Question

Let f(x)=πcosx+x².The value of c∈(0,π) where f attains its local maximum/minumum is

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

Answer 1

The Correct Option is B

Step 1: Find the derivative of \( f(x) \): \[ f(x) = \pi \cos x + x^2 \] \[ f'(x) = -\pi \sin x + 2x \]

Step 2: Find critical points by setting \( f'(x) = 0 \): \[ -\pi \sin x + 2x = 0 \] \[ \pi \sin x = 2x \] \[ \sin x = \frac{2x}{\pi} \]

Step 3: Test the given options in \( (0, \pi) \):

Option (A) \( x = \frac{\pi}{3} \): \[ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \approx 0.866 \] \[ \frac{2x}{\pi} = \frac{2(\pi/3)}{\pi} = \frac{2}{3} \approx 0.666 \] Not equal

Option (B) \( x = \frac{\pi}{2} \): \[ \sin\left(\frac{\pi}{2}\right) = 1 \] \[ \frac{2x}{\pi} = \frac{2(\pi/2)}{\pi} = 1 \] This satisfies the equation!

Option (C) \( x = 3\pi \): Outside the interval \( (0, \pi) \), so invalid

Option (D) \( x = \pi \): At endpoint, not in open interval \( (0, \pi) \)

Option (E) \( x = \pi \): Same as (D), invalid

Step 4: Verify it's an extremum by second derivative test: \[ f''(x) = -\pi \cos x + 2 \] At \( x = \frac{\pi}{2} \): \[ f''\left(\frac{\pi}{2}\right) = -\pi \cos\left(\frac{\pi}{2}\right) + 2 = 2 > 0 \] This indicates a local minimum at \( x = \frac{\pi}{2} \).

Conclusion: The correct value where \( f \) attains a local extremum in \( (0, \pi) \) is \(\boxed{B}\) (\(\frac{\pi}{2}\)).

Answer 2

To find the critical points of the function \( f(x) = \pi \cos x + x^2 \), we first need to find the derivative of the function. \[ f'(x) = -\pi \sin x + 2x \] Now, set the derivative equal to zero to find the critical points: \[ -\pi \sin x + 2x = 0 \] \[ \pi \sin x = 2x \] This equation cannot be solved algebraically, so we need to estimate the value of \( x \) within the interval \( (0, \pi) \) where this holds true. Let's examine the behavior of the function: - At \( x = \frac{\pi}{2} \), we have: \[ \pi \sin \left( \frac{\pi}{2} \right) = \pi \] and \[ 2 \times \frac{\pi}{2} = \pi \] Thus, \( x = \frac{\pi}{2} \) is a solution where the derivative is zero. This suggests a potential local extremum at \( x = \frac{\pi}{2} \). Now, to determine whether this is a maximum or minimum, we can check the second derivative: \[ f''(x) = -\pi \cos x + 2 \] At \( x = \frac{\pi}{2} \), we get: \[ f''\left( \frac{\pi}{2} \right) = -\pi \cos \left( \frac{\pi}{2} \right) + 2 = 2 \] Since the second derivative is positive, the function has a local minimum at \( x = \frac{\pi}{2} \).