Let $m$ and $n$ be the numbers of real roots of the quadratic equations $x ^ { 2 } - 12 x + [ x ] + 31 = 0$ and $x ^ { 2 } - 5 | x + 2 | - 4 = 0$ respectively, where $[ x ]$ denotes the greatest integer $\leq x$. Then $m ^ { 2 } + m n + n ^ { 2 }$ is equal to

Let $S = \left\{ \alpha : \log _ { 2 } \left( 9 ^ { 2 \alpha - 4 } + 13 \right) - \log _ { 2 } \left( \frac { 5 } { 2 } \cdot 3 ^ { 2 \alpha - 4 } + 1 \right) = 2 \right\}$. Then the maximum value of $\beta$ for which the equation $\mathrm { x } ^ { 2 } - 2 \left( \sum _ { \alpha \in s } \alpha \right) ^ { 2 } \mathrm { x } + \sum _ { a \in s } ( \alpha + 1 ) ^ { 2 } \beta = 0$ has real roots, is $\_\_\_\_$ .

If the value of real number $\alpha > 0$ for which $x^{2} - 5\alpha x + 1 = 0$ and $x^{2} - \alpha x - 5 = 0$ have a common real roots is $\frac{3}{\sqrt{2\beta}}$ then $\beta$ is equal to $\_\_\_\_$

The number of real solutions of the equation $3\left(x^2 + \frac{1}{x^2}\right) - 2\left(x + \frac{1}{x}\right) + 5 = 0$ is
(1) 4
(2) 0
(3) 3
(4) 2

The equation $x ^ { 2 } - 4 x + [ x ] + 3 = x [ x ]$, where $[ x ]$ denotes the greatest integer function, has:
(1) exactly two solutions in $( - \infty , \infty )$
(2) no solution
(3) a unique solution in $( - \infty$, 1)
(4) a unique solution in $( - \infty , \infty )$

Let $q$ be the maximum integral value of $p$ in $[0, 10]$ for which the roots of the equation $x^{2} - px + \frac{5}{4}p = 0$ are rational. Then the area of the region $\left\{(x, y): 0 \leq y \leq (x - q)^{2},\, 0 \leq x \leq q\right\}$ is
(1) 243
(2) 25
(3) $\frac{125}{3}$
(4) 164

For $0 < c < b < a$, let $( a + b - 2 c ) x ^ { 2 } + ( b + c - 2 a ) x + ( c + a - 2 b ) = 0$ and $\alpha \neq 1$ be one of its root. Then, among the two statements (I) If $\alpha \in ( -1, 0)$, then $b$ cannot be the geometric mean of $a$ and $c$. (II) If $\alpha \in (0, 1)$, then $b$ may be the geometric mean of $a$ and $c$.
(1) Both (I) and (II) are true
(2) Neither (I) nor (II) is true
(3) Only (II) is true
(4) Only (I) is true

Let $x = \frac { m } { n } \left( m , n \right.$ are co-prime natural numbers) be a solution of the equation $\cos \left( 2 \sin ^ { - 1 } x \right) = \frac { 1 } { 9 }$ and let $\alpha , \beta ( \alpha > \beta )$ be the roots of the equation $m x ^ { 2 } - n x - m + n = 0$. Then the point $( \alpha , \beta )$ lies on the line
(1) $3 x + 2 y = 2$
(2) $5 x - 8 y = - 9$
(3) $3 x - 2 y = - 2$
(4) $5 x + 8 y = 9$

Let $\int _ { \alpha } ^ { \log _ { e } 4 } \frac { \mathrm {~d} x } { \sqrt { \mathrm { e } ^ { x } - 1 } } = \frac { \pi } { 6 }$. Then $\mathrm { e } ^ { \alpha }$ and $\mathrm { e } ^ { - \alpha }$ are the roots of the equation : (1) $x ^ { 2 } + 2 x - 8 = 0$ (2) $x ^ { 2 } - 2 x - 8 = 0$ (3) $2 x ^ { 2 } - 5 x + 2 = 0$ (4) $2 x ^ { 2 } - 5 x - 2 = 0$

