Question

Let f(x)=cosx for \(0\leq x \leq\frac{\pi}{3}\). Then the value of c which satisfies the conclusion of the Mean Value Theorem for the function f on \([0,\frac{\pi}{3}]\) is equal to

• A. \(sin^{-1}(\frac{3}{2\pi})\)
• B. \(sin^{-1}(\frac{1}{3\pi})\)
• C. \(sin^{-1}(\frac{\pi}{12})\)
• D. \(sin^{-1}(\frac{1}{6\pi})\)
• E. \(sin^{-1}(\frac{\pi}{4})\)

Answer 1

The Correct Option is A

The Mean Value Theorem (MVT) states that if a function \( f(x) \) is continuous on the closed interval \( [a, b] \) and differentiable on the open interval \( (a, b) \), then there exists at least one point \( c \in (a, b) \) such that: \[ f'(c) = \frac{f(b) - f(a)}{b - a} \] Here, \( f(x) = \cos(x) \) on the interval \( [0, \frac{\pi}{3}] \). Therefore, we need to compute the derivative of \( f(x) \) and apply the MVT.
Step 1: Compute \( f(0) \) and \( f\left( \frac{\pi}{3} \right) \) \[ f(0) = \cos(0) = 1 \] \[ f\left( \frac{\pi}{3} \right) = \cos\left( \frac{\pi}{3} \right) = \frac{1}{2} \]
Step 2: Apply the Mean Value Theorem We apply the MVT to the given interval \( [0, \frac{\pi}{3}] \): \[ f'(c) = \frac{f\left( \frac{\pi}{3} \right) - f(0)}{\frac{\pi}{3} - 0} \] Substitute the values of \( f(0) \) and \( f\left( \frac{\pi}{3} \right) \): \[ f'(c) = \frac{\frac{1}{2} - 1}{\frac{\pi}{3}} = \frac{-\frac{1}{2}}{\frac{\pi}{3}} = -\frac{3}{2\pi} \]
Step 3: Differentiate \( f(x) = \cos(x) \) The derivative of \( f(x) = \cos(x) \) is: \[ f'(x) = -\sin(x) \] So, we have: \[ -\sin(c) = -\frac{3}{2\pi} \] \[ \sin(c) = \frac{3}{2\pi} \]
Step 4: Solve for \( c \) Finally, solving for \( c \): \[ c = \sin^{-1} \left( \frac{3}{2\pi} \right) \]

The correct option is (A) : \(sin^{-1}(\frac{3}{2\pi})\)

Answer 2

We are given the function \(f(x) = \cos x\) for \(0 \leq x \leq \frac{\pi}{3}\). We want to find the value of \(c\) that satisfies the conclusion of the Mean Value Theorem (MVT) for the function \(f\) on the interval \([0, \frac{\pi}{3}]\).

The Mean Value Theorem states that if a function \(f(x)\) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one point \(c\) in the interval \((a, b)\) such that \[f'(c) = \frac{f(b) - f(a)}{b - a}\]

In our case, \(f(x) = \cos x\), \(a = 0\), and \(b = \frac{\pi}{3}\). The derivative of \(f(x)\) is: \[f'(x) = -\sin x\]

Now, we calculate \(f(a)\) and \(f(b)\): \[f(0) = \cos(0) = 1\] \[f\left(\frac{\pi}{3}\right) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\]

The right side of the MVT equation is: \[\frac{f(b) - f(a)}{b - a} = \frac{\frac{1}{2} - 1}{\frac{\pi}{3} - 0} = \frac{-\frac{1}{2}}{\frac{\pi}{3}} = -\frac{3}{2\pi}\]

According to the MVT, there exists a \(c\) in \((0, \frac{\pi}{3})\) such that \(f'(c) = -\frac{3}{2\pi}\). So, \[-\sin c = -\frac{3}{2\pi}\] \[\sin c = \frac{3}{2\pi}\]

Therefore, \(c = \sin^{-1}\left(\frac{3}{2\pi}\right)\).