WBJEE · Maths · Definite Integration
If \(f(x)=\frac{e^x}{1+e^x}, l_1=\int_{f(-a)}^{f(a)} x g(x(1-x)) d x\) and \(l_2=\int_{f(-a)}^{f(a)} g(x(1-x)) d x\), then the value of \(\frac{l_2}{l_1}\) is
- A -1
- B -3
- C 2
- D 1
Answer & Solution
Correct Answer
(C) 2
Step-by-step Solution
Detailed explanation
Hint: \(f(a)+f(-a)=\frac{e^a}{1+e^a}+\frac{e^{-a}}{1+e^{-a}}=1\) Let \(f(-a)=t \therefore f(a)=1-t\) Now,…
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