Question

For the function, $f\left(x\right)=e^{x}cos x, \, x\in \left[0 , \, 2 \pi \right]$ , 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)

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

So it is minima at $x=\pi $