Let function $f:R\to R$ be defined by $f(x)=2x+sinx$
 for $x\in R$, then f is
A. One-one and onto
B. one-one but NOT onto
C. onto but NOT one-one
D. Neither one-one nor onto

Hint:One-to-one functions are continuous and differentiable functions whose derivative is always positive (> >0) or consistently negative ( <0). We calculate the derivative f, and if we can identify whether it is always positive or negative, we can say that the function f(x) is one-to-one and for onto, an element x is linked to every element y by the function f.


Formula used:Derivative of $\sin x=\dfrac{\text{d}(\sin x)}{\text{d}x}=\cos x$


Complete step by step solution:Given,function $f(x)=2 x+\sin x, f:R\to R$
$\Rightarrow f^{\prime}(x)=2+\cos x \\$
$\Rightarrow f^{\prime}(x)>0, \quad \forall x \in R$
This shows that f(x) is one-one, as $f(x)$ is strictly increasing.
Because it is the case where $f(x)$ increases with each $x \in R$, so all values between $(-\infty and \infty)$ are accepted by $f(x)$.
$\therefore Range of f(x) \in R.$
As a result, $f(x)$ is one-to-one and onto.


Thus, Option (A) is correct.

Note:The alternative way to solve this question is by checking the necessary conditions for the function to be one -one and onto. For a one-to-one function, we should add the requirement that every image in the range has a distinct pre-image in the domain. An element x is linked to every element y by the function f known as the onto function.
For one-one:
$f\left(x_1\right)=f\left(x_2\right) $
$\Rightarrow 2 x_1+\sin x_1=2 x_2+\sin x_2$
It is possible only when $x_1=x_2$
So, f(x) is one - one
For onto: Range of f(x)=R \Rightarrow Range = co-domain
Thus, f(x) is onto.
Hence, it is one-one and onto.