Question

If $||x-3|-4|=3$, then there exist

• A. exactly $2$ distinct values of $x$ in $[1,5]$
• B. exactly $2$ distinct values of $x$ in $[-5,0]$
• C. exactly $2$ distinct values of $x$ in $[1,10]$
• D. exactly $4$ distinct values of $x$ in $[-5,10]$

Answer 1

Answer: (A), (D)

The correct options are
A exactly $2$ distinct values of $x$ in $[1,5]$
D exactly $4$ distinct values of $x$ in $[-5,10]$

$\text{Case}1:|x-3|-4\ge 0\text{i.e.,}x\in (-\infty ,-1]\cup [7,\infty )$
Then $|x-3|-4=3$
$\Rightarrow |x-3|=7$
$\Rightarrow x=10$ or $x=-4$
$\therefore x=-4,10\cdots \left(1\right)$

$\text{Case}2:|x-3|-4<0\text{i.e.,}x\in (-1,7)$
Then $|x-3|-4=-3$
$\Rightarrow |x-3|=1$
$\Rightarrow x=2,4\cdots \left(2\right)$

Hence, from $\left(1\right)$ and $\left(2\right)$, $x=-4,2,4,10$