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Let be defined by f(x)=\begin{cases}x^{2}, & x\le 2\\ 2x-1, & x>2\\end{cases} Is one‑to‑one on ?

easy MCQ 1 mark Class 12 · Mathematics · Relations and Functions
  1. Yes
  2. No, because the two pieces are not injective
  3. No, because the two pieces overlap
  4. No, because the function is not defined at
Answer
No, because the two pieces overlap

The correct answer is No, because the two pieces overlap — 5. Thus two distinct inputs give the same output; is not injective.

Step-by-step solution

  1. 1. Assume and .
  2. 2. Take which lies in the first piece: .
  3. 3. Solve for the second piece: .
  4. 4. Compute , equal to .
  5. 5. Thus two distinct inputs give the same output; is not injective.
Common mistake: Thinking each piece is injective implies the whole function is injective

Concept tested: Injectivity of piecewise functions

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