# The value of $c$ from the lagrange’s mean value theorem for which \[\] in $\left[ {1,5} \right]$ is

A). 5

B). 1

C). $\sqrt {15} $

D). None of these

Last updated date: 19th Mar 2023

•

Total views: 310.5k

•

Views today: 6.89k

Answer

Verified

310.5k+ views

Hint- Here, we will be finding the required unknown by equating the value of $f'\left( c \right)$ obtained through differentiating the given function with that obtained with the help of lagrange’s mean value theorem.

Given function is $f\left( x \right) = \sqrt {25 - {x^2}} $

This function is continuous for all the values of $x \in \left[ {1,5} \right]$because the above function is defined for all of these values.

Now let us differentiate the given function with respect to $x$, we get

$

f'\left( x \right) = \dfrac{{d\left[ {{{\left( {25 - {x^2}} \right)}^{\dfrac{1}{2}}}} \right]}}{{dx}} = \left( {\dfrac{1}{2}} \right){\left( {25 - {x^2}} \right)^{\dfrac{1}{2} - 1}}\left( { - 2x} \right) = - x{\left( {25 - {x^2}} \right)^{ - \dfrac{1}{2}}} \\

\Rightarrow f'\left( x \right) = - \dfrac{x}{{\sqrt {\left( {25 - {x^2}} \right)} }}{\text{ }} \to {\text{(1)}} \\

$

Clearly, the above function is also differentiable for all the values of $x \in \left[ {1,5} \right]$because the above function is defined for set of all these values.

So, according to lagrange’s mean value theorem if a function is continuous as well as differentiable for $x \in \left[ {a,b} \right]$, there exists $c \in \left[ {1,5} \right]$ such that $f'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}}$ .

$\therefore f'\left( c \right) = \dfrac{{f\left( 5 \right) - f\left( 1 \right)}}{{5 - 1}} = \dfrac{{\sqrt {25 - {5^2}} - \sqrt {25 - {1^2}} }}{4} = \dfrac{{0 - \sqrt {24} }}{4} = - \dfrac{{\sqrt 6 }}{2}$

By using equation (1), we have

$f'\left( c \right) = - \dfrac{c}{{\sqrt {\left( {25 - {c^2}} \right)} }} \Rightarrow - \dfrac{{\sqrt 6 }}{2} = - \dfrac{c}{{\sqrt {\left( {25 - {c^2}} \right)} }}$

Squaring both sides, the above equation becomes

\[

\Rightarrow {\left( { - \dfrac{{\sqrt 6 }}{2}} \right)^2} = {\left( { - \dfrac{c}{{\sqrt {\left( {25 - {c^2}} \right)} }}} \right)^2} \Rightarrow \dfrac{6}{4} = \dfrac{{{c^2}}}{{\left( {25 - {c^2}} \right)}} \Rightarrow \dfrac{3}{2} = \dfrac{1}{{\left( {\dfrac{{25}}{{{c^2}}} - 1} \right)}} \Rightarrow \left( {\dfrac{{25}}{{{c^2}}} - 1} \right) = \dfrac{2}{3} \\

\Rightarrow \left( {\dfrac{{25}}{{{c^2}}}} \right) = \dfrac{2}{3} + 1 \Rightarrow \left( {\dfrac{{25}}{{{c^2}}}} \right) = \dfrac{5}{3} \Rightarrow \left( {\dfrac{{{c^2}}}{{25}}} \right) = \dfrac{3}{5} \Rightarrow {c^2} = \dfrac{{3 \times 25}}{5} = 15 \\

\Rightarrow c = \sqrt {15} {\text{ }} \in \left[ {1,5} \right] \\

\\

\]

Therefore, the value of $c$ is \[\sqrt {15} {\text{ }}\].

Hence, option C is correct.

Note- In these types of problems where lagrange’s mean value theorem is used, at the end it is ensured that the value of $c$ obtained should lie in the interval of $x$.

Given function is $f\left( x \right) = \sqrt {25 - {x^2}} $

This function is continuous for all the values of $x \in \left[ {1,5} \right]$because the above function is defined for all of these values.

