Definite Integrals

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Let $f (x)$ be a continuous function satisfying $f (x) f (a - x) = 1$ . Then $\int_{0}^{a} \frac{1}{1 + f ( x )} d x$ equals:

Select the correct option:

$a$

$a /2$

$0$

$2 a$

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About This Question

Subject: mathematics
Chapter: integral calculus
Topic: definite integrals
Difficulty: Medium
Year: 2025

Solution

Correct Answer:

$a /2$

Let $I = \int_{0}^{a} \frac{1}{1 + f ( x )} d x$ . Using the property $\int_{0}^{a} g (x) d x = \int_{0}^{a} g (a - x) d x$ : $I = \int_{0}^{a} \frac{1}{1 + f ( a - x )} d x$ .

Given $f (x) f (a - x) = 1$ , we imply $f (a - x) = 1/ f (x)$ . Substitute this back: $I = \int_{0}^{a} \frac{1}{1 + 1/ f ( x )} d x = \int_{0}^{a} \frac{f ( x )}{f ( x ) + 1} d x$ .

Now add the original $I$ and the new form: $2 I = \int_{0}^{a} \frac{1}{1 + f ( x )} d x + \int_{0}^{a} \frac{f ( x )}{1 + f ( x )} d x$ $2 I = \int_{0}^{a} \frac{1 + f ( x )}{1 + f ( x )} d x = \int_{0}^{a} 1 d x = a$ . Thus, $I = a /2$ .

This medium difficulty mathematics question is from the chapter integral calculus, covering the topic of definite integrals. It appeared in the 2025 exam.

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