CAT 2025

Slot 2

QA

Question & Solution

1. Concept Used

• Topic: Domain of Composite Functions
• Formula: $$h(x) = f(g(x)) + g(f(x)), \quad \text{where } f(x) = \frac{x}{2x-1}, \quad g(x) = \frac{x}{x-1}$$

2. Calculation

To find the domain of $$h(x) = f(g(x)) + g(f(x))$$, we must identify every value of $$x$$ that causes any component — the inner functions, the outer functions applied to them, or the resulting expressions — to be undefined.

Step 1: Individual function restrictions.

$$f(x) = \frac{x}{2x-1} \text{ is undefined when } 2x - 1 = 0 \Rightarrow x = \frac{1}{2}$$

$$g(x) = \frac{x}{x-1} \text{ is undefined when } x - 1 = 0 \Rightarrow x = 1$$

So, $$x = \frac{1}{2}$$ and $$x = 1$$ are already excluded from the domain.

Step 2: Compute $$f(g(x))$$ and find its restrictions.

Substitute $$g(x) = \frac{x}{x-1}$$ into $$f$$:

$$f(g(x)) = \frac{g(x)}{2 \cdot g(x) - 1} = \frac{\dfrac{x}{x-1}}{2 \cdot \dfrac{x}{x-1} - 1}$$

Simplify the denominator:

$$2 \cdot \frac{x}{x-1} - 1 = \frac{2x}{x-1} - \frac{x-1}{x-1} = \frac{2x - (x-1)}{x-1} = \frac{x+1}{x-1}$$

So:

$$f(g(x)) = \frac{\dfrac{x}{x-1}}{\dfrac{x+1}{x-1}} = \frac{x}{x+1}$$

This simplified expression $$\frac{x}{x+1}$$ is undefined when $$x + 1 = 0 \Rightarrow x = -1$$.

Also, the computation itself required $$g(x)$$ to be defined ($$x eq 1$$) and the denominator $$2g(x)-1 eq 0$$, which we already captured via $$x eq -1$$.

Step 3: Compute $$g(f(x))$$ and find its restrictions.

Substitute $$f(x) = \frac{x}{2x-1}$$ into $$g$$:

$$g(f(x)) = \frac{f(x)}{f(x) - 1} = \frac{\dfrac{x}{2x-1}}{\dfrac{x}{2x-1} - 1}$$

Simplify the denominator:

$$\frac{x}{2x-1} - 1 = \frac{x - (2x-1)}{2x-1} = \frac{1-x}{2x-1}$$

So:

$$g(f(x)) = \frac{\dfrac{x}{2x-1}}{\dfrac{1-x}{2x-1}} = \frac{x}{1-x}$$

This expression is undefined when $$1 - x = 0 \Rightarrow x = 1$$, which is already excluded.

Step 4: Collect all excluded values.

From $$g(x)$$ undefined: $$x = 1$$

From $$f(x)$$ undefined: $$x = \frac{1}{2}$$

From $$f(g(x)) = \frac{x}{x+1}$$ undefined: $$x = -1$$

From $$g(f(x)) = \frac{x}{1-x}$$ undefined: $$x = 1$$ (already listed)

Therefore, $$h(x)$$ is undefined at $$x \in \left{-1,\ \frac{1}{2},\ 1\right}$$.

---

3. Solution

Answer = Option 1 ✅

The domain of $$h(x) = f(g(x)) + g(f(x))$$ is all real numbers except $$-1,\ \dfrac{1}{2},\ \text{and}\ 1$$.