Question

Let $f\left(x\right)=x+2|x+1|+2|x-1|$. If $f\left(x\right)=k$ has exactly one real solution, then the value of $k$ is

• A. $3$
• B. $0$
• C. $2$
• D. $4$

Answer 1

The correct option is A $3$

$f\left(x\right)=\{\begin{array}{cc}x-2(x+1)-2(x-1),& x<-1\\ x+2(x+1)-2(x-1),& -1\le x\le 1\\ x+2(x+1)+2(x-1),& x>1\end{array}$

$f\left(x\right)=\{\begin{array}{cc}-3x,& x<-1\\ x+4,& -1\le x\le 1\\ 5x,& x>1\end{array}$


$f\left(x\right)=k$ has exactly one real solution, if $y=k$ touches the graph of $y=f\left(x\right)$ where $y=f\left(x\right)$ attains the minimum value.
The minimum value of $f$ is $3$.
Hence, $k=3$