Question

The value of ‘a’ for which the function $\left( a+2 \right){{x}^{3}}-3a{{x}^{2}}+9ax-1$ decreases monotonically for all real x, is

Answer 1

For solving this question you should know about the decreasing function and its properties. The decreasing functions have a good quality that their first derivative is always greater than or equal to zero. So, then we will find the values of roots of this and thus we will find the answer.

Complete step-by-step solution:
According to the question it is asked to us to find the value for ‘a’, for which the function $\left( a+2 \right){{x}^{3}}-3a{{x}^{2}}+9ax-1$ decreases monotonically.
Here, the given function is $\left( a+2 \right){{x}^{3}}-3a{{x}^{2}}+9ax-1$ the decreasing functions can be written as:
$f'\left( x \right)\ge 0$
So, the value of $f'\left( x \right)$ for this is:
$\Rightarrow \dfrac{d}{dx}\left( f\left( x \right) \right)=\dfrac{d}{dx}\left[ \left( a+2 \right){{x}^{3}}-3a{{x}^{2}}+9ax-1 \right]$
$\Rightarrow f'\left( x \right)=3\left( a+2 \right){{x}^{2}}-6ax+9a$
If $f'\left( x \right)\ge 0$
So, it can be written as
$3\left( a+2 \right){{x}^{2}}-6ax+9a\ge 0$
If we solve this getting the value of ‘a’ then,
$3\left( \left( a+2 \right){{x}^{2}}-2ax+3a \right)\ge 0$
Or

& D\le 0 \\\ & \because D={{b}^{2}}-4ac \\\ \end{aligned}$$ So, $$4{{a}^{2}}-12a\left( a+2 \right)\ge 0$$ $$\begin{aligned} & 4a\left( a-3\left( a+2 \right) \right)\ge 0 \\\ & 4a\left( -6-2a \right)\ge 0 \\\ & 8a\left( -3-a \right)\ge 0 \\\ \end{aligned}$$ Or it can be written as $$8a\left( 3+a \right)\le 0$$ Hence, $$a>0$$ and $$a+3<0$$ OR $$a<0$$ and $$a+3>0$$ Hence, $$a\in \left( -\infty ,-3 \right]$$ So, this is the final value for ‘a’. **Note:** While solving these types of questions you have to find that the roots of any quadratic equation can be solved by the $${{b}^{2}}-4ac$$. And if the question asks for the values of any variable in this, then we will solve it by the help of this formula and find the value.