If the function f is given by f(x)=x^3-3(a-2) | vedclass

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(C) Given $f(x)=x^3-3(a-2)x^2+3ax+7$. Since $f(x)$ is increasing in $(0,1]$ and decreasing in $[1,5)$,$f(x)$ must have a local maximum at $x=1$. Thus,

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$7$

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(C) Given $f(x)=x^3-3(a-2)x^2+3ax+7$. Since $f(x)$ is increasing in $(0,1]$ and decreasing in $[1,5)$,$f(x)$ must have a local maximum at $x=1$. Thus,$f'(1)=0$. $f'(x)=3x^2-6(a-2)x+3a$. Substituting $x=1$: $3(1)^2-6(a-2)(1)+3a=0$. $3-6a+12+3a=0 \Rightarrow -3a+15=0 \Rightarrow a=5$. Now,substitute $a=5$ into $f(x)$: $f(x)=x^3-3(5-2)x^2+3(5)x+7 = x^3-9x^2+15x+7$. We need to solve $\frac{f(x)-14}{(x-1)^2}=0$. $f(x)-14 = x^3-9x^2+15x+7-14 = x^3-9x^2+15x-7$. By synthetic division or polynomial division,$x^3-9x^2+15x-7 = (x-1)^2(x-7)$. So,$\frac{(x-1)^2(x-7)}{(x-1)^2} = 0 \Rightarrow x-7=0 \Rightarrow x=7$.

If the function $f$ is given by $f(x)=x^3-3(a-2)x^2+3ax+7$,for some $a \in R$,is increasing in $(0,1]$ and decreasing in $[1,5)$,then a root of the equation $\frac{f(x)-14}{(x-1)^2}=0$ $(x \neq 1)$ is

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