IIT-JEE 2008 · Paper 2 · Mathematics Q2

MathematicsSingle correct+3 / -1Question 2 of 66

Let the function $g:(-\infty,\infty)\to\left(-\frac\pi2,\frac\pi2\right)$ be given by $g(u)=2\tan^{-1}(e^u)-\frac\pi2$ . Then, $g$ is

even and is strictly increasing in

(0,\infty)

odd and is strictly decreasing in

(-\infty,\infty)

odd and is strictly increasing in

(-\infty,\infty)

neither even nor odd, but is strictly increasing in

(-\infty,\infty)

Answer and worked solution

Official answer: C

Odd. $\tan^{-1}e^{-u}=\frac\pi2-\tan^{-1}e^u$ , so
$g(-u)=2\left(\frac\pi2-\tan^{-1}e^u\right)-\frac\pi2=-g(u).$

Increasing. $g'(u)=\dfrac{2e^u}{1+e^{2u}}>0$ for all $u$ .

(C)

IIT-JEE 2008 · Paper 2, Mathematics question 2