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
Answer
Verified366k+ 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 $x$.
Last updated date: 25th Sep 2023
•
Total views: 366k
•
Views today: 4.66k
Recently Updated Pages
