Question 174
A função $f(x) = x^2 - 4x + 3$ tem vértice no ponto
(A) $(2, -1)$ (B) $(2, 1)$ (C) $(-2, -1)$ (D) $(-2, 1)$ (E) $(1, 0)$

A função $f(x) = x^2 - 4x + 3$ tem vértice no ponto
(A) $(1, 0)$ (B) $(2, -1)$ (C) $(3, 0)$ (D) $(2, 1)$ (E) $(4, 3)$

The interior of a cup was generated by the rotation of a parabola around an axis $z$, as shown in the figure.
The real function that expresses the parabola, in the Cartesian plane of the figure, is given by the law $f(x) = \frac{3}{2}x^{2} - 6x + C$, where $C$ is the measure of the height of the liquid contained in the cup, in centimetres. It is known that the point $V$, in the figure, represents the vertex of the parabola, located on the $x$ axis.
Under these conditions, the height of the liquid contained in the cup, in centimetres, is
(A) 1. (B) 2. (C) 4. (D) 5. (E) 6.

The Body Mass Index (BMI) can be considered a practical, easy and inexpensive alternative for direct measurement of body fat. Its value can be obtained by the formula $\text{BMI} = \frac{\text{Mass}}{(\text{Height})^{2}}$, in which mass is in kilograms and height is in meters. Children naturally begin life with a high body fat index, but become thinner as they age, so scientists created a BMI especially for children and young adults, from two to twenty years of age, called BMI by age.
The graph shows the BMI by age for boys.
A mother decided to calculate the BMI of her son, a ten-year-old boy, with 1.20 m height and $30.92 \mathrm{~kg}$.
To be in the range considered normal for BMI, the minimum and maximum values that this boy needs to lose weight, in kilograms, should be, respectively,
(A) 1.12 and 5.12.
(B) 2.68 and 12.28.
(C) 3.47 and 7.47.
(D) 5.00 and 10.76.
(E) 7.77 and 11.77.

The parabola $y = x^2 - 6x + 8$ has vertex at:
(A) $(2, 0)$
(B) $(3, -1)$
(C) $(3, 1)$
(D) $(4, 0)$
(E) $(6, 8)$

3. Execute the flowchart shown in Figure 1. If the input is $n = 3$, then the output is
A. $\frac { 6 } { 7 }$
B. $\frac { 3 } { 7 }$
C. $\frac { 8 } { 9 }$
D. $\frac { 4 } { 9 }$ [Figure]
(4) If variables $x , y$ satisfy the constraint conditions $\left\{ \begin{array} { c } x + y \geq - 1 \\ 2 x - y \leq 1 \\ y \leq 1 \end{array} \right.$, then the minimum value of $z = 3 x - y$ is
(A) $- 7$
(B) $- 1$
(C) $1$
(D) $2$
(5) Let the function $f ( x ) = \ln ( 1 + x ) - \ln ( 1 - x )$. Then $f ( x )$ is
(A) an odd function and increasing on $( 0,1 )$
(B) an odd function and decreasing on $(0,1)$
(C) an even function and increasing on $( 0,1 )$
(D) an even function and decreasing on $(0,1)$ (6) Given that the expansion of $\left( \sqrt { \mathrm { x } } - \frac { \mathrm { a } } { \sqrt { \mathrm { x } } } \right) ^ { 5 }$ contains a term with $\mathrm { x } ^ { \frac { 3 } { 2 } }$ whose coefficient is 30, then $\mathrm { a } =$
(A) $\sqrt { 3 }$
(B) $- \sqrt { 3 }$
(C) $6$
(D) $- 6$ (7) In the square shown in Figure 2, 10000 points are randomly thrown. The estimated number of points falling in the shaded region (where curve C is the density curve of the normal distribution $N ( 0,1 )$) is
[Figure]
Figure 2
(A) $2386$
(B) $2718$
(C) $3413$
(D) $4772$
Attachment: If $X \sim N \left( \mu , \sigma ^ { 2 } \right)$, then
$$\begin{aligned} & P ( \mu - \sigma < x \leq \mu + \sigma ) = 0.6826 \\ & P ( \mu - 2 \sigma < x \leq \mu + 2 \sigma ) = 0.9544 \end{aligned}$$
(8) Points $\mathrm { A } , \mathrm { B } , \mathrm { C }$ move on the circle $\mathrm { x } ^ { 2 } + \mathrm { y } ^ { 2 } = 1$, and $\mathrm { AB } \perp \mathrm { BC }$. If point P has coordinates $( 2,0 )$, then the maximum value of $| \overrightarrow { P A } + \overrightarrow { P B } + \overrightarrow { P C } |$ is
A. $6$
B. $7$
C. $8$
D. $9$ (9) The graph of the function $f ( x ) = \sin 2 x$ is shifted to the right by $\varphi \left( 0 < \varphi < \frac { \pi } { 2 } \right)$ units to obtain the graph of function $g ( x )$. If for $x _ { 1 }$ and $x _ { 2 }$ satisfying $\left| f \left( x _ { 1 } \right) - g \left( x _ { 2 } \right) \right| = 2$, we have $\left| x _ { 1 } - x _ { 2 } \right| _ { \min } = \frac { \pi } { 3 }$, then $\varphi =$
A. $\frac { 5 \pi } { 12 }$
B. $\frac { \pi } { 3 }$
C. $\frac { \pi } { 4 }$
D. $\frac { \pi } { 6 }$

