Question

The values of $\mathrm{a}$ for which the function $(\mathrm{a}+2){\mathrm{x}}^3-3{\mathrm{ax}}^2+9\mathrm{ax}-1$ decreases monotonically throughout for all real $\mathrm{x}$, are

• A. $\mathrm{a}<-2$
• B. $\mathrm{a}>-2$
• C. $-3<\mathrm{a}<0$
• D. $-\infty <\mathrm{a}\le 3$

Answer 1

The correct option is D

$-\infty <\mathrm{a}\le 3$

Explanation for the correct option:

Given: $(a+2)x^3-3a{x}^2+9ax-1$

The function is monotonically decreasing if $f\text{'}\left(x\right)\le 0$ for all $\mathrm{x}\in \mathrm{ℝ}$

$$\begin{gathered} f\text{'}\left(x\right)=3x^2\left(a+2\right)-6ax+9a \\ f\text{'}\left(x\right)\le 0 \\ 3x^2\left(a+2\right)-6ax+9a\le 0 \\ {x}^2(a+2)-2ax+3a\le 0 \end{gathered}$$

Find the roots of the above quadratic equation

$$\begin{gathered} \Rightarrow x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ \Rightarrow x=\frac{2a\pm \sqrt{4a^2-4\times \left(a+2\right)\times 3a}}{2(a+2)} \\ \Rightarrow x=\frac{2a\pm \sqrt{-8a^2-24a}}{2(a+2)} \\ \Rightarrow x=\frac{2a+2\sqrt{-2a^2-6a}}{2\left(a+2\right)} \end{gathered}$$

Roots area real if $-2{\mathrm{a}}^2-6\mathrm{a}\ge 0$

$$\begin{gathered} \Rightarrow {\mathrm{a}}^2+3\mathrm{a}\le 0 \\ \Rightarrow \mathrm{a}(\mathrm{a}+3)\le 0 \\ \Rightarrow -\infty <\mathrm{a}\le -3 \end{gathered}$$

Hence option D is the correct answer.