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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
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$\mathrm{f}(\mathrm{x})=\left\{\begin{array}{cc}x^2-4 x+3, & \text { if } x < 2 \\ x-3, & \text { if } x \geq 2\end{array}\right.$
Then the number of real numbers $x$ for which $f(x)=8$ is
Solution:
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Verified Answer
The correct answer is:
2
$f(x)=8$ ...(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 $=2$
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 $=2$
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