For the function f(x) = e^{cos x}, Rolle's th | vedclass

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(A) Rolle's theorem states that for a function $f(x)$ to be applicable on an interval $[a, b]$, it must satisfy three conditions: $1$. $f(x)$ is conti

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applicable when $\frac{\pi}{2} \leq x \leq \frac{3\pi}{2}$

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(A) Rolle's theorem states that for a function $f(x)$ to be applicable on an interval $[a, b]$, it must satisfy three conditions: $1$. $f(x)$ is continuous on $[a, b]$. $2$. $f(x)$ is differentiable on $(a, b)$. $3$. $f(a) = f(b)$. For $f(x) = e^{\cos x}$, the function is continuous and differentiable everywhere. We check the condition $f(a) = f(b)$ for the given options: For option $A$: $f(\frac{\pi}{2}) = e^{\cos(\pi/2)} = e^0 = 1$ and $f(\frac{3\pi}{2}) = e^{\cos(3\pi/2)} = e^0 = 1$. Since $f(\frac{\pi}{2}) = f(\frac{3\pi}{2})$, Rolle's theorem is applicable on the interval $[\frac{\pi}{2}, \frac{3\pi}{2}]$.

For the function $f(x) = e^{\cos x}$, Rolle's theorem is

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