Q33 / 75 · JEE Main 2025 (28 Jan Shift 1) · Mathematics
Two number $\mathrm{k}_1$ and $\mathrm{k}_2$ are randomly chosen from the set of natural numbers. Then, the probability that the value of $\mathrm{i}^{\mathrm{k}_1}+\mathrm{i}^{\mathrm{k}_2},(\mathrm{i}=\sqrt{-1})$ is non-zero, equals
- A$\frac{1}{2}$
- Correct answerB$\frac{3}{4}$
- C$\frac{1}{4}$
- D$\frac{2}{3}$
Solution
Let $k_1=4 \lambda_1+r_1, r_1 \in\{0,1,2,3\}$
$k_2=4 \lambda_2+r_2$
$\begin{aligned}
& (i)^{k_1}+(i)^{k_2}=(i)^{r_1}+(i)^{r_2} \\
& (i)^{r_1} \in\{1, i,-1,-i\}
\end{aligned}$
$\text { Zero } \Rightarrow 1,(-1) \text { pair } \quad \Rightarrow\left\{\begin{array}{cc}
1, & -1 \\
i, & -i \\
-i,+i \\
-1, & 1
\end{array}\right\}$
$i,(-i)$ pair
$\begin{aligned} & \text { Zero probability }=\frac{4}{{ }^4 C_1 \cdot{ }^4 C_1}=\frac{1}{4} \\ & \text { Probability (non-zero) }=1-\frac{1}{4}=\frac{3}{4}\end{aligned}$
Hence, the correct answer is option: 'B'.