Important Question for class 12 Maths Academic team of entrancei prepared selected questions for class 12 Maths download your free copy of class 12 Maths questions from entrancei .do find NCERT solutions for class 12 Maths 1.
The general solution of the trigonometrical equation sin x + cos x = 1, for n = 0, ± 1, ± 2,… is given by 1 p 2 (D) none of these
(A) x = 2np
(B) x = 2np +
(C) x = np + (–1)n (p/4) – p/4 Sol.
2.
Sol.
3.
Sol.
4.
We write the given equation as
( 2)
sin x +
1
( 2)
cos x =
1
( 2)
or cos(p/4) sin x + sin (p/4) cos x = sin (p/4) or sin (x + p/4) = sin (p/4) \ x + p/4 = np + (–1)n (p/4) or x = np+ (–1)n (p./4) – (p/4) when n = 0, ± 1, ± 2,… The number of all possible triplets (a1, a2, a3) such that a1 + a2 cos 2x + a3 sin2 x = 0 for all x is (A) zero (B) one (C) three (D) infinite (D). We write the given relation as a1 + a2 cos 2x + a3 ×
1 (1 - cos 2x ) = 0 2
or æç a1 + 1 a3 ö÷ + æç a2 - 1 a3 ö÷ cos 2x = 0 2 ø è 2 ø è The above relation will hold for all x if 1 1 a1 + a3 = 0 and a2 - a3 = 0 2 2 Choosing a3 = k, kÎR, we get, a1 = –k/2, a2 = k/2. Hence the solution set is (–k/2, k/2, k), where k is any real number. Thus the number of solutions is infinite. The solution set of (2cos x – 1) (3 + 2cos x) = 0 in the interval 0 £ x £ 2p is (A) {p/3} (B) {p/3, 5p/3} (C) {p/3, 5p/3, cos–1 (–3/2) (D) none of these (B). We have 2cos x – 1 = 0 or 3 + 2cos x = 0 But 3 + 2cos x = 0 gives cos x = –3/2 which is not possible. From 2cos x – 1 = 0, we get cos x = 1/2 whence x = p/3, 2p – p/3 i.e. 5p/3 in the interval 0 £ x £ 2p. Thus the solution set in the given interval is {p/3, 5p/3}. The value of q lying between q = 0 and q = p/2 and satisfying the equation 1 + cos2 q cos q cos2 q 2
sin2 q
4 sin 4q
1 + sin q 4 sin 4q = 0 is 2 sin q 1 + 4 sin 4q 2
11p (B) p 24 3 5p p (C) (D) 24 24 1 + cos2 q sin2 q 4 sin 4q æ By R3 ® R3 - R 2 and ö (A). -1 1 0 ç ÷ è R2 ® R2 - R1 ø 0 -1 1
(A)
Sol.
1
Entrancei
Applying C1 ® C1 + C2 , we get 2 sin2 q D= 0 1 -1 -1
4 sin 4q Þ sin 4q = - 1 = sin æç - p ö÷ 0 2 è 6ø 1
pö n æ 4q = np + ( -1) ç - ÷ , n Î I è 6ø 7p 11p pù é q= or for q Î ê0, 24 24 2 úû ë
5.
Sol.
If 1 + sin q + sin2 q +… to ¥ = 4 + 2 3 0 < q < p, q ¹ p/2, then (A) q = p/6 (C) q = p/6 or p/6
(B) q = p/3 (D) q = p/3 or 2p/3
(D). The left hand side of the given equation is an infinite G.P. with common ratio sin q. Since 0 < q < p, q ¹ p/2, we have 0 < sin q < 1. We now sum up the infinite G.P. so that the given equation becomes 1 = 4+2 3 (1 - sin q ) or 1 - sin q =
1
=
(4 + 2 3 )
4-2 3 4-2 3 = 16 - 12 4
or 1 - sin q = 1 - 3 2
\ sin q =
3 p = sin 2 3
\ q = p/3 or 2p/3 in the interval 0 < q < p. 6.
The value of q satisfying the equation cosq + Ö3 sinq = 0, in 0 £ q £ p, is p 6 2p (C) 6
5p 6 4p (D) 6
(A)
Sol.
7.
Sol.
8.
Sol.
(B)
(B). The given equation is 1 cos q + 3 sin q = 0 2
2 5p Þ cos(q – p/3) = 0 = cosp/2 Þ q = 6
The least value of a for which the equation
4 1 + sin x 1 - sin x
(A) 9 (C) 8 (A). The given equation is a sin2 x – (3 + a)sin x + 4 = 0. It has real roots if (3 + a)2 – 16a ³ 0 Þ (a – 9) (a – 1) ³ 0 Þ a £ 1 or a ³ 9. If a = 1 Þ sin x = – 2. Hence a = 9. If tan2q.sec2q < 0, then q lies in the internal (A) (0, p/2) (C) (p, 3p/2) (B). tan 2q sec 2q =
Entrancei
sin 2q <0 cos2 2q
= a has at least one solution for x Î (0, p/2) is
(B) 4 (D) 1
(B) (p/2, p) (D) (3p/2, 2p)
Þ sin 2q < 0 Þ 9.
p <q<p 2
If tan2q tanq = – 3, then q is equal to (A) np +
p 3
(B) np ±
(C) 2np ± p
(D) none of these
3
Sol.
2 (B). tan2q tanq = – 3 Þ 2 tan 2q = –3
1 - tan q
Þ
Entrancei
tan2q
p 3
= 3 Þ tanq = ±
3.