= Solution
At $-a$, the squared prefactor tends to $4$, so the oscillatory discontinuity again remains. Near zero,
$$
f_2(x)=\left(1-\frac{2a}{x}+\frac{a^2}{x^2}\right)[x+O(x^3)]
=\frac{a^2}{x}-2a+O(x).
$$
It is unbounded near zero, despite its assigned zero value there, so it is neither continuous nor <differentiable> at zero.
At $a$, put $x=a+h$. Then
$$
|f_2(a+h)|\leq\frac{h^2}{(a+h)^2},\qquad
\left|\frac{f_2(a+h)-f_2(a)}h\right|\leq\frac{|h|}{(a+h)^2}\longrightarrow0.
$$
Thus \b[$f_2$ is continuous and <differentiable> at $x=\pi/2$, with $f_2'(\pi/2)=0$]. After translating the point to zero, the estimate shows how <quadratic damping restores differentiability at the origin>. The only discontinuities and nondifferentiability points are \b[$x=-\pi/2$ and $x=0$]; the <function> is <smooth> away from these two points and $\pi/2$. At $\pi/2$, its <derivative> exists but is not continuous.
Indeed, for $x\ne a$ near $a$, put $B(x)=(x-a)^2/x^2$. Then $f_2'=B'\sin(\tan x)+B\sec^2x\cos(\tan x)$. Here $B'\to0$ and $B\sec^2x\to1/a^2$, so the <derivative> retains oscillations. Existence of the <derivative> at a point does not require <continuity> of nearby <derivatives>.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2011/paper-1-oscillation-n2.png]
{title=Quadratic suppression giving a zero derivative at the positive pole, and an unbounded discontinuity at zero}
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