Find the maximum value of the real number $k$ such that the graphs of $y = \sqrt { x + 3 }$ and $y = \sqrt { 1 - x } + k$ intersect. [4 points]

9. Let real numbers $x , y$ satisfy $\left\{ \begin{array} { c } 2 x + y \leq 10 , \\ x + 2 y \leq 14 , \\ x + y \geq 6 , \end{array} \right.$ Then the maximum value of $x y$ is
(A) $\frac { 25 } { 2 }$
(B) $\frac { 49 } { 2 }$
(C) 12
(D) 16

14. Given that real numbers $x , y$ satisfy $x ^ { 2 } + y ^ { 2 } \leq 1$ , then the maximum value of $| 2 x + y - 4 | + | 6 - x - 3 y |$ is $\_\_\_\_$ .

Let $x, y, z \in \mathbf{R}$ and $x + y + z = 1$.
(1) Find the minimum value of $(x-1)^2 + (y+1)^2 + (z+1)^2$;
(2) If $(x-2)^2 + (y-1)^2 + (z-a)^2 \geqslant \frac{1}{3}$ holds for all $x, y, z$ satisfying $x + y + z = 1$, prove that $a \leqslant -3$ or $a \geqslant -1$.

If $x _ { 1 } > x _ { 2 } > \cdots > x _ { 10 }$ are real numbers, what is the least possible value of $$\left( \frac { x _ { 1 } - x _ { 10 } } { x _ { 1 } - x _ { 2 } } \right) \left( \frac { x _ { 1 } - x _ { 10 } } { x _ { 2 } - x _ { 3 } } \right) \cdots \left( \frac { x _ { 1 } - x _ { 10 } } { x _ { 9 } - x _ { 10 } } \right) ?$$ (A) $10 ^ { 10 }$
(B) $10 ^ { 9 }$
(C) $9 ^ { 9 }$
(D) $9 ^ { 10 }$

The range of values that the function $$f ( x ) = \frac { x ^ { 2 } + 2 x + 4 } { 2 x ^ { 2 } + 4 x + 9 }$$ takes as $x$ varies over all real numbers in the domain of $f$ is:
(A) $\frac { 3 } { 7 } < f ( x ) \leq \frac { 1 } { 2 }$
(B) $\frac { 3 } { 7 } \leq f ( x ) < \frac { 1 } { 2 }$
(C) $\frac { 3 } { 7 } < f ( x ) \leq \frac { 4 } { 9 }$
(D) $\frac { 3 } { 7 } \leq f ( x ) \leq \frac { 1 } { 2 }$

8. If $\mathrm { a } , \mathrm { b } , \mathrm { c } , \mathrm { d }$ are positive real numbers such that $\mathrm { a } + \mathrm { b } + \mathrm { c } + \mathrm { d } = 2$, then $\mathrm { M } = ( \mathrm { a } + \mathrm { b } ) ( \mathrm { c } +$ d) satisfies the relation :
(A) $0 \leq M \leq 1$
(B) $1 \leq \mathrm { M } \leq 2$
(C) $2 \leq M \leq 3$
(D) $3 \leq M \leq 4$

Let $x$ and $y$ be real numbers which satisfy
$$3x^2 + 2xy + 3y^2 = 32. \tag{1}$$
Then we are to find the ranges of the values of $x + y$ and $xy$.
First, we set
$$a = x + y. \tag{2}$$
By eliminating $y$ from (1) and (2), we obtain the quadratic equation in $x$
$$\mathbf{A}x^2 - \mathbf{B}ax + \mathbf{C}a^2 - 32 = 0.$$
Since $x$ is a real number, we have
$$\mathbf{DE} \leqq a \leqq \mathbf{F}.$$
Next, we set
$$b = xy. \tag{4}$$
From (1), (2) and (4) we obtain
$$b = \frac{\mathbf{G}}{\mathbf{H}}a^2 - \mathbf{I}.$$
Hence from (3) and (5) we have
$$\mathbf{JK} \leqq b \leqq \mathbf{L}.$$

