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The polynomial equation $x^{3}-3 a x^{2}+\left(27 a^{2}+9\right) x+2016=0$ has -
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Verified Answer
The correct answer is:
exactly one real root for any real a
$$
\begin{aligned}
\mathrm{f}^{\prime}(\mathrm{x}) &=3 \mathrm{x}^{2}-6 \mathrm{ax}+27 \mathrm{a}^{2}+9 \\
&=3\left[\mathrm{x}^{2}-2 \mathrm{ax}+9 \mathrm{a}^{2}+3\right]=3\left((\mathrm{x}-\mathrm{a})^{2}+8 \mathrm{a}^{2}+3\right)
\end{aligned}
$$
$\therefore f^{\prime}(x)$ is $+$ ve for $x \in R$ so $f(x)$ is monotonic $\uparrow$ for $\mathrm{x} \in \mathrm{R}$.
\begin{aligned}
\mathrm{f}^{\prime}(\mathrm{x}) &=3 \mathrm{x}^{2}-6 \mathrm{ax}+27 \mathrm{a}^{2}+9 \\
&=3\left[\mathrm{x}^{2}-2 \mathrm{ax}+9 \mathrm{a}^{2}+3\right]=3\left((\mathrm{x}-\mathrm{a})^{2}+8 \mathrm{a}^{2}+3\right)
\end{aligned}
$$
$\therefore f^{\prime}(x)$ is $+$ ve for $x \in R$ so $f(x)$ is monotonic $\uparrow$ for $\mathrm{x} \in \mathrm{R}$.
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