Question

If $y = m\log x + n{x^2} + x$ has its extreme values at $x = 2$ and $x = 1$, then $2m + 10n$ is equal to
${\text{A}}{\text{. }} - 1 \\\ {\text{B}}{\text{. }} - 4 \\\ {\text{C}}{\text{. }} - 2 \\\ {\text{D}}{\text{. 1}} \\\ {\text{E}}{\text{. }} - 3 \\\$

Answer 1

Hint- Here, we will be proceeding by differentiating the given function with respect to x and then putting $\dfrac{{dy}}{{dx}} = 0$ and afterwards substituting the given extreme values of x to obtain the two equation in terms of m and n only and then we will solve them.
Given, $y = m\log x + n{x^2} + x{\text{ }} \to {\text{(1)}}$ which is a function in terms of variable x.

It is also given that the function represented in equation (1) has its extreme values at $x = 2$ and $x = 1$.
As we know that extreme values of any function y are obtained by finding $\dfrac{{dy}}{{dx}}$ and then putting $\dfrac{{dy}}{{dx}} = 0$.
By differentiating the function given by equation (1) both sides with respect to x, we get
$\dfrac{{dy}}{{dx}} = \dfrac{{d\left[ {m\log x + n{x^2} + x} \right]}}{{dx}} = \dfrac{{d\left( {m\log x} \right)}}{{dx}} + \dfrac{{d\left( {n{x^2}} \right)}}{{dx}} + \dfrac{{dx}}{{dx}}$
Since, m and n are constants (i.e., independent of variable x) so we can take them out of the differentiation.

$$\Rightarrow \dfrac{{dy}}{{dx}} = m\left[ {\dfrac{{d\left( {\log x} \right)}}{{dx}}} \right] + n\left[ {\dfrac{{d\left( {{x^2}} \right)}}{{dx}}} \right] + 1 = m\left[ {\dfrac{1}{x}} \right] + n\left[ {2x} \right] + 1 \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{m}{x} + 2nx + 1 \\\$$

Put $\dfrac{{dy}}{{dx}} = 0$ for the extreme values of the variable x, we get

$$\Rightarrow 0 = \dfrac{m}{x} + 2nx + 1 \\\ \Rightarrow 0 = \dfrac{{m + 2n{x^2} + x}}{x} \\\ \Rightarrow 0 = m + 2n{x^2} + x \\\ \Rightarrow 2n{x^2} + x + m = 0{\text{ }} \to {\text{(2)}} \\\$$

So, we are having a quadratic equation in variable x which means there will be two extreme values of x. These extreme values are $x = 2$ and $x = 1$. These two given extreme values should satisfy the quadratic equation given by equation (2).
Put x=2 in equation (2), we get

$$\Rightarrow 2n{\left( 2 \right)^2} + 2 + m = 0 \\\ \Rightarrow 8n + 2 + m = 0{\text{ }} \to {\text{(3)}} \\\$$

Put x=1 in equation (2), we get

$$\Rightarrow 2n{\left( 1 \right)^2} + 1 + m = 0 \\\ \Rightarrow 2n + 1 + m = 0{\text{ }} \to {\text{(4)}} \\\$$

In order to obtain the values of constants m and n we will subtract equation (4) from equation (3), we have

$$\Rightarrow 8n + 2 + m - \left( {2n + 1 + m} \right) = 0 - 0 \\\ \Rightarrow 8n + 2 + m - 2n - 1 - m = 0 \\\ \Rightarrow 6n + 1 = 0 \\\ \Rightarrow 6n = - 1 \\\ \Rightarrow n = \dfrac{{ - 1}}{6} \\\$$

Put $n = \dfrac{{ - 1}}{6}$ in equation (4) to obtain the value of m, we have

$$\Rightarrow 2\left[ {\dfrac{{ - 1}}{6}} \right] + 1 + m = 0 \\\ \Rightarrow \dfrac{{ - 1}}{3} + 1 + m = 0 \\\ \Rightarrow m = \dfrac{1}{3} - 1 = \dfrac{{1 - 3}}{3} \\\ \Rightarrow m = \dfrac{{ - 2}}{3} \\\$$

Put $m = \dfrac{{ - 2}}{3}$ and $n = \dfrac{{ - 1}}{6}$ to find the value for the expression $2m + 10n$, we get
$2m + 10n = 2\left[ {\dfrac{{ - 2}}{3}} \right] + 10\left[ {\dfrac{{ - 1}}{6}} \right] = \dfrac{{ - 4}}{3} - \dfrac{5}{3} = \dfrac{{ - 4 - 5}}{3} = \dfrac{{ - 9}}{3} \\\ \Rightarrow 2m + 10n = - 3 \\\$
Hence, option E is correct.

Note- At the extreme values for any function y, $\dfrac{{dy}}{{dx}} = 0$. Also, if $\dfrac{{{d^2}y}}{{d{x^2}}} > 0$ that means at this extreme value, minima occurs (i.e., we will be getting the minimum value of y corresponding to this extreme value of x) and if $\dfrac{{{d^2}y}}{{d{x^2}}} < 0$ that means at this extreme value, maxima occurs (i.e., we will be getting the maximum value of y corresponding to this extreme value of x).