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CAT 2025Slot 2QAQues & Sol

AlgebraMedium

Question

Let $f(x)=\frac{x}{(2x-1)}$ and $g(x)=\frac{x}{(x-1)}$. Then the domain of the funtion $h(x)=f(g(x))+g(f(x))$ is all real numbers except

Options

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 given equation.

3. Solution

Answer = Option 1 ✅

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