# For the map $f\left( x \right) = x\sin \pi x$, $x>0$, points at which $f'(x)$ vanishes

Let $$f\left( x \right) = x\sin \pi x$$, $$x>0$$. Then for all natural number n, $$f'(x)$$ vanishes at

(A) A unique point in the interval $$(n,n+\frac{1}{2})$$

(B) A unique point in the interval $$(n+\frac{1}{2},n+1)$$

(C) A unique point in the interval $$(n,n+1)$$

(D) two points in the interval $$(n,n+1)$$

My approach is as follow

$$f'\left( x \right) = \pi x\cos \pi x + \sin \pi x = 0$$

$$\Rightarrow - \tan \pi x = \pi x$$

From here onward how do I proceed further

• Focus first on the equation $\tan \theta=\theta$ and use the intermediate value theorem. Commented Nov 5, 2020 at 12:29
• The derivative of $\tan \theta$ is $\frac{1}{1 + \theta^2} \le 1$. Commented Nov 5, 2020 at 12:30

$$\sin\pi x$$ keeps a constant sign in all unit intervals $$(n,n+1)$$, and the derivative changes sign once. We can reject $$D$$ and accept $$C$$, but the choice between $$A$$ and $$B$$ is unsure. To decide, we can consider the sign of
$$f'(n)f'\left(n+\frac12\right)=\pi n\cos \pi n \sin \pi\left(n+\frac12\right)>0.$$