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Applications of Derivatives — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsApplications of Derivatives4 Marks
Question
Verify Rolle's theorem for the function $f(x)=e^x(\sin x-\cos x)$ on $\left[\frac{\pi}{4}, \frac{5 \pi}{4}\right]$.
✓

Answer

Given that,
$
f(x)=e^x(\sin x-\cos x)
$
We know that $e^x, \sin x$ and $\cos x$ are continuous and differentiable on their domains. Therefore $f(x)$ is continuous and differentiable on $\left[\frac{\pi}{4}, \frac{5 \pi}{4}\right]$ and $\left(\frac{\pi}{4}, \frac{5 \pi}{4}\right)$ respectively.
Let $a=\frac{\pi}{4}$, and $b=\frac{5 \pi}{4}$
For $x=a=\frac{\pi}{4}$ from (I) we get,
$f(a)=f\left(\frac{\pi}{4}\right)=e^{\frac{\pi}{4}}\left[\sin \left(\frac{\pi}{4}\right)-\cos \left(\frac{\pi}{4}\right)\right]=e^{\frac{\pi}{4}}\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right)=0$
For $x=b=\left(\frac{5 \pi}{4}\right)$ from (I) we get,
$f(a)=f\left(\frac{5 \pi}{4}\right)=e^{\frac{5 \pi}{4}}\left[\sin \left(\frac{5 \pi}{4}\right)-\cos \left(\frac{5 \pi}{4}\right)\right]=e^{\frac{5 \pi}{4}}\left(-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)=0$
$\therefore \quad f(a)=f(b) \quad$ i.e. $f\left(\frac{\pi}{4}\right)=f\left(\frac{5 \pi}{4}\right)$.
All the conditions of Rolle's theorem are satisfied.
To get the value of $c$, we should have $f^{\prime}(c)=0$ for some $c \in\left[\frac{\pi}{4}, \frac{5 \pi}{4}\right]$
Differentiate (I) w.r.t. $x$.
$f^{\prime}(x)=e^x(\cos x+\sin x)+(\sin x-\cos x) e^x=2 e^x \sin x$
Now, for $x=c, f^{\prime}(c)=0 \Rightarrow 2 e^c \sin c=0$. As $e^c \neq 0$ for any $c \in R$ $\sin c=0 \Rightarrow c=0, \pm \pi, \pm 2 \pi, \pm 3 \pi, \ldots$
It is clearly seen that $\pi \in\left[\frac{\pi}{4}, \frac{5 \pi}{4}\right] \therefore c=\pi$
Thus Rolle's theorem is verified.

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