JEE MATHEMATICS MCQ: FUNCTIONS

1 Easy

1 of 1

Application +4 | -1

\mathrm{f}:\left[\frac{\pi}{2}, \frac{3 \pi}{2}\right] \rightarrow[-1,1]

is defined by

f(x)=\sin x

, then

f^{-1}(x)

is given by

View answer and explanation

Correct option: C

\pi-\sin ^{-1} \mathrm{x}

The correct answer is C (

\pi-\sin ^{-1} \mathrm{x}

) because the inverse function is found by solving for x in terms of y and considering the restricted domain of the original function.

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f:(-\infty, 2] \longrightarrow(-\infty, 4]

f(x)=x(4-x)

f^{-1}(x)

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f(g(x))=(\sin \sqrt{x})^2

f:\left( {2,\;3} \right) \to \left( {0,\;1} \right)

f\left( x \right) = x - \left[ x \right]

{f^{ - 1}}\left( x \right)

f: R \rightarrow R

f(x)=x^4

f(x)=(x+1)^2-1(x \geq-1)

S=\left\{x: f(x)=f^{-1}(x)\right\}