Question

Let $f\left(x\right)=|x-1|+|x+1|$ and $f\left(x\right)=k$ has atleast one real solution, then which of the following is correct

• A. $k\in [1,\infty )$
• B. $k\in [2,\infty )$
• C. $k=-1$
• D. $k=0$

Answer 1

The correct option is B $k\in [2,\infty )$

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


Clearly, $f\left(x\right)$ is continuous in $R$ and Range of $f\left(x\right)$ is $[2,\infty )$
From $I.V.T.$: $f\left(x\right)=k$ will have atleast one solution, if $k\in \text{Range of}f\left(x\right)$
$\therefore k\in [2,\infty )$