Question

If Rolle’s theorem for $f\left(x\right)=e^x\left(sinx-cosx\right)$ is verified on $[\pi /4$, $5\pi /4]$, then the value of $c$ is

• A. $\pi /3$
• B. $\pi /2$
• C. $3\pi /4$
• D. $\pi$

Answer 1

Answer: (D)

Given, $f(x)=e^x(sinx=cosx)$
on differentiating both sides w.r.t. , $x_1$ we get
$f^{\prime }(x)=e^x\frac{d}{dx}(sinx-cosx)+(sinx-cosx)\frac{d}{dx}\left(e^x\right)$
[by using product rule of derivative]
$=e^x(cosx+sinx)+(sinx-cosx)e^x$
$=2e^{x}sinx$
We know that, if Rolle's theorem is verified,
then their exist $c\in \left(\frac{\pi }{4},\frac{5\pi }{4}\right)$, such that $f^{\prime }(c)=0$
$\therefore 2e^{c}sinc=0$
$\Rightarrow sinc=0$
$\Rightarrow c=\frac{\pi }{2}\in \left(\frac{\pi }{4},\frac{5\pi }{4}\right)$