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Functions — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
The domain of the function y=2x−12x3+3x2+x+sin⁡−1(log⁡2x)y=\sqrt\dfrac{2x-1}{2x^{3}+3x^{2}+x}+\sqrt{\sin^{-1}(\log_{2}x)} is:
A(12,∞)\left(\dfrac{1}{2},\infty\right)
B(12,2]\left(\dfrac{1}{2},2\right]
C[1,2][1,2]correct
D(1,∞)(1,\infty)
Solution
Solution: Step 1: Condition (i): x>0x>0 (for log⁡2x\log_{2}x to be defined). Condition (ii): sin⁡−1(log⁡2x)\sin^{-1}(\log_{2}x) requires −1≤log⁡2x≤1  ⇒  2−1≤x≤21-1\le\log_{2}x\le 1\;\Rightarrow\;2^{-1}\le x\le 2^{1}, so x∈[12,2]…(2)x\in\left[\dfrac{1}{2},2\right]\quad\ldots(2) Condition (iii): sin⁡−1(log⁡2x)≥0\sin^{-1}(\log_{2}x)\ge 0 (needed for outer square root): log⁡2x≥sin⁡0=0  ⇒  x≥1…(3)\log_{2}x\ge\sin 0=0\;\Rightarrow\;x\ge 1\quad\ldots(3) Condition (iv): 2x−12x3+3x2+x≥0\dfrac{2x-1}{2x^{3}+3x^{2}+x}\ge 0. Factor denominator: x(2x+1)(x+1)x(2x+1)(x+1). For x>0x>0, denominator >0>0. So need numerator ≥0\ge 0: 2x−1≥0  ⇒  x≥12…(4)2x-1\ge 0\;\Rightarrow\;x\ge\dfrac{1}{2}\quad\ldots(4) Also denominator ≠0\neq 0: x≠0,−12,−1x\neq 0,-\dfrac{1}{2},-1 (all satisfied for x≥12>0x\ge\dfrac{1}{2}>0). Step 2: Combining (1), (2), (3), (4): x∈[1,2]x\in[1,2]. Answer: (C) [1,2][1,2]
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