Let $f : \mathbb { R } \rightarrow \mathbb { R }$ be a function defined by $f ( x ) = ( 2 + 3 a ) x ^ { 2 } + \left( \frac { a + 2 } { a - 1 } \right) x + b , a \neq 1$. If $f ( x + \mathrm { y } ) = f ( x ) + f ( \mathrm { y } ) + 1 - \frac { 2 } { 7 } x \mathrm { y }$, then the value of $28 \sum _ { i = 1 } ^ { 5 } | f ( i ) |$ is
(1) 545
(2) 715
(3) 735
(4) 675

The number of solutions of the equation $\left(\frac{9}{x} - \frac{9}{\sqrt{x}} + 2\right)\left(\frac{2}{x} - \frac{7}{\sqrt{x}} + 3\right) = 0$ is:
(1) 2
(2) 3
(3) 1
(4) 4

The product of all the rational roots of the equation $\left(x^2 - 9x + 11\right)^2 - (x-4)(x-5) = 3$, is equal to
(1) 14
(2) 21
(3) 28
(4) 7

Let $\alpha _ { \theta }$ and $\beta _ { \theta }$ be the distinct roots of $2 x ^ { 2 } + ( \cos \theta ) x - 1 = 0 , \theta \in ( 0,2 \pi )$. If m and M are the minimum and the maximum values of $\alpha _ { \theta } ^ { 4 } + \beta _ { \theta } ^ { 4 }$, then $16 ( M + m )$ equals :
(1) 24
(2) 25
(3) 17
(4) 27

Q1 Consider the equation
$$(x-1)^2 = |3x-5|.$$
(1) Among all solutions of equation (1), the solutions satisfying $x \geqq \frac{5}{3}$ are $x = \mathbf{A}$ and $x = \mathbf{B}$, where $\mathbf{A} < \mathbf{B}$.
(2) Equation (1) has a total of $\mathbf{C}$ solutions. When the minimum one is denoted by $\alpha$, the integer $m$ satisfying $m-1 < \alpha \leqq m$ is $\mathbf{DE}$.

Consider the following quadratic equations in $x$
$$x^2+2x-15=0 \tag{1}$$ $$2x^2+3x+a^2+12a=0 \tag{2}$$
Let us denote the two solutions of (1) by $\alpha$ and $\beta$ ($\alpha < \beta$). We are to find the range of values which $a$ in (2) can take, in order that (2) has two real solutions $\gamma$ and $\delta$ and they satisfy
$$\alpha < \gamma < \beta < \delta.$$
(1) $\alpha = \mathbf{AB}$ and $\beta = \mathbf{C}$.
(2) When we set $b = a^2+12a$, from the condition $\alpha < \gamma$ we have
$$b > \mathbf{DEF}$$
and from the condition $\gamma < \beta < \delta$ we have
$$b < \mathbf{GHI}.$$
Hence the range of the values which $a$ can take is
$$\mathbf{JK} < a < \mathbf{LM}, \quad \mathbf{NO} < a < \mathbf{PQ},$$
where $\mathrm{JK} < \mathrm{NO}$.

Q1 Consider the equation
$$(x-1)^2 = |3x-5|.$$
(1) Among all solutions of equation (1), the solutions satisfying $x \geqq \frac{5}{3}$ are $x = \mathbf{A}$ and $x = \mathbf{B}$, where $\mathbf{A} < \mathbf{B}$.
(2) Equation (1) has a total of $\mathbf{C}$ solutions. When the minimum one is denoted by $\alpha$, the integer $m$ satisfying $m-1 < \alpha \leqq m$ is $\mathbf{DE}$.

