1. Let $x$ and $y$ be real numbers, satisfying
$$
x+y=1, x^{4}+y^{4}=\frac{7}{2} \text {. }
$$

Then the value of $x^{2}+y^{2}$ is ( ).
(A) 2
(B) 3
(C) 4
(D) 5
-1 . A.
Let $x^{2}+y^{2}=a$. Then
$$
\begin{array}{l}
x y=\frac{(x+y)^{2}-\left(x^{2}+y^{2}\right)}{2}=\frac{1-a}{2} . \\
\text { Also } x^{4}+y^{4}=\left(x^{2}+y^{2}\right)^{2}-2 x^{2} y^{2} \\
=a^{2}-2\left(\frac{1-a}{2}\right)^{2}=\frac{1}{2} a^{2}+a-\frac{1}{2}=\frac{7}{2} \\
\Rightarrow a^{2}+2 a-1=7 \\
\Rightarrow a=2 \text { or }-4 .
\end{array}
$$

Since $a>0$, then $x^{2}+y^{2}=2$.