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Solution:

Given:- Function $f(x)=x-\sin x$

Theorem:- Let $f$ be a differentiable real function defined on an open interval $(a, b)$.

(i) If $f^{\prime}(x)>0$ for $a l l x \in(a, b)$, then $f(x)$ is increasing on $(a, b)$

(ii) If $f^{\prime}(x)<0$ for all $x \in(a, b)$, then $f(x)$ is decreasing on $(a, b)$

Algorithm:-

(i) Obtain the function and put it equal to $f(x)$

(ii) Find $f^{\prime}(x)$

(iii) Put $f^{\prime}(x)>0$ and solve this inequation.

For the value of $x$ obtained in (ii) $f(x)$ is increasing and for remaining points in its domain it is decreasing.

Here we have,

$f(x)=x-\sin x$

$\Rightarrow \mathrm{f}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}-\sin \mathrm{x})$

$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=1-\cos \mathrm{x}$

Now, as given

$x \in R$

$\Rightarrow-1<\cos x<1$

$\Rightarrow-1>\cos x>0$

$\Rightarrow f^{\prime}(x)>0$

hence, Condition for $f(x)$ to be increasing

Thus $f(x)$ is increasing on interval $x \in R$