Let $a$ be a constant. For the two functions in $x$
$$\begin{aligned} & f ( x ) = 2 x ^ { 2 } + x + a - 2 \\ & g ( x ) = - 4 x - 5 \end{aligned}$$
we are to find the real values of $x$ for which $f ( x ) = g ( x )$ and also find the values of the two functions there.
(1) For each of $\mathbf { N } , \mathbf { O }$ and $\mathbf { P }$ in the following statements, choose the appropriate condition from (0) $\sim$ (8) below.
When $\mathbf { N }$, there are two real values of $x$ for which $f ( x ) = g ( x )$. When $\mathbf { O }$, there is only one real value of $x$ for which $f ( x ) = g ( x )$. When $\mathbf{P}$, there is no real value of $x$ for which $f ( x ) = g ( x )$.
(0) $a > \frac { 1 } { 8 }$
(1) $a = \frac { 17 } { 8 }$
(2) $a = \frac { 1 } { 6 }$
(3) $a < \frac { 1 } { 6 }$
(4) $a < \frac { 17 } { 8 }$
(5) $a < \frac { 1 } { 8 }$ (6) $a > \frac { 1 } { 6 }$ (7) $a = \frac { 1 } { 8 }$ (8) $a > \frac { 17 } { 8 }$
(2) When N, the values of $x$ for which $f ( x ) = g ( x )$ are $\frac { - \mathrm { Q } \pm \sqrt { \mathrm { R } - \mathbf { S } a } } { \mathbf{T} }$, and the values of the functions there are $\mp \sqrt { \mathbf { U } - \mathbf { V } a }$.
When O, the value of $x$ for which $f ( x ) = g ( x )$ is $- \frac { \mathrm { W } } { \mathrm { X} }$, and the value of the functions there is $\mathbf{Y}$.
(3) Consider the case where $f ( x ) = g ( x )$ and the absolute value of these functions there is greater than or equal to 3. The condition for this case is that $a \leqq - \mathbf { Z }$.

Q1 A quadratic function $y = ax^2 + bx + \frac{3}{a}$ satisfies the following two conditions:
(i) $y$ is maximized at $x = 3$,
(ii) the value of $y$ at $x = 1$ is 2.
We are to find the values of $a$ and $b$.
Using conditions (i) and (ii), we obtain the following relationships between $a$ and $b$:
$$\left\{ \begin{aligned} b & = \mathbf{AB}a \\ \mathbf{C} & = a + b + \frac{\mathbf{D}}{a}. \end{aligned} \right.$$
From these two equalities, we have the equation
$$\mathbf{E}a^2 + \mathbf{F}a - \mathbf{G} = 0$$
and hence
$$a = \mathbf{HI}, \quad b = \mathbf{J}.$$
Thus the maximum value of this function is $\mathbf{K}$.

Q1 A quadratic function $y = ax^2 + bx + \frac{3}{a}$ satisfies the following two conditions:
(i) $y$ is maximized at $x = 3$,
(ii) the value of $y$ at $x = 1$ is 2.
We are to find the values of $a$ and $b$.
Using conditions (i) and (ii), we obtain the following relationships between $a$ and $b$:
$$\left\{ \begin{aligned} b & = \mathbf{AB}a \\ \mathbf{C} & = a + b + \frac{\mathbf{D}}{a}. \end{aligned} \right.$$
From these two equalities, we have the equation
$$\mathbf{E}a^2 + \mathbf{F}a - \mathbf{G} = 0$$
and hence
$$a = \mathbf{HI}, \quad b = \mathbf{J}.$$
Thus the maximum value of this function is $\mathbf{K}$.

