Question

Number of roots of equation $3\left|x\right|-|2-\left|x\right||=1$ is

• A. 0
• B. 2
• C. 4
• D. none of these

Answer 1

Answer: (A)

0

$3|x|-|2-|x||=1\{\begin{array}{c}3x-(2-|x|)=1x\ge 2\\ -3x(2-|x|)=1x<2\end{array}$

$=\{\begin{array}{c}-3x-2+|x|=1x\ge 2\\ -3x+2-|x|=1x<2\end{array}$

$=\{\begin{array}{c}3x+|x|=3x\ge 2\\ -3x-|x|=-1x<2\end{array}$

$=\{\begin{array}{c}3x+|x|=3x\ge 2\\ 3x+|x|=1x<2\end{array}$

$=\{\begin{array}{c}3x+x=3x\ge 2\\ 3x-x=3x<2\end{array}$

$\{\begin{array}{c}3x+x=1x\ge 2\\ 3x-x=1x<2\end{array}$

$=\{4x=3,2x=3,4x=1,2x=1$

$\therefore x=\frac{3}{4},x=\frac{3}{2},x=\frac{1}{4},x=\frac{1}{2}$

For $x=\frac{3}{4},\frac{3}{4}-(2-\frac{3}{4})=\frac{3}{4}-\frac{5}{4}=\frac{-2}{4}\ne 1$

For $x=\frac{3}{2},\frac{9}{2}-(2-\frac{3}{2})=\frac{9}{2}-\frac{1}{2}=\frac{8}{2}\ne 1$

For $x=\frac{1}{4},\frac{3}{4}-(2-\frac{1}{4})=\frac{3}{4}-\frac{7}{4}=-\frac{4}{4}\ne 1$

For $x=\frac{1}{2},\frac{3}{2}-(2-\frac{1}{2})=\frac{3}{2}-\frac{3}{2}=0\ne 1$

Hence these exist no solution

No. of roots $=0$