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

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(A) Given $f(x) = 5 - |x - 2|$. The maximum value of $f(x)$ occurs when $|x - 2| = 0$,so $\alpha = 2$.Given $g(x) = |x + 1|$. The minimum value of $g(

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$\frac{1}{2}$

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(A) Given $f(x) = 5 - |x - 2|$. The maximum value of $f(x)$ occurs when $|x - 2| = 0$,so $\alpha = 2$.Given $g(x) = |x + 1|$. The minimum value of $g(x)$ occurs when $x + 1 = 0$,so $\beta = -1$.The limit is to be evaluated at $x \rightarrow -\alpha \beta = - (2)(-1) = 2$.We need to calculate $\lim _{x \rightarrow 2} \frac{(x - 1)(x^2 - 5x + 6)}{(x^2 - 6x + 8)}$.Factorizing the expressions: $x^2 - 5x + 6 = (x - 2)(x - 3)$ and $x^2 - 6x + 8 = (x - 2)(x - 4)$.Substituting these into the limit: $\lim _{x \rightarrow 2} \frac{(x - 1)(x - 2)(x - 3)}{(x - 2)(x - 4)}$.Canceling the common factor $(x - 2)$: $\lim _{x \rightarrow 2} \frac{(x - 1)(x - 3)}{x - 4}$.Evaluating the limit: $\frac{(2 - 1)(2 - 3)}{2 - 4} = \frac{(1)(-1)}{-2} = \frac{-1}{-2} = \frac{1}{2}$.

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 $\lim _{x \rightarrow-\alpha \beta} \frac{(x-1)(x^2-5x+6)}{(x^2-6x+8)}$ is equal to

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