CUET · MATHS · PYQ PAPER 2023
Match List - I with List - II. Evaluate the integrals.
| \(List - I\) | \(List - II\) |
| (A) \(\int_0^{\pi / 2} \frac{\sin x-\cos x}{1+\sin x \cos x} d x\) | \((I)\) \(2 \pi\) |
| (B) \(\int_0^{1 / 2} \frac{d x}{\sqrt{x-x^2}}\) | \((II)\) \(\frac{\pi}{4}\) |
| (C) \(\int_{-\pi}^\pi x \sin x d x\) | \((III)\) \(0\) |
| (D) \(\int_0^1 \frac{1}{1+x^2} d x\) | \((IV)\) \(\frac{\pi}{2}\) |
- A \((A)-(I), (B)-(IV), (C)-(II), (D)-(III)\)
- B \((A)-(III), (B)-(IV), (C)-(I), (D)-(II)\)
- C \((A)-(III), (B)-(II), (C)-(I), (D)-(IV)\)
- D \((A)-(I), (B)-(III), (C)-(IV), (D)-(II)\)
Answer & Solution
Correct Answer
(B) \((A)-(III), (B)-(IV), (C)-(I), (D)-(II)\)
Step-by-step Solution
Detailed explanation
(A) \(I = \int_0^{\pi / 2} \frac{\sin x-\cos x}{1+\sin x \cos x} d x\) \(I = \int_0^{\pi / 2} \frac{\sin (\pi/2 - x)-\cos (\pi/2 - x)}{1+\sin (\pi/2 - x) \cos (\pi/2 - x)} d x = \int_0^{\pi / 2} \frac{\cos x-\sin x}{1+\cos x \sin x} d x = -I\) \(2I = 0 \Rightarrow I = 0\) (B)…
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