Skip to content
Exercise 1.5 · Q19

Q.The function f:[0,2π]→[−1,1]f:[0,2\pi]\to[-1,1] defined by f(x)=sin⁡xf(x)=\sin x is

(1) one-to-one
(2) onto
(3) bijection
(4) cannot be defined
Puducherry TnboardTextbookSubjectiveImportance★★★★★
63% · 66/104 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 (One-to-one?). sin⁡(0)=sin⁡(π)=sin⁡(2π)=0\sin(0)=\sin(\pi)=\sin(2\pi)=0 -- multiple distinct inputs share the same output. Not one-to-one.

Step 2 (Onto?). As xx ranges over [0,2π][0,2\pi], sin⁡x\sin x attains every value from its minimum −1-1 (at x=3π/2x=3\pi/2) to its maximum 11 (at x=π/2x=\pi/2), covering all of [−1,1][-1,1] continuously. Onto its stated co-domain [−1,1][-1,1]. …

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.