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 Church of Saint Francis of Assisi, a modernist architectural work by Oscar Niemeyer, located at Pampulha Lake, in Belo Horizonte, has parabolic vaults. Figure 2 provides a front view of one of the vaults, with hypothetical measurements to simplify the calculations.
What is the measure of the height H, in meters, indicated in Figure 2?
(A) $\frac{16}{3}$
(B) $\frac{31}{5}$
(C) $\frac{25}{4}$
(D) $\frac{25}{3}$
(E) $\frac{75}{2}$

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

Let $A$, $B$ and $C$ be unknown constants. Consider the function $f(x)$ defined by
$$\begin{aligned} f(x) &= Ax^2 + Bx + C \text{ when } x \leq 0 \\ &= \ln(5x + 1) \text{ when } x > 0 \end{aligned}$$
Write the values of the constants $A$, $B$ and $C$ such that $f''(x)$, i.e., the double derivative of $f$, exists for all real $x$. If this is not possible, write ``not possible''. If some of the constants cannot be uniquely determined, write ``not unique'' for each such constant.

15. Given the function $\mathrm { f } ( \mathrm { x } ) = \left\{ \begin{array} { l l } x ^ { 3 } , & \mathrm { x } \leq \mathrm { a } , \\ \mathrm { x } ^ { 2 } , & \mathrm { x } > \mathrm { a } , \end{array} \right.$ if there exists a real number $b$ such that the function $\mathrm { g } ( \mathrm { x } ) = \mathrm { f } ( \mathrm { x } ) - \mathrm { b }$ has exactly two zeros, then the range of values for $a$ is $\_\_\_\_$.
III. Solution Questions: This section has 6 questions, for a total of 75 points. Show your work, proofs, or calculation steps in your answers.

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 $x, y, z \in \mathbf{R}$ 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 $x , y$ satisfy the constraints $\left\{ \begin{array} { l } x + y \geqslant 0 , \\ 2 x - y \geqslant 0 , \\ x \leqslant 1 , \end{array} \right.$ then the maximum value of $z = 3 x + 2 y$ is $\_\_\_\_$ .

If $x , y$ satisfy the constraints $\left\{ \begin{array} { l } x - 3 y \leqslant - 1 \\ x + 2 y \leqslant 9 \\ 3 x + y \geqslant 7 \end{array} \right.$, then the maximum value of $z = 2 x - y$ is \_\_\_\_

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 $k \in \mathbb { R } \backslash \{ 9 \}$.

Throughout this problem, $I = ]-1, +\infty[$, and $f(x) = \int_0^{\pi/2} (\sin(t))^x \mathrm{~d}t$.
An application $h$ from a non-trivial interval $J$ of $\mathbf{R}$ to $\mathbf{R}$ is said to be log-convex if, and only if, it takes values in $\mathbf{R}_+^*$ and $\ln \circ h$ is convex on $J$.
Verify that $f$ is an application from $I$ to $\mathbf{R}$ that is log-convex.

Throughout this problem, $I = ]-1, +\infty[$, and $f(x) = \int_0^{\pi/2} (\sin(t))^x \mathrm{~d}t$. We call $\tilde{f}$ the application from $\mathbf{R}_+$ to $\mathbf{R}$, defined by: $$\forall x \in \mathbf{R}^+, \tilde{f}(x) = \ln(f(2x))$$
Suppose here that $x \in \mathbf{R}_+^*$, $(n,p) \in (\mathbf{N}^*)^2$ and $x \leq p$. Verify that $$\tilde{f}(n) - \tilde{f}(n-1) \leq \frac{\tilde{f}(n+x) - \tilde{f}(n)}{x} \leq \frac{\tilde{f}(n+p) - \tilde{f}(n)}{p}$$ and that $(\tilde{f}(n+x) - \tilde{f}(n))$ has a limit as $n$ tends to $+\infty$.

