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
B
x26x+9=0
C
x214x+49=0
D
x2+6x+9=0

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$${x}^{2} - 10x + 21 = 0$$Mathematics

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