JEE Main 2025 (Online) 28th January Morning Shift

Explanation $i^{k_1} + i^{k_2} \ne 0 \Rightarrow i^{k_1} \rightarrow 4$ option for $i, -1, -i, 1$ Total cases $ \Rightarrow 4 \times 4 = 16$ Unfovourble cases $ \Rightarrow i^{k_1} + i^{k_2} = 0$ $ \{\begin{array}{c}1, -1 \\ -1, 1 \\ i, -i \\ -i, i\end{array}\}$ 4 Cases $\Rightarrow$ Probability $ = \frac{16-4}{16} = \frac{3}{4}$

Explanation Probability distribution $x = 0$, $p = \frac{^7C_2}{^{10}C_2} = \frac{42}{90}$ $x = 1$, $p = \frac{^7C_1 \times ^3C_1}{^{10}C_2} = \frac{42}{90}$ $x = 2$, $p = \frac{^3C_2}{^{10}C_2} = \frac{6}{90}$ Now, $\mu = \sum x_i p_i = \frac{42}{90} + \frac{12}{90} = \frac{54}{90}$ $\sigma^2 = \sum p_i x_i^2 - \mu^2 = \frac{42}{90} + \frac{24}{90} - (\frac{54}{90})^2$ $\Rightarrow \frac{66}{90} - (\frac{54}{90})^2$ $\sigma^2 \Rightarrow \frac{28}{75}$

Explanation The given relation is $R = {(x, y) : x, y ∈ Z$ and $x + y$ is even $}$. Reflexive: For any $x ∈ Z$, $x + x = 2x$, which is even. So, $(x, x) ∈ R$. Thus, $R$ is reflexive. Symmetric: If $(x, y) ∈ R$, then $x + y$ is even. Since $x + y = y + x$, $y + x$ is also even. So, $(y, x) ∈ R$. Thus, $R$ is symmetric. Transitive: If $(x, y) ∈ R$ and $(y, z) ∈ R$, then $x + y$ is even and $y + z$ is even. Then $(x + y) + (y + z) = x + 2y + z$ is even. Since $2y$ is even, it follows that $x + z$ is even. So, $(x, z) ∈ R$. Thus, $R$ is transitive. Since $R$ is reflexive, symmetric, and transitive, it is an equivalence relation.