Let S=N \ 0\ . Define a relation R from S to R by : R=\ (x, y): _ e y=x _ e ((2)/(5)), x ∈ ~S, y ∈ R\ Then, the sum of all the elements in the range of R is equal to :

Correct answer: D. (5)/(3) aligned & S=\ 0,1,2,3 \ & _y y= _e((2)/(5)) & y=((2)/(5))^x aligned Required Sum =1+((2)/(5))^1+((2)/(5))^2+((2)/(5))^3+ . .-=(1)/(1-(2)/(5))=(5)/(3)

JEE Main 202529 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 29 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Sets and Relations

JEE Main 2025 · 29 January · Evening Shift · Mathematics · Sets and Relations

Let S=N \ 0\ . Define a relation R from S to R by : R=\ (x, y): _ e...

Let

S=\mathbf{N} \cup\{0\}

. Define a relation

R

from

S

\mathbf{R}

by :

\mathrm{R}=\left\{(x, y): \log _{\mathrm{e}} y=x \log _{\mathrm{e}}\left(\frac{2}{5}\right), x \in \mathrm{~S}, y \in \mathbf{R}\right\}

Then, the sum of all the elements in the range of

R

is equal to :

Options

A $\frac{10}{9}$
B $\frac{3}{2}$
C $\frac{5}{2}$
D $\frac{5}{3}$

Correct answer

D. $\frac{5}{3}$

Step-by-step solution

\begin{aligned} & S=\{0,1,2,3 \ldots \ldots\} \\ & \log _y y=\log _e\left(\frac{2}{5}\right) \\ & \Rightarrow y=\left(\frac{2}{5}\right)^x\end{aligned}

Required

\text{ Sum }=1+\left(\frac{2}{5}\right)^1+\left(\frac{2}{5}\right)^2+\left(\frac{2}{5}\right)^3+\ldots . .-=\frac{1}{1-\frac{2}{5}}=\frac{5}{3}

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