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TARGET : JEE(Main+Advanced) 2021

NURTURE COURSE
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PHASE-ALL

RACE # 31 MATH EM ATI CS

Straight Objective Type
1. If sin q and sec q(0 < q < p/2)are the roots of the equation 2x2 + kx + 1 = 0, then the value of k is equal to-

7 2 7 5 7 5 7 5
(A) - (B) (C) •(D) -
5 5 2 5
Ans.(D)
1 1
sin q.sec q = Þ tan q =
2 2
1 5
Þ sin q = ,sec q =
5 2

k æ 1 5ö
- = sin q + sec q Þ k =–2 çç + ÷÷
2 è 5 2 ø

-7 5
=
5
MULTIPLE CORRECT CHOICE TYPE]
2. If 2cos q + sin q = 1 , then the value of 4cos q + 3sin q is equal to
7
•(A) 3 (B) –5 •(C) (D) –4
5
Ans.(A,C)
2cosq =1– sinq Þ 2 1 - sin 2 q = 1 – sinq
3
Þ sin q = - ,1
5
4 cosq + 3sinq = 2(2 cosq + sinq) + sinq
7
= 2 + sinq = ,3
5

17 5
3. If sec A = and cosecB = , then sec(A + B) can have the value equal to -
8 4

85 85 85 85
•(A) •(B) – •(C) – •(D)
36 36 84 84

Ans. (A,B,C,D)

8 15
cos A = and sinA = ±
17 17

Maths / R # 31 1 /1
 4 3
sin B = Þ cos B = ±
4 5

cos(A + B) = cosA cosB – sinAsinB

8 3 15 4 8 3 15 4
= . - . or - . - .
17 5 17 5 17 5 17 5

8 3 15 4 8 3 15 4
or . + . or - . + .
17 5 17 5 17 5 17 5

36 84 84 36
=- or - or or
85 85 85 85
[MATRIX TYPE]
4. Column-I Column-II
(A) The number of value/s of x satisfying the relation (P) 0
1
(tanx + cotx)2 = if x Î [0,2p]
sin x.cos x
tan(4q - f) p
(B) If 17sinf = sin(8q – 3f) and = (Q) –1
tan(4q - 2f) q
where p & q are co-prime numbers, then the value of (p–q) is (R) 1
a
2
sin x + a - 2
2 2

(C) The values of a for which the equation = (S) 2

1 - tan x
2
cos2x
has solution can be
Ans. (A) P; (B) R; (C) Q,R,S
1 1
(A) 2 2
=
sin x cos x sin x.cos x
sinx ¹ 0, cos x ¹ 0
sinx cosx = 1
Þ sin2x = 2
No solution.
sin(8q - 3q) 17 - 1 sin(8q - 3f) - sin f
(B) 17 = Þ =
sin f 17 + 1 sin(q - 3f) + sin f

tan(4q - f) 9
Þ =
tan(4q - 2f) 8
2
(C) a = -1
2

sin 2 x
Þ a2 Î [1,¥) Þ a Î (–¥,–1] È [1,¥)
5. Column-I Column-II
sin 54° cos 54°
(A) - has the value equal to (P) 1
sin18° cos18°
cos3 9° - cos 27° sin 3 9° + sin 27°
(B) + has the value equal to (Q) 2
cos 9° sin 9°
1 1 1
(C) - -
log 2 (1/ 6) log 3 (1/ 6) log 4 (1/ 6) is equal to (R) 3

Maths / R # 31 2 /1
 (D) sin(410° – A) cos(400° + A) + cos(410° – A) sin(400° + A) (S) 4
has the value equal to
2sin 3°.sin1°
(E) is less than
cos 2 1° - cos 2 2°
(F) The value of tan 16° tan42° tan44° tan 76° is equal to
[Ans. (A) Q; (B) R; (C) P; (D) P; (E) R,S ; (F) P]
sin 36°
(A) =2
sin18° cos18°
cos3 9° - 4 cos3 9° + 3cos 9° sin 3 9° + 3sin 9° - 4 sin 3 9°
(B) cos 9°
+
sin 9°
= 3 – 3 cos29° + 3 – 3sin29° = 3
æ 2 ö
(C) log1/ 6 ç ÷ =1
è 3´ 4 ø
(D) sin(50°–A) cos(40°+A) + cos(50°–A) sin(40° +A)
= sin90° = 1
2(cos 2° - cos 4°)
(E) =2
cos 2° - cos 4°
(F) (tan16°. tan44°.tan76°) tan42°
= tan48° cot48° = 1

[INTEGER TYPE]
6. If sin 2a = 4sin 2b , show that 5tan(a - b) = 3tan(a + b) [4]
Ans.
sin 2a
=4
sin 2b
Applying componendo & dividendo,
sin 2a - sin 2b 4 - 1
=
sin 2a + sin 2b 4 + 1
2sin(a - b) cos(a + b) 3
Þ =
2sin(a + b) cos(a - b) 5
Þ 5 tan(a - b) = 3 tan(a + b)
Hence proved.
7. If sinx + cosx + tanx + cotx + secx + cosecx = 7 then sin2x = a - b 7 , where a, b Î N.
Find the value of (a + b) [Ans. 30] [7]
Ans.30
sinx + cosx = t - 2£t£ 2
1 + sin2x = t2 Þ sin2x = t2 – 1
2 2t
t+ 2 + 2 =7
t -1 t -1
t3 – t + 2 + 2t = 7t 2 – 7
Þ t3 – 7t2 + t + 9 = 0
(t + 1) (t2 – 8t + 9) = 0
Þ t = –1, 4 - 7, 4 + 7 (rejected)

Maths / R # 31 3 /1
 sin2x = 0 (rejected Q out of domain)
sin2x = 22 -8 7
a + b = 22 + 8 = 30
p 5
8. If 0 < x < and cosx + sinx = , find the numerical value of cosx – sinx. [5]
4 4
7
[Ans. y = ]
4
7
Ans. y =
4
25
(cos x + sin x) 2 =
16
9
Þ cos x sin x =
32
9 7
(cosx – sinx)2 = 1 – =
16 16

é p ù
cos x - sin x =
7 ê 0<x < 4 ú
4 ê ú
ë\ cos x > sin x û

Maths / R # 31 4 /1
Maths / R # 31 5 /1
Maths / R # 31 6 /1