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Question

Let $$f(x)=\begin{cases}x^{2}-1,0 < x < 2\\2x+3,2 \leq x < 3\end{cases},$$ 
The quadratic equation whose roots are $$\displaystyle \lim_{x\rightarrow 2^{-}}f(x)$$ and $$\displaystyle \lim_{x\rightarrow 2^{+}}f(x)$$ is


A
x210x+21=0
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B
x26x+9=0
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C
x214x+49=0
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D
x2+6x+9=0
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Solution

The correct option is A $$x^{2}-10x+21=0$$
For $$0 < x < 2$$,
$$f(x) =  {x}^{2} - 1 $$
$$\lim_{ x \rightarrow {2}^{-}}   f(x) = 3 $$
For, $$2 < x < 3$$ 
$$f(x) = 2x + 3$$
$$\lim_{ x \rightarrow {2}^{+}}   f(x) = 7 $$
Thus the equation with roots, 3 and 7 is
$$ {x}^{2} - 10x + 21 = 0 $$

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