A function f: R → [-1, 1] defined by f(x) = sin x, ∀ x ∈ R, where R is the subset of real numbers is one-one and onto if R is the interval:

Question

A function f: R → [-1, 1] defined by f(x) = sin x, ∀ x ∈ R, where R is the subset of real numbers is one-one and onto if R is the interval: (A) [0, 2 π] (B) [- (π/2) , (π/2) ] (C) [-π, π] (D) [0, π]

Tardigrade Admin · Accepted Answer

We know that f (x) is said to be one-one If f (x1) = f (x2) ⇒ x1= x2 f (x) is said to be onto if f (x) is always increasing. ∴ x ∈ [- (π/2) , (π/2) ] (∵ f (x) = sin x)

Q. A function $f : R \to [- 1, 1]$ defined by $f (x) = sin x, \forall x \in R$ , where R is the subset of real numbers is one-one and onto if R is the interval:

$[0, 2 π]$

$[- \frac{π}{2}, \frac{π}{2}]$

$[- π, π]$

$[0, π]$

We know that f (x) is said to be one-one
If $f (x_{1}) = f (x_{2}) \Rightarrow x_{1} = x_{2}$
f (x) is said to be onto if f (x) is always increasing. $∴ x \in [- \frac{π}{2}, \frac{π}{2}] (∵ f (x) = sin x)$

Q. A function f:R→[−1,1] defined by f(x)=sinx,∀x∈R, where R is the subset of real numbers is one-one and onto if R is the interval:

[0,2π]

[−2π​,2π​]

[−π,π]

[0,π]