Skip to content
Exercise 1.5 · Q17

Q.The rule f(x)=x2f(x)=x^2 is a bijection if the domain and the co-domain are given by

(1) R, RR,\ R
(2) R, (0,∞)R,\ (0,\infty)
(3) (0,∞), R(0,\infty),\ R
(4) [0,∞), [0,∞)[0,\infty),\ [0,\infty)
Puducherry TnboardTextbookSubjectiveImportance★★★★★
62% · 64/104 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Step 1. Check each option. (1) R,RR,R: x2x^2 is not injective on RR (e.g. f(1)=f(−1)f(1)=f(-1)) -- fails.

Step 2. (2) R,(0,∞)R,(0,\infty): still not injective on RR -- fails (and 00 is unreachable in (0,∞)(0,\infty) anyway, so not onto either).

Step 3. (3) (0,∞),R(0,\infty),R: injective on (0,∞)(0,\infty) (strictly increasing there), but the range is (0,∞)(0,\infty), which does NOT cover all of the stated co-domain RR (misses 00 and every negative) -- not onto, fails. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.