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
$$a = \mathbf { H } , \quad b = \mathbf { I } .$$

Let us simultaneously throw three dice which are different in size and denote the number on the large, medium and small dice by $x , y$ and $z$, respectively.
Let $A$ be the event where $x = y = z$; let $B$ be the event where $x + y + z = 6$; let $C$ be the event where $x + y = z$.
(1) The numbers of outcomes in event $A$ is $\mathbf { J }$, in event $B$ is $\mathbf { K } \mathbf { L }$, and in event $C$ is $\mathbf { M N }$.
(2) The numbers of outcomes in event $A \cap B$ is $\mathbf { O }$, in event $B \cap C$ is $\mathbf { P }$, and in event $C \cap A$ is $\mathbf { Q }$.
(3) The probability of event $B \cup C$ is
$$P ( B \cup C ) = \frac { \mathbf { R S } } { \mathbf { T U V } } .$$

Consider the expression in $x$
$$P = | x - 1 | + | x - 2 | + | x - a | .$$
We are to find the range of real numbers $a$ such that the value of $P$ is minimized at $x = a$.
First, let us note that the inequality
$$| x - 1 | + | x - 2 | + | x - a | \geqq | x - 1 | + | x - 2 |$$
always holds, and is an equality in the case $x = a$.
When we set
$$y = | x - 1 | + | x - 2 | ,$$
we have
$$y = \begin{cases} - \mathbf { A } & x + \mathbf { B } \\ \mathbf { D } & ( x < \mathbf { C } ) \\ \mathbf { F } & ( \mathbf { C } \leqq x \leqq \mathbf { E } ) \\ \mathbf { G } & ( \mathbf { E } < x ) . \end{cases}$$
When we consider the graph of (1), we see that the minimum value of $y$ is $\mathbf { H }$ and $y$ takes the value $\mathbf{H}$ at every $x$ satisfying $\mathbf{I} \leqq x \leqq \mathbf{J}$.
Thus, for every $a$ satisfying $\mathbf{K} \leqq a \leqq \mathbf{L}$, the value of $P$ is minimized at $x = a$ and its value there is $\mathbf{M}$.

Let $a$ and $b$ be natural numbers such that the greatest common divisor of $a$ and $b$ is 3. We are to find the natural numbers $a$ and $b$ such that
$$3 a - 2 b = \ell + 3$$
is satisfied, where $\ell$ is the least common multiple of $a$ and $b$.
When we set $a = 3 p$ and $b = 3 q$, the natural numbers $p$ and $q$ are mutually prime (co-prime), and hence $\ell = \mathbf { N } p q$. Thus using $p$ and $q$, the equality (1) can be transformed to
$$p q - \mathbf { O } p + \mathbf { P } q + \mathbf { Q } = 0 .$$
This can be further transformed to
$$( p + \mathbf { R } ) ( q - \mathbf { S } ) = - \mathbf { S } \mathbf { T } .$$
Among the pairs of integers $p$ and $q$ which satisfy this equation, the pair such that both $a$ and $b$ are natural numbers is
$$p = \mathbf { U } , \quad q = \mathbf { V } ,$$
which gives
$$a = \mathbf { W X } , \quad b = \mathbf { Y } .$$

For each of $\mathbf { A } \sim \mathbf { M }$ in the following statements, choose the correct answer from among (0) ~ (9) at the bottom of this page.
We are to solve the following simultaneous inequalities
$$\left\{ \begin{aligned} x ^ { 2 } - 2 x < 3 & \cdots \cdots \cdots (1)\\ a x ^ { 2 } - a x - x + 1 > 0 , & \cdots \cdots \cdots (2) \end{aligned} \right.$$
where $0 < a < 1$.
When we solve (1), we have
$$\mathbf { A } < x < \mathbf { B } .$$
Next, when we transform (2), we have
$$( a x - \mathbf { C } ) ( x - \mathbf { D } ) > 0 .$$
Hence, noting that $0 < a < 1$, we see that the solution of (2) is
$$x < \mathbf { E } \text { or } \mathbf { F } < x .$$
Thus, when $0 < a \leqq \mathbf { G }$, the solution of the simultaneous inequalities is
$$\mathbf { H } < x < \mathbf { I }$$
and when $\mathbf { G } < a < 1$, the solution is
$$\mathbf { J } < x < \mathbf { K } \text { or } \mathbf { L } < x < \mathbf { M }$$
where $\mathbf { K } < \mathbf { M }$.
(0) 0
(1) 1
(2) 2
(3) 3
(4) $- 1$
(5) $\frac { 1 } { 2 }$ (6) $\frac { 1 } { 3 }$ (7) $\frac { 1 } { a }$ (8) $\frac { 2 } { a }$ (9) $\frac { 3 } { a }$

