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If $f: \mathbb{R} \rightarrow \mathbb{R}$ is defined by $f(\mathrm{x})=2 \mathrm{x}+\sin \mathrm{x}, \mathrm{x} \in \mathrm{R}$, then $f$ is
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The correct answer is:
one-one and onto
$$
\begin{aligned}
& f(x)=2 x+\sin x \\
& f^{\prime}(x)=2+\cos x>0
\end{aligned}
$$
$\therefore f(x)$ is one-one.
$\because \quad \forall y \in f(x)$, there exist some $x$ as there is a polynomial function $2 x$.
$\therefore f(x)$ is onto.
\begin{aligned}
& f(x)=2 x+\sin x \\
& f^{\prime}(x)=2+\cos x>0
\end{aligned}
$$
$\therefore f(x)$ is one-one.
$\because \quad \forall y \in f(x)$, there exist some $x$ as there is a polynomial function $2 x$.
$\therefore f(x)$ is onto.
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