Q42 / 160 · AP EAMCET 2016 · Mathematics

If $f(x)=|x-1|+|x-2|+|x-3|, 2 < x < 3$, then $f$ is

  • A
    an onto function but not one-one
  • B
    one-one function but not onto
  • Correct answer
    C
    a bijection
  • D
    neither one-one nor onto

Solution

We have, $\begin{aligned} & f(x)=|x-1|+|x-2|+|x-3| \\ & \Rightarrow f(x)=\left\{\begin{array}{cc} 6-3 x, & x < 1 \\ 4-x, & 1 < x < 2 \\ x, & 2 < x < 3 \\ 3 x-6, & x>3 \end{array}\right. \end{aligned}$ For $2 < x < 3$, then $f(x)=x$ So, $f(x)$ is one-one and onto function. It implies that $f(x)$ is a bijection.

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

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