Question

$f(x)=(a+2){{x}^{3}}-3a{{x}^{2}}+9ax-1$ Decreases for all real value of $x$ is
(a). $(-\infty ,-3)$
(b). $(-\infty ,0)$
(c). $(-3,0)$
(d). $(-3,\infty )$

Answer 1

Hint:Given that f(x) is decreasing, so would take its derivative $f'(x)\le 0$

Complete step by step solution:
Given that:$f(x)=(a+2){{x}^{3}}-3a{{x}^{2}}+9ax-1$
                  Taking derivative with respect to $x$
      $f'(x)=3(a+2){{x}^{2}}-6ax+9a$
As given that, $f(x)$ is decreasing for all real values of $x$.
Therefore, $f'(x)$
$f'(x)<0$ for all $x\in R$

     $\begin{align}
  & 3(a+2){{x}^{2}}-6x+9a<0 \\
 & (a+2){{x}^{2}}-2ax+3a<0 \\
\end{align}$
     $(a+2)$ and $4{{a}^{2}}-4\times (a+2)\times 3a<0$ [ because $a{{x}^{2}}+bx+c<0$ for all$x\in R$ ]
                                 [$\Rightarrow a<0$ ]
$\begin{align}
  & \Rightarrow a+2<0and4{{a}^{2}}-4\times (a+2)\times 3a<0 \\
 & \Rightarrow a<-2and{{a}^{2}}-3{{a}^{2}}-6a<0 \\
 & \Rightarrow a<-2and-2{{a}^{2}}-6a<0 \\
 & \Rightarrow a<-2and-2a(a+3)<0 \\
 & \\
\end{align}$
     Now, $\begin{align}
  & -2a(a+3)<0 \\
 & \Rightarrow a(a+3)>0 \\
 & \\
\end{align}$
                         $\Rightarrow a<-3$ or $a>0$

  $\begin{align}
  & a\in (-\infty ,-3)\cup (0,\infty ) \\
 & \\
\end{align}$
$a<-2$ and $-2a(a+3)<0$
Therefore, $a<-2$ and $\begin{align}
  & a\in (-\infty ,-3)\cup (0,\infty ) \\
 & \Rightarrow a\in (-\infty ,-3) \\
\end{align}$

Hence, $f(x)$ decreases for all $x\in R$, if all $a\in (-\infty ,-3)$ hence, option A is correct.

Option B: $(-\infty ,0)$ is not valid as since $a<-3$ or $a>0$is the solution we received and values of $a$ don’t lies in $(-\infty ,0)$
Option C: $(-3,0)$ is not valid since $a<-3$ or $a>0$is the solution we received and values of $a$ don’t lies in $(-3,0)$
Option D: $(-3,\infty )$ is not valid since $a<-3$ or $a>0$is the solution we received and values of $a$ don’t lies in $(-3,\infty )$

Note: () is known as the open bracket and we can define and assume that R belongs to (-2, 2), then all values between -2 & +2 except -2 & 2.