CUET · MATHS · PYQ PAPER 2025
The value of the definite integral \(I=\int_{-1}^1 \frac{1}{1+\sqrt{e^x}} d x\) is :
- A \(0\)
- B 1
- C 2
- D 3
Answer & Solution
Correct Answer
(B) 1
Step-by-step Solution
Detailed explanation
\(f(x) = \frac{1}{1+\sqrt{e^x}}\) \(f(-x) = \frac{1}{1+\sqrt{e^{-x}}} = \frac{\sqrt{e^x}}{1+\sqrt{e^x}}\) \(f(x) + f(-x) = \frac{1}{1+\sqrt{e^x}} + \frac{\sqrt{e^x}}{1+\sqrt{e^x}} = \frac{1+\sqrt{e^x}}{1+\sqrt{e^x}} = 1\) \(I = \int_{-1}^1 f(x) dx = \int_0^1 [f(x) + f(-x)] dx\)…
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