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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Aicharya · Accepted Answer

$\begin{aligned} & |x-2| \geq 0 \ & \Rightarrow-|x-2| \leq 0 \ & \Rightarrow 5-|x-2| \leq 5 \end{aligned}$ Maximum value of $\mathrm{f}(x)$ is 5 . $\begin{aligned} 	herefore \quad & 5-|x-2|=5 \ & \Rightarrow|x-2|=0 \ & \Rightarrow x=2 \ & \Rightarrow \alpha=2 \ & |x+1| \geq 0 \end{aligned}$ Minimum value of $\mathrm{g}(x)$ is 0 . $\begin{aligned} 	herefore \quad & |x+1|=0 \ & \Rightarrow x=-1 \ & \Rightarrow \beta=-1 \end{aligned}$ $\begin{aligned} 	herefore \quad & \lim _{x ightarrow-\alpha \beta} \frac{(x-1)\left(x^2-5 x+6ight)}{x^2-6 x+8} \ = & \lim _{x ightarrow 2} \frac{(x-1)(x-2)(x-3)}{(x-2)(x-4)} \ = & \lim _{x ightarrow 2} \frac{(x-1)(x-3)}{x-4} \ = & \frac{(1)(-1)}{-2} \ = & \frac{1}{2} \end{aligned}$

Let $\mathrm{f}(x)=5-|x-2|$ and $\mathrm{g}(x)=|x+1|, x \in \mathrm{R}$ If $\mathrm{f}(x)$ attains maximum…

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