If the function $f(x)=x^3+2 p x^2+27 x+16$ is strictly increasing for all $x \in R$, then the range of $p$ is

If the function $f(x)=x^3+2 p x^2+27 x+16$ is strictly increasing for all $x \in R$, then the range of $p$ is
  1. $\left(-\infty, \frac{-9}{2}\right) \cup\left(\frac{9}{2}, \infty\right)$
  2. $(-\infty,-9) \cup(9, \infty)$
  3. $\left(\frac{-9}{2}, \frac{9}{2}\right)$
  4. $(-9,9)$

Solution

$ \begin{aligned} & f(x)=x^3+2 p x^2+27 x+16 \\ & f^{\prime}(x)=3 x^2+4 p x+27 \end{aligned} $ $f(x)$ is strictly increasing function, so $f^{\prime}(x)>0$ $ \Rightarrow \quad 3 x^2+4 p x+27>0 $ Possible only when its $D < 0$ $ \begin{aligned} (4 p)^2-4(3)(27) & < 0 \\ 16 p^2-4 \cdot 81 & < 0 \\ p^2-\frac{81}{4} & < 0 \\ \Rightarrow \quad\left(p+\frac{9}{2}\right)\left(p-\frac{9}{2}\right) & < 0 \\ \Rightarrow \quad p & \in\left(-\frac{9}{2}, \frac{9}{2}\right) . \end{aligned} $

Asked in: AP EAMCET 2018 (24 Apr Shift 1)

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