Let function f:R→R be defined by f(x)=2x+sinx for x∈R, then f is [2002]
(1)
Given: f(x)=2x+sinx, x∈R
⇒ f'(x)=2+cosx.
Now, -1≤cosx≤1
⇒ 1≤2+cosx≤3
∴ f'(x)>0, ∀ x∈R
⇒ f(x) is strictly increasing and therefore one-one.
Also, limx→∞f(x)=∞ and limx→-∞f(x)=-∞
∴ Range of f(x)=R=domain of f(x)⇒f(x) is onto.
Hence, f(x) is one-one and onto.