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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 $(n,n+1)$

The official answer is B&C

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

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

1 Answer 1

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$$\sin\pi x$$ 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

$$f'(n)f'\left(n+\frac12\right)=\pi n\cos \pi n \sin \pi\left(n+\frac12\right)>0.$$

You can conclude.

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