JEE Main 2022 · 29 Jun · Shift 1 · Mathematics Q15

← JEE Main 2022 · 29 Jun · Shift 1

MathematicsSingle correct+4 / -1Question 15 of 90

The domain of the function $\cos^{-1}\left(\dfrac{2\sin^{-1}\left(\frac1{4x^2-1}\right)}\pi\right)$ is :

\mathbf R-\left\{-\dfrac12,\dfrac12\right\}

(-\infty,-1]\cup[1,\infty)\cup\{0\}

\left(-\infty,\dfrac{-1}2\right)\cup\left(\dfrac12,\infty\right)\cup\{0\}

\left(-\infty,\dfrac{-1}{\sqrt2}\right]\cup\left[\dfrac1{\sqrt2},\infty\right)\cup\{0\}

Answer and worked solution

Official answer: D

Conditions. The outer $\cos^{-1}$ is fine whenever the inner $\sin^{-1}$ exists. Need $\left|\frac1{4x^2-1}\right|\le1$ , i.e. $|4x^2-1|\ge1$ : $x^2\ge\frac12$ or $x=0$ .

(D)

Topics:Inverse trigonometric functions Domain and range

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JEE Main 2022 · 29 Jun · Shift 1, Mathematics question 15