If ${{\text{S}}_1}$ and ${{\text{S}}_2}$ are respectively the sets of local minimum and local maximum points of the functions $f\left( x \right) = 9{x^4} + 12{x^3} - 36{x^2} + 25,x \in R$, then write the elements contained by the two sets ${{\text{S}}_1}$ and ${{\text{S}}_2}$.
$
{\text{A}}{\text{. }}{{\text{S}}_1} = \left\{ { - 2,1} \right\};{{\text{S}}_2} = \left\{ 0 \right\} \\
{\text{B}}{\text{. }}{{\text{S}}_1} = \left\{ { - 2,0} \right\};{{\text{S}}_2} = \left\{ 1 \right\} \\
{\text{C}}{\text{. }}{{\text{S}}_1} = \left\{ { - 2} \right\};{{\text{S}}_2} = \left\{ {0,1} \right\} \\
{\text{D}}{\text{. }}{{\text{S}}_1} = \left\{ { - 1} \right\};{{\text{S}}_2} = \left\{ {0,2} \right\} \\
$
Last updated date: 25th Mar 2023
•
Total views: 306.3k
•
Views today: 6.84k
Answer
306.3k+ views
Hint- Here, we will proceed by differentiating the given function once and then putting \[f'\left( x \right) = 0\] in order to obtain the values of x where local maxima and local minima can occur according to the sign of \[f''\left( x \right)\].
Complete step-by-step answer:
The given function in x is $f\left( x \right) = 9{x^4} + 12{x^3} - 36{x^2} + 25{\text{ }} \to {\text{(1)}},x \in R$
It is also given that ${{\text{S}}_1}$ corresponds to the set of values of x where local minima occurs and ${{\text{S}}_2}$ corresponds to the set of values of x where local maxima occurs.
Let us differentiate the given function with respect to x, we get
\[
\Rightarrow \dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}} = \dfrac{d}{{dx}}\left[ {9{x^4} + 12{x^3} - 36{x^2} + 25} \right] \\
\Rightarrow f'\left( x \right) = \dfrac{d}{{dx}}\left[ {9{x^4}} \right] + \dfrac{d}{{dx}}\left[ {12{x^3}} \right] - \dfrac{d}{{dx}}\left[ {36{x^2}} \right] + \dfrac{d}{{dx}}\left[ {25} \right] \\
\Rightarrow f'\left( x \right) = 9\dfrac{d}{{dx}}\left[ {{x^4}} \right] + 12\dfrac{d}{{dx}}\left[ {{x^3}} \right] - 36\dfrac{d}{{dx}}\left[ {{x^2}} \right] + 0 \\
\Rightarrow f'\left( x \right) = 9\left( {4{x^3}} \right) + 12\left( {3{x^2}} \right) - 36\left( {2x} \right) \\
\Rightarrow f'\left( x \right) = 36{x^3} + 36{x^2} - 72x \\
\Rightarrow f'\left( x \right) = 36x\left( {{x^2} + x - 2} \right){\text{ }} \to {\text{(2)}} \\
\]
Since, we know that local maxima or local minima are the points where local maximum and local minimum values will be occurring. At local maxima and local minima, for any function f(x) the necessary condition is \[f'\left( x \right) = 0\].
By putting \[f'\left( x \right) = 0\] in equation (2), we get
\[
\Rightarrow 0 = 36x\left( {{x^2} + x - 2} \right) \\
\Rightarrow x\left( {{x^2} + x - 2} \right) = 0 \\
\Rightarrow x\left( {{x^2} - x + 2x - 2} \right) = 0 \\
\Rightarrow x\left[ {x\left( {x - 1} \right) + 2\left( {x - 1} \right)} \right] = 0 \\
\Rightarrow x\left( {x - 1} \right)\left( {x + 2} \right) = 0 \\
\]
From the above equation, we have
x=0 or $
x - 1 = 0 \\
\Rightarrow x = 1 \\
$ or $
x + 2 = 0 \\
\Rightarrow x = - 2 \\
$
So, the points where maxima or minima can occur are x=0,1,-2
Also we know that for any function f(x) to attain local maxima at a point x=a, \[f''\left( a \right) < 0\] and for this function f(x) to attain local minima at a point x=b, \[f''\left( b \right) > 0\].
