JEE Main 2025 (Online) 3rd April Evening Shift

Explanation To find $P(X \geq 3)$, we first determine the constant $k$ using the total probability for $X$. The probability $P(X = x)$ is given by: $P(X= x) = k(x+1) \cdot 3^{-x}, x= 0,1,2,3,\dots$ The total probability must equal 1: $s = \sum_{x=0}^{\infty} k(x+1) \cdot 3^{-x}$ Calculating that series: $s = k3^0 + 2\frac{k}{3} + 3\frac{k}{3^2} + \dots$ Therefore, dividing the series by 3: $\frac{s}{3} = \frac{k}{3} + 2\frac{k}{3^2} + \dots$ Subtracting these: $s - \frac{s}{3} = k + \frac{k}{3} + \frac{k}{3^2} + \dots$ The resulting series is a geometric series: $2\frac{s}{3} = k(1 + \frac{1}{3} + \frac{1}{3^2} + \dots )$ The sum of the infinite geometric series is: $2\frac{s}{3} = k \cdot \frac{1}{1 - \frac{1}{3}} = \frac{3k}{2}$ Equating: $s = \frac{9k}{4} = 1$ Thus, solving for $k$: $k = \frac{4}{9}$ Next, compute $P(X \geq 3)$: $P(X \geq 3) = 1 - (P(X=0) + P(X=1) + P(X=2))$ Calculating these: $P(X=0) = k = \frac{4}{9}$ $P(X=1) = 2\frac{k}{3} = \frac{8}{27}$ $P(X=2) = 3\frac{k}{9} = \frac{4}{27}$ Adding these probabilities: $P(X=0) + P(X=1) + P(X=2) = \frac{4}{9} + \frac{8}{27} + \frac{4}{27} = \frac{8}{9}$ Finally, calculate $P(X \geq 3)$: $P(X \geq 3) = 1 - \frac{8}{9} = \frac{1}{9}$

Explanation The relation $R$ is defined such that $y = \max(x, 1)$. Therefore, we find the following pairs: $x = -2 \implies y = \max(-2, 1) = 1$, so $(-2, 1) \in R$. $x = -1 \implies y = \max(-1, 1) = 1$, so $(-1, 1) \in R$. $x = 0 \implies y = \max(0, 1) = 1$, so $(0, 1) \in R$. $x = 1 \implies y = \max(1, 1) = 1$, so $(1, 1) \in R$. $x = 2 \implies y = \max(2, 1) = 2$, so $(2, 2) \in R$. $x = 3 \implies y = \max(3, 1) = 3$, so $(3, 3) \in R$. Thus, $R = \{(-2, 1), (-1, 1), (0, 1), (1, 1), (2, 2), (3, 3)\}$, and $l = 6$. To make $R$ reflexive, we need to add $(-2, -2), (-1, -1), (0, 0)$. Thus, $m = 3$. To make $R$ symmetric, we need to add $(1, -2), (1, -1), (1, 0)$. Thus, $n = 3$. Therefore, $l + m + n = 6 + 3 + 3 = 12$.