4. If variables $x , y$ satisfy the constraint conditions $\left\{ \begin{array} { c } x + y \geq 1 , \\ y - x \leq 1 , \text { then } z = 2 x - y \text { has a minimum value of } \\ x \leq 1 , \end{array} \right.$
A. $-1$
B. 0
C. 1
D. 2

17. (11 points)
A student uses the ``five-point method'' to sketch the graph of $f(x) = A\sin(\omega x + \varphi)$ (where $\omega > 0$, $|\varphi| < \frac{\pi}{2}$) over one period and creates a table with partial data filled in as follows:
\begin{tabular}{ | c | c | c | c | c | c | } \hline $\omega x + \varphi$ & 0 & $\frac{\pi}{2}$ & $\pi$ & $\frac{3\pi}{2}$ & $2\pi$ \hline $x$ & & $\frac Thus $\overrightarrow { P B } \cdot \overrightarrow { D E } = 0$, that is, $P B \perp D E$. Since $E F \perp P B$ and $D E \cap E F = E$, we have $P B \perp$ plane $D E F$. Because $\overrightarrow { P C } = ( 0,1 , - 1 ) , \overrightarrow { D E } \cdot \overrightarrow { P C } = 0$, then $D E \perp P C$, so $D E \perp$ plane $P B C$. From $D E \perp$ plane $P B C$ and $P B \perp$ plane $D E F$, we know that all four faces of tetrahedron $B D E F$ are right triangles, that is, tetrahedron $B D E F$ is an orthocentric tetrahedron, with right angles on its four faces being $\angle D E B , \angle D E F , \angle E F B , \angle D F B$ respectively.
[Figure]
Solution diagram 1 for Question 19
[Figure]
Solution diagram 2 for Question 19
(II) Since $P D \perp$ plane $A B C D$, we have $\overrightarrow { D P } = ( 0,0,1 )$ is a normal vector to plane $A B C D$; From (I), $P B \perp$ plane $D E F$, so $\overrightarrow { B P } = ( - \lambda , - 1,1 )$ is a normal vector to plane $D E F$. If the dihedral angle between plane $D E F$ and plane $A B C D$ is $\frac { \pi } { 3 }$, then $\cos \frac { \pi } { 3 } = \left| \frac { \overrightarrow { B P } \cdot \overrightarrow { D P } } { | \overrightarrow { B P } | \cdot | \overrightarrow { D P } | } \right| = \left| \frac { 1 } { \sqrt { \lambda ^ { 2 } + 2 } } \right| = \frac { 1 } { 2 }$, solving gives $\lambda = \sqrt { 2 }$. Therefore $\frac { D C } { B C } = \frac { 1 } { \lambda } = \frac { \sqrt { 2 } } { 2 }$. Thus when the dihedral angle between plane $D E F$ and plane $A B C D$ is $\frac { \pi } { 3 }$, we have $\frac { D C } { B C } = \frac { \sqrt { 2 } } { 2 }$.

18. (This question is worth 15 points)
Given the function $f ( x ) = x ^ { 2 } + ax + b$ ( $a , b \in \mathbb{R}$ ), let $M ( a , b )$ denote the maximum value of $| f ( x ) |$ on the interval $[ - 1 , 1 ]$ . (I) Prove that when $| a | \geq 2$ , $M ( a , b ) \geq 2$ ; (II) When $a , b$ satisfy $M ( a , b ) \leq 2$ , find the maximum value of $| a | + | b |$ .

