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If $\mathrm{f}(x)=x^3+\mathrm{b} x^2+\mathrm{c} x+\mathrm{d}$ and $0 < \mathrm{b}^2 < \mathrm{c}$, then in $(-\infty, \infty)$
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Verified Answer
The correct answer is:
$\mathrm{f}(x)$ is strictly increasing function.
$$
\begin{aligned}
& \mathrm{f}(x)=x^3+\mathrm{b} x^2+\mathrm{c} x+\mathrm{d} \\
\therefore \quad & \mathrm{f}^{\prime}(x)=3 x^2+2 \mathrm{~b} x+\mathrm{c}
\end{aligned}
$$
Now its discriminant $=4\left(b^2-3 c\right)$ $\Rightarrow 4\left(\mathrm{~b}^2-\mathrm{c}\right)-8 \mathrm{c} < 0$, as $\mathrm{b}^2 < \mathrm{c}$ and $\mathrm{c}>0$
$\Rightarrow \mathrm{f}^{\prime}(x)>0$ for all $x \in \mathrm{R}$
$\Rightarrow f$ is strictly increasing on $R$.
\begin{aligned}
& \mathrm{f}(x)=x^3+\mathrm{b} x^2+\mathrm{c} x+\mathrm{d} \\
\therefore \quad & \mathrm{f}^{\prime}(x)=3 x^2+2 \mathrm{~b} x+\mathrm{c}
\end{aligned}
$$
Now its discriminant $=4\left(b^2-3 c\right)$ $\Rightarrow 4\left(\mathrm{~b}^2-\mathrm{c}\right)-8 \mathrm{c} < 0$, as $\mathrm{b}^2 < \mathrm{c}$ and $\mathrm{c}>0$
$\Rightarrow \mathrm{f}^{\prime}(x)>0$ for all $x \in \mathrm{R}$
$\Rightarrow f$ is strictly increasing on $R$.
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