Q112 / 200 · WBJEE 2013 · Mathematics

The minimum value of the function $f(x)=2|x-1|+|x-2|$ is

  • A
    0
  • Correct answer
    B
    1
  • C
    2
  • D
    3

Solution

Given, $$ \begin{array}{l} \quad f(x)=2|x-1|+|x-2| \\ \Rightarrow \quad f(x)=\left\{\begin{array}{c} -2(x-1)-(x-2), x < 1 \\ 2(x-1)-(x-2), 1 \leq x < 2 \\ 2(x-1)+(x-2), x \geq 2 \\ -3 x+4, \quad x < 1 \\ x_{1} \\ 3 x-4, \quad 1 \leq x < 2 \\ x \geq 2 \end{array}\right. \end{array} $$ $f^{\prime}(x)=\left\{\begin{array}{cc}-3 & x < 1 \\ 1 & 1 \leq x < 2 \\ 3 & x \geq 2\end{array}\right.$ So, $f(x)$ will be minimum at $x=1$ and the minimum value is 1 .

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

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