Sets, Relations & Functions
Functions
nta_abhyas_2025
Grade 11
Question:
Let a function $f: (4, \infty) \to (0, \infty)$ defined as $f(x) = \frac{x^2}{x^2 - 1}$, then $f$ is
injective & surjective
injective but surjective
injective but not surjective
neither injective nor surjective
Step-by-Step Solution
Key Concept: Determine surjectivity by comparing the range of the function with its codomain.
For $f(x) = |1 - \frac{1}{x-2}|$ with domain $x > 2$: as $x \to 2^+$, we have $\frac{1}{x-2} \to \infty$, so $|1 - \frac{1}{x-2}| \to \infty$. As $x \to \infty$, we have $\frac{1}{x-2} \to 0$, so $|1 - \frac{1}{x-2}| \to 1$. The minimum value is $0$, achieved when $1 - \frac{1}{x-2} = 0$, i.e., $x = 3$. Therefore the range is $[0, \infty)$, which equals the codomain, making the function surjective but not injective.
Correct Answer: 2