JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Definite Integrals

Consider the function

f : (- \infty, \infty) \to (- \infty, \infty)

defined by

f (x) = \frac{x ^{2} - a x + 1}{x ^{2} + a x + 1}, 0 < a < 2.

Let

g (x) = \int_{0}^{e^{x}} \frac{f ^{'} ( t )}{1 + t ^{2}} d t

Which of the following is true?

(A) $g^{'} (x)$ is positive on $(- \infty, 0)$ and negative on $(0, \infty)$
(B) $g^{'} (x)$ is negative on $(- \infty, 0)$ and positive on $(0, \infty)$ ✓ Correct
(C) $g^{'} (x)$ changes sign on both $(- \infty, 0)$ and $(0, \infty)$
(D) $g^{'} (x)$ does not change sign on $(- \infty, \infty)$

Answer: (B)

Solution

By Leibniz rule,

g^{'} (x) = \frac{f ^{'} ( e ^{x} ) \cdot e ^{x}}{1 + e ^{2 x}}

. Since

f^{'} (t) = \frac{2 a ( t ^{2} - 1 )}{( t ^{2} + a t + 1 ) ^{2}}

, we have

f^{'} (e^{x}) < 0

for

x < 0

(since

e^{x} < 1

) and

f^{'} (e^{x}) > 0

for

x > 0

(since

e^{x} > 1

). Thus

g^{'} (x) < 0

(- \infty, 0)

and

g^{'} (x) > 0

(0, \infty)

JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Single correct option