WBJEE · Maths · Definite Integration
Let \(f\) be a non-constant continuous function for all \(x \geq 0\). Let \(f\) satisfy the relation \(f(x) f(a-x)=1\) for some \(a \in R^{+}\). Then, \(I=\int_{0}^{a} \frac{d x}{1+f(x)}\) is equal to
- A \(a\)
- B \(\frac{a}{4}\)
- C \(\frac{a}{2}\)
- D \(f(a)\)
Answer & Solution
Correct Answer
(C) \(\frac{a}{2}\)
Step-by-step Solution
Detailed explanation
\(I=\int_{0}^{a} \frac{d x}{1+f(x)}\) \(=\int_{0}^{a} \frac{1}{1+f(a-x)} d x\) \(\quad\left[\because \int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\right]\) \(=\int_{0}^{a} \frac{d x}{1+\frac{1}{f(x)}} \quad[\because f(x) f(a-x)=1]\) \(I=\int_{0}^{a} \frac{f(x)}{f(x)+1} d x\)…
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