JEE Advanced · Mathematics · 25. AOD
Let the function \(g:(-\infty, \infty) \rightarrow\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) be given by \(g(u)=2 \tan ^{-1}\left(e^u\right)-\frac{\pi}{2}\). Then, \(g\) is
- A
even and is strictly increasing in \((0, \infty)\)
- B
odd and is strictly decreasing in \((-\infty, \infty)\)
- C
odd and is strictly increasing in \((-\infty, \infty)\)
- D
Neither even nor odd, but is strictly increasing in \((-\infty, \infty)\)
Answer & Solution
Correct Answer
(C)
odd and is strictly increasing in \((-\infty, \infty)\)
Step-by-step Solution
Detailed explanation
\(\therefore \quad g(u)=2 \tan ^{-1}\left(e^u\right)-\frac{\pi}{2}\)
for \(\quad u \in(-\infty, \infty)\)
and
\[
\begin{aligned}
g(-u) & =2 \tan ^{-1}\left(e^{-u}\right)-\frac{\pi}{2} \\
& =2\left(\cot ^{-1}\left(e^u\right)\right)-\frac{\pi}{2} \\
& =2\left(\frac{\pi}{2}-\tan ^{-1}\left(e^u\right)\right)-\frac{\pi}{2} \\
& =\frac{\pi}{2}-2 \tan ^{-1}\left(e^u\right)=-g(u)
\end{aligned}
\]
\(\therefore \quad g(-u)=-g(u)\)
\(\Rightarrow g(u)\) is an odd function.
for \(\quad u \in(-\infty, \infty)\)
and
\[
\begin{aligned}
g(-u) & =2 \tan ^{-1}\left(e^{-u}\right)-\frac{\pi}{2} \\
& =2\left(\cot ^{-1}\left(e^u\right)\right)-\frac{\pi}{2} \\
& =2\left(\frac{\pi}{2}-\tan ^{-1}\left(e^u\right)\right)-\frac{\pi}{2} \\
& =\frac{\pi}{2}-2 \tan ^{-1}\left(e^u\right)=-g(u)
\end{aligned}
\]
\(\therefore \quad g(-u)=-g(u)\)
\(\Rightarrow g(u)\) is an odd function.
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