Question

$f\left(x\right)=e^x\cos x$ in $[0,2\pi ]$ has maximum slope at

• A. $\frac{3\pi }{4}$
• B. $0$
• C. $\pi$
• D. $2\pi$

Answer 1

The correct option is D $2\pi$

$f\left(x\right)=e^x\cos x$

$\therefore$ $f^{\prime }\left(x\right)=f\left(x\right)=e^x$($\cos x-sinx)$

$f^{\prime \prime }\left(x\right)=-2e^{x}sinx$

$f^{\prime \prime \prime }\left(x\right)=-2e^x(\cos x+\sin x)$

lf $f\left(x\right)$ is maxlmum

${\varphi }^{\prime }\left(x\right)=f^{\prime \prime }\left(x\right)=0$ and ${\varphi }^{\prime \prime }\left(x\right)=f^{\prime \prime \prime }\left(x\right)<0$

${\varphi }^{\prime }\left(x\right)=0\Rightarrow sinx=0\Rightarrow x=0,\pi$ or 2 $\pi$

${\varphi }^{\prime \prime }\left(x\right)<0$ when $x=0$ or 2 $\pi$

But $\varphi \left(0\right)=f\left(0\right)=1$ and$\varphi ($ 2 $\pi )=e^{2\pi }>1$

Hence the maximum slope is at $x=2\pi$