For a 0,if the function f(x) = 2x^3 - 9ax^2 | vedclass

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(C) Given,$f(x) = 2x^3 - 9ax^2 + 12a^2x + 1$. To find the critical points,we calculate the derivative: $f'(x) = 6x^2 - 18ax + 12a^2$. Setting $f'(x) =

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$2$

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(C) Given,$f(x) = 2x^3 - 9ax^2 + 12a^2x + 1$. To find the critical points,we calculate the derivative: $f'(x) = 6x^2 - 18ax + 12a^2$. Setting $f'(x) = 0$,we get $6(x^2 - 3ax + 2a^2) = 0$,which factors as $6(x - a)(x - 2a) = 0$. Thus,the critical points are $x = a$ and $x = 2a$. We find the second derivative: $f''(x) = 12x - 18a$. For $x = a$,$f''(a) = 12a - 18a = -6a$. Since $a > 0$,$f''(a) For $x = 2a$,$f''(2a) = 12(2a) - 18a = 6a$. Since $a > 0$,$f''(2a) > 0$,so $f(x)$ has a minimum at $q = 2a$. Given the condition $p^2 = q$,we substitute the values: $a^2 = 2a$. Since $a > 0$,we can divide by $a$ to get $a = 2$.

For $a > 0$,if the function $f(x) = 2x^3 - 9ax^2 + 12a^2x + 1$ attains its maximum value at $p$ and minimum value at $q$ such that $p^2 = q$,then $a =$

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