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← JEE Main 2022 · 27 Jul · Shift 2
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JEE Main 2022 · 27 Jul · Shift 2, Mathematics question 1

The domain of the function f(x)=sin⁡−1[2x2−3]+log⁡2(log⁡12(x2−5x+5))f(x)=\sin^{-1}\left[2x^2-3\right]+\log_2\left(\log_{\frac12}\left(x^2-5x+5\right)\right), where [t][t] is the greatest integer function, is :

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Answer and worked solution

Official answer: C

First term. −1≤[2x2−3]≤1-1\le\left[2x^2-3\right]\le1 needs 1≤x2<521\le x^2<\frac52.

Second term. log⁡12(x2−5x+5)>0\log_{\frac12}\left(x^2-5x+5\right)>0 needs 0<x2−5x+5<10<x^2-5x+5<1. That gives 1<x<5−521<x<\frac{5-\sqrt5}2 or 5+52<x<4\frac{5+\sqrt5}2<x<4.

Together only (1,5−52)\left(1,\frac{5-\sqrt5}2\right) survives, since 5−52≈1.38<2.5\frac{5-\sqrt5}2\approx1.38<\sqrt{2.5}.

(C)

Topics:Domain and rangeInverse trigonometric functionsLogarithms
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