20. (12 marks) (I) Let the daily production quantities of products $A$ and $B$ be $x$ and $y$ respectively, with corresponding profit $z$. Then we have
$$\left\{ \begin{array} { l } 2 x + 1.5 y \leq W \\ x + 1.5 y \leq 12 \\ 2 x - y \geq 0 \\ x \geq 0 , \quad y \geq 0 \end{array} \right.$$
The objective function is $z = 1000x + 1200y$.
[Figure]
Solution diagram 1 for Question 20
[Figure]
Solution diagram 2 for Question 20
[Figure]
Solution diagram 3 for Question 20
When $W = 12$, the planar region represented by (1) is shown in Figure 1, with three vertices $A ( 0,0 ) , B ( 2.4,4.8 ) , C ( 6,0 )$ respectively. Rewriting $z = 1000 x + 1200 y$ as $y = - \frac { 5 } { 6 } x + \frac { z } { 1200 }$, when $x = 2.4 , y = 4.8$, the line $l : y = - \frac { 5 } { 6 } x + \frac { z } { 1200 }$ has maximum $y$-intercept, maximum profit $Z = z _ { \max } = 2.4 \times 1000 + 4.8 \times 1200 = 8160$. When $W = 15$, the planar region represented by (1) is shown in Figure 2, with three vertices $A ( 0,0 ) , B ( 3,6 ) , C ( 7.5,0 )$ respectively. Rewriting $z = 1000 x + 1200 y$ as $y = - \frac { 5 } { 6 } x + \frac { z } { 1200 }$, when $x = 3 , y = 6$, the line $l : y = - \frac { 5 } { 6 } x + \frac { z } { 1200 }$ has maximum $y$-intercept, maximum profit $Z = z _ { \max } = 3 \times 1000 + 6 \times 1200 = 10200$. When $W = 18$, the planar region represented by (1) is shown in Figure 3, with four vertices $A ( 0,0 ) , B ( 3,6 ) , C ( 6,4 ) , D ( 9,0 )$ respectively. Rewriting $z = 1000 x + 1200 y$ as $y = - \frac { 5 } { 6 } x + \frac { z } { 1200 }$, when $x = 6 , y = 4$, the line $l : y = - \frac { 5 } { 6 } x + \frac { z } { 1200 }$ has maximum $y$-intercept, maximum profit $Z = z _ { \max } = 6 \times 1000 + 4 \times 1200 = 10800$. Thus the distribution of maximum profit $Z$ is

$Z$	8160	10200	10800
$P$	0.3	0.5	0.2

Therefore, $E ( Z ) = 8160 \times 0.3 + 10200 \times 0.5 + 10800 \times 0.2 = 9708$. (II) From (I), the probability that daily maximum profit exceeds 10000 yuan is $p _ { 1 } = P ( Z > 10000 ) = 0.5 + 0.2 = 0.7$. By the binomial distribution, the probability that at least one day out of 3 days has maximum profit exceeding 10000 yuan is $p = 1 - \left( 1 - p _ { 1 } \right) ^ { 3 } = 1 - 0.3 ^ { 3 } = 0.973$.

Let $a, b, c$ be real numbers such that $a = a^2 + b^2 + c^2$. What is the smallest possible value of $b$?
(A) 0
(B) $-1$
(C) $-\frac{1}{4}$
(D) $-\frac{1}{2}$.

If non-zero real numbers $b$ and $c$ are such that $\min f ( x ) > \max g ( x )$, where $f ( x ) = x ^ { 2 } + 2 b x + 2 c ^ { 2 }$ and $g ( x ) = - x ^ { 2 } - 2 c x + b ^ { 2 } , ( x \in R )$; then $\left| \frac { c } { b } \right|$ lies in the interval
(1) $( \sqrt { 2 } , \infty )$
(2) $\left[ \frac { 1 } { 2 } , \frac { 1 } { \sqrt { 2 } } \right)$
(3) $\left( 0 , \frac { 1 } { 2 } \right)$
(4) $\left[ \frac { 1 } { \sqrt { 2 } } , \sqrt { 2 } \right]$

