The domain of the function f(x)=sin^-1[2x^2-3]+log_2 (log_frac12(x^2-5x+5) ) , where [t] is the greatest integer function, is :

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JEE Main 2022 (Online) 27th July Evening ShiftQuestion 1

The domain of the function $f(x)=\sin^{-1}[2x^2-3]+\log_2\left(\log_{\frac12}(x^2-5x+5)\right)$, where $[t]$ is the greatest integer function, is :

Sets, Relations and Function

$\left(-\sqrt{\frac52},\frac{5-\sqrt5}{2}\right)$$\left(\frac{5-\sqrt5}{2},\frac{5+\sqrt5}{2}\right)$$\left(1,\frac{5-\sqrt5}{2}\right)$✓ Correct answer$\left[1,\frac{5+\sqrt5}{2}\right)$

Solution

$\sin^{-1}[2x^2-3]$ requires $[2x^2-3]\in\{-1,0,1\}$, i.e. $-1\le2x^2-3<2\Rightarrow1\le x^2<\tfrac52$. $\log_2(\log_{1/2}u)$ with $u=x^2-5x+5$ requires $\log_{1/2}u>0\iff0<u<1$. $u>0\Rightarrow x<\tfrac{5-\sqrt5}{2}$ or $x>\tfrac{5+\sqrt5}{2}$; $u<1\Rightarrow1<x<4$. So $x\in(1,\tfrac{5-\sqrt5}2)\cup(\tfrac{5+\sqrt5}2,4)$. With $1\le x^2<2.5$ ($|x|<1.58$) only $(1,\tfrac{5-\sqrt5}{2})$ survives. Option C.

Sets, Relations and Function - JEE Mathematics question (JEE Main 2022 (Online) 27th July Evening Shift)

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