Throughout this problem, $I = ]-1, +\infty[$, and $f(x) = \int_0^{\pi/2} (\sin(t))^x \mathrm{~d}t$. An application $h$ from a non-trivial interval $J$ of $\mathbf{R}$ to $\mathbf{R}$ is said to be log-convex if, and only if, it takes values in $\mathbf{R}_+^*$ and $\ln \circ h$ is convex on $J$. The functional equation referred to as (1) is: $$\forall x \in I, (x+1)f(x) = (x+2)f(x+2)$$
Conclude that $f$ is the unique application from $I$ to $\mathbf{R}$, which is log-convex, which satisfies (1) and such that $$f(0) = \frac{\pi}{2}$$

An application $h$ from a non-trivial interval $J$ of $\mathbf{R}$ to $\mathbf{R}$ is said to be log-convex if, and only if, it takes values in $\mathbf{R}_+^*$ and $\ln \circ h$ is convex on $J$.
Does there exist an application $h$, from $\mathbf{R}$ to $\mathbf{R}$ and log-convex, satisfying $$\forall t \in \mathbf{R}, (t+T)h(t) = (t+2T)h(t+2T)?$$

104. The graph of $y = -x^2 + 2x + 5$ is shifted 3 units to the right along the positive $x$-axis, then 2 units to the negative $y$-axis. What is the new vertex of the parabola?
(1) $(3,4)$ (2) $(2,5)$ (3) $(3,5)$ (4) $(2,6)$

102-- The minimum value of the function $y = mx^2 - 12x + 5m - 1$ for $m = 2$ is the axis of symmetry of the parabola. What is $x$?
(1) $x=2$ (2) $x=2.5$ (3) $x=3$ (4) $x=3.5$

Let $A B C D$ be a unit square. Four points $E , F , G$ and $H$ are chosen on the sides $A B , B C , C D$ and $D A$ respectively. The lengths of the sides of the quadrilateral $E F G H$ are $\alpha , \beta , \gamma$ and $\delta$. Which of the following is always true?
(A) $1 \leq \alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 } + \delta ^ { 2 } \leq 2 \sqrt { 2 }$
(B) $2 \sqrt { 2 } \leq \alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 } + \delta ^ { 2 } \leq 4 \sqrt { 2 }$
(C) $2 \leq \alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 } + \delta ^ { 2 } \leq 4$
(D) $\sqrt { 2 } \leq \alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 } + \delta ^ { 2 } \leq 2 + \sqrt { 2 }$

Let $S$ be the square formed by the four vertices $( 1,1 ) , ( 1 , - 1 ) , ( - 1,1 )$, and $( - 1 , - 1 )$. Let the region $R$ be the set of points inside $S$ which are closer to the centre than to any of the four sides. Find the area of the region $R$.

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) $-\frac{1}{2}$.

Let $K$ be the set of all points $(x , y)$ such that $| x | + | y | \leq 1$. Given a point $A$ in the plane, let $F _ { A }$ be the point in $K$ which is closest to $A$. Then the points $A$ for which $F _ { A } = ( 1,0 )$ are
(A) all points $A = ( x , y )$ with $x \geq 1$.
(B) all points $A = ( x , y )$ with $x \geq y + 1$ and $x \geq 1 - y$.
(C) all points $A = ( x , y )$ with $x \geq 1$ and $y = 0$.
(D) all points $A = ( x , y )$ with $x \geq 0$ and $y = 0$.

17. Let $f ( x ) = ( 1 + b 2 ) x 2 + 2 b x + 1$ and let $m ( b )$ be the minimum value of $f ( x )$. As $b$ varies, the range of $m ( b )$ is:
(A) $[ 0,1 ]$
(B) $[ 0,1 / 2 ]$
(C) $[ 1 / 2,1 ]$
(D) $[ 0,1 ]$

Let $\mathbb { R } ^ { 2 }$ denote $\mathbb { R } \times \mathbb { R }$. Let
$$S = \left\{ ( a , b , c ) : a , b , c \in \mathbb { R } \text { and } a x ^ { 2 } + 2 b x y + c y ^ { 2 } > 0 \text { for all } ( x , y ) \in \mathbb { R } ^ { 2 } - \{ ( 0,0 ) \} \right\}$$
Then which of the following statements is (are) TRUE?
(A) $\left( 2 , \frac { 7 } { 2 } , 6 \right) \in S$
(B) If $\left( 3 , b , \frac { 1 } { 12 } \right) \in S$, then $| 2 b | < 1$.
(C) For any given $( a , b , c ) \in S$, the system of linear equations
$$\begin{aligned} & a x + b y = 1 \\ & b x + c y = - 1 \end{aligned}$$
has a unique solution.
(D) For any given $( a , b , c ) \in S$, the system of linear equations
$$\begin{aligned} & ( a + 1 ) x + b y = 0 \\ & b x + ( c + 1 ) y = 0 \end{aligned}$$
has a unique solution.