If the function f(x) = x^3 - 3(a - 2)x^2 + 3a | vedclass

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(C) Given $f(x) = x^3 - 3(a - 2)x^2 + 3ax + 7$. The derivative is $f'(x) = 3x^2 - 6(a - 2)x + 3a$. Since the function changes behavior at $x = 1$,we h

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$7$

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(C) Given $f(x) = x^3 - 3(a - 2)x^2 + 3ax + 7$. The derivative is $f'(x) = 3x^2 - 6(a - 2)x + 3a$. Since the function changes behavior at $x = 1$,we have $f'(1) = 0$. $3(1)^2 - 6(a - 2)(1) + 3a = 0$ $3 - 6a + 12 + 3a = 0$ $-3a + 15 = 0 \Rightarrow a = 5$. Substituting $a = 5$ into $f(x)$,we get $f(x) = x^3 - 3(5 - 2)x^2 + 3(5)x + 7 = x^3 - 9x^2 + 15x + 7$. Now,solve $\frac{f(x) - 14}{(x - 1)^2} = 0$ for $x 
eq 1$. $f(x) - 14 = 0 \Rightarrow x^3 - 9x^2 + 15x + 7 - 14 = 0$. $x^3 - 9x^2 + 15x - 7 = 0$. Since $x = 1$ is a root,we divide by $(x - 1)^2$: $(x - 1)^2(x - 7) = 0$. The roots are $x = 1$ and $x = 7$. Since $x 
eq 1$,the root is $x = 7$.

If the function $f(x) = x^3 - 3(a - 2)x^2 + 3ax + 7$,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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