JEE Mathematics Sets, Relations, and Functions Q #1006
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4 Marks 2025 JEE Main 2025 (Online) 29th January Evening Shift MCQ SINGLE
Let $S = \mathbb{N} \cup \{0\}$. Define a relation R from S to $\mathbb{R}$ by: $R = \{(x, y) : \log_e y = x \log_e (\frac{2}{5}), x \in S, y \in \mathbb{R}\}$. Then, the sum of all the elements in the range of $R$ is equal to:
(A) $\frac{3}{2}$
(B) $\frac{10}{9}$
(C) $\frac{5}{2}$
(D) $\frac{5}{3}$
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Correct Answer: D
Explanation
Given $S = \{0, 1, 2, 3, ...\}$. Also, $\log_e y = x \log_e (\frac{2}{5})$.
This implies $y = (\frac{2}{5})^x$.
Since $x \in S$, $x$ can take values $0, 1, 2, 3, ...$.
The required sum is $1 + (\frac{2}{5})^1 + (\frac{2}{5})^2 + (\frac{2}{5})^3 + ... = \frac{1}{1 - \frac{2}{5}} = \frac{1}{\frac{3}{5}} = \frac{5}{3}$.

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