For real numbers p and q, define f(x)=lim n→∞(x^2n+2+(px+q))/(x^2n+1) for every real x, and suppose f is continuous on mathbb R.

For $|x|<1$, $x^{2n}\to0$ so $f(x)=px+q$; for $|x|>1$, dividing by $x^{2n}$ gives $f(x)=x^{2}$; at $x=\pm1$ the expression equals $\frac{1+(\pm p+q)}{2}$. Continuity at $x=1$ requires $\frac{1+p+q}{2}=1=p+q$, and at $x=-1$ requires $-p+q=1$. Solving, $p=0$ and $q=1$ (A true). So $f(x)=1$ on $(-1,1)$ and $f(x)=x^{2}$ for $|x|\ge1$. At $x=1$ the left derivative is $0$ while the right derivative is $2x=2$, so $f$ is not differentiable at $1$ (B false), and by symmetry the same kink appears at $-1$ with slopes $0$ and $-2$ (D false). The integral is $\int_{-1}^{1}1\,dx+2\int_{1}^{2}x^{2}dx=2+2\cdot\frac73=\frac{20}{3}$ (C true). Numerical quadrature of the piecewise function reproduces $\frac{20}{3}$. Answer: A, C.