Question ID: #552
Let $S = N \cup \{0\}$. 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:
- (1) $\frac{3}{2}$
- (2) $\frac{5}{2}$
- (3) $\frac{10}{9}$
- (4) $\frac{5}{3}$
Solution:
The relation is given by $\log_{e}y = x \log_{e}(\frac{2}{5})$.
Using log properties, $\log_{e}y = \log_{e}(\frac{2}{5})^x$.
$\Rightarrow y = (\frac{2}{5})^x$.
The set $S = \{0, 1, 2, 3, \dots\}$ (since $S = N \cup \{0\}$).
The range of $R$ consists of values of $y$ for each $x \in S$:
Range $= \{ (\frac{2}{5})^0, (\frac{2}{5})^1, (\frac{2}{5})^2, \dots \}$
Range $= \{ 1, \frac{2}{5}, \frac{4}{25}, \dots \}$
We need the sum of these elements, which forms an infinite Geometric Progression (GP) with first term $a = 1$ and common ratio $r = \frac{2}{5}$.
Sum $S_{\infty} = \frac{a}{1 – r}$
$= \frac{1}{1 – \frac{2}{5}}$
$= \frac{1}{\frac{3}{5}} = \frac{5}{3}$.
Ans. (4)
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