Verified Solution • 2025–26 Session moderate

QJEEM25Ma9ab2: JEE Main 2025 Mathematics - Class 11 JEEM25Ma9ab2 Solution: Let . Define a relation R from S to by : Then, the sum of all…

Let S=N∪{0}\mathrm{S}=\mathbf{N} \cup\{0\} . Define a relation R from S to R\mathbf{R} by :

R={(x,y):log⁡ey=xlog⁡e(25),x∈ S,y∈R}.\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 RR is equal to :

Multiple Choice Options

  • A 32\frac{3}{2}
  • B 109\frac{10}{9}
  • C 52\frac{5}{2}
  • D 53\frac{5}{3}

Step-by-Step Verified Solution

amp;S={0,1,2,3…..}amp;log⁡ey=log⁡e(25)amp;⇒y=(25)x\begin{aligned}& S=\{0,1,2,3 \ldots . .\} \\& \log _{\mathrm{e}} \mathrm{y}=\log _{\mathrm{e}}\left(\frac{2}{5}\right) \\& \Rightarrow \mathrm{y}=\left(\frac{2}{5}\right)^{\mathrm{x}}\end{aligned}

JEE Main 2025 (Online) 29th January Evening Shift Mathematics - Sets and Relations Question 11 English Explanation

Required

 Sum =1+(25)1+(25)2+(25)3+…..−=11−25=53\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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