Question

How do you determine the values of c that satisfy the mean value theorem on the interval $\left[ {\dfrac{\pi }{2},\dfrac{{3\pi }}{2}} \right]$ for $f\left( x \right) = \sin \left( {\dfrac{x}{2}} \right)$ ?

Answer 1

Hint : In order to solve the above question, we will use the mean value theorem which states that if a function is continuous on the closed interval $\left[ {a,b} \right]$ and differentiable in the open interval $\left( {a,b} \right)$ such that $f\left( a \right) = f\left( b \right)$ then $f'\left( x \right) = 0$ for some $c$ in $\left[ {a,b} \right]$ . Using this concept, we will solve the above sum.
Formula used:
To solve the above question, we will be using the mean value theorem.
$f'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}}$ .

Complete step by step solution:
We are given: $f\left( x \right) = \sin \left( {\dfrac{x}{2}} \right)$ is continuous in $\left[ {\dfrac{\pi }{2},\dfrac{{3\pi }}{2}} \right]$ .
Also, $f\left( x \right)$ is differential in $\left( {\dfrac{\pi }{2},\dfrac{{3\pi }}{2}} \right)$ .
So, there must exist $c$ on $\left( {\dfrac{\pi }{2},\dfrac{{3\pi }}{2}} \right)$ such that
$f'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}}$ .
On equating the values of $a$ and $b$ , we get,
$\Rightarrow f'\left( c \right) = \dfrac{{f\left( {\dfrac{{3\pi }}{2}} \right) - f\left( {\dfrac{\pi }{2}} \right)}}{{\dfrac{{3\pi }}{2} - \dfrac{\pi }{2}}}$ .
Now, $f\left( {\dfrac{{3\pi }}{2}} \right) = \sin \left( {\dfrac{{3\pi }}{4}} \right)$
$= \dfrac{{\sqrt 2 }}{2}$ .
And,
$f\left( {\dfrac{\pi }{2}} \right) = \sin \left( {\dfrac{\pi }{4}} \right)$
$= \dfrac{{\sqrt 2 }}{2}$ .
From this, we get that,

f'\left( c \right) = \dfrac{{f\left( {\dfrac{{3\pi }}{2}} \right) - f\left( {\dfrac{\pi }{2}} \right)}}{\pi } \\\ \Rightarrow f'\left( c \right) = \dfrac{{\dfrac{{\sqrt 2 }}{2} - \dfrac{{\sqrt 2 }}{2}}}{\pi } \\\ $$ , Now. $$f'\left( c \right) = 0$$ But, $$f'\left( x \right) = \left( {\dfrac{1}{2}} \right)\cos \left( {\dfrac{x}{2}} \right)$$ , $$f'\left( c \right) = \left( {\dfrac{1}{2}} \right)\cos \left( {\dfrac{c}{2}} \right)$$ $$ = 0$$ From this we get, $$ \Rightarrow \cos \left( {\dfrac{c}{2}} \right) = 0$$ , This can be written as $$\dfrac{c}{2} = \dfrac{\pi }{2}$$ , as $${\cos ^{ - 1}}0 = \dfrac{\pi }{2}$$ . Therefore, we get, $$c = \pi $$ . So, from this we found the value of $$c$$ , that is, $$c = \pi $$ . **So, the correct answer is $$c = \pi $$.** **Note** : While solving sums similar to the one given above, you need to remember Rolle’s mean value theorem in which he stated that if a function is continuous on the closed interval $$\left[ {a,b} \right] $$ and differentiable in the open interval $$\left( {a,b} \right)$$ such that $$f\left( a \right) = f\left( b \right)$$ then $$f'\left( x \right) = 0$$ for some $$c$$ in $$\left[ {a,b} \right] $$ . Also, you need to remember the formula to calculate $$f'\left( c \right)$$ , which is, $$f'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}}$$ . In questions similar to this, you also need to remember the values of trigonometric functions, for example, in this question we used the value of $${\cos ^{ - 1}}0$$ which is equal to $$\dfrac{\pi }{2}$$ .