CAT 2025 Slot 2 · Quant · Question 4

A value of x is excluded if it breaks either composite, so collect the bad points of each and take the union.

Each composite can fail in two ways:

the inner function undefined, or

the outer function undefined at the inner value.

In f\left( g\left( x \right) \right), the inner function g needs x-1\ne 0, so x\ne 1.

The outer function f breaks when its input is \frac{1}{2}, so we need \frac{x}{x-1}\ne \frac{1}{2}, that is 2x\ne x-1, giving x\ne -1.

In g\left( f\left( x \right) \right), the inner function f needs 2x-1\ne 0, so x\ne \frac{1}{2}.

The outer function g breaks when its input is 1, so we need \frac{x}{2x-1}\ne 1, that is x\ne 2x-1, giving x\ne 1.

Taking the union, the excluded values are -1, \frac{1}{2} and 1.

Shortcut. Options 2, 3 and 4 all omit -1 or add a value that no step can produce. Since x=\frac{1}{2} and x=1 are obvious from the denominators alone, the only real question is whether a fourth bad point exists — and testing x=-1 in g gives \frac{1}{2}, exactly the value f cannot take. That single check settles it.