Now let us differentiate the given function with respect to $x$, we get

$

f'\left( x \right) = \dfrac{{d\left[ {{{\left( {25 - {x^2}} \right)}^{\dfrac{1}{2}}}} \right]}}{{dx}} = \left( {\dfrac{1}{2}} \right){\left( {25 - {x^2}} \right)^{\dfrac{1}{2} - 1}}\left( { - 2x} \right) = - x{\left( {25 - {x^2}} \right)^{ - \dfrac{1}{2}}} \\

\Rightarrow f'\left( x \right) = - \dfrac{x}{{\sqrt {\left( {25 - {x^2}} \right)} }}{\text{ }} \to {\text{(1)}} \\

$

Clearly, the above function is also differentiable for all the values of $x \in \left[ {1,5} \right]$because the above function is defined for set of all these values.

So, according to lagrange’s mean value theorem if a function is continuous as well as differentiable for $x \in \left[ {a,b} \right]$, there exists $c \in \left[ {1,5} \right]$ such that $f'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}}$ .

$\therefore f'\left( c \right) = \dfrac{{f\left( 5 \right) - f\left( 1 \right)}}{{5 - 1}} = \dfrac{{\sqrt {25 - {5^2}} - \sqrt {25 - {1^2}} }}{4} = \dfrac{{0 - \sqrt {24} }}{4} = - \dfrac{{\sqrt 6 }}{2}$

By using equation (1), we have

$f'\left( c \right) = - \dfrac{c}{{\sqrt {\left( {25 - {c^2}} \right)} }} \Rightarrow - \dfrac{{\sqrt 6 }}{2} = - \dfrac{c}{{\sqrt {\left( {25 - {c^2}} \right)} }}$

Squaring both sides, the above equation becomes

\[

\Rightarrow {\left( { - \dfrac{{\sqrt 6 }}{2}} \right)^2} = {\left( { - \dfrac{c}{{\sqrt {\left( {25 - {c^2}} \right)} }}} \right)^2} \Rightarrow \dfrac{6}{4} = \dfrac{{{c^2}}}{{\left( {25 - {c^2}} \right)}} \Rightarrow \dfrac{3}{2} = \dfrac{1}{{\left( {\dfrac{{25}}{{{c^2}}} - 1} \right)}} \Rightarrow \left( {\dfrac{{25}}{{{c^2}}} - 1} \right) = \dfrac{2}{3} \\

\Rightarrow \left( {\dfrac{{25}}{{{c^2}}}} \right) = \dfrac{2}{3} + 1 \Rightarrow \left( {\dfrac{{25}}{{{c^2}}}} \right) = \dfrac{5}{3} \Rightarrow \left( {\dfrac{{{c^2}}}{{25}}} \right) = \dfrac{3}{5} \Rightarrow {c^2} = \dfrac{{3 \times 25}}{5} = 15 \\

\Rightarrow c = \sqrt {15} {\text{ }} \in \left[ {1,5} \right] \\

\\

\]

Therefore, the value of $c$ is \[\sqrt {15} {\text{ }}\].

Hence, option C is correct.

Note- In these types of problems where lagrange’s mean value theorem is used, at the end it is ensured that the value of $c$ obtained should lie in the interval of $x$.

Recently Updated Pages

If ab and c are unit vectors then left ab2 right+bc2+ca2 class 12 maths JEE_Main

A rod AB of length 4 units moves horizontally when class 11 maths JEE_Main

Evaluate the value of intlimits0pi cos 3xdx A 0 B 1 class 12 maths JEE_Main

Which of the following is correct 1 nleft S cup T right class 10 maths JEE_Main

What is the area of the triangle with vertices Aleft class 11 maths JEE_Main

KCN reacts readily to give a cyanide with A Ethyl alcohol class 12 chemistry JEE_Main

Trending doubts

What was the capital of Kanishka A Mathura B Purushapura class 7 social studies CBSE

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

Tropic of Cancer passes through how many states? Name them.

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

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