Question: The explanation of the banking angle of the road is said to be that one part of the normal force exerted by the road on a moving object neutralizes the object's weight and another part is providing the centripetal force necessary to turn the object in a circular path.
So the normal force is greater than the weight of the object. What is the source of this force? Why is this force greater than the weight?

Answer: 
It should be mentioned that the mathematical condition for the validity of the WKB approximation for the 1D [TISE](https://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation) can be neatly formulated as
$$\left| \frac{d\lambda(x)}{dx} \right|~\ll~2\pi, \tag{46.6}$$
where
$$ \lambda(x) ~=~\frac{h}{p(x)}, \qquad p(x)~=~\sqrt{2m|E-V(x)|} ,$$
is the [de Broglie wavelength](https://en.wikipedia.org/wiki/Matter_wave), cf. Ref. [LL].

References:

* [LL] L.D. Landau & E.M. Lifshitz, *QM,* Vol. 3, [2nd](https://archive.org/details/ost-physics-landaulifshitz-quantummechanics) & 3rd ed, 1981; $\S46$.
The basic assumption for the WKB approximation is that the potential $V(x)$ should vary slowly in space, compared to the wave function $\Psi(x)$. The more precise way to express this is
$$\left| \frac{V''(x)}{V'(x)}\right| \ll \left|\frac{\Psi''(x)}{\Psi'(x)}\right|.$$
Usually, the wave function has a wave-like profile of the form $\sim e^{ikx}$, where $k$ is the wave vector defined locally and is related to wavelength through $\lambda=\frac{2\pi}{k}$, and therefore
$$\left|\frac{\Psi''(x)}{\Psi'(x)}\right| \sim \frac{1}{\lambda}.$$
So
$$\left|\frac{V''(x)}{V'(x)}\right|\ll \frac{1}{\lambda},$$
which is required for WKB to be valid. Approximately we can write,
$$k\approx \sqrt{2m[E-V(x)]}$$
At points where $V(x)=E$ (the classical turning points) the local wavelength becomes large, and the inequality cannot be satisfied. So special care should be given to these points and matching the WKB solutions on both sides of these points is required.