Question

7. The function ƒ : [0,2π] → [−1,1] defined by f(x) = sin x is (i) one-to-one (ii) onto (iii) bijection (iv) cannot be defined​

Answer 1

Answer:

Many One Onto Function

Step-by-step explanation:

If we analyse the graph of sinx from x=0 to x=2 pi we see it perfectly fits the given codomain in the question and thus onto, but for two different values of x we get same f(x) and thus many one function.

So it is a many one onto function

Alternative:

Solving algebraically , we can differentiate sinx and we get cos x and set it to zero we get x=pi/2 and 3pi/2  in the domain of [0,2pi] we can see cos x is not monotonically increasing or decreasing and thus the nature of sinx is many one in the domain [0,2pi] and from [0,2pi] we get sin x in [-1,1] so the range perfectly coincides with the given CoDomain so Onto function

So it is a many one onto function