Question

Number of solution(s) of the equation $x^2-5x\text{sgn}(x^2-9)+6=0$ is
(Here, $\text{sgn}$ denotes the signum function)

• A. $0$
• B. $2$
• C. $3$
• D. $1$

Answer 1

The correct option is D $1$

We know that
$\text{sgn}x=\{\begin{array}{c}1x>0\\ 0x=0\\ -1x<0\end{array}$

$\text{(i)}{x}^2-9>0$
i.e., $x\in (-\infty ,-3)\cup (3,\infty )$
Then $x^2-5x+6=0$
$\Rightarrow (x-2)(x-3)=0\Rightarrow x=2,3$
No solution as $x\in (-\infty ,-3)\cup (3,\infty )$

$\text{(ii)}{x}^2-9=0$
i.e., $x=-3,3$
Then $x^2+6=0$
No real roots.

$\text{(iii)}{x}^2-9<0$
i.e., $x\in (-3,3)$
Then $x^2+5x+6=0$
$\Rightarrow (x+2)(x+3)=0\Rightarrow x=-2[\because x\in (-3,3\left)\right]$

$\therefore x=-2$ is the only solution.
Hence, only one solution is possible.