GATE CS 2015 Set 1 Q.15 (CS Q.5): Sets, Relations, Functions, Partial Orders and Lattices

If $g(x) = 1 - x$ and $h(x) = \dfrac{x}{x-1}$ , then $\dfrac{g(h(x))}{h(g(x))}$ is:

$\dfrac{h(x)}{g(x)}$

$\dfrac{-1}{x}$

$\dfrac{g(x)}{h(x)}$

$\dfrac{x}{(1-x)^2}$

Official answer and solution

Solution

Substitute the definitions of $g(x)$ and $h(x)$ into numerator and denominator and simplify the rational expression.

Given $g(x) = 1 - x$ and $h(x) = \frac{x}{x-1}$ .

g(h(x)) = 1 - h(x) = 1 - \frac{x}{x-1} = \frac{(x-1) - x}{x-1} = \frac{-1}{x-1} = \frac{1}{1-x}.

h(g(x)) = \frac{g(x)}{g(x) - 1} = \frac{1 - x}{(1 - x) - 1} = \frac{1 - x}{-x} = \frac{x - 1}{x}.

\frac{g(h(x))}{h(g(x))} = \frac{\frac{-1}{x-1}}{\frac{x-1}{x}} = \frac{-x}{(x-1)^2}.

\frac{h(x)}{g(x)} = \frac{\frac{x}{x-1}}{1-x} = \frac{x}{(x-1)(1-x)} = \frac{-x}{(x-1)^2}.

Both expressions are identical. Thus, $\frac{g(h(x))}{h(g(x))} = \frac{h(x)}{g(x)}$ (option A).

Final answer

Common trap

Making an algebraic sign error when simplifying the denominator $h(g(x)) = (1-x)/(-x)$ .

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

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