Question

The value of C in Rolles's theorem for the function \(f(x)=e^xsinx,x\epsilon[0,\pi]\),is :

• A. \(\frac{\pi}{6}\)
• B. \(\frac{\pi}{4}\)
• C. \(\frac{\pi}{2}\)
• D. \(\frac{3\pi}{4}\)

Answer 1

The Correct Option is D

Given the function \(f(x) = e^x \sin x\) on the interval \([0, \pi]\), we are to find the value of \(c\) that satisfies Rolles's Theorem.

Rolles's Theorem states that if a function \(f\) is continuous on \([a, b]\), differentiable on \((a, b)\), and \(f(a) = f(b)\), then there exists at least one \(c \in (a, b)\) such that \(f'(c) = 0\).

First, we verify the conditions of the theorem:

• Continuity: \(f(x) = e^x \sin x\) is continuous on \([0, \pi]\) since both \(e^x\) and \(\sin x\) are continuous.
• Differentiability: \(f(x)\) is differentiable on \((0, \pi)\) as \(e^x\) and \(\sin x\) are differentiable.
• Equal values: Calculate \(f(0)\) and \(f(\pi)\):

  \(f(0) = e^0 \sin 0 = 1 \times 0 = 0\)
  \(f(\pi) = e^\pi \sin \pi = e^\pi \times 0 = 0\)

  Both \(f(0)\) and \(f(\pi)\) are 0, satisfying the condition \(f(a) = f(b)\).

Now, find \(f'(x)\):

\(f'(x) = \frac{d}{dx}(e^x \sin x) = e^x \sin x + e^x \cos x\) using the product rule.

Set \(f'(x) = 0\):
\(e^x (\sin x + \cos x) = 0\)

Since \(e^x \neq 0\), we solve:

\(\sin x + \cos x = 0\)

Or equivalently:

\(\tan x = -1\)

The general solutions for \(\tan x = -1\) are \(x = \frac{3\pi}{4} + n\pi\).

In the interval \((0, \pi)\), the solution is:

\(x = \frac{3\pi}{4}\)

Thus, the value of \(c\) in the interval \((0, \pi)\) that satisfies Rolles's Theorem is \(\frac{3\pi}{4}\).