Let f(x)=5-|x-2| and g(x)=|x+1|, x in R. If f | vedclass

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(A) Given $f(x) = 5 - |x - 2|$. Since $|x - 2| \geq 0$,the maximum value of $f(x)$ is $5$,which occurs when $|x - 2| = 0$,i.e.,$x = 2$. Thus,$\alpha =

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$\frac{1}{2}$

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(A) Given $f(x) = 5 - |x - 2|$. Since $|x - 2| \geq 0$,the maximum value of $f(x)$ is $5$,which occurs when $|x - 2| = 0$,i.e.,$x = 2$. Thus,$\alpha = 2$.Given $g(x) = |x + 1|$. Since $|x + 1| \geq 0$,the minimum value of $g(x)$ is $0$,which occurs when $|x + 1| = 0$,i.e.,$x = -1$. Thus,$\beta = -1$.We need to evaluate $\lim _{x \rightarrow -\alpha \beta} \frac{(x - 1)(x^2 - 5x + 6)}{x^2 - 6x + 8}$.Since $-\alpha \beta = -(2)(-1) = 2$,the limit is $\lim _{x \rightarrow 2} \frac{(x - 1)(x - 2)(x - 3)}{(x - 2)(x - 4)}$.Canceling the common factor $(x - 2)$,we get $\lim _{x \rightarrow 2} \frac{(x - 1)(x - 3)}{x - 4}$.Substituting $x = 2$,we get $\frac{(2 - 1)(2 - 3)}{2 - 4} = \frac{(1)(-1)}{-2} = \frac{-1}{-2} = \frac{1}{2}$.

Let $f(x)=5-|x-2|$ and $g(x)=|x+1|, x \in R$. If $f(x)$ attains its maximum value at $\alpha$ and $g(x)$ attains its 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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