Question

If \( \int f(x)\cos x \, dx = \frac{1}{2}\{f(x)\}^2 + c \), then \( f\left(\frac{\pi}{2}\right) \) is

• A. \( c \)
• B. \( \frac{\pi}{2} + c \)
• C. \( c + 1 \)
• D. \( 2\pi + c \)
• E. \( c + 2 \)

Hint:

When a function is inside an integral identity, differentiate both sides to recover a differential equation for the function.

Answer 1

The Correct Option is C

Concept: This problem involves identifying a function from a given integral identity. The key idea is that if: \[ \int f(x)\cos x \, dx = \frac{1}{2}[f(x)]^2 + c \] then differentiating both sides with respect to \(x\) helps us retrieve a differential equation involving \(f(x)\).

Step 1: Differentiate both sides with respect to \(x\).
Left-hand side becomes: \[ \frac{d}{dx}\left(\int f(x)\cos x \, dx\right) = f(x)\cos x \] Right-hand side becomes: \[ \frac{d}{dx}\left(\frac{1}{2}[f(x)]^2 + c\right) = f(x)\cdot f'(x) \]

Step 2: Equate both derivatives: \[ f(x)\cos x = f(x)f'(x) \]

Step 3: Simplify equation.
Assuming \(f(x) \neq 0\), divide both sides by \(f(x)\): \[ f'(x) = \cos x \]

Step 4: Integrate both sides to find \(f(x)\): \[ f(x) = \int \cos x \, dx = \sin x + C \]

Step 5: Substitute into original relation to determine constant consistency.
Plugging \(f(x) = \sin x + C\) into RHS: \[ \frac{1}{2}(\sin x + C)^2 + c \] Differentiating confirms the identity holds for constant \(C=0\). Thus: \[ f(x) = \sin x \]

Step 6: Evaluate at \(x = \frac{\pi}{2}\): \[ f\left(\frac{\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) = 1 \]

Step 7: Hence: \[ f\left(\frac{\pi}{2}\right) = c + 1 \]