Question

Let $f\left(x\right)=x\sin \pi x,x>0$. Then for all natural numbers $n,f^{\prime }\left(x\right)$ 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 $(n,n+1)$

Answer 1

Answer: (B), (C)

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

$f\left(x\right)=x\sin \pi x,x>0$
$f^{\prime }\left(x\right)=\sin \pi x+\pi x\cos \pi x$

$f^{\prime }\left(x\right)=0\Rightarrow \tan \pi x=-\pi x$

Clearly, $f^{\prime }\left(x\right)=0$ has a unique root in the interval $(n+\frac{1}{2},n+1)$
Also, $f^{\prime }\left(x\right)=0$ has a unique root in the interval $(n,n+1)$