Let us consider the real numbers $x , y , t$ and $u$ satisfying the following four conditions:
$$\begin{aligned} & y \geqq | x | \\ & x + y = t \\ & x ^ { 2 } + y ^ { 2 } = 12 \\ & x ^ { 3 } + y ^ { 3 } = u \end{aligned}$$
We are to find the ranges of values which $t$ and $u$ can take.
(1) From (1) and (3), we see that the point $( x , y )$ is located on the arc which is a quadrant of the circle having its center at the origin and the radius $\mathbf { A }$. Moreover, the coordinates of the end points of this arc are
$$( \sqrt { \mathbf { C } } , \sqrt { \mathbf { CD } } ) \text { and } ( - \sqrt { \mathbf { C } } , \sqrt { \mathbf { D } } ) .$$
From this and (2), we also see that the range of values which $t$ can take is
$$\mathbf { E } \leqq t \leqq \mathbf { F } . \mathbf { G } .$$
(2) Next, from (2) and (3), we have
$$x y = \frac { \mathbf { H } } { \mathbf { I } } \left( t ^ { 2 } - \mathbf { J K } \right)$$
and further, using (4) we also have
$$u = \frac { \mathbf { L } } { \mathbf{L}} \left( \mathbf{NO} \, t - t ^ { 3 } \right)$$
Hence, since
$$\frac { d u } { d t } = \frac { \mathbf { P } } { \mathbf { Q } } \left( \mathbf { RS } - t ^ { 2 } \right)$$
the range of values which $u$ can take under the condition (5) is
$$\mathbf { TL } \leqq u \leqq \mathbf { UV } \sqrt { \mathbf{ W } } .$$

Consider two squares as in the figure to the right. Let the coordinates of their vertexes be
$$\begin{array} { l l } \mathrm { A } ( 2 t , 0 ) , \quad \mathrm { B } ( 0,2 t ) , & \mathrm { C } ( - 2 t , 0 ) , \quad \mathrm { D } ( 0 , - 2 t ) , \\ \mathrm { P } \left( 4 - t ^ { 2 } , 4 - t ^ { 2 } \right) , & \mathrm { Q } \left( - 4 + t ^ { 2 } , 4 - t ^ { 2 } \right) , \\ \mathrm { R } \left( - 4 + t ^ { 2 } , - 4 + t ^ { 2 } \right) , & \mathrm { S } \left( 4 - t ^ { 2 } , - 4 + t ^ { 2 } \right) , \end{array}$$
where $0 < t < 2$. Denote the areas of the two squares ABCD and PQRS by $S _ { 1 }$ and $S _ { 2 }$, respectively.
Then we have
$$S _ { 1 } = \mathbf { M } t ^ { 2 } \text { and } S _ { 2 } = \mathbf { N } \left( t ^ { 2 } - \mathbf { O } \right) ^ { 2 } .$$
(1) $S _ { 1 } + S _ { 2 }$ is minimized at $t = \sqrt { \mathbf { P } }$, and the minimum value is $\mathbf { Q } \mathbf { R }$.
(2) For $\mathbf { W }$ and $\mathbf { X }$ below, choose the correct answer from among (0) $\sim$ (9), and for the other $\square$, enter the correct numbers.
We are to find the range of $t$ such that $S _ { 1 } < S _ { 2 }$. If $S _ { 1 } < S _ { 2 }$, then $t$ satisfies the inequality
$$t ^ { 4 } - \mathbf { ST } t ^ { 2 } + \mathbf { UV } > 0 .$$
From the above inequality, a condition on $t ^ { 2 }$ is $\mathbf { W }$. Hence, $S _ { 1 } < S _ { 2 }$ if and only if $t$ satisfies $\mathbf { X }$.
(0) $t ^ { 2 } < 4$ or $6 < t ^ { 2 }$ (1) $4 < t ^ { 2 } < 6$ (2) $t ^ { 2 } < 2$ or $8 < t ^ { 2 }$ (3) $2 < t ^ { 2 } < 8$ (4) $t ^ { 2 } \neq 4$ (5) $0 < t < 2$ (6) $0 < t < \sqrt { 2 }$ (7) $\sqrt { 2 } < t < 2$ (8) $2 < t < \sqrt { 6 }$ (9) $t \neq 2$

For points $(x, y)$ on the boundary of the bounded region between the parabola $y = x ^ { 2 }$ and the line $y = 2 - x$, what is the maximum value of the expression $\mathbf { x } ^ { \mathbf { 2 } } + \mathbf { y } ^ { \mathbf { 2 } }$?
A) 25
B) 20
C) 17
D) 13
E) 10