If $x^2 - y^2 + 2hxy + 2gx + 2fy + c = 0$ is the locus of a point, which moves such that it is always equidistant from the lines $x + 2y + 7 = 0$ and $2x - y + 8 = 0$, then the value of $g + c + h - f$ equals
(1) 14
(2) 6
(3) 8
(4) 29

Let $a$ be a constant. Consider the quadratic function in $x$
$$y = 2 x ^ { 2 } + a x + 3 .$$
Suppose that the vertex of the graph of (1) is in the first (upper right-hand) quadrant.
(1) The range of values which $a$ can take is
$$\mathbf { A B } \sqrt { \mathbf { C } } < a < \mathbf { D } ,$$
and the least integer $a$ satisfying this inequality is $\mathbf{EF}$.
(2) Let $a = \mathrm { EF }$ in (1). Let
$$y = 2 x ^ { 2 } + p x + q$$
be the equation of the graph which is obtained by translating the graph of (1) by $- \frac { 1 } { n }$ in the $x$-direction and by $\frac { 6 } { n ^ { 2 } }$ in the $y$-direction. Then
$$p = \frac { \mathbf { G } } { n } - \mathbf { H } , \quad q = \frac { \mathbf { I } } { n ^ { 2 } } - \frac { \mathbf { J } } { n } + \mathbf { K } .$$
(3) The total number of natural numbers $n$ for which $p$ in (2) is an integer is $\mathbf { L }$. Among these $n$, consider the ones such that the value of $q$ is also an integer. Then $\mathbf { M }$ is the value of the $n$ for which $q$ takes the minimum value $\mathbf { N }$.

Consider two quadratic functions
$$\begin{aligned} & y = 2 x ^ { 2 } + 3 a x + 4 b \tag{1}\\ & y = b x ^ { 2 } + c x + d \tag{2} \end{aligned}$$
whose graphs are mutually symmetric with respect to the origin.
(1) From the symmetry with respect to the origin we see that
$$b = \mathbf { AB } , \quad c = \mathbf { C } a , \quad d = \mathbf { D } .$$
Hence (2) can be reduced to
$$y = \mathbf{AB} x ^ { 2 } + \mathbf{C} a x + \mathbf{D} . \tag{3}$$
(2) Let $0 < a < 1$, and consider the graph of (3).
When the range of values of $x$ is $0 \leqq x \leqq \frac { 3 } { 2 }$, the range of values of $y$ in (3) is
$$\frac { \mathbf { E } } { \mathbf { F } } a ^ { 2 } - \frac { \mathbf { G } } { \mathbf { H } } \leq y \leqq \frac { \mathbf { I } } { \mathbf{J}} a ^ { 2 } + \mathbf { K }$$
(3) For any value of $a$, the vertex of the graph of (3) is on the graph of the quadratic function
$$y = \mathbf { L } x ^ { 2 } + \mathbf { M } .$$

Let $a$ and $b$ be constants where $a > 0$. Translate the graph of the quadratic function
$$y = 4x^2 + 2ax + b$$
by $a$ in the $x$-direction and by $1 - 7a$ in the $y$-direction. If this graph passes through the point $(0, 4)$, we have
$$b = \mathbf{AB}\, a^2 + \mathbf{C}\, a + \mathbf{D},$$
and the quadratic function representing the graph resulting from these translations is
$$y = \mathbf{E}\, x^2 - \mathbf{F}\, ax + \mathbf{G}.$$
When the graph of quadratic function (1) is tangent to the $x$-axis, we have $a = \frac{\mathbf{H}}{\mathbf{I}}$, and the $x$-coordinate of the point of tangency is $x = \mathbf{J}$.

Let $a$ and $b$ be constants where $a > 0$. Translate the graph of the quadratic function
$$y = 4x^2 + 2ax + b$$
by $a$ in the $x$-direction and by $1 - 7a$ in the $y$-direction. If this graph passes through the point $(0, 4)$, we have
$$b = \mathbf{AB}\, a^2 + \mathbf{C}\, a + \mathbf{D}$$
and the quadratic function representing the graph resulting from these translations is
$$y = \mathbf{E}\, x^2 - \mathbf{F}\, ax + \mathbf{G}.$$
When the graph of quadratic function (1) is tangent to the $x$-axis, we have $a = \frac{\mathbf{H}}{\mathbf{I}}$, and the $x$-coordinate of the point of tangency is $x = \mathbf{J}$.

