Q41 / 160 · TS EAMCET 2016 · Mathematics

If $f(x)=x^2-2 x+4$, then the set of values of $x$ satisfying $f(x-1)=f(x+1)$ is

  • A
    $\{-1\}$
  • B
    $\{-1,1\}$
  • Correct answer
    C
    $\{1\}$
  • D
    $\{1,2\}$

Solution

Given, $f(x)=x^2-2 x+4$ Now, $f(x-1)=f(x+1)$ $$ \begin{aligned} & \Rightarrow \quad(x-1)^2-2(x-1)+4=(x+1)^2-2(x+1)+4 \\ & \Rightarrow \quad x^2-2 x+1-2 x+2+4 \\ & =x^2+2 x+1-2 x-2+4 \\ & \Rightarrow x^2-4 x+7=x^2+3 \\ & \Rightarrow \quad \quad-4 x=-4 \\ & \Rightarrow \quad x=1 \\ & \therefore \quad \quad x=\{1\} \end{aligned} $$

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

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