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