find the friction, dude.

$$ At the point where the particle makes angle $a$ with the center of the arc w.r.to horizontal, velocity be $r\omega$ . the FBD of the particle is given. form this , $N\quad =\quad mg\times \cos { (a) } +mr{ \omega }^{ 2 }$ as $f=xN$ , tangential acceleration ${ a }_{ t }$ is ${ a }_{ t }\quad =\quad g(\cos { (a) } -x\sin { (a) } )-xr\omega ^{ 2 }\quad =\quad r\frac { d\omega }{ dt } \quad =\quad r\omega \frac { d\omega }{ da }$ let $k\quad =\quad \frac { r{ \omega }^{ 2 } }{ 2g }$ . then the above equation can be written as $\cos { (a) } -x\sin { (a) } =\quad \frac { dk }{ da } \quad +\quad 2xk$ Now apply maths:multiply both sides by ${e}^{2xa}$ - $(\cos { (a) } -x\sin { (a) } ){ e }^{ 2xa }\quad =\quad \frac { dk }{ da } { e }^{ 2xa }\quad +\quad 2xk{ e }^{ 2xa }\quad =\quad \frac { d }{ da } (x{ e }^{ 2xa })$ $(\cos { (a) } -x\sin { (a) } ){ e }^{ 2xa }\quad da\quad =\quad d(x{ e }^{ 2xa })$ integrate from $a=0$ to $a=\pi/2$ . $\int _{ 0 }^{ \pi /2 }{ (\cos { (a) } -x\sin { (a) } ){ e }^{ 2xa }\quad da } \quad =\quad \int _{ 0 }^{ \pi /2 }{ d(x{ e }^{ 2xa }) } \quad =\quad 0$ . solving this, we get the answer.

The provided options are wrong , the correct answer should be $3x = (1-2x^2) \times e^{\pi x}$