Question

The slope of the curve $y = e^x \cos x$, $x \in (-\pi, \pi)$ is maximum at:

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

Hint:

To find the maximum of any quantity (like slope), always treat that quantity as a new function and find where its derivative is zero.

Answer 1

The Correct Option is D

Concept: The slope of the curve is given by $f'(x)$. To find where the slope is maximum, we need to maximize $f'(x)$, which means finding where its derivative (the second derivative of the original function, $f''(x)$) is zero.

Step 1: Find the slope function $f'(x)$.
Using the product rule on $y = e^x \cos x$: \[ f'(x) = e^x \cos x - e^x \sin x = e^x(\cos x - \sin x) \]

Step 2: Find the derivative of the slope $f''(x)$.
Differentiate $f'(x)$ to find critical points for the slope: \[ f''(x) = e^x(\cos x - \sin x) + e^x(-\sin x - \cos x) \] \[ f''(x) = e^x(\cos x - \sin x - \sin x - \cos x) = -2e^x \sin x \]

Step 3: Solve for $f''(x) = 0$.
\[ -2e^x \sin x = 0 \quad \Rightarrow \quad \sin x = 0 \] Within the interval $(-\pi, \pi)$, the only solution is $x = 0$.

Step 4: Check for maximum slope.
At $x = 0$, $f'(0) = e^0(\cos 0 - \sin 0) = 1(1 - 0) = 1$. Comparing with endpoints or other values shows this is the maximum slope in the vicinity of the origin.