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.