Question:
Let a function \(f : (0, \infty) \to (0, \infty)\) be defined by \(f(x) = \left|1 - \frac{1}{x}\right|\). Then f is :
Let a function \(f : (0, \infty) \to (0, \infty)\) be defined by \(f(x) = \left|1 - \frac{1}{x}\right|\). Then f is :
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Check f(2) and f(2/3) for injectivity, then compare the actual range of f with the stated codomain (0, infinity); note f(1) = 0.
Updated On: Nov 5, 2026
- not injective but it is surjective
- neiter injective nor surjective
- injective only
- both injective as well as surjective
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The Correct Option is B
Solution and Explanation
Injectivity: \(f(2) = \left|1 - \tfrac12\right| = \tfrac12\) and \(f(\tfrac23) = \left|1 - \tfrac32\right| = \tfrac12\). Two different inputs give the same output, so f is not injective.
Surjectivity: for \(x \in (0,1)\), \(f(x) = \frac1x - 1\) takes all values in \((0, \infty)\); for \(x \ge 1\), \(f(x) = 1 - \frac1x \in [0, 1)\). In particular \(f(1) = 0\). So the range of f is \([0, \infty)\), which is not the stated codomain \((0, \infty)\). By the intended reading, f is not onto.
Hence f is neither injective nor surjective.
Answer: (2) neiter injective nor surjective
Surjectivity: for \(x \in (0,1)\), \(f(x) = \frac1x - 1\) takes all values in \((0, \infty)\); for \(x \ge 1\), \(f(x) = 1 - \frac1x \in [0, 1)\). In particular \(f(1) = 0\). So the range of f is \([0, \infty)\), which is not the stated codomain \((0, \infty)\). By the intended reading, f is not onto.
Hence f is neither injective nor surjective.
Answer: (2) neiter injective nor surjective
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