IIT JEE Advanced 2009 Paper 2

(A) We have f ′(x) > 0, ∀x ∈(0,π /2). Therefore,
f(0) < 0 and f(π/2) > 0
Hence, there is so one solution.
(B) Let us consider that (a, b, c) is direction ratio of the intersected line. Therefore,
ak + 4b + c = 0
4a + kb + 2c = 0

a

8−k

=

b

4−2k

=

c

k2−16

We need to have
2(8 − k) + 2(4 − 2k) + (k2 − 16) = 0
⇒ k = 2, 4.
(C) Let us consider
f (x) = x + 2 + x +1 + x −1 + x − 2
Therefore, k can take values: 2, 3, 4, 5.

(D) ∫

dy

y+1

= ∫ dx
⇒ f (x) = 2ex−1
⇒ f (ln2) = 3