JEE MATHEMATICS MCQ: FUNCTIONS

1 Easy

1 of 1

Understanding +4 | -1

Let

f:(-1,1) \rightarrow B

, be a function defined by

f(x)=\tan ^{-1} \frac{2 x}{1-x^2}

, then

f

is both one- one and onto when

B

is the interval

View answer and explanation

Correct option: B

\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

Option B

\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

is correct because the range of

f(x) = \tan^{-1} \frac{{2x}}{{1-x^2}}

for

x \in (-1, 1)

(-\frac{\pi}{2}, \frac{\pi}{2})

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