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If $f(x)=\left\{\begin{array}{cl}x^2-1, & 0 < x < 2 \\ 2 x+3, & 2 \leq x < 3\end{array}\right.$,the quadratic equation whose roots are $\lim _{x \rightarrow 2^{-}} f(x)$ and $\lim _{x \rightarrow 2^{+}} f(x)$ is
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The correct answer is:
$x^2-10 x+21=0$
$f(x)= \begin{cases}x^2-1, & 0 < x < 2 \\ 2 x+3, & 2 \leq x < 3\end{cases}$
Now, $\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}} x^2-1=\lim _{h \rightarrow 0}\left[(2-h)^2-1\right]$ $=\lim _{h \rightarrow 0}\left[4+h^2-4 h-1\right]=4+0-0-1=3$
and $\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}}(2 x+3)=\lim _{h \rightarrow 0} 2(2+h)+3$ $=\lim _{h \rightarrow 0} 4+2 h+3=4+0+3=7$
Therefore, the quadratic equation whose roots are 3 and 7 is given by $x^2-(3+7) x+(3 \times 7)=0$ which is $x^2-10 x+21=0$
Now, $\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}} x^2-1=\lim _{h \rightarrow 0}\left[(2-h)^2-1\right]$ $=\lim _{h \rightarrow 0}\left[4+h^2-4 h-1\right]=4+0-0-1=3$
and $\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}}(2 x+3)=\lim _{h \rightarrow 0} 2(2+h)+3$ $=\lim _{h \rightarrow 0} 4+2 h+3=4+0+3=7$
Therefore, the quadratic equation whose roots are 3 and 7 is given by $x^2-(3+7) x+(3 \times 7)=0$ which is $x^2-10 x+21=0$
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