= Solution
{c}
Since $T_2=2x^2-1$, take $x_1=-1/\sqrt2$, $x_2=1/\sqrt2$. Exactness for $T_0$ and $T_1$ requires $A+B=\pi$ and $(-A+B)/\sqrt2=0$. Thus
$$
\boxed{x_{1,2}=\mp\frac1{\sqrt2},\qquad A=B=\frac\pi2.}
$$
For $T_2$, both node values and its weighted integral vanish. For $T_3=4x^3-3x$, oddness and symmetric nodes likewise give zero on both sides. Therefore $R(T_j)=0$ for $j=0,1,2,3$. These four <Chebyshev polynomials> have successive degrees zero through three and form a <basis> of the degree-at-most-three polynomial space. By <linearity>, the rule is exact on that whole space: this is two-node <Gauss–Chebyshev quadrature>.
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