Question

If \( f(x) = \int_0^x t(\sin x - \sin t)dt \) then :

• A. \( f'''(x) - f''(x) = \cos x - 2x\sin x \)
• B. \( f'''(x) + f'(x) = \cos x - 2x\sin x \)
• C. \( f'''(x) + f''(x) = \sin x \)
• D. \( f'''(x) + f''(x) - f'(x) = \cos x \)

Hint:

Always extract terms containing 'x' outside the integral before differentiating. If you see a term in the derivative that matches an earlier order derivative, substitute it immediately to find the required differential equation. \

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to differentiate the given integral function up to the third order. Note that \(x\) is part of the integrand, so we must separate it before differentiating.

Step 2: Key Formula or Approach:
Use the Leibniz Rule for differentiation under the integral sign: \[ \frac{d}{dx} \int_0^x g(x, t)\,dt = g(x, x) + \int_0^x \frac{\partial}{\partial x} g(x, t)\,dt \]

Step 3: Detailed Explanation:
Rewrite \( f(x) \): \[ f(x) = \sin x \int_0^x t\,dt - \int_0^x t \sin t\,dt = \frac{x^2}{2} \sin x - \int_0^x t \sin t\,dt \] Differentiating once: \[ f'(x) = \left[ x \sin x + \frac{x^2}{2} \cos x \right] - [x \sin x] = \frac{x^2}{2} \cos x \] Differentiating a second time: \[ f''(x) = x \cos x - \frac{x^2}{2} \sin x \] Differentiating a third time: \[ f'''(x) = (\cos x - x \sin x) - \left( x \sin x + \frac{x^2}{2} \cos x \right) \] \[ f'''(x) = \cos x - 2x \sin x - \frac{x^2}{2} \cos x \] Observing that \( \frac{x^2}{2} \cos x = f'(x) \), we substitute: \[ f'''(x) = \cos x - 2x \sin x - f'(x) \] \[ f'''(x) + f'(x) = \cos x - 2x \sin x \]

Step 4: Final Answer:
The correct relation is \( f'''(x) + f'(x) = \cos x - 2x \sin x \).