Question

For the function, $f\left(x\right)=e^{x}cosx,x\in \left[0,2\pi \right]$ , the slope of the tangent at any point on the curve of the function is minimum at

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

Answer 1

Answer: (A)

$f\left(x\right)=e^{x}cosx$
$f^{{}^{\prime }}\left(x\right)=-e^{x}sinx+e^{x}cosx$
$f^{{}^{\prime }}\left(x\right)=e^x\left(cosx-sinx\right)$
Let $g\left(x\right)=e^x\left[cosx-sinx\right]$ is the slope of the tangent to the curve, then,
$g^{{}^{\prime }}\left(x\right)=e^x\left[-sinx-cosx\right]+e^x\left[cosx-sinx\right]$
$=e^x\left[-sinx-cosx+cosx-sinx\right]$
$g^{{}^{\prime }}\left(x\right)=-2e^{x}sinx\Rightarrow x=0,\pi ,2\pi$

So it is minima at $x=\pi$