# 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\} \\

$

Answer

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362.1k+ 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.

Last updated date: 29th Sep 2023

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