Question

The number of integral values of $\alpha$ for which the equation $|x-1|-2|x-3|+|x+2|={\alpha }^2-2\alpha +1$ has atleast one solution, is

• A. 1
• B. 2
• C. 3
• D. 5

Answer 1

Answer: (D)

$f(x)=|x-1|-2|x-3|+|x+2|$

$\therefore$ Range of $f(x)$ is $[-7,7]$
$f(x)={\alpha }^2-2\alpha +1$ will have a solution if ${\alpha }^2-2\alpha +1\le 7$
$(\alpha -1{)}^2\le 7$
$-\sqrt{7}+1\le \alpha \le \sqrt{7}+1$
$\therefore$ number of integral values of $\alpha$ is 5