Let the function g:(-∞, ∞) arrow (-(π)/(2), (π)/(2)) be given by g(u)=2 ^ -1 (e^u)-(π)/(2) .Then, g is

Correct answer: C. odd and is strictly increasing in (-∞, ∞) Given that g(u)=2 ^ -1 (e^u)-(π)/(2) g(-u)=2 ^ -1 (e^ -u )-(π)/(2)=2 ^ -1 ((1)/(e^u))-(π)/(2)=2 ^ -1 (e^u)-(π)/(2)=2[(π)/(2)- ^ -1 (e^u)]-(π)/(2)=π-2 ^ -1 (e^u)-(π)/(2)=(π)/(2)-2 ^ -1 (e^u)=-g(u) g is an odd function. Also g^ (u)= 2 e^u 1+e^ 2 u >0, u ∈(-∞, ∞) g is strictly increasing on (-∞, ∞) .

BITSAT 2024Paper 1

Exam: BITSAT
Year: 2024
Paper: Paper 1
Subject: Mathematics
Chapter: Functions

BITSAT 2024 · Paper 1 · Mathematics · Functions

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

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

Options

Aeven and is strictly increasing in $(0, \infty)$
Bodd and is strictly decreasing in $(-\infty, \infty)$
Codd and is strictly increasing in $(-\infty, \infty)$
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

Given that

g(u)=2 \tan ^{-1}\left(e^u\right)-\frac{\pi}{2}\therefore \quad g(-u)=2 \tan ^{-1}\left(e^{-u}\right)-\frac{\pi}{2}=2 \tan ^{-1}\left(\frac{1}{e^u}\right)-\frac{\pi}{2}=2 \cot ^{-1}\left(e^u\right)-\frac{\pi}{2}=2\left[\frac{\pi}{2}-\tan ^{-1}\left(e^u\right)\right]-\frac{\pi}{2}=\pi-2 \tan ^{-1}\left(e^u\right)-\frac{\pi}{2}=\frac{\pi}{2}-2 \tan ^{-1}\left(e^u\right)=-g(u)\therefore \quad g

is an odd function. Also

g^{\prime}(u)=\frac{2 e^u}{1+e^{2 u}}>0, \forall u \in(-\infty, \infty)\therefore g

is strictly increasing on

(-\infty, \infty)

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