Question

For the function $f(x) = {e^{cos x}}$, Rolle’s theorem is
$1)$applicable when $\dfrac{\pi }{2} \leqslant x \leqslant \dfrac{{3\pi }}{2}$
$2)$applicable when $0 \leqslant x \leqslant \dfrac{\pi }{2}$
$3)$applicable when $0 \leqslant x \leqslant \pi$
$4)$applicable when $\dfrac{\pi }{4} \leqslant \;x \leqslant \dfrac{\pi }{2}$

Answer 1

We have to find the interval for which the function $f(x) = {e^{cos x}}$ follows Rolle’s law . We solve this question using the concept of Rolle’s theorem . The conditions which a function should satisfy to follow Rolle’s Theorem . We should also know about the concept of continuity and differentiability of a function.

Complete step-by-step solution:
Given : $f(x) = {e^{cos x}}$
We know that the conditions of a Rolle’s theorem:
If $f\left( x \right)$ is a real valued function defined on closed interval $\left[ {a,b} \right]$ such that
\left( 1 \right)$$$$f\left( x \right) is continuous in $\left[ {a,b} \right]$
\left( 2 \right)$$$$f\left( x \right) is differentiable in $(a,b)$
\left( 3 \right)$$$$f\left( a \right) = f\left( b \right)
Then there must must exist one real number such that $c \in \left( {a,b} \right)$ such that $f'\left( c \right) = 0.$
So , if a function follows these conditions then the function follows Rolle’s theorem .
Now ,
We know that every exponential function is continuous and differentiable for all the values of $x$.
So ,
The given function follows the first two conditions of Rolle’s theorem.
Now we have to check the third condition i.e $.f\left( a \right)$ should be equal to $f\left( b \right)$.
So ,
${e^{(cosa)}} = {e^{(cosb)}}$ Taking log both sides , we get
$cosa = cosb$——-(1)
So , the interval which follows Rolle’s Theorem should satisfy the condition in equation (1)
Now let us take all the limits as given in the question .
Let $x \in [\dfrac{\pi }{2},\dfrac{{3\pi }}{2}]$
So, on comparing the interval values , we get $a = \dfrac{\pi }{2}$ and $b = \dfrac{{3\pi }}{2}$
Now , using the values of cos function
$cos\left( {\dfrac{\pi }{2}} \right) = 0$
$cos\left( {\dfrac{{3\pi }}{2}} \right) = 0$
Hence ,
$cos\left( {\dfrac{\pi }{2}} \right) =$ $cos\left( {\dfrac{{3\pi }}{2}} \right)$
Thus , this satisfies the third condition for Rolle’s theorem .
Hence the interval for which Rolle’s theorem is applicable is $[\dfrac{\pi }{2}{\text{ ,}}\dfrac{{3\pi }}{2}]$ .
Thus , the correct option is $\left( 1 \right)$.

Note: A function is continuous at $x = c$ if the function is defined at $x = c$ and if the value of the function at $x = c$ equals the limit of the function at $x = c$ . If f is not continuous at $c$ , we say f is discontinuous at $c$ and $c$ is called a point of discontinuity of f .
In simple terms a function is said to be continuous at a point $x = c$ if the
$\mathop {\lim }\limits_{x \to {c^ - }} f\left( x \right) = {\text{ }}\mathop {\lim }\limits_{x \to {c^ + }} f\left( x \right) = f\left( c \right)$
A function f(x) is said to be differentiable at $x = c$ if
$\mathop {\lim }\limits_{h \to 0} {\text{ }}\dfrac{{\left[ {f\left( {a - h} \right) - f\left( a \right)} \right]}}{{\left( { - h} \right)}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left[ {f\left( {a + h} \right) - f\left( a \right)} \right]}}{h}$.