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

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(C) Given,$f(x)=2 x^{3}-9 a x^{2}+12 a^{2} x+1$. The critical points are found by setting $f^{\prime}(x)=0$. $f^{\prime}(x) = 6x^{2} - 18ax + 12a^{2}

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$2$

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(C) Given,$f(x)=2 x^{3}-9 a x^{2}+12 a^{2} x+1$. The critical points are found by setting $f^{\prime}(x)=0$. $f^{\prime}(x) = 6x^{2} - 18ax + 12a^{2} = 6(x^{2} - 3ax + 2a^{2}) = 6(x-a)(x-2a) = 0$. So,the critical points are $x=a$ and $x=2a$. We find the second derivative $f^{\prime \prime}(x) = 12x - 18a$. For a maximum at $p$,$f^{\prime \prime}(p) For a minimum at $q$,$f^{\prime \prime}(q) > 0 \Rightarrow 12q - 18a > 0 \Rightarrow q > \frac{3}{2}a$. Comparing the critical points $a$ and $2a$ with these conditions,we get $p=a$ and $q=2a$. Given $p^{2}=q$,we substitute the values: $a^{2} = 2a$. $a^{2} - 2a = 0 \Rightarrow a(a-2) = 0$. Thus,$a=0$ or $a=2$. If $a=0$,$f(x)=2x^{3}+1$,which is a strictly increasing function and has no local maxima or minima. Therefore,$a=2$ is the only valid solution.

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

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