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 \[\dfrac{\pi }{4} \leqslant \;x \leqslant \dfrac{\pi }{2}\]

Hint: 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
\[\mathop {\lim }\limits_{h \to 0} {\text{ }}\dfrac{{\left[ {f\left( {a - h} \right) - f\left( a \right)} \right]}}{{\left( { - h} \right)}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left[ {f\left( {a + h} \right) - f\left( a \right)} \right]}}{h}\].