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

Correct Answer is : odd and is strictly increasing in (−∞,∞)(-∞, ∞)(−∞,∞).

even and is strictly increasing in (0, ∞).

odd and is strictly decreasing in (-∞, ∞).

odd and is strictly increasing in (-∞, ∞).

neither even nor odd, but is strictly increasing in (-∞, ∞).

Correct Answer is : odd and is strictly increasing in (−∞,∞)(-∞, ∞)(−∞,∞).

Test Index

MHTCET 16 May 2024 Shift 1 Paper

Question : 137 of 150

Marks: +1, -0

Let the function

g:(-\infty, \infty) \rightarrow\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

g(u)=2 \tan ^{-1}\left(e^{u}\right)-\frac{\pi}{2}

g

Solution:

Concept:

Check whether

g

is odd/even using

g(-u)

, and test monotonicity using the derivative.

Explanation:

Given

g(u)=2\tan^{-1}(e^u)-\frac{\pi}{2}

For

x>0

, we have

\tan^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{2}-\tan^{-1}x

g(-u)=2\tan^{-1}(e^{-u})-\frac{\pi}{2}

This equals

2\left(\frac{\pi}{2}-\tan^{-1}(e^u)\right)-\frac{\pi}{2}

Simplifying gives

\frac{\pi}{2}-2\tan^{-1}(e^u)=-g(u)

Hence

g

is an odd function.

Now differentiate:

g'(u)=\frac{2e^u}{1+e^{2u}}

Since

e^u>0

and

1+e^{2u}>0

for all real

u

, we get

g'(u)>0

for all

u\in(-\infty,\infty)

Therefore

g

is strictly increasing on

(-\infty,\infty)

Answer:

g

is odd and strictly increasing in

(-\infty,\infty)

, so the correct option is C.

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