If $f: \mathbf{R} \rightarrow \mathbf{R}$ be defined by $f(x)=x+2|x+1|+2|x-1|$, then the element in the co-domain, which has unique pre image in the domain is

• Correct answer
  A

  3
• B

  1
• C

  2
• D

  5

Solution

We have, $\begin{aligned} & f(x)=x+2|x+1|+2|x-1| \\ & f(x)=\left\{\begin{array}{cc} x+2(-x-1)+2(-x+1), & x \leq-1 \\ x+2(x+1)+2(-x+1), & -1 < x \leq 1 \\ x+2(x+1)+2(x-1), & x \geq 1 \end{array}\right. \\ & f(x)=\left\{\begin{array}{cc} -3 x, & x \leq-1 \\ x+4, & -1 < x < 1 \\ 5 x, & x \geq 1 \end{array}\right. \end{aligned}$ Graph of $f(x)$ is
Clearly, 3 has unique image in the domain.

Hence, the correct answer is option: 'A'.