Question

The values of ‘a’ for which the function $(a + 2){x^3} - 3a{x^2} + 9ax - 1$ decreases monotonically throughout for all real $x$ are
A. $a < \- 2$
B. $a > - 2$
C. $- 3 < a < 0$
D. $- \infty < a \leqslant - 3$

Answer 1

Hint : The derivative of a function can be at times used to determine whether a function is increasing or decreasing on any interval in its domain. If $f'(x) > 0$ in an interval I, then the function is said to be increasing on I and if $f'(x) < 0$ in an interval I, then the function is said to be decreasing on I.

Complete step-by-step answer :
If $f'(x) > 0$ then $f$ is increasing on the interval, and if $f'(x) < 0$ then $f$ is decreasing on the interval.
Following steps are involved in the process of finding the intervals of increasing and decreasing function:
Firstly, differentiate the given function with respect to the constant variable.
Then solve $f'(x) = 0$ .
After solving the equation of the first derivative and finding the points of discontinuity we get the open intervals with the value of $x$ , through which the sign of the intervals can be taken into consideration.
If the sign of the interval in their first derivative form gives more than $0$ then the function is said to be increasing in nature, while if the sign of the intervals in their first derivative form gives less than $0$ then the function is said to be decreasing in nature.
Finally, we get increasing as well as decreasing intervals of the function.
We are given the function $f(x) = (a + 2){x^3} - 3a{x^2} + 9ax - 1$
Taking derivative on both the sides with respect to $x$ we get ,
$f'(x) = 3(a + 2){x^2} - 6ax + 9a$
For the function to be monotonically decreasing $f'(x) \leqslant 0$ $\forall x \in \mathbb{R}$
Therefore $3(a + 2){x^2} - 6ax + 9a \leqslant 0\forall x \in \mathbb{R}$
Or we can say that $(a + 2){x^2} - 2ax + 3a \leqslant 0\forall x \in \mathbb{R}$
This is a quadratic equation in terms of $x$ .
Therefore for the function to be monotonically decreasing, discriminant $\leqslant 0$
i.e. $\sqrt {{b^2} - 4ac} = \sqrt {{{( - 2a)}^2} - 4(a + 2)(3a)} \leqslant 0$
Therefore we get $- 8{a^2} - 24a \leqslant 0$
Therefore we get $a \geqslant 0$ and $a \leqslant - 3$
Therefore we get $- \infty < a \leqslant - 3$
Therefore option (4) is the correct answer.
So, the correct answer is “Option D”.

Note : If $f'(x) > 0$ then $f$ is increasing on the interval, and if $f'(x) < 0$ then $f$ is decreasing on the interval. The derivative of a function can be at times used to determine whether a function is increasing or decreasing on any interval in its domain.