Let $a \neq 0$. Let $G$ be a curve which is symmetric with respect to the origin $(0,0)$ to the graph of the quadratic function in $x$
$$y = ax^2 - 4x - 4a. \tag{1}$$
(1) The coordinates of the vertex of the graph of (1) are
$$\left( \frac{\mathbf{A}}{a}, -\frac{\mathbf{B}}{a} - 4a \right).$$
(2) Among the following choices, the quadratic function whose graph is $G$ is $\square$ C. (0) $y = ax^2 + 4x + 4a$
(1) $y = ax^2 + 4x - 4a$
(2) $y = ax^2 - 4x + 4a$
(3) $y = -ax^2 + 4x + 4a$
(4) $y = -ax^2 - 4x + 4a$
(5) $y = -ax^2 - 4x - 4a$
(3) The curve $G$ intersects the graph of the quadratic function (1) at the two points
$$(\mathrm{DE},\ \mathrm{F}) \text{ and } (\mathrm{G},\ \mathrm{HI}).$$
(4) Let $a = 2$. Then over the interval $\mathrm{DE} \leqq x \leqq \square$, the maximum and the minimum values of the quadratic function whose graph is $G$ are $\square$ JK and $\square$ LM, respectively.

Let $a \neq 0$. Let $G$ be a curve which is symmetric with respect to the origin $(0,0)$ to the graph of the quadratic function in $x$
$$y = ax^2 - 4x - 4a. \tag{1}$$
(1) The coordinates of the vertex of the graph of (1) are
$$\left( \frac{\mathbf{A}}{a}, -\frac{\mathbf{B}}{a} - 4a \right).$$
(2) Among the following choices, the quadratic function whose graph is $G$ is $\square$ C. (0) $y = ax^2 + 4x + 4a$
(1) $y = ax^2 + 4x - 4a$
(2) $y = ax^2 - 4x + 4a$
(3) $y = -ax^2 + 4x + 4a$
(4) $y = -ax^2 - 4x + 4a$
(5) $y = -ax^2 - 4x - 4a$
(3) The curve $G$ intersects the graph of the quadratic function (1) at the two points
$$(\mathrm{DE},\ \mathrm{F}) \text{ and } (\mathrm{G},\ \mathrm{HI}).$$
(4) Let $a = 2$. Then over the interval $\mathrm{DE} \leqq x \leqq \square\mathrm{G}$, the maximum and the minimum values of the quadratic function whose graph is $G$ are JK and LM, respectively.

Suppose we have a quadratic function $y = a x ^ { 2 } + b x + c$ in $x$ which satisfies the following conditions 【＊】：
【＊】 When $x = - 1$ ，then $y = - 8$ and when $x = 3$ ，then $y = 16$ ．Further，in the interval $- 1 \leqq x \leqq 3$ ，the value of $y$ increases with the increase of the value of $x$ ．
We are to find the conditions which $a , b$ and $c$ must satisfy．
From【＊】，it follows that $b$ and $c$ can be expressed in terms of $a$ as
$$\begin{aligned} & b = \mathbf { A B } a + \mathbf { A } \\ & c = \mathbf { D E } a - \mathbf { F } . \end{aligned}$$
Hence，the axis of symmetry of the graph of this quadratic function has the equation
$$x = \mathbf { G } - \frac { \mathbf { H } } { a } .$$
Thus $a , b$ and $c$ must satisfy the relationships（1）and（2），and furthermore
$$0 < a \leqq \frac { \mathbf { I } } { \mathbf { J } } \quad \text { or } \frac { \mathbf { K L } } { \mathbf { M } } \leqq a < 0 .$$

Suppose we have a quadratic function $y = a x ^ { 2 } + b x + c$ in $x$ which satisfies the following conditions 【＊】：
【＊】 When $x = - 1$ ，then $y = - 8$ and when $x = 3$ ，then $y = 16$ ．Further，in the interval $- 1 \leqq x \leqq 3$ ，the value of $y$ increases with the increase of the value of $x$ ．
We are to find the conditions which $a , b$ and $c$ must satisfy．
From【＊】，it follows that $b$ and $c$ can be expressed in terms of $a$ as
$$\begin{aligned} & b = \mathbf { A B } a + \mathbf { C } \\ & c = \mathbf { D E } a - \mathbf { F } . \end{aligned}$$
Hence，the axis of symmetry of the graph of this quadratic function has the equation
$$x = \mathbf { G } - \frac { \mathbf { H } } { a } .$$
Thus $a , b$ and $c$ must satisfy the relationships（1）and（2），and furthermore
$$0 < a \leqq \frac { \mathbf { I } } { \mathbf { J } } \quad \text { or } \frac { \mathbf { K L } } { \mathbf { M } } \leqq a < 0 .$$

