Question

The number of real roots of the equation $x^2-12|x|+20=0$ is $P.$ Then the values of $a$ for which the equation $||x-2|+a|=P$ can have four distinct solutions, is

• A. $a\le -4$
• B. $a<-4$
• C. $a\le 2$
• D. $a<-2$

Answer 1

The correct option is B $a<-4$

$x^2-12|x|+20=0\Rightarrow |x{|}^2-12|x|+20=0\Rightarrow (|x|-10)(|x|-2)=0\Rightarrow |x|=10,|x|=2\Rightarrow x=\pm 10,x=\pm 2\Rightarrow P=4$

Now, $||x-2|+a|=4$

Case $\left(i\right)$ $|x-2|+a=4$
$\Rightarrow |x-2|=4-a$
It has $2$ distinct solutions, if
$4-a>0\Rightarrow a<4\dots \left(1\right)$

Case $\left(ii\right)$ $|x-2|+a=-4$
$\Rightarrow |x-2|=-4-a$
It has $2$ distinct solutions, if
$-4-a>0\Rightarrow a<-4\dots \left(2\right)$

So, equation $||x-2|+a|=4$ will have four distinct solutions, when $\left(1\right)$ and $\left(2\right)$ both satisfy simultaneously.
Hence, final range of $a$ is $\left(1\right)\cap \left(2\right)$
$\Rightarrow a<-4$