Question

Let the function $f\left(x\right)=\left|x + 1\right|$ . The number of values of $x\in \left[- 2,2\right]$ for which $f\left(x - 3\right)$ , $f\left(x - 1\right)$ and $f\left(x + 1\right)$ are in the arithmetic progression is

• A. $0$
• B. $1$
• C. $2$
• D. Infinite

Answer 1

Answer: (C)

$f\left(x\right)=\left|x + 1\right|\Rightarrow f\left(x - 3\right)=\left|x - 2\right|$
$f\left(x - 1\right)=\left|x\right|$ , $f\left(x + 1\right)=\left|x + 2\right|$
Given $f\left(x - 3\right),f\left(x - 1\right),f\left(\right.x+1\left.\right)$ are in A.P.
$\Rightarrow 2\left|x\right|=\left|x - 2\right|+\left|x + 2\right|$

Case $1$ : $x < -2$
$-x+2-x-2=-2x\Rightarrow 0=0$
$\Rightarrow x < -2$
Case $2$ : $-2\leq x < 0$
$-x+2+x+2=-2x\Rightarrow x=-2$
$\Rightarrow x=-2$
Case $3$ : $0\leq x < 2$
$-x+2+x+2=2x\Rightarrow x=2$ $\Rightarrow $ no solution
Case $4$ : $x\geq 2$
$x-2+x+2=2x\Rightarrow 0=0$
$\Rightarrow x\geq 2$
Taking union, we get $x\leq -2$ or $x\geq 2$
But. $x\in \left[- 2,2\right]$
Taking intersection, we get,
$x=-2$ and $2$