Question

The number of roots of the equation, $x\sin x=1$ in the interval $0<x<2\pi$ is :

• A. $0$
• B. $1$
• C. $2$
• D. $4$

Answer 1

The correct option is C $2$

let take $f\left(x\right)=x\sin x-1$

$f^{\prime }\left(x\right)=\sin x+x\cos x=0,x=\pi$ in interval (so at $x=\pi$ there must be maxima or minima),

$f\left(\pi \right)=\pi \sin \pi -1<0$ (negative)

$f\left(\frac{\pi }{2}\right)>0$ (so there must be a root in interval $(\pi ,\frac{\pi }{2})$

$f\left(\frac{3\pi }{2}\right)>0$, (there must be a root in interval $(0,\frac{\pi }{2})$
So total two root exist.