In the figure to the right, let $\angle \mathrm { XOY } = 60 ^ { \circ }$, and let OZ be the half-line (ray) which bisects $\angle \mathrm { XOY }$. In addition, the points A and B on the half-lines OX and OY satisfy $\mathrm { OA } = \mathrm { OB } = 1$.
Let $\mathrm { P } , \mathrm { Q }$ and R be moving points on $\mathrm { OX } , \mathrm { OZ }$ and OY that start simultaneously from $\mathrm { A } , \mathrm { O }$ and B, moving in the direction away from point O at the speeds of $1 , \sqrt { 3 }$ and 2 units per second.
We are to find the moment at which the three points $\mathrm { P } , \mathrm { Q }$ and R are arranged on a straight line by considering the area of the triangle PQR.
First, the lengths of $\mathrm { OP } , \mathrm { OQ }$ and OR at $t$ seconds after the start are
$$\mathrm { OP } = t + \mathbf { A } , \quad \mathrm { OQ } = \sqrt { \mathbf { B } } t , \quad \mathrm { OR } = \mathbf { C } t + \mathbf { D } .$$
At this time the areas of the triangles are
$$\begin{aligned} \triangle \mathrm { OPQ } & = \frac { \sqrt { \mathbf { E } } t ( t + \mathbf { F } ) } { 4 } , \\ \triangle \mathrm { ORQ } & = \frac { \sqrt { \mathbf { G } } t ( \mathbf { H } t + \mathbf { I } ) } { 4 } , \\ \triangle \mathrm { OPR } & = \frac { \sqrt { \mathbf { G } } ( t + \mathbf { K } ) ( \mathbf { L } t + \mathbf { M } ) } { 4 } \end{aligned}$$
Hence we obtain
$$\triangle \mathrm { PQR } = \frac { \sqrt { \mathbf { N } } } { 4 } \left| - t ^ { 2 } + t + \mathbf { O } \right| .$$
So, to find the moment such that the three points $\mathrm { P } , \mathrm { Q }$ and R are arranged on a straight line, we should find the case where
$$t ^ { 2 } - t - \mathbf { O } = \mathbf { P } .$$
Thus the time required is
$$t = \frac { \mathbf { Q } + \sqrt { \mathbf { R } } } { \mathbf { S } } \text{ (seconds).}$$

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
$$a = \mathbf { H } , \quad b = \mathbf { I } .$$

Let us simultaneously throw three dice which are different in size and denote the number on the large, medium and small dice by $x , y$ and $z$, respectively.
Let $A$ be the event where $x = y = z$; let $B$ be the event where $x + y + z = 6$; let $C$ be the event where $x + y = z$.
(1) The numbers of outcomes in event $A$ is $\mathbf { J }$, in event $B$ is $\mathbf { K } \mathbf { L }$, and in event $C$ is $\mathbf { M N }$.
(2) The numbers of outcomes in event $A \cap B$ is $\mathbf { O }$, in event $B \cap C$ is $\mathbf { P }$, and in event $C \cap A$ is $\mathbf { Q }$.
(3) The probability of event $B \cup C$ is
$$P ( B \cup C ) = \frac { \mathbf { R S } } { \mathbf { T U V } } .$$

Given two points $\mathrm { A } ( 1 , - 1,0 )$ and $\mathrm { B } ( - 2,1,2 )$ in a coordinate space with the origin O, let us set $\overrightarrow { \mathrm { OA } } = \vec { a } , \overrightarrow { \mathrm { OB } } = \vec { b }$.
(1) First, we are to find the value of $t$ at which $| \vec { a } + t \vec { b } |$ is minimized. Since
$$| \vec { a } + t \vec { b } | ^ { 2 } = \mathbf { A } t ^ { 2 } - \mathbf { B } t + \mathbf { C }$$
$| \vec { a } + t \vec { b } |$ is minimized at $t = \frac { \mathbf { D } } { \mathbf { E } }$, and its minimum value is $\mathbf { F }$.
(2) Next, the vectors $\vec { c }$ which are orthogonal to the vectors $\vec { a }$ and $\vec { b }$ can be represented as
$$\vec { c } = s ( \mathbf { G } , \mathbf { H } , 1 )$$
where $s$ is a non-zero real number. Now, let C and D be the points such that $\overrightarrow { \mathrm { OC } } = ( \mathbf { G } , \mathbf { H } , 1 )$ and $\overrightarrow { \mathrm { OD } } = 3 \vec { a } + \vec { b }$. Since $\angle \mathrm { CBD } = \frac { \pi } { \mathbf { I } }$, the area of the triangle BCD is $\frac { \mathbf { J } \sqrt { \mathbf { K } } } { \mathbf { L } }$.

