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.

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 $| a | + | b |$ .

Give a term for a function $h$ defined on $\mathbb { R }$ such that the term $\sqrt { h ( x ) }$ is defined exactly for $x \in [ - 2 ; 4 ]$. Explain the reasoning underlying your answer.
The figure shows a section of the graph $G$ of the function $f$ defined on $\mathbb { R } \backslash \{ - 3 \}$ with $f ( x ) = x - 3 + \frac { 5 } { x + 3 }$. $G$ has exactly one minimum point $T$. [Figure]
(1a) [4 marks] The lines with equations $x = - 3$ and $y = x - 3$ have special significance for $G$. Draw both lines into the figure and state this significance. Also give the coordinates of the intersection point of the two lines.
(1b) [3 marks] Calculate the coordinates of the intersection point of $G$ with the y-axis. Justify based on the given term of $f$ that $G$ lies above the line with equation $y = x - 3$ for $x > - 3$.
(1c) [3 marks] Prove that $f ( x ) = \frac { x ^ { 2 } - 4 } { x + 3 }$ holds by appropriately transforming the term $x - 3 + \frac { 5 } { x + 3 }$, and justify that $f$ has exactly the zeros $- 2$ and $2$.
(1d) [5 marks] Determine by calculation a term for the first derivative function $f ^ { \prime }$ of $f$ and calculate the x-coordinate of $T$.
(1e) [3 marks] Determine from the figure an approximate value for the integral $\int _ { - 2 } ^ { 2 } f ( x ) \mathrm { dx }$.
Consider the integral function $J : x \mapsto \int _ { - 2 } ^ { x } f ( t ) \mathrm { dt }$ defined on ] $- 3 ; + \infty$ [. (1f) [6 marks] [0pt] Justify that the function $F : x \mapsto \frac { 1 } { 2 } x ^ { 2 } - 3 x + 5 \cdot \ln ( x + 3 )$ defined on ] $- 3 ; + \infty \left[ \right.$ is an antiderivative of $f$ for $x > - 3$. Use this to show that $\lim _ { x \rightarrow - 3 } J ( x ) = - \infty$ holds, and interpret this statement geometrically.
(1g) [3 marks] Justify without further calculation that $J$ has at least two zeros.
Consider the family of functions $f _ { k } : x \mapsto \frac { x ^ { 2 } - k } { x + 3 }$ defined on $\mathbb { R } \backslash \{ - 3 \}$ with $k \in \mathbb { R } \backslash \{ 9 \}$. The graph of $f _ { k }$ is denoted by $G _ { k }$. The function $f$ from task 1 is thus the function $f _ { 4 }$ of this family.
(2a) [4 marks] Give the number of zeros of $f _ { k }$ as a function of $k$ and justify that the function $f _ { 0 }$ of the family has a zero without sign change.
For the first derivative function of $f _ { k }$, we have $f _ { k } ^ { \prime } ( x ) = \frac { x ^ { 2 } + 6 x + k } { ( x + 3 ) ^ { 2 } }$. (2b) [2 marks] Justify that $G _ { k }$ has no extrema for $k > 9$.
The tangent to $G _ { k }$ at the point $\left( 0 \mid f _ { k } ( 0 ) \right)$ is denoted by $t _ { k }$.
(2c) [3 marks] Show that $t _ { k }$ has slope $\frac { k } { 9 }$, and determine the value of $k$ for which $t _ { k }$ is perpendicular to the line with equation $y = x - 3$.
(2d) [4 marks] Give an equation of $t _ { k }$ and assess the following statement: There exists a point that lies on $t _ { k }$ for all $k \in \mathbb { R } \backslash \{ 9 \}$.

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

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 in $x$
$$y = a x ^ { 2 } + b x + c .$$
The function (1) takes its maximum value 16 at $x = 1$, its graph intersects the $x$-axis at two points, and the length of the segment connecting those two points is 8. We are to find the values of $a$, $b$ and $c$.
From the conditions, (1) can be represented as
$$y = a ( x - \mathbf { A } ) ^ { 2 } + \mathbf { B } \mathbf { C }$$
and the coordinates of the two points at which the graph of (1) and the $x$-axis intersect are
$$( - \mathbf { D } , 0 ) , \quad ( \mathbf { E } , 0 ) .$$
Thus we obtain $a = \mathbf { F G }$. Hence we have
$$b = \mathbf { H } , \quad c = \mathbf { I J } .$$

Consider the quadratic function in $x$
$$y = a x ^ { 2 } + b x + c .$$
The function (1) takes its maximum value 16 at $x = 1$, its graph intersects the $x$-axis at two points, and the length of the segment connecting those two points is 8. We are to find the values of $a$, $b$ and $c$.
From the conditions, (1) can be represented as
$$y = a ( x - \mathbf { A } ) ^ { 2 } + \mathbf { B } \mathbf { C } ,$$
and the coordinates of the two points at which the graph of (1) and the $x$-axis intersect are
$$( - \mathbf { D } , 0 ) , \quad ( \mathbf { E } , 0 ) .$$
Thus we obtain $a = \mathbf { F G }$. Hence we have
$$b = \mathbf { H } , \quad c = \mathbf { I J } .$$

Let us consider the quadratic function
$$f ( x ) = \frac { 1 } { 4 } x ^ { 2 } - ( 2 a - 1 ) x + a ,$$
where $a$ is a real number.
(1) The coordinates of the vertex of the graph of $y = f ( x )$ are
$$\left( \mathbf { A } a - \mathbf { B } , - \mathbf { C } a ^ { 2 } + \mathbf { D } a - \mathbf { E } \right) .$$
(2) The range of $a$ such that the graph of $y = f ( x )$ and the $x$-axis intersect at two different points, A and B , is
$$a < \frac { \mathbf { F } } { \mathbf{G} } \text { or } \mathbf { H } < a .$$
(3) The range of $a$ such that the $x$-coordinates of both points A and B in (2) are greater than or equal to 0 and less than or equal to 6 is
$$\mathbf { I } < a \leqq \frac { \mathbf { J K } } { \mathbf { L M } } .$$

