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A function f:[0,π/2]→R given by f(x)=sinx is*ontoone to one functionbijectivenot one to one

Question

A function f:[0,π/2]→R given by f(x)=sinx is

  • onto
  • one to one function
  • bijective
  • not one to one
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Solution

The function f(x) = sinx from [0, π/2] to R is both one-to-one (injective) and onto (surjective), which makes it a bijective function.

Here's why:

  1. One-to-One (Injective): A function is said to be one-to-one, or injective, if it never maps distinct inputs to the same output. In other words, if f(a) = f(b), then a must equal b. For the function f(x) = sinx in the interval [0, π/2], this is true. In this interval, the sine function is increasing, so it will not map distinct inputs to the same output.

  2. Onto (Surjective): A function is said to be onto, or surjective, if every element in the output space is

This problem has been solved

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