Q41 / 160 · TS EAMCET 2023 (14 May Shift 2) · Mathematics

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function defined by $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{cc}x^2-4 x+3, & \text { if } x < 2 \\ x-3, & \text { if } x \geq 2\end{array}\right.$ Then the number of real numbers $x$ for which $f(x)=8$ is

  • A
    1
  • Correct answer
    B
    2
  • C
    3
  • D
    4

Solution

$f(x)=8$ ...(i) For $x < 2$ $\begin{aligned} & f(x)=x^2-4 x+3=8 \\ & x^2-4 x-5=0 \\ & (x-5)(x+1)=0 \\ & x=-1 ; 5\end{aligned}$ $\because x < 2$ $\therefore x=5$ not possible and for $x \geq 2$ $f(x)=x-3=8 \Rightarrow x=11$ $\therefore$ Number of solutions $=2$

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

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