MAH-CET 2026 — Mathematics PYQ

Step 1: Simplify the Logarithmic Relation

We are given the governing equation for the relation:

$${\log }_{e}y=x{\log }_e\left(\frac{2}{5}\right)$$

Using the logarithmic power property, $a{\log }_b(c)={\log }_b(c^a)$, we can rewrite the right side:

$${\log }_{e}y={\log }_e{\left(\frac{2}{5}\right)}^x$$

Taking the antilog on both sides (removing the logarithm):

$$y={\left(\frac{2}{5}\right)}^x$$

Step 2: Identify the Range of the Relation

The range of a relation is the set of all possible output values ($y$). We are given that $x\in S$, where $S=\mathbb{N}\cup \{0\}=\{0,1,2,3,\dots \}$.

Let's find the values of $y$ by substituting the values of $x$:

• For $x=0⟹y={\left(\frac{2}{5}\right)}^0=1$

• For $x=1⟹y={\left(\frac{2}{5}\right)}^1=\frac{2}{5}$

• For $x=2⟹y={\left(\frac{2}{5}\right)}^2={\left(\frac{2}{5}\right)}^2$

• For $x=3⟹y={\left(\frac{2}{5}\right)}^3={\left(\frac{2}{5}\right)}^3$

Thus, the range set is:

$$\text{Range}=\left\{1,\frac{2}{5},{\left(\frac{2}{5}\right)}^2,{\left(\frac{2}{5}\right)}^3,\dots \right\}$$

Step 3: Calculate the Sum of the Elements

We need to find the sum of all elements in this range:

$$\text{Sum}=1+\frac{2}{5}+{\left(\frac{2}{5}\right)}^2+{\left(\frac{2}{5}\right)}^3+\dots \infty$$

This is an infinite Geometric Progression (G.P.) where:

• The first term ($a$) = $1$

• The common ratio ($r$) = $\frac{2}{5}$

Since the absolute value of the common ratio |r| = \left|\frac{2}{5}\right| < 1, we can use the formula for the sum of an infinite geometric series:

$$S_{\infty }=\frac{a}{1-r}$$

Substituting our values into the formula:

$$\text{Sum}=\frac{1}{1-\frac{2}{5}}$$

$$\text{Sum}=\frac{1}{\frac{5-2}{5}}$$

$$\text{Sum}=\frac{1}{\frac{3}{5}}=\frac{5}{3}$$

Correct Answer:

(d) $\frac{5}{3}$