IPMAT Indore 2022 · Quantitative Ability · Question 32

Step 1 — split on the sign of x. The definition of \left| x \right| gives two cases.

If x>0: \left| x \right|=x, so

f\left( x \right)=\frac{x}{x}=1

If x<0: \left| x \right|=-x, so

f\left( x \right)=\frac{x}{-x}=-1

Step 2 — note the excluded point. x=0 is explicitly removed from the domain (the expression would be 0/0), so 0 is not in the range.

Step 3 — collect. The function takes exactly two values:

\{1, -1\}

This is the signum function, \text{sgn}\left( x \right), restricted to x\ne 0.

Answer: Option 2 — \{1,-1\}

Shortcut. Recognise \frac{x}{\left| x \right|} as the sign of x. It reports +1 for positives and -1 for negatives, and nothing else.

The only trap is whether 0 belongs. It does not: the question states x\ne 0, so options (c) and (d) are out. And since negative x is permitted, (a) is incomplete.

A useful related fact: \frac{\left| x \right|}{x} gives the same set, and x=\left| x \right|\text{sgn}\left( x \right) for all x — worth having when a question mixes moduli and signs.