Question:

The point in the interval [0, 2π], where $f(x)=eˣsin~x$ has maximum slope, is

Show Hint

The point in the interval [0, 2π], where $f(x)=e
Updated On: Apr 15, 2026
  • $\pi/4$
  • $\pi/2$
  • $\pi$
  • $3\pi/2$
Show Solution
collegedunia
Verified By Collegedunia

The Correct Option is B

Solution and Explanation

Step 1: Concept
Slope is given by $f'(x)$. To maximize the slope, find the critical points of $f'(x)$ by setting $f''(x) = 0$.
Step 2: Analysis
$f'(x) = e^x(sin~x + cos~x)$. $f''(x) = e^x(sin~x + cos~x + cos~x - sin~x) = 2e^x cos~x$.
Step 3: Evaluation
Set $f''(x) = 0 \Rightarrow 2e^x cos~x = 0 \Rightarrow cos~x = 0 \Rightarrow x = \pi/2$ in the given interval.
Step 4: Conclusion
Check $f'''(x)$ at $x = \pi/2$: $f'''(x) = 2e^x(cos~x - sin~x)$. At $x = \pi/2$, $f'''(\pi/2) = -2e^{\pi/2}<0$. Thus, slope is maximum at $x = \pi/2$.
Final Answer: (b)
Was this answer helpful?
0
0