The maximum value of \(\rm y=\tan^{-1}\frac{1-x}{1+x}\) on [0, 1] is:

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AAI ATC Junior Executive 21 Feb 2023 Shift 1 Official Paper

Answer 1

Given:

\(\rm y=\tan^{-1}\frac{1-x}{1+x}\) on [0,1]

Concept:

Use formula \(\rm tan(A-B)=\frac{tanA-tanB}{1+tanA\cdot tanB}\)

Calculation:

Let \(\rm x=tanA\implies A=tan^{-1}x\)

Let \(\rm f(x)=\tan^{-1}\frac{1-x}{1+x}\)

\(\rm \implies f(x)=\tan^{-1}({\frac{1-tanA}{1+tanA}})\)

\(\rm \implies f(x)=\tan^{-1}({tan(\frac{\pi}{4}-A)})\)

\(\rm \implies f(x)=\frac{\pi}{4}-A\)

\(\rm \implies f(x)=\frac{\pi}{4}-tan^{-1}x\)

In [0,1] f(x) is maximum when tan^-1x is minimum then

At x=0

\(\rm \implies f(0)=\frac{\pi}{4}-tan^{-1}(0)\)

\(\rm \implies f(0)=\frac{\pi}{4}\) (Maximum)

Hence the option (1) is correct.

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