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Let $\mathrm{f}(\mathrm{x})=5-|\mathrm{x}-2|$ and $\mathrm{g}(\mathrm{x})=|\mathrm{x}+1|, \mathrm{x} \in \mathrm{R}$. If $\mathrm{f}(\mathrm{x})$ attains maximum value at $\alpha$ and $g(x)$ attains minimum value at $\beta$, then $\lim _{x \rightarrow-\alpha \beta} \frac{(x-1)\left(x^2-5 x+6\right)}{\left(x^2-6 x+8\right)}$ is equal to
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The correct answer is:
$\frac{1}{2}$
$\mathrm{f}(\mathrm{x})=5-|\mathrm{x}-2|$ is maximum at $\mathrm{x}=2 \Rightarrow \alpha=2$
$\mathrm{g}(\mathrm{x})=|\mathrm{x}+1|$ is minimum at $\mathrm{x}=-1 \Rightarrow \beta=-1$
Now $\lim _{x \rightarrow-\alpha \beta} \frac{(x-1)\left(x^2-5 x+6\right)}{\left(x^2-6 x+8\right)}=\lim _{x \rightarrow-(2)(-1)} \frac{(x-1)(x-2)(x-3)}{(x-2)(x-4)}$
$\begin{aligned} & =\lim _{x \rightarrow 2} \frac{(x-1)(x-3)}{x-4} \\ & =\frac{(2-1)(2-3)}{2-4} \\ & =\frac{1}{2}\end{aligned}$
$\mathrm{g}(\mathrm{x})=|\mathrm{x}+1|$ is minimum at $\mathrm{x}=-1 \Rightarrow \beta=-1$
Now $\lim _{x \rightarrow-\alpha \beta} \frac{(x-1)\left(x^2-5 x+6\right)}{\left(x^2-6 x+8\right)}=\lim _{x \rightarrow-(2)(-1)} \frac{(x-1)(x-2)(x-3)}{(x-2)(x-4)}$
$\begin{aligned} & =\lim _{x \rightarrow 2} \frac{(x-1)(x-3)}{x-4} \\ & =\frac{(2-1)(2-3)}{2-4} \\ & =\frac{1}{2}\end{aligned}$
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