Let us consider the quadratic function
$$f ( x ) = \frac { 1 } { 4 } x ^ { 2 } - ( 2 a - 1 ) x + a$$
where $a$ is a real number.
(1) The coordinates of the vertex of the graph of $y = f ( x )$ are
$$\left( \mathbf { A } a - \mathbf { B } , - \mathbf { C } a ^ { 2 } + \mathbf { D } a - \mathbf { E } \right) .$$
(2) The range of $a$ such that the graph of $y = f ( x )$ and the $x$-axis intersect at two different points, A and B , is
$$a < \frac { \mathbf { F } } { \mathbf{G} } \quad \text { or } \mathbf { H } < a .$$
(3) The range of $a$ such that the $x$-coordinates of both points A and B in (2) are greater than or equal to 0 and less than or equal to 6 is
$$\mathbf { I } < a \leqq \frac { \mathbf { J K } } { \mathbf { L M } } .$$

$f(x)$ is a quadratic function in $x$. The graph of $y = f(x)$ passes through the point $(1, -1)$ and has a turning point at $(-1, 3)$.
Find an expression for $f(x)$.

The curve $S$ has equation
$$y = px^2 + 6x - q$$
where $p$ and $q$ are constants.
$S$ has a line of symmetry at $x = -\frac{1}{4}$ and touches the $x$-axis at exactly one point.
What is the value of $p + 8q$?
A $6$
B $18$
C $21$
D $25$
E $38$

Let $0 < x _ { 1 } < x _ { 2 }$. A function f defined on the set of real numbers as
$$f ( x ) = \left( x - x _ { 1 } \right) \left( x - x _ { 2 } \right)$$
The parabola represented by this function intersects the axes at different points A and B in the rectangular coordinate plane as shown in the figure.
The distances from points A and B to the origin are equal, and this parabola takes its minimum value when $x = \frac { 3 } { 5 }$. Accordingly, what is the ratio $\frac { \mathbf { x } _ { \mathbf { 2 } } } { \mathbf { x } _ { \mathbf { 1 } } }$?
A) 2
B) 3
C) 4
D) 5
E) 6

Ayşe, who wants to set aside a portion of her notebook for history class, folds the page she wants to set aside more easily to find, as shown in the figure, from the upper right corner of the page.
In this notebook with rectangular pages, each line on the pages is parallel to the top edge of the page.
Accordingly, what is x in degrees?
A) 50 B) 52 C) 54 D) 56 E) 58

Three right triangles are positioned as follows: their hypotenuses lie on the same line and one vertex of each coincides.
Given that each of the blue angles measures $115^{\circ}$, what is the measure of the yellow angle in degrees?
A) 100 B) 105 C) 110 D) 115 E) 120

A glazier using his extendable ladder to reach a window 6 meters high from the ground positions the ladder 4.8 meters away from a garden wall 3.6 meters high, as shown in Figure 1, and extends the ladder to touch the wall and reach the bottom of the window. The glazier then brings the ladder into the garden and, as shown in Figure 2, leans one end against the wall and extends it to the bottom of the window.
When the glazier positions the ladder as shown in Figure 2, he extends it 3.5 meters less than when he positions it as shown in Figure 1. Accordingly, what is the thickness of the wall in meters?
A) 0.5 B) 0.6 C) 0.7 D) 0.8 E) 0.9

In an application used to adjust the sound level of a music program, consisting of 100 equal units in the shape of a right triangle, the appearance of the application when the sound level is 60 units is given in Figure 1.
When the sound level is increased to 70 units as shown in Figure 2, the area of the green right triangle increases by 260 square units.
Accordingly, what is the height marked with ? in the appearance of the application in units?
A) 30 B) 35 C) 40 D) 45 E) 50

Three square-shaped tables with perimeters of 12, 16, and 28 units are given in Figure 1. These three tables are combined as shown in Figure 2 with no gaps between them to create a new table.
Accordingly, what is the perimeter length of the new table created in units?
A) 42 B) 46 C) 48 D) 52 E) 54

An isosceles trapezoid-shaped cardboard with a height of 30 units has an upper base length of 20 units. When this cardboard is cut along a line parallel to the lower base, reducing the height by 12 units, it is observed that the lower base length decreases by 6 units.
Accordingly, what is the area of the new cardboard obtained in square units?
A) 363 B) 385 C) 441 D) 450 E) 464

The interior angle measure of a regular n-sided polygon is calculated as $\frac{(n-2) \cdot 180^{\circ}}{n}$.
Six identical isosceles trapezoid-shaped mirrors, each with a perimeter of 28 units and shown in Figure 1, are combined as shown in Figure 2 with no gaps between them and all mirrors visible. In the resulting figure, the sum of the perimeter lengths of the red regular hexagon and the blue regular hexagon is 96 units.
Accordingly, what is the area of one of the mirrors used in square units?
A) $18\sqrt{3}$ B) $24\sqrt{3}$ C) $28\sqrt{3}$ D) $30\sqrt{3}$ E) $36\sqrt{3}$

Let $a$ and $b$ be real numbers. The function $f$ defined on the set of real numbers as
$$f(x) = \begin{cases} x^{2} - ax + 6 & , x \leq a \\ 2x + a & , a < x \leq b \\ 11 - 2x + b & , x > b \end{cases}$$
is continuous on its domain.
Accordingly, what is the product $a \cdot b$?
A) 4 B) 6 C) 8 D) 10 E) 12