Question

Suppose the cubic ${x^3} - px + q$has three distinct real roots where and. Then which one of the following holds?
1)The cubic has a maxima at both $\sqrt {\dfrac{p}{3}}$and$- \sqrt {\dfrac{p}{3}}$
2)The cubic has a minima at both$\sqrt {\dfrac{p}{3}}$and$- \sqrt {\dfrac{p}{3}}$
3)The cubic has a minima at $\sqrt {\dfrac{p}{3}}$and maxima at$- \sqrt {\dfrac{p}{3}}$
4)The cubic has a maxima at $\sqrt {\dfrac{p}{3}}$and minima at$- \sqrt {\dfrac{p}{3}}$

Answer 1

Hint : This question is based on cubic equations and analysis of its roots using the concept of differentiation. The concept of minima and maxima using differentiation is used in this question. For maxima and minima, we need to find the critical points using,
And then, find if the critical point is maxima or minima by checking the sign off at that crucial point, $f'(x) = 0$
It is positive; then the point is a minima $f''(x) > 0$
If it is negative then the point is a maxima$f''(x) < 0$.

Complete step-by-step answer :
Now, we are given the cubic function as shown below
$\Rightarrow f(x) = {x^3} - px + q$,$p > 0$and $q < 0$
Now, let us differentiate the function as shown
$\Rightarrow f'(x) = 3{x^2} - p$
Now, let us find the critical points using the below condition
$\Rightarrow f'(x) = 0$
$\Rightarrow 3{x^2} - p = 0$
Equating the differentiated function to zero, to find the critical points as shown below
$\Rightarrow x = \pm \sqrt {\dfrac{p}{3}}$
Now, again differentiating the value of the function to find the nature of the critical point
$\Rightarrow f''(x) = 6x$
Now, by putting $x = + \sqrt {\dfrac{p}{3}}$we get,
$\Rightarrow f''(\sqrt {\dfrac{p}{3}} ) = 6\sqrt {\dfrac{p}{3}}$
As $p > 0$ the R.H.S. is positive, as shown below,
$\Rightarrow f''(\sqrt {\dfrac{p}{3}} ) > 0$,
Therefore, we have a minimum $x = + \sqrt {\dfrac{p}{3}}$.
Similarly, repeating the process for $x = - \sqrt {\dfrac{p}{3}}$we get,
$\Rightarrow f''( - \sqrt {\dfrac{p}{3}} ) = - 6(\sqrt {\dfrac{p}{3}} )$
As $p > 0$the R.H.S is negative, as shown below
$\Rightarrow f''( - \sqrt {\dfrac{p}{3}} ) < 0$
Thus, $x = - \sqrt {\dfrac{p}{3}}$we have a maxima.
Thus, option(3) is the correct answer.
So, the correct answer is “Option 3”.

Note : This question involves basic concepts like differentiation and analysis of cubic equations. One should be well versed with these concepts to solve this question. Do not commit calculation mistakes, and be sure of the final answer.