1. Given the equation about $x$
$$
(1-2 k) x^{2}-2 \sqrt{k+1} x-1=0
$$

has real roots. Then the range of values for $k$ is ( ).
(A) $k \geqslant 2$
(B) $k \geqslant-1$, and $k \neq \frac{1}{2}$
(C) $-1 \leqslant k \leqslant 2$, and $k \neq \frac{1}{2}$
(D) $-1 \leqslant k \leqslant 2$
- 1. D.

When $1-2 k=0 \Rightarrow k=\frac{1}{2}$, the original equation must have one real root.

When $k \neq \frac{1}{2}$, by $k+1 \geqslant 0$, and the discriminant of the original equation $\Delta=(-2 \sqrt{k+1})^{2}+4(1-2 k) \geqslant 0$, solving yields $-1 \leqslant k \leqslant 2$, and $k \neq \frac{1}{2}$.
In summary, $-1 \leqslant k \leqslant 2$.