Question

A real differentiable function \(f\) satisfies \(f(x)+f(y)+2xy=f(x+y)\). Given \(f''(0)=0\), then \[ \int_0^{\pi/2} f(\sin x)\,dx = \]

• A. \(0\)
• B. \( \frac{\pi}{4} \)
• C. \( \frac{\pi}{2} \)
• D. \( \pi \)

Hint:

Recognize identity pattern: \( (x+y)^2 \).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We use differentiation to determine the functional form of \(f(x)\) and then solve the definite integral.
Step 2: Detailed Explanation:
1. Find \(f(x)\):
Differentiate \(f(x) + f(y) + 2xy = f(x + y)\) with respect to \(y\):
\[ f'(y) + 2x = f'(x + y) \]
At \(y = 0\): \(f'(0) + 2x = f'(x)\).
Integrate both sides:
\[ f(x) = x^{2} + f'(0)x + c \]
Since \(f(x) + f(y) + 2xy = f(x + y)\), setting \(x=0, y=0\) gives \(2f(0) = f(0) \implies f(0) = 0\). Thus, \(c = 0\).
Also, \(f''(x) = 2\). The given \(f''(0) = 0\) in the question appears to be a typo or misinterpreted; given the standard functional form, we assume \(f(x) = x^{2}\) (as per the provided solution logic).
2. Evaluate the integral:
\[ I = \int_{0}^{\pi/2} f(\sin x) dx = \int_{0}^{\pi/2} \sin^{2} x dx \]
Using the property \(\int_{0}^{a} g(x) dx = \int_{0}^{a} g(a - x) dx\):
\[ I = \int_{0}^{\pi/2} \cos^{2} x dx \]
Summing the two forms:
\[ 2I = \int_{0}^{\pi/2} (\sin^{2} x + \cos^{2} x) dx = \int_{0}^{\pi/2} 1 dx = \pi/2 \]
\[ I = \pi/4 \]
Step 3: Final Answer:
The integral is \(\pi/4\).