Given, ,$$f(x)=x \cdot \sin \pi x, x>0$$

$$\Rightarrow \quad f^{\prime}(x)=1 \cdot \sin \pi x+x \cdot \pi \cos \pi x$$

Apply $$f^{\prime}(x)=0$$

$$\begin{aligned} & \Rightarrow \sin \pi x+\pi x \cos \pi x=0 \\ & \Rightarrow \sin \pi x=-\pi x \cos \pi x \\ & \Rightarrow \frac{\sin \pi x}{\cos \pi x}=-\pi x \\ & \Rightarrow \tan \pi x=-\pi x \end{aligned}$$

Draw the graph of $$y=\tan \pi x$$ and $$y=-\pi x$$ for $$x>0$$

It is clear that $$y=\tan \pi x$$ and $$y=-\pi x$$ intersect at a unique point if $$\frac{1}{2} < x < 1$$ as $$\frac{3}{2} < x < 2$$ as $$\frac{5}{2} < x < 3$$ or.....

So, $$y=\tan \pi x$$ and $$y=-\pi x$$ intersect at a unique point is $$x \in\left(n+\frac{1}{2}, n+1\right)$$ as $$(n, n+1)$$.

Hints:

The number of point of intersection of graphs of $$y=f(x)$$ and $$y=g(x)$$ is equal to the number of solution of $$f(x)=g(x)$$