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JEE Advanced 2008Paper 2
Exam
JEE Advanced
Year
2008
Paper
Paper 2
Subject
Mathematics
Chapter
Application of Derivatives

JEE Advanced 2008 · Paper 2 · Mathematics · Application of Derivatives

Let the function g:(-∞, ∞) arrow(-(π)/(2), (π)/(2)) be given by...

Let the function g:(−∞,∞)→(−π2,π2)g:(-\infty, \infty) \rightarrow\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) be given by g(u)=2tan⁡−1(eu)−π2g(u)=2 \tan ^{-1}\left(e^u\right)-\frac{\pi}{2}. Then, gg is

Options

  1. Aeven and is strictly increasing in (0,∞)(0, \infty)
  2. Bodd and is strictly decreasing in (−∞,∞)(-\infty, \infty)
  3. Codd and is strictly increasing in (−∞,∞)(-\infty, \infty)
  4. DNeither even nor odd, but is strictly increasing in (−∞,∞)(-\infty, \infty)

Correct answer

C. odd and is strictly increasing in (−∞,∞)(-\infty, \infty)

Step-by-step solution

∴g(u)=2tan⁡−1\therefore \quad g(u)=2 \tan ^{-1}\left(e^u\right)-\frac{\pi}{2}forfor\quad u \in(-\infty, \infty)andandg(−u)=2tan⁡−1(e−u)−π2=2(cot⁡−1(eu))−π2=2(π2−tan⁡−1(eu))−π2=π2−2tan⁡−1(eu)=−g(u)\begin{aligned}g(-u) & =2 \tan ^{-1}\left(e^{-u}\right)-\frac{\pi}{2}\\ & =2 \left(\cot ^{-1}\left(e^u\right)\right)-\frac{\pi}{2}\\ & =2 \left(\frac{\pi}{2}-\tan ^{-1}\left(e^u\right)\right)-\frac{\pi}{2}\\ & =\frac{\pi}{2}-2 \tan ^{-1}\left(e^u\right)=-g(u) \end{aligned}\therefore \quad g(-u)=-g(u)\Rightarrow g(u)$ is an odd function.
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