Consider the quadratic function
$$y = - x ^ { 2 } - a x + 3 .$$
(1) If $a > 0$ and the maximum value of function (1) is 7 , then $a = \square$. In this case, the equation of the axis of symmetry of the graph is $x = \mathbf { B C }$, and the $x$-coordinates of the points of intersection of this graph and the $x$-axis are $\mathbf { D E } \pm \sqrt { \mathbf { F } }.$
(2) If the curve obtained by translating the graph of function (1) by 2 in the $x$-direction and by $-3$ in the $y$-direction passes through $( - 3 , - 5 )$, then $a =$ $\square$ G.

Consider the quadratic function
$$y = - x ^ { 2 } - a x + 3 .$$
(1) If $a > 0$ and the maximum value of function (1) is 7 , then $a = \square$. In this case, the equation of the axis of symmetry of the graph is $x = \mathbf { B C }$, and the $x$-coordinates of the points of intersection of this graph and the $x$-axis are $\mathbf { D E } \pm \sqrt { \mathbf { F } }.$
(2) If the curve obtained by translating the graph of function (1) by 2 in the $x$-direction and by $-3$ in the $y$-direction passes through $( - 3 , - 5 )$, then $a =$ $\square$ G.

Consider the two parabolas $$\begin{aligned} \ell : & & y = ax^2 + 2bx + c \\ m : & & y = (a+1)x^2 + 2(b+2)x + c + 3. \end{aligned}$$ Four points A, B, C and D are assumed to be in the relative positions shown in the figure. One of the two parabolas passes through the three points A, B and C, and the other one passes through the three points B, C and D.
(1) The parabola passing through the three points A, B and C is $\mathbf{A}$. Here, for $\mathbf{A}$ choose the correct answer from (0) or (1), just below. (0) parabola $\ell$ (1) parabola $m$
(2) Since both parabolas $\ell$ and $m$ pass through the two points B and C, the $x$-coordinates of B and C are the solutions of the quadratic equation $$x^2 + \mathbf{B}x + \mathbf{C} = 0.$$ Hence, the $x$-coordinate of point B is $\mathbf{DE}$, and the $x$-coordinate of point C is $\mathbf{FG}$.
(3) In particular, we are to find the values of $a$, $b$ and $c$ when $\mathrm{AB} = \mathrm{BC}$ and $\mathrm{CO} = \mathrm{OD}$.
Since the two points C and D are symmetric with respect to the $y$-axis, we have $b = \mathbf{H}$. On the other hand, since $\mathrm{AB} = \mathrm{BC}$, the straight line $x = \mathbf{IJ}$ is the axis of symmetry of $\mathbf{A}$. Hence we have $a = -\frac{\mathbf{K}}{\mathbf{L}}$. And we have $c = \frac{\mathbf{M}}{\mathbf{L}}$.

Let $a$ and $b$ be real numbers where $0 < b < 7$. Let us consider the maximum value $M$ and the minimum value $m$ of the quadratic function
$$f ( x ) = x ^ { 2 } - 6 x + a$$
over the interval $b \leqq x \leqq 7$.
The function $f ( x )$ can be represented as
$$f ( x ) = ( x - \mathbf { A } ) ^ { 2 } + a - \mathbf { B } .$$
(1) For each of $\mathbf { C }$ ~ $\mathbf { G }$ in the following statements, choose the correct answer from among (0) ~ (9) below.
We are to find $M$ and $m$. There are two cases.
(i) When $0 < b \leqq \mathbf { C }$, then
$$M = \mathbf { D } , \quad m = \mathbf { E } .$$
(ii) When $\mathbf { C } < b < 7$, then
$$M = \mathbf { F } , \quad m = \mathbf { G } .$$
(0) 0
(1) 1
(2) 2
(3) 3
(4) $a - 6$
(5) $a + 7$ (6) $a + 8$ (7) $a - 9$ (8) $b ^ { 2 } - 6 b + a$ (9) $b ^ { 2 } + 6 b + a$
(2) In the case that $M = 13$ and $m = 1$, we have
$$a = \mathbf { H } , \quad b = \mathbf { I } .$$