Let S=x[-6,3]--2,2: |x+3|-1|x|-20 and T=x Z: x^2-7|x|+90.

Question

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

Accepted Answer

For $T$: $|x|^2-7|x|+9\le0 \implies |x|\in\left[\dfrac{7-\sqrt{13}}2,\dfrac{7+\sqrt{13}}2\right]\approx[1.7,5.3]$. Integer values of $|x|$: $2,3,4,5$, so $x\in\{\pm2,\pm3,\pm4,\pm5\}$.

Checking each against the inequality defining $S$ (piecewise analysis of $\dfrac{|x+3|-1}{|x|-2}\ge0$ over $[-6,3]$), the values that also lie in $S$ are $x=3,-4,-5$.

$$n(S\cap T)=3.$$

Solutions · 1