Question

If $f\left( x \right) = {x^3} + 4{x^2} + ax + 5$ is a monotonically decreasing function of $x$ in the largest possible interval $\left( { - 2, - \dfrac{2}{3}} \right)$, then the value if $a$ is
A) $4$
B) $2$
C) $- 1$
D) None of these

Answer 1

The given function is monotonically decreasing in the given interval, which means that its derivative is negative in this interval. So, for solving this question, we will first differentiate the given function. Since the given interval is the largest possible interval in which the given function can monotonically decrease, this means that the derivative at the end points must be equal to zero. Therefore, on equating the derivative of the function equal to zero at any of the end points, we will get the required value of $a$.

Complete step by step solution:
The function given in the question is $f\left( x \right) = {x^3} + 4{x^2} + ax + 5$.
Differentiating both sides with respect to $x$, we get
$\Rightarrow f'\left( x \right) = 3{x^2} + 8x + a$……………………….$\left( 1 \right)$
According to the question, the given function $f\left( x \right)$ is monotonically decreasing in the interval $\left( { - 2, - \dfrac{2}{3}} \right)$. This means that the slope of $f\left( x \right)$ is negative in this interval. Further, it is also given that the interval $\left( { - 2, - \dfrac{2}{3}} \right)$ is the largest possible interval in which the given function can monotonically decrease. Therefore, the slope, or the derivative of the given function is equal to zero at the end points of the given interval, which means that
$f'\left( { - 2} \right) = 0$………………………….$\left( 2 \right)$
Substituting $x = - 2$ in equation $\left( 1 \right)$, we get
$f'\left( { - 2} \right) = 3{\left( { - 2} \right)^2} + 8\left( { - 2} \right) + a$
Now substituting equation $\left( 2 \right)$ in above equation, we get
$\Rightarrow 0 = 12 - 16 + a$
$\Rightarrow a - 4 = 0$
Adding $4$ on both the sides, we finally get
$\Rightarrow a = 4$

Hence, the correct answer is option A.

Note:
Although we used the left end point $x = - 2$ of the given interval to get the value of $a$, but we can also use the right end point $x = - \dfrac{2}{3}$ at which the derivative to zero. On equating the derivative to zero at $x = - \dfrac{2}{3}$ also, we will get the same value of $a$. Differentiation is a way of finding the derivative of a function or the rate of change of a function with respect to a particular variable. Integration is the inverse of differentiation and hence it is called antiderivative.