Let $a$ and $b$ be real numbers, where $a > 0$. Consider the two quadratic functions
$$f ( x ) = 2 x ^ { 2 } - 4 x + 5 , \quad g ( x ) = x ^ { 2 } + a x + b .$$
We are to find the values of $a$ and $b$ when the function $g ( x )$ satisfies the following two conditions.
(i) The minimum value of $g ( x )$ is 8 less than the minimum value of $f ( x )$.
(ii) There exists only one $x$ which satisfies $f ( x ) = g ( x )$.
Since the minimum value of $f ( x )$ is $\mathbf { A }$, from condition (i), we derive the equality
$$b = \frac { a ^ { 2 } } { \mathbf { B } } - \mathbf { C } \text {. }$$
Hence the equation from which we can find the $x$ satisfying $f ( x ) = g ( x )$ is
$$x ^ { 2 } - ( a + \mathbf { D } ) x - \frac { a ^ { 2 } } { \mathbf { E } } + \mathbf { F G } = 0 .$$
Thus, since $a > 0$, from condition (ii) we obtain
$$a = \mathbf { H } , \quad b = \mathbf { I J } .$$
In this case, the $x$ satisfying $f ( x ) = g ( x )$ is $\square \mathbf{ K }$.

Let $a$ and $b$ be real numbers, where $a > 0$. Consider the two quadratic functions
$$f ( x ) = 2 x ^ { 2 } - 4 x + 5 , \quad g ( x ) = x ^ { 2 } + a x + b .$$
We are to find the values of $a$ and $b$ when the function $g ( x )$ satisfies the following two conditions.
(i) The minimum value of $g ( x )$ is 8 less than the minimum value of $f ( x )$.
(ii) There exists only one $x$ which satisfies $f ( x ) = g ( x )$.
Since the minimum value of $f ( x )$ is $\mathbf{A}$, from condition (i), we derive the equality
$$b = \frac { a ^ { 2 } } { \mathbf { B } } - \mathbf { C } \text {. }$$
Hence the equation from which we can find the $x$ satisfying $f ( x ) = g ( x )$ is
$$x ^ { 2 } - ( a + \mathbf { D } ) x - \frac { a ^ { 2 } } { \mathbf { E } } + \mathbf { F G } = 0 .$$
Thus, since $a > 0$, from condition (ii) we obtain
$$a = \mathbf { H } , \quad b = \mathbf { I J } .$$
In this case, the $x$ satisfying $f ( x ) = g ( x )$ is $\mathbf { K }$.

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 $a$ be a constant other than 0 . Let
$$\begin{aligned} & f ( x ) = x ^ { 2 } + 2 a x - 4 a - 12 , \\ & g ( x ) = a x ^ { 2 } + 2 x - 4 a + 4 . \end{aligned}$$
(1) When the solutions of $f ( x ) = 0$ and the solutions of $g ( x ) = 0$ coincide, $a$ is $\mathbf { M N }$, and their solutions are $x = \mathbf { O P }$ and $x = \mathbf { Q }$.
(2) $g ( x ) = 0$ has just one solution when $a = \frac { \mathbf { R } } { \mathbf { S } }$, and in this case the solution is $x =$ $\mathbf{TU}$.
(3) The range of $a$ such that $f ( x ) < g ( x )$ for all $x$ is $\mathbf{VW}$.

Let $a, b, c$ be nonzero real numbers, and the two roots of the quadratic equation $ax^2 + bx + c = 0$ both lie between 1 and 3. Select the equation whose two roots must lie between 4 and 5.
(1) $a(x-2)^2 + b(x-2) + c = 0$
(2) $a(x+2)^2 + b(x+2) + c = 0$
(3) $a(2x-7)^2 + b(2x-7) + c = 0$
(4) $a\left(\frac{x+7}{2}\right)^2 + b\left(\frac{x+7}{2}\right) + c = 0$
(5) $a(3x-11)^2 + b(3x-11) + c = 0$

A sales station sells three types of mobile phones: A, B, and C. The profit per unit is 100 yuan for type A, 400 yuan for type B, and 240 yuan for type C. Last year, $A, B, C$ units of types A, B, C were sold respectively, with an average profit of 260 yuan per unit. It is also known that the average profit for selling types A and B together ($A + B$ units) is 280 yuan per unit. The ratio of the quantities of the three types of mobile phones sold last year is $A : B : C =$ (13－1)：(13－2)：(13－3) (expressed as a ratio of integers in lowest terms)