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CUET · MATHS · PYQ PAPER 2025

\(\int \frac{\sqrt{1-x}}{\sqrt{1+x}} d x=\alpha \sqrt{1-x^2}+\beta \sin ^{-1} x+C\), where \(C\) is an arbitrary constant, then which of the following are TRUE?
(A) \(\alpha=1\)
(B) \(\alpha=-1\)
(C) \(\beta=1\)
(D) \(\beta=-1\)
Choose the correct answer from the options given below :

  1. A (B) and (D) only
  2. B (A) and (C) only
  3. C (A) and (D) only
  4. D (B) and (C) only
Verified Solution

Answer & Solution

Correct Answer

(B) (A) and (C) only

Step-by-step Solution

Detailed explanation

\(\int \frac{\sqrt{1-x}}{\sqrt{1+x}} d x = \int \frac{\sqrt{1-x}}{\sqrt{1+x}} \cdot \frac{\sqrt{1-x}}{\sqrt{1-x}} d x\) \(= \int \frac{1-x}{\sqrt{1-x^2}} d x\) \(= \int \frac{1}{\sqrt{1-x^2}} d x - \int \frac{x}{\sqrt{1-x^2}} d x\) \(= \sin^{-1} x - (-\sqrt{1-x^2}) + C\)…
From CUET
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