By differentiating the equation (2) both sides with respect to x, we get
\[
\Rightarrow \dfrac{d}{{dx}}\left[ {f'\left( x \right)} \right] = \dfrac{d}{{dx}}\left[ {36x\left( {{x^2} + x - 2} \right)} \right] \\
\Rightarrow f''\left( x \right) = 36\dfrac{d}{{dx}}\left[ {{x^3} + {x^2} - 2x} \right] \\
\Rightarrow f''\left( x \right) = 36\left[ {\dfrac{d}{{dx}}\left[ {{x^3}} \right] + \dfrac{d}{{dx}}\left[ {{x^2}} \right] - 2\dfrac{{dx}}{{dx}}} \right] \\
\Rightarrow f''\left( x \right) = 36\left[ {3{x^2} + 2x - 2} \right]{\text{ }} \to {\text{(3)}} \\
\]
Put x=0 in equation (3), we get
\[ \Rightarrow f''\left( 0 \right) = 36\left[ {3{{\left( 0 \right)}^2} + 2\left( 0 \right) - 2} \right] = - 72\]
Clearly, \[f''\left( 0 \right) < 0\] so x=0 is a point of local maxima i.e., corresponding to point x=0, the given function f(x) has local maximum value. So, x=0 is a value in the set ${{\text{S}}_2}$.
Put x=1 in equation (3), we get
\[ \Rightarrow f''\left( 1 \right) = 36\left[ {3{{\left( 1 \right)}^2} + 2\left( 1 \right) - 2} \right] = 36\left[ 3 \right] = 108\]
Clearly, \[f''\left( 1 \right) > 0\] so x=1 is a point of local minima i.e., corresponding to point x=1, the given function f(x) has local minimum value. So, x=1 is a value in the set ${{\text{S}}_1}$.
Put x=-2 in equation (3), we get
\[ \Rightarrow f''\left( -2 \right) = 36\left[ {3{{\left( { - 2} \right)}^2} + 2\left( { - 2} \right) - 2} \right] = 36\left[ 6 \right] = 216\]
Clearly, \[f''\left( -2 \right) > 0\] so x=-2 is a point of local minima i.e., corresponding to point x=-2, the given function f(x) has local minimum value. So, x=-2 is a value in the set ${{\text{S}}_1}$.
So, set ${{\text{S}}_1} = \left\{ { - 2,1} \right\};{{\text{S}}_2} = \left\{ 0 \right\}$
Hence, option A is correct.
Note- In this particular problem, we have used second derivative test i.e., if \[f''\left( a \right) < 0\], then x=a is a point of local maxima and if \[f''\left( a \right) > 0\], then x=a is a point of local minima. Also, if \[f''\left( a \right) = 0\]occurs then x=a is a point of inflection. Here, the local maximum value of function f(x) is obtained by substituting x=0 in the function and the local minimum values of f(x) is obtained by substituting x=-2 and x=1 in the function.
Complete step-by-step answer:
The given function in x is $f\left( x \right) = 9{x^4} + 12{x^3} - 36{x^2} + 25{\text{ }} \to {\text{(1)}},x \in R$
It is also given that ${{\text{S}}_1}$ corresponds to the set of values of x where local minima occurs and ${{\text{S}}_2}$ corresponds to the set of values of x where local maxima occurs.
Let us differentiate the given function with respect to x, we get
\[
\Rightarrow \dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}} = \dfrac{d}{{dx}}\left[ {9{x^4} + 12{x^3} - 36{x^2} + 25} \right] \\
\Rightarrow f'\left( x \right) = \dfrac{d}{{dx}}\left[ {9{x^4}} \right] + \dfrac{d}{{dx}}\left[ {12{x^3}} \right] - \dfrac{d}{{dx}}\left[ {36{x^2}} \right] + \dfrac{d}{{dx}}\left[ {25} \right] \\
\Rightarrow f'\left( x \right) = 9\dfrac{d}{{dx}}\left[ {{x^4}} \right] + 12\dfrac{d}{{dx}}\left[ {{x^3}} \right] - 36\dfrac{d}{{dx}}\left[ {{x^2}} \right] + 0 \\
\Rightarrow f'\left( x \right) = 9\left( {4{x^3}} \right) + 12\left( {3{x^2}} \right) - 36\left( {2x} \right) \\
\Rightarrow f'\left( x \right) = 36{x^3} + 36{x^2} - 72x \\
\Rightarrow f'\left( x \right) = 36x\left( {{x^2} + x - 2} \right){\text{ }} \to {\text{(2)}} \\
\]
Since, we know that local maxima or local minima are the points where local maximum and local minimum values will be occurring. At local maxima and local minima, for any function f(x) the necessary condition is \[f'\left( x \right) = 0\].
