JEE Main 2022 · 28 Jul · Shift 2 · Mathematics Q61

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MathematicsSingle correct+4 / -1Question 61 of 90

Let $S=\left\{x\in[-6,3]-\{-2,2\}:\dfrac{|x+3|-1}{|x|-2}\ge0\right\}$ and $T=\left\{x\in\mathbf Z:x^2-7|x|+9\le0\right\}$ . Then the number of elements in $S\cap T$ is :

Answer and worked solution

Official answer: D

T. $x^2-7|x|+9\le0$ for $|x|\in[1.70,5.30]$ , so $T=\{\pm2,\pm3,\pm4,\pm5\}$ .

S. With $|x|>2$ the numerator must be $\ge0$ : $[-6,-4]\cup(2,3]$ . With $|x|<2$ it must be $\le0$ , which needs $-4\le x\le-2$ : impossible. So $S=[-6,-4]\cup(2,3]$ .

Common $-5,-4,3$ . (D)

Topics:Inequalities and modulus Sets and operations

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JEE Main 2022 · 28 Jul · Shift 2, Mathematics question 61