . ๐ป. ๐ OR
(iii) ๐ก๐๐๐
2 =
๐ ๐๐๐
1+๐๐๐ ๐
๐ . ๐ป. ๐ =sin ๐
1 + ๐๐๐ ๐
=2๐ ๐๐
๐
2๐๐๐
๐
2
1+ ๐๐๐ 2๐
2 โ ๐๐๐ 2๐
2
โต sin ๐ = 2๐ ๐๐๐
2๐๐๐
๐
2 & ๐๐๐ ๐ = ๐๐๐ 2 ๐
2 โ ๐๐๐ 2 ๐
2
=2๐ ๐๐
๐
2๐๐๐
๐
2
1โ๐๐๐ 2๐
2 + ๐ ๐๐2๐
2
=2๐ ๐๐
๐
2๐๐๐
๐
2
๐ ๐๐2๐
2 + ๐ ๐๐2๐
2
=2๐ ๐๐
๐
2๐๐๐
๐
2
2๐ ๐๐2๐
2
=๐๐๐
๐
2
๐ ๐๐๐
2
= ๐ก๐๐๐
2= ๐ฟ. ๐ป. ๐
Q.7(C) QUESTION: Solve: ๐ฅ2 + ๐ฆ2 = 34, ๐ฅ๐ฆ + 15 = 0 SOLUTION:
๐ฅ2 + ๐ฆ2 = 34
๐ฅ๐ฆ + 15 = 0
๐ฅ๐ฆ = โ15
๐ฆ = โ15
๐ฅ ----(1)
2 โน ๐ฅ2 + โ15
๐ฅ
2
= 34 ๐๐ ๐๐๐ (1)
๐ฅ2 +225
๐ฅ2 = 34
ร ๐๐๐ ๐๐ฆ ๐ฅ2 , ๐ค๐ ๐๐๐ก
๐ฅ4 + 225 = 34๐ฅ2
๐ฅ4 โ 34๐ฅ2 + 225 = 0
๐ฅ4 โ 25๐ฅ2 โ 9๐ฅ2 + 225 = 0
๐ฅ2 ๐ฅ2 โ 25 โ 9 ๐ฅ2 โ 25 = 0
๐ฅ2 โ 25 ๐ฅ2 โ 9 = 0
Either: ๐ฅ2 โ 25 = 0
๐ฅ2 = 25 โน ๐ฅ = ยฑ5
Either: ๐ฅ2 โ 9 = 0
๐ฅ2 = 9 โน ๐ฅ = ยฑ3
(1) โน ๐ฆ = โ15
๐ฅ
๐ฅ = 5
๐ฆ = โ15
5
๐ฆ = โ3, ๐ฅ = 5
๐ฅ = โ5
๐ฆ = โ15
โ5
๐ฆ = 3, ๐ฅ = โ5
๐ฅ = 3
๐ฆ = โ15
3
๐ฆ = โ5, ๐ฅ = 3
๐ฅ = โ3
๐ฆ = โ15
โ3
๐ฆ = 5, ๐ฅ = โ3
๐. ๐ = 5, โ3 , โ5,3 , 3, โ5 , (โ3,5)