Solution

If the rider uses minimum speed possible, the friction from the surface is maximum possible, i.e. $f=\mu N$. As the vertical component of the total force is zero, we have

$$f\sin\alpha-N\cos\alpha-mg=0\qquad\Rightarrow\qquad N(\mu\sin\alpha-\cos\alpha)=mg.$$

The horizontal component of the total force serves as the centripetal force, so

$$f\cos\alpha+N\sin\alpha=\frac{mv^2}{R\sin\alpha}\qquad\Rightarrow\qquad N(\mu\cos\alpha+\sin\alpha)=\frac{mv^2}{R\sin\alpha}.$$

Eliminating $N$ from both equations and solving for $v$, we have

$$v=\sqrt{\frac{gR\sin\alpha(\mu\cos\alpha+\sin\alpha)}{\mu\sin\alpha-\cos\alpha}}=\sqrt{\frac{gR\sin\alpha(\mu+\tan\alpha)}{\mu\tan\alpha-1}},$$

which is possible when $\tan\alpha>1/\mu$.