Let S = mathbb{N} cup {0}. Define a relation | vedclass

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(D) Given $S = \{0, 1, 2, 3, \dots\}$.From the relation $\log_e y = x \log_e \left(\frac{2}{5}\right)$,we have $\log_e y = \log_e \left(\left(\frac{2}

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$\frac{5}{3}$

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(D) Given $S = \{0, 1, 2, 3, \dots\}$.From the relation $\log_e y = x \log_e \left(\frac{2}{5}\right)$,we have $\log_e y = \log_e \left(\left(\frac{2}{5}\right)^x\right)$.This implies $y = \left(\frac{2}{5}\right)^x$.Since $x \in S$,the range of $R$ is the set of values $y = \left(\frac{2}{5}\right)^x$ for $x = 0, 1, 2, \dots$.Range $= \left\{ \left(\frac{2}{5}\right)^0, \left(\frac{2}{5}\right)^1, \left(\frac{2}{5}\right)^2, \dots \right\} = \left\{ 1, \frac{2}{5}, \left(\frac{2}{5}\right)^2, \dots \right\}$.This is an infinite geometric progression with first term $a = 1$ and common ratio $r = \frac{2}{5}$.The sum of the elements in the range is $S_{\infty} = \frac{a}{1 - r} = \frac{1}{1 - \frac{2}{5}} = \frac{1}{\frac{3}{5}} = \frac{5}{3}$.

List-$I$	List-$II$
$A$. Range of $\sec ^{-1}\left[1+\cos ^2 x\right]$, where $[.]$ denotes the greatest integer function	$I$. Odd function
$B$. Domain of $f(x)$ where $f\left(x+\frac{1}{x}\right)=x^2+\frac{1}{x^2}$	$II$. $\left\{0, \frac{1}{2}\right\}$
$C$. $f(x+y)=f(x)+f(y) ; f(1)=5$	$III$. $\left\{\sec ^{-1} 5, \sec ^{-1} 4\right\}$
$D$. $\sin ^{-1} x-\cos ^{-1} x+\sin ^{-1}(1-x)=0 \Rightarrow x \in$	$IV$. $R$
	$V$. $\left\{\sec ^{-1} 1, \sec ^{-1} 2\right\}$

Let $S = \mathbb{N} \cup \{0\}$. Define a relation $R$ from $S$ to $\mathbb{R}$ by: $R = \{(x, y) : \log_e y = x \log_e \left(\frac{2}{5}\right), x \in S, y \in \mathbb{R}\}$. Then,the sum of all the elements in the range of $R$ is equal to

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