JEE MATHEMATICS MCQ: FUNCTIONS

1 Easy

1 of 1

Application +4 | -1

The domain of the function

f(x)=\sin ^{-1}\left(\log _2\left(\frac{1}{2} x^2\right)\right)

is :

View answer and explanation

Correct option: C

[-2,-1] \cup[1,2]

The correct answer is C ([-2,-1]

\cup

[1,2]) because the argument of the arcsine function must be between -1 and 1, leading to the inequality

1 \le x^2 \le 4

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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\}