11.B. Given the linear function $y_{1}=2 x$, and the quadratic function $y_{2}=x^{2}+1$. Does there exist a quadratic function $y_{3}=a x^{2}+b x+c$, whose graph passes through the point $(-5,2)$, and for any real number $x$, the function values $y_{1} 、 y_{2} 、 y_{3}$ satisfy $y_{1} \leqslant y_{3} \leqslant y_{2}$? If it exists, find the expression of the function $y_{3}$; if not, explain the reason.
$11 . B$. There exists a quadratic function that satisfies the conditions.
Since $y_{1}-y_{2}=2 x-\left(x^{2}+1\right)$
$$
=-x^{2}+2 x-1=-(x-1)^{2} \leqslant 0 \text {, }
$$

Therefore, for any real number $x$, $y_{1} \leqslant y_{2}$ always holds.

Also, the graph of the quadratic function $y_{3}=a x^{2}+b x+c$ passes through the point $(-5,2)$, so
$25 a-5 b+c=2$.
When $x=1$, we have $y_{1}=y_{2}=2, y_{3}=a+b+c$.
For any real number $x$, $y_{1} \leqslant y_{3} \leqslant y_{2}$ always holds, thus
$$
2 \leqslant a+b+c \leqslant 2 .
$$

Hence, $a+b+c=2$.
From equations (1) and (2), we get $b=4 a, c=2-5 a$.
Therefore, $y_{3}=a x^{2}+4 a x+(2-5 a)$.
When $\dot{y}_{1} \leqslant y_{3}$, we have
$$
2 x \leqslant a x^{2}+4 a x+(2-5 a) \text {, }
$$

which simplifies to $a x^{2}+(4 a-2) x+(2-5 a) \geqslant 0$.
Thus, the quadratic function
$$
y=a x^{2}+(4 a-2) x+(2-5 a)
$$

is non-negative for all real numbers $x$. Hence,
$$
\left\{\begin{array}{l}
a>0, \\
(4 a-2)^{2}-4 a(2-5 a) \leqslant 0,
\end{array}\right.
$$

which simplifies to $\left\{\begin{array}{l}a>0, \\ (3 a-1)^{2} \leqslant 0 .\end{array}\right.$
Therefore, $a=\frac{1}{3}$.
When $y_{3} \leqslant y_{2}$, we have
$$
a x^{2}+4 a x+(2-5 a) \leqslant x^{2}+1 \text {, }
$$

which simplifies to
$$
(1-a) x^{2}-4 a x+(5 a-1) \geqslant 0 \text {. }
$$

Thus, the quadratic function
$$
y=(1-a) x^{2}-4 a x+(5 a-1)
$$

is non-negative for all real numbers $x$. Hence,
$$
\left\{\begin{array}{l}
1-a>0, \\
(-4 a)^{2}-4(1-a)(5 a-1) \leqslant 0,
\end{array}\right.
$$

which simplifies to $\left\{\begin{array}{l}a<1, \\ (3 a-1)^{2} \leqslant 0 .\end{array}\right.$
Therefore, $a=\frac{1}{3}$.
In summary, $a=\frac{1}{3}, b=\frac{4}{3}, c=\frac{1}{3}$.
Thus, there exists a quadratic function $y_{3}=\frac{1}{3} x^{2}+\frac{4}{3} x+\frac{1}{3}$, such that for any real number $x$, $y_{1} \leqslant y_{3} \leqslant y_{2}$ always holds.