Differentiation and its Application: Let g(x)=int_0^e^x(f'(t))/(1+t^2) dt. Which of the following is true? | JEE Advanced 2008 Paper 2

Comprehension · Questions 1–1

Read the passage given below and answer the questions that follow.

Consider the function $f:(-\infty,\infty)\to(-\infty,\infty)$ defined by $$f(x)=\frac{x^2-ax+1}{x^2+ax+1},\quad 0<a<2.$$

JEE Advanced 2008 Paper 2Question 1

Let $$g(x)=\int_0^{e^x}\frac{f'(t)}{1+t^2}\,dt.$$ Which of the following is true?

Differentiation and its Application

$g'(x)$ is positive on $(-\infty,0)$ and negative on $(0,\infty)$$g'(x)$ is negative on $(-\infty,0)$ and positive on $(0,\infty)$✓ Correct answer$g'(x)$ changes sign on both $(-\infty,0)$ and $(0,\infty)$$g'(x)$ does not change sign on $(-\infty,\infty)$

Solution

Concept. Here $f(x)=\dfrac{x^2-ax+1}{x^2+ax+1}$ with $0<a<2$, and $f'(x)=\dfrac{2a(x^2-1)}{(x^2+ax+1)^2}$. Differentiate $g$ with the chain rule; the sign of $g'$ is the sign of $f'(e^x)$. Formulas used. (1) $\dfrac{d}{dx}\displaystyle\int_0^{u(x)}h(t)\,dt=h(u(x))\,u'(x)$Step 1 - $g'$. By (1), $$g'(x)=\frac{f'(e^x)\,e^x}{1+e^{2x}}.$$ Step 2 - Sign. $e^x>0$, so $g'$ has the sign of $f'(e^x)$, i.e. of $e^{2x}-1$: negative for $x<0$, positive for $x>0$. Answer: (B)

Differentiation and its Application - JEE Mathematics question (JEE Advanced 2008 Paper 2)

More from Differentiation and its Application