Question

If $f(x)$ is a twice differentiable function and $f^{\prime}(0)=0$, then $\int_{0}^{\pi/2}(f(x)+f^{\prime\prime}(x))\cos x dx=$

• A. $f\left(\frac{\pi}{2}\right)$
• B. $f^{\prime}\left(\frac{\pi}{2}\right)$
• C. 1
• D. 0

Hint:

For integrals pairing functions with their derivatives (like $f+f^{\prime\prime}$), look for cancellation patterns using integration by parts.

Answer 1

The Correct Option is A

Step 1: Concept
We split the expression into two separate definite integrals and apply integration by parts to the term containing the second derivative: $\int u dv = uv - \int v du$.

Step 2: Meaning
Let's evaluate $\int_{0}^{\pi/2} f^{\prime\prime}(x)\cos x dx$. Take $u = \cos x \implies du = -\sin x dx$ and $dv = f^{\prime\prime}(x)dx \implies v = f^{\prime}(x)$: $\left[f^{\prime}(x)\cos x\right]_{0}^{\pi/2} - \int_{0}^{\pi/2} f^{\prime}(x)(-\sin x) dx = (0 - f^{\prime}(0)) + \int_{0}^{\pi/2} f^{\prime}(x)\sin x dx$. Since $f^{\prime}(0) = 0$, this becomes $\int_{0}^{\pi/2} f^{\prime}(x)\sin x dx$.

Step 3: Analysis
Now apply integration by parts a second time to $\int_{0}^{\pi/2} f^{\prime}(x)\sin x dx$. Take $u = \sin x \implies du = \cos x dx$ and $dv = f^{\prime}(x)dx \implies v = f(x)$: $\left[f(x)\sin x\right]_{0}^{\pi/2} - \int_{0}^{\pi/2} f(x)\cos x dx = \left(f\left(\frac{\pi}{2}\right)\cdot 1 - f(0)\cdot 0\right) - \int_{0}^{\pi/2} f(x)\cos x dx = f\left(\frac{\pi}{2}\right) - \int_{0}^{\pi/2} f(x)\cos x dx$.

Step 4: Conclusion
Substitute this back into the full original integral expression: $\int_{0}^{\pi/2} f(x)\cos x dx + \left(f\left(\frac{\pi}{2}\right) - \int_{0}^{\pi/2} f(x)\cos x dx\right) = f\left(\frac{\pi}{2}\right)$. This beautifully cancels out the integral terms, leaving exactly option (A).

Final Answer: (A)