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

Given, \(f(x)=2 x^{3}-9 a x^{2}+12 a^{2} x+1\) attains maximum and minimum at \(p\) and \(q\) respectively. \(\therefore\)
\(
f^{\prime}(p)=0, f^{\prime}(q)=0
\)
Now,
\(
\begin{array}{c}
f^{\prime \prime}(p) < 0 \text { and } f^{\prime \prime}(q)>0 \\
f^{\prime}(p)=0
\end{array}
\)
and
\(
\begin{array}{r}
f^{\prime}(q)=0 \\
6 p^{2}-18 a p+12 a^{2}=0 \\
6 q^{2}-18 a q+12 a^{2}=0
\end{array}
\)
and
\(
\begin{array}{ll}
\Rightarrow & p^{2}-3 a p+2 a^{2}=0 \\
\text { and } & q^{2}-3 a q+2 a^{2}=0
\end{array}
\)
\(p=a, 2 a, q=a, 2 a\)
\(\begin{array}{lrl}\Rightarrow & p=a, 2 a, q=a, \\ \text { Now, } & f^{\prime \prime}(p) < 0\end{array}\)
\(\Rightarrow \quad 12 p-18 a < 0\)
\(\Rightarrow \quad p < \frac{3}{2} a\)
and \(f^{\prime \prime}(q)>0 \Rightarrow 12 q-18 a>0\)
\(\Rightarrow \quad q>\frac{3}{2} a\)
From Eqs. (i), (ii) and (iii), we get
\(\begin{aligned} p &=a, q=2 a \\ \text { Now, } \quad p^{2} &=q \end{aligned}\)
\(
\begin{array}{lc}
\Rightarrow & a^{2}=2 a \\
\Rightarrow & a=0,2
\end{array}
\)
But for \(a=0, f(x)=2 x^{3}+1\) which does not
attains a maximum or minimum for any value of \(x\). Hence, \(a=2\).