Absolute Value

Let $S=\left\{x\in \left[-6,3\right]-\left\{-2,2\right\}:\frac{\left|x+3\right|-1}{\left|x\right|-2}\ge 0\right\}$ and $T=\left\{x\in Z:x^2-7\left|x\right|+9\le 0\right\}$, then the number of elements in $S\cap T$ is

Solution set of inequality $\left|\frac{2}{x-4}\right|>1$ is

If the solution set of the inequality $\left|a+3x\right|\le 6$ is $\left[\frac{-8}{3},\frac{4}{3}\right]$, then the value of $a$ is equal to

The set of all $x$ satisfying the inequality $\left|3x+4\right|\le 7$ is

Solve inequality: $\frac{\left|x-1\right|-2}{x}\le 0$.

If $\left|a{x}^2+bx+c\right|\le 1$ for all $x$ in $\left[0, 1\right]$, then which of the following is NOT correct

The value of $x$ for which $\left|x+3\right|>\left|2x-1\right|$ is

The number of solution(s) the equation $\left| x – 1\right| + \left|x – 2\right|+\left|x – 3\right|+\left|x – 4\right|= 3$ is 

If $\left| {\sin }^2\mathrm{x}+ 17 – {\mathrm{x}}^2\right|= \left|16 – {\mathrm{x}}^2\right| +\left| 2 {\sin }^2\mathrm{x}\right| + \left|{\cos }^2\mathrm{x}\right|$, then $\mathrm{x}$ lies in
,

$\left|\frac{x}{x+1}\right|<{\mathrm{10}}^{-4}$ holds if: