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TS EAMCET · Maths · Functions

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

  1. A 1
  2. B 2
  3. C 3
  4. D 4
Verified Solution

Answer & Solution

Correct Answer

(B) 2

Step-by-step Solution

Detailed explanation

\(f(x)=8\) ...(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 \(=2\)