Let us consider the solutions to the equation in the complex number $z$
$$z ^ { 4 } = - 324 \quad \cdots (1)$$
and the solutions to the equation in the complex number $z$
$$z ^ { 4 } = t ^ { 4 } \quad \cdots (2)$$
where $t$ is a positive real number.
(1) To find the solutions to (1), let us set
$$z = r ( \cos \theta + i \sin \theta ) \quad ( r > 0,0 < \theta \leqq 2 \pi )$$
Then
$$z ^ { 4 } = r ^ { \mathbf { M } } ( \cos \mathbf { N } \theta + i \sin \mathbf { N } \theta ) .$$
The values of $r$ and $\theta$ such that this expression is equal to $-324$ are
$$\begin{aligned} & r = \mathbf { O } \sqrt { \mathbf { P } } , \\ & \theta = \frac { \mathbf { Q } } { \mathbf { R } } \pi , \frac { \mathbf { S } } { \mathbf { R } } \pi , \frac { \mathbf { T } } { \mathbf { R } } \pi , \frac { \mathbf { U } } { \mathbf { R } } \pi , \end{aligned}$$
where $\mathbf { Q } < \mathbf { S } < \mathbf { T } < \mathbf { U }$.
(2) There are $\mathbf { V }$ solutions to equation (2), and these solutions are dependent on $t$. Now, consider any one of the solutions to (1) and any one of the solutions to (2), and let $d$ be the distance on the complex number plane between these two solutions. Then, over the interval $0 < t \leqq 4$, the minimum value of $d$ is $\mathbf { W }$ and the maximum value is $\sqrt { \mathbf { X Y } }$.

Let $f ( x ) = x ^ { 4 } + 2 x ^ { 3 } - 12 x ^ { 2 } + 4$. We are to find the values of $p$ such that we can draw three tangents to the curve $y = f ( x )$ from the point $\mathrm { P } ( 0 , p )$ on the $y$-axis.
(i) The equation of the tangent to the curve $y = f ( x )$ at the point $( t , f ( t ) )$ is
$$y = \left( \mathbf { A } t ^ { 3 } + \mathbf { B } t ^ { 2 } - \mathbf { C D } t \right) x - \mathbf { E } t ^ { 4 } - \mathbf { F } t ^ { 3 } + \mathbf { G H } t ^ { 2 } + \mathbf { I }$$
The condition under which this straight line passes through the point $\mathrm { P } ( 0 , p )$ is that
$$p = - \mathbf { J } t ^ { 4 } - \mathbf { K } t ^ { 3 } + \mathbf { L M } t ^ { 2 } + \mathbf { N }$$
(ii) For $\mathbf { O }$ and $\mathbf { S }$ in the following statements, choose either (0) or (1) and for the other blanks, enter the correct number. (0) local minimum
(1) local maximum
When the right side of (1) is set to $g ( t )$, the function $g ( t )$ takes a $\mathbf{O}$ at $t = \mathbf { P Q }$ and $t = \mathbf { R }$. On the other hand, $g ( t )$ takes a $\mathbf { S }$ at $t = \mathbf { T }$.
Hence the values of $p$ such that we can draw three tangents to the curve $y = f ( x )$ from the point $\mathrm { P } ( 0 , p )$ are
$$p = \mathbf { U } \text { and } p = \mathbf { V } ,$$
where $\mathbf { U } < \mathbf { V }$.

The coordinates $( x , y )$ of a moving point P are given by the following functions in time $t$:
$$\begin{aligned} & x = 4 t - \sin 4 t \\ & y = 4 - \cos 4 t \end{aligned}$$
(1) The derivatives of $x$ and $y$ with respect to $t$ are
$$\begin{aligned} \frac { d x } { d t } & = \mathbf { A } ( \mathbf { A } - \mathbf { B } \cos 4 t ) \\ \frac { d y } { d t } & = \mathbf { C } \sin 4 t . \end{aligned}$$
Hence we have
$$\left( \frac { d x } { d t } \right) ^ { 2 } + \left( \frac { d y } { d t } \right) ^ { 2 } = \mathbf { D E } \sin ^ { 2 } \mathbf { F } t$$
(2) As the point P moves from the time $t = 0$ to the time $t = 2 \pi$, its speed $v$ is maximized a total of $\mathbf { G }$ times. Let us denote by $t _ { 0 }$ the moment of the first time the speed is maximized and the moment of the last time it is maximized by $t _ { 1 }$. Then
$$t _ { 0 } = \frac { \mathbf { H } } { \mathbf { I } } \pi , \quad t _ { 1 } = \frac { \mathbf { J } } { \mathbf { I } } \pi$$
and the maximum speed is $v = \mathbf { L }$.
(3) For $t _ { 0 }$ and $t _ { 1 }$ in (2), the distance that point P moves during the period from $t = t _ { 0 }$ to $t = t _ { 1 }$ is $\mathbf{MN}$.