Question

Let $f\left(x\right)=\sin x-x\cos x,x\in R$

If $\alpha$ is the least positive value for which $\tan \alpha =\alpha ,$ then the area bounded by $y=f\left(x\right)$, x-axis, $x=0$ and $x=2\pi$ is

• A. $4$
• B. $4(1-\cos \alpha )$
• C. $4(1+\cos \alpha )$
• D. None of these

Answer 1

The correct option is D None of these

Required area
$={\int }_0^{\alpha }f\left(x\right)dx-{\int }_{\alpha }^{2\pi }f\left(x\right)dx$
$=-(2\cos x+x\sin x{)}_0^{\alpha }+(2\cos x+x\sin x{)}_{\alpha }^{2\pi }$
$=-(2\cos \alpha +\alpha \sin \alpha )+\left(2\right)+\left(2\right)-(2\cos \alpha +\alpha \sin \alpha )$
$=4-2(2\cos \alpha +\alpha \sin \alpha )$
$=4-4\cos \alpha -2{\alpha }^2.\cos \alpha$ (as $\sin \alpha =\alpha \cos \alpha$)
$=4-2(2+{\alpha }^2)\cos \alpha$