Page 31 - ISC-12

8Jai Baba Ki ISC Mathematics – Class XII by Gupta–Bansal
f(x) π
(C) lim = . (D) There is an x ∈ R such that (g ◦ f)(x) = 1.
x→0 g(x) 6

[JEE 2015]
−→ −→ −→ √
~
~
~
33. Let PQR be a triangle. Let ~a = QR, b = RP and ~c = PQ. If |~a| = 12, |b| = 4 3 and b · ~c = 24,
then which of the following is (are) true?
|~c| 2 |~c| 2
(A) − |~a| = 12. (B) + |~a| = 30.
2 2
√
~
~
(C) |~a × b + ~c × ~a| = 48 3. (D) ~a · b = −72.
[JEE 2015]

6 4
34. If α = 3 sin −1 and β = 3 cos −1 , where the inverse trigonometric functions take only the
11 9
principal values, then the correct option(s) is(are) .

(A) cos β > 0. (B) sin β < 0. (C) cos (α + β) > 0. (D) cos α < 0.

[JEE 2015]

35. The option(s) with the values of a and L that satisfy the following equation is(are)
4π
Z
4
6
t
e (sin at + cos at) dt
0
π
Z = L?
t
6
4
e (sin at + cos at) dt
0
e 4π − 1 e 4π + 1
(A) a = 2, L = . (B) a = 2, L = .
π
e − 1 e + 1
π
e 4π − 1 e 4π + 1
(C) a = 4, L = . (D) a = 4, L = .
π
π
e − 1 e + 1
[JEE 2015]
36. Let f, g : [−1, 2] → R be continuous functions which are twice differentiable on the interval (−1, 2).
Let the values of f and g at the points −1, 0 and 2 be as given in the following table:
x = −1 x = 0 x = 2
f(x) 3 6 0
g(x) 0 1 −1

00
In each of the intervals (−1, 0) and (0, 2) the function (f − 3g) never vanishes. Then the correct
statement(s) is(are)

0
0
(A) f (x) − 3g (x) = 0 has exactly three solutions in (−1, 0) ∪ (0, 2).
0
0
(B) f (x) − 3g (x) = 0 has exactly one solution in (−1, 0).
0
0
(C) f (x) − 3g (x) = 0 has exactly one solution in (0, 2).
0
0
(D) f (x) − 3g (x) = 0 has exactly two solutions in (−1, 0) and exactly two solutions in (0, 2).
[JEE 2015]