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Question: The value of \(c\) from the lagrange’s mean value theorem for which \[\] in \(\left[ {1,5} \right]\)...

The value of cc from the lagrange’s mean value theorem for which $$$$ in [1,5]\left[ {1,5} \right] is
A). 5
B). 1
C). 15\sqrt {15}
D). None of these

Explanation

Solution

Hint- Here, we will be finding the required unknown by equating the value of f′(c)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(x)=25−x2f\left( x \right) = \sqrt {25 - {x^2}}
This function is continuous for all the values of x∈[1,5]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 xx, we get
f′(x)=d[(25−x2)12]dx=(12)(25−x2)12−1(−2x)=−x(25−x2)−12 ⇒f′(x)=−x(25−x2) →(1)  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∈[1,5]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∈[a,b]x \in \left[ {a,b} \right], there exists c∈[1,5]c \in \left[ {1,5} \right] such that f′(c)=f(b)−f(a)b−af'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}} .
∴f′(c)=f(5)−f(1)5−1=25−52−25−124=0−244=−62\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′(c)=−c(25−c2)⇒−62=−c(25−c2)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

⇒(−62)2=(−c(25−c2))2⇒64=c2(25−c2)⇒32=1(25c2−1)⇒(25c2−1)=23 ⇒(25c2)=23+1⇒(25c2)=53⇒(c225)=35⇒c2=3×255=15 ⇒c=15 ∈[1,5]   \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 cc is 15 \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 cc obtained should lie in the interval of xx.