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Let $\mathrm{f}(x)=5-|x-2|$ and $\mathrm{g}(x)=|x+1|, x \in \mathrm{R}$ If $\mathrm{f}(x)$ attains maximum value at $\alpha$ and $\mathrm{g}(x)$ attains minimum value at $\beta$, then $\lim _{x \rightarrow-\alpha \beta} \frac{(x-1)\left(x^2-5 x+6\right)}{x^2-6 x+8}$ is equal to
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Verified Answer
The correct answer is:
$\frac {1}{2}$
$\begin{aligned}
& |x-2| \geq 0 \\
& \Rightarrow-|x-2| \leq 0 \\
& \Rightarrow 5-|x-2| \leq 5
\end{aligned}$
Maximum value of $\mathrm{f}(x)$ is 5 .
$\begin{aligned}
\therefore \quad & 5-|x-2|=5 \\
& \Rightarrow|x-2|=0 \\
& \Rightarrow x=2 \\
& \Rightarrow \alpha=2 \\
& |x+1| \geq 0
\end{aligned}$
Minimum value of $\mathrm{g}(x)$ is 0 .
$\begin{aligned}
\therefore \quad & |x+1|=0 \\
& \Rightarrow x=-1 \\
& \Rightarrow \beta=-1
\end{aligned}$
$\begin{aligned}
\therefore \quad & \lim _{x \rightarrow-\alpha \beta} \frac{(x-1)\left(x^2-5 x+6\right)}{x^2-6 x+8} \\
= & \lim _{x \rightarrow 2} \frac{(x-1)(x-2)(x-3)}{(x-2)(x-4)} \\
= & \lim _{x \rightarrow 2} \frac{(x-1)(x-3)}{x-4} \\
= & \frac{(1)(-1)}{-2} \\
= & \frac{1}{2}
\end{aligned}$
& |x-2| \geq 0 \\
& \Rightarrow-|x-2| \leq 0 \\
& \Rightarrow 5-|x-2| \leq 5
\end{aligned}$
Maximum value of $\mathrm{f}(x)$ is 5 .
$\begin{aligned}
\therefore \quad & 5-|x-2|=5 \\
& \Rightarrow|x-2|=0 \\
& \Rightarrow x=2 \\
& \Rightarrow \alpha=2 \\
& |x+1| \geq 0
\end{aligned}$
Minimum value of $\mathrm{g}(x)$ is 0 .
$\begin{aligned}
\therefore \quad & |x+1|=0 \\
& \Rightarrow x=-1 \\
& \Rightarrow \beta=-1
\end{aligned}$
$\begin{aligned}
\therefore \quad & \lim _{x \rightarrow-\alpha \beta} \frac{(x-1)\left(x^2-5 x+6\right)}{x^2-6 x+8} \\
= & \lim _{x \rightarrow 2} \frac{(x-1)(x-2)(x-3)}{(x-2)(x-4)} \\
= & \lim _{x \rightarrow 2} \frac{(x-1)(x-3)}{x-4} \\
= & \frac{(1)(-1)}{-2} \\
= & \frac{1}{2}
\end{aligned}$
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