CUET · MATHS · PYQ PAPER 2023
The value of \(\int_1^2 \frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}} d x\) is:
- A 3
- B \(\frac{3}{2}\)
- C \(\frac{1}{2}\)
- D 2
Answer & Solution
Correct Answer
(C) \(\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
Let \(I = \int_1^2 \frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}} d x\). Using \(\int_a^b f(x) dx = \int_a^b f(a+b-x) dx\), we have \(I = \int_1^2 \frac{\sqrt{3-x}}{\sqrt{3-(3-x)}+\sqrt{3-x}} d x = \int_1^2 \frac{\sqrt{3-x}}{\sqrt{x}+\sqrt{3-x}} d x\).…
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