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 \\
$
Last updated date: 26th Mar 2023
•
Total views: 308.7k
•
Views today: 7.85k
Answer
308.7k+ views
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).
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).
Recently Updated Pages
Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts
Name the Largest and the Smallest Cell in the Human Body ?

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

A ball impinges directly on a similar ball at rest class 11 physics CBSE

Lysosomes are known as suicidal bags of cell why class 11 biology CBSE

Two balls are dropped from different heights at different class 11 physics CBSE

A 30 solution of H2O2 is marketed as 100 volume hydrogen class 11 chemistry JEE_Main

A sample of an ideal gas is expanded from 1dm3 to 3dm3 class 11 chemistry CBSE
