Match List - I with List - II. | List - I | List - II | |---|---| | (A) If A and B are mutually exclusive events, then $P(A \cup B) =$ | (I) $\frac{P(A \cap B)}{P(B)}, P(B) \neq 0$ | | (B) If A and B are independent events, then $P(A \cap B) =$ | (II) $\frac{P(A \cap B)}{P(A)}, P(A) \neq 0$ | | (C) If A and B are two events of a sample space of an experiment, then $P(A/B) =$ | (III) $P(A) \cdot P(B)$ | | (D) If A and B are two events of a sample space of an experiment, then $P(B/A) =$ | (IV) $P(A) + P(B)$ | Choose the correct answer from the options given below :

Let $h = \frac{\pi}{2} - x$. Then $\cos x = \sin h$ and $\pi - 2x = 2h$. So $\lim_{x \to \pi/2} \frac{k\cos x}{\pi - 2x} = \lim_{h \to 0} \frac{k \sin h}{2h} = \frac{k}{2}$. For continuity: $\frac{k}{2} = 3$, so $k = 6$.

Match List - I with List - II. | List - I | List - II | |---|---| | (A) If A… - CUET UG 2023 Mathematics PYQ

CUET UG 2023 — Mathematics Algebra

2023mcqeasy

Match List - I with List - II.

List - I	List - II
(A) If A and B are mutually exclusive events, then $P(A \cup B) =$	(I) $\frac{P(A \cap B)}{P(B)}, P(B) \neq 0$
(B) If A and B are independent events, then $P(A \cap B) =$	(II) $\frac{P(A \cap B)}{P(A)}, P(A) \neq 0$
(C) If A and B are two events of a sample space of an experiment, then $P(A/B) =$	(III) $P(A) \cdot P(B)$
(D) If A and B are two events of a sample space of an experiment, then $P(B/A) =$	(IV) $P(A) + P(B)$

Choose the correct answer from the options given below :

Official previous-year question

Held on 30 May 2023 · Verified 13 Jul 2026.

Options

A
(A)-(IV), (B)-(III), (C)-(I), (D)-(II)
B
(A)-(III), (B)-(IV), (C)-(I), (D)-(II)
C
(A)-(II), (B)-(III), (C)-(IV), (D)-(I)
D
(A)-(I), (B)-(II), (C)-(III), (D)-(IV)

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Solution

Let $h = \frac{\pi}{2} - x$ . Then $\cos x = \sin h$ and $\pi - 2x = 2h$ . So $\displaystyle \lim_{x \to \pi/2} \frac{k\cos x}{\pi - 2x} = \lim_{h \to 0} \frac{k \sin h}{2h} = \frac{k}{2}$ . For continuity: $\frac{k}{2} = 3$ , so $k = 6$ .

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