If the function f given by f(x) = x^3 - 3(a - | 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 $f(x)$ is increasing in $(0, 1]$ and decreasing

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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 $f(x)$ is increasing in $(0, 1]$ and decreasing in $[1, 5)$,the function must have a local maximum at $x = 1$.Thus,$f'(1) = 0$.$f'(1) = 3(1)^2 - 6(a - 2)(1) + 3a = 3 - 6a + 12 + 3a = 15 - 3a = 0$.Solving for $a$,we get $a = 5$.Substituting $a = 5$ into $f(x)$,we get $f(x) = x^3 - 3(5 - 2)x^2 + 3(5)x + 7 = x^3 - 9x^2 + 15x + 7$.We need to solve $\frac{f(x) - 14}{(x - 1)^2} = 0$,which implies $f(x) - 14 = 0$ for $x 
eq 1$.$x^3 - 9x^2 + 15x + 7 - 14 = x^3 - 9x^2 + 15x - 7 = 0$.Since $x = 1$ is a root of $f(x) - 14 = 0$,we can divide by $(x - 1)^2$:$x^3 - 9x^2 + 15x - 7 = (x - 1)^2(x - 7) = 0$.Thus,the root of the equation $\frac{f(x) - 14}{(x - 1)^2} = 0$ is $x = 7$.

If the function $f$ 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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