By putting \[f'\left( x \right) = 0\] in equation (2), we get
\[
\Rightarrow 0 = 36x\left( {{x^2} + x - 2} \right) \\
\Rightarrow x\left( {{x^2} + x - 2} \right) = 0 \\
\Rightarrow x\left( {{x^2} - x + 2x - 2} \right) = 0 \\
\Rightarrow x\left[ {x\left( {x - 1} \right) + 2\left( {x - 1} \right)} \right] = 0 \\
\Rightarrow x\left( {x - 1} \right)\left( {x + 2} \right) = 0 \\
\]
From the above equation, we have
x=0 or $
x - 1 = 0 \\
\Rightarrow x = 1 \\
$ or $
x + 2 = 0 \\
\Rightarrow x = - 2 \\
$
So, the points where maxima or minima can occur are x=0,1,-2
Also we know that for any function f(x) to attain local maxima at a point x=a, \[f''\left( a \right) < 0\] and for this function f(x) to attain local minima at a point x=b, \[f''\left( b \right) > 0\].
By differentiating the equation (2) both sides with respect to x, we get
\[
\Rightarrow \dfrac{d}{{dx}}\left[ {f'\left( x \right)} \right] = \dfrac{d}{{dx}}\left[ {36x\left( {{x^2} + x - 2} \right)} \right] \\
\Rightarrow f''\left( x \right) = 36\dfrac{d}{{dx}}\left[ {{x^3} + {x^2} - 2x} \right] \\
\Rightarrow f''\left( x \right) = 36\left[ {\dfrac{d}{{dx}}\left[ {{x^3}} \right] + \dfrac{d}{{dx}}\left[ {{x^2}} \right] - 2\dfrac{{dx}}{{dx}}} \right] \\
\Rightarrow f''\left( x \right) = 36\left[ {3{x^2} + 2x - 2} \right]{\text{ }} \to {\text{(3)}} \\
\]
Put x=0 in equation (3), we get
\[ \Rightarrow f''\left( 0 \right) = 36\left[ {3{{\left( 0 \right)}^2} + 2\left( 0 \right) - 2} \right] = - 72\]
Clearly, \[f''\left( 0 \right) < 0\] so x=0 is a point of local maxima i.e., corresponding to point x=0, the given function f(x) has local maximum value. So, x=0 is a value in the set ${{\text{S}}_2}$.
Put x=1 in equation (3), we get
\[ \Rightarrow f''\left( 1 \right) = 36\left[ {3{{\left( 1 \right)}^2} + 2\left( 1 \right) - 2} \right] = 36\left[ 3 \right] = 108\]
Clearly, \[f''\left( 1 \right) > 0\] so x=1 is a point of local minima i.e., corresponding to point x=1, the given function f(x) has local minimum value. So, x=1 is a value in the set ${{\text{S}}_1}$.
Put x=-2 in equation (3), we get
\[ \Rightarrow f''\left( -2 \right) = 36\left[ {3{{\left( { - 2} \right)}^2} + 2\left( { - 2} \right) - 2} \right] = 36\left[ 6 \right] = 216\]
Clearly, \[f''\left( -2 \right) > 0\] so x=-2 is a point of local minima i.e., corresponding to point x=-2, the given function f(x) has local minimum value. So, x=-2 is a value in the set ${{\text{S}}_1}$.
So, set ${{\text{S}}_1} = \left\{ { - 2,1} \right\};{{\text{S}}_2} = \left\{ 0 \right\}$
Hence, option A is correct.
Note- In this particular problem, we have used second derivative test i.e., if \[f''\left( a \right) < 0\], then x=a is a point of local maxima and if \[f''\left( a \right) > 0\], then x=a is a point of local minima. Also, if \[f''\left( a \right) = 0\]occurs then x=a is a point of inflection. Here, the local maximum value of function f(x) is obtained by substituting x=0 in the function and the local minimum values of f(x) is obtained by substituting x=-2 and x=1 in the function.
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
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

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

Write a letter to the Principal of your school to plead class 10 english CBSE
