If the function f(x)=2x^{3}-9ax^{2}+12a^{2}x+ | vedclass

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(A) Given the function $f(x)=2x^{3}-9ax^{2}+12a^{2}x+1$. First, find the derivative $f'(x)$: $f'(x) = 6x^{2}-18ax+12a^{2}$. Set $f'(x)=0$ to find crit

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$2$

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(A) Given the function $f(x)=2x^{3}-9ax^{2}+12a^{2}x+1$. First, find the derivative $f'(x)$: $f'(x) = 6x^{2}-18ax+12a^{2}$. Set $f'(x)=0$ to find critical points: $6(x^{2}-3ax+2a^{2})=0 \Rightarrow 6(x-a)(x-2a)=0$. So, the critical points are $x=a$ and $x=2a$. Now, find the second derivative $f''(x)$: $f''(x) = 12x-18a$. Check the nature of the critical points: 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 local minimum at $q=2a$. Given the condition $p^{2}=q$, substitute the values: $a^{2} = 2a$. Since $a>0$, we can divide by $a$: $a = 2$.

If the function $f(x)=2x^{3}-9ax^{2}+12a^{2}x+1$ where $a>0$ attains its maximum and minimum at $p$ and $q$ respectively such that $p^{2}=q$, then $a$ is equal to

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