JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Quadratic Equations and Inequalities

First, solve for set $S$: The inequality $\frac{|x+3|-1}{|x|-2}\ge 0$ requires analyzing sign changes. The numerator $|x+3|-1=0$ when $x=-4$ or $x=-2$, and the denominator $|x|-2=0$ when $x=\pm 2$ (excluded from domain). Testing intervals in $[-6,3]-\{-2,2\}$ gives the solution set.

For set $T$: $x^2-7|x|+9\le 0$. Let $u=|x|$, so $u^2-7u+9\le 0$. Using the quadratic formula: $u=\frac{7\pm \sqrt{49-36}}{2}=\frac{7\pm \sqrt{13}}{2}$. Since $\sqrt{13}\approx 3.6$, we get $u\in [1.7,5.3]$ approximately. For integer $x$, $|x|\in \{2,3,4,5\}$, giving $T=\{-5,-4,-3,-2,2,3,4,5\}$.

Intersecting with $S$ (after detailed analysis): $S\cap T=\{-5,-4,3\}$, which has 3 elements.