JEE Main 2022 (Online) 28th July Evening ShiftQuestion 1
Let $S=\left\{x\in[-6,3]-\{-2,2\}:\frac{|x+3|-1}{|x|-2}\ge0\right\}$ and $T=\{x\in\mathbb{Z}:x^2-7|x|+9\le0\}$. Then the number of elements in $S\cap T$ is :
Sets, Relations and Function
Solution
$S$: need numerator and denominator of the same sign. Integers of $T$ are tested directly below. $T$: $x^2-7|x|+9\le0\Rightarrow |x|\in\left[\frac{7-\sqrt{13}}2,\frac{7+\sqrt{13}}2\right]\approx[1.70,5.30]$, so integers $\pm2,\pm3,\pm4,\pm5$. Check in $S$: $x=-5$: $\frac{1}{3}\ge0$ yes; $x=-4$: $\frac{0}{2}=0$ yes; $x=-3$: $\frac{-1}{1}<0$ no; $x=\pm2$ excluded; $x=2$ excluded; $x=3$: $\frac{5}{1}\ge0$ yes. So $S\cap T=\{-5,-4,3\}$, 3 elements. Option (D).