Consider the quadratic function
$$y = 3 x ^ { 2 } - 6 .$$
(1) Suppose that the graph obtained by a parallel translation of the graph of $y = 3 x ^ { 2 } - 6$ passes through the two points $( 1,5 )$ and $( 4,14 )$. The quadratic function of this graph is
$$y = \mathbf { A } x ^ { 2 } - \mathbf { B C } x + \mathbf { D E } .$$
This graph is the parallel translation of the graph of $y = 3 x ^ { 2 } - 6$ by $\mathbf{F}$ in the $x$-direction and by $\mathbf { G }$ in the $y$-direction.
(2) The quadratic function having the graph which is symmetric to the graph of $y = 3 x ^ { 2 } - 6$ with respect to the straight line $y = c$ is
$$y = - \mathbf { H } x ^ { 2 } + \mathbf { I } c + \mathbf { J } .$$
When the graphs of the two quadratic functions (1) and (2) have just one common point, it follows that $c = \mathbf { K }$, and the coordinates of the common point are ( $\mathbf { L } , \mathbf { M }$ ).

We have four white cards, three red cards and three black cards. A different number is written on each of the ten cards.
(1) Choose two of the ten cards and put one in box A, and one in box B. There are $\mathbf{NO}$ ways of putting two cards in the two boxes.
(2) There are $\mathbf { P Q }$ ways of choosing two cards of the same color, and $\mathbf { R S }$ ways of choosing two cards of different colors.
Next, put the ten cards in a box and take out one card and without returning it to the box, take out second card.
(3) The probability that the two cards taken out have the same color is $\square\mathbf{ T UV}$
(4) The probability that the color of the first card taken out is white or red, and the color of the second card taken out is red or black is $\frac { \mathbf { W X } } { \mathbf { Y } \mathbf { Y } }$.

Consider the real numbers $a$ and $b$ such that the equation in $x$
$$| x - 3 | + | x - 6 | = a x + b \tag{1}$$
has a solution.
Set the left side of (1) as $y = | x - 3 | + | x - 6 |$. This can be represented without using the absolute value signs in the following way.
$$\begin{array} { l l } \text { If } x < \mathbf { A } , & \text { then } y = - \mathbf { B } x + \mathbf { C } ; \\ \text { if } \mathbf { A } \leqq x < \mathbf { D } , & \text { then } y = \mathbf { E } ; \\ \text { if } \mathbf { D } \leqq x , & \text { then } y = \mathbf { F } x - \mathbf { G } . \end{array}$$
Next, let us consider the common point(s) of the graph of this function and the straight line $y = a x + b$ on the $x y$-plane. Then we see the following:
(i) If $a = 1$, then the range of the values of $b$ such that (1) has one or more solutions is
$$b \geqq \mathbf { H I } .$$
(ii) If $b = 6$, then the range of the values of $a$ such that (1) has two different solutions is
$$\mathbf { J K } < a < \mathbf { L }.$$

Consider two squares as in the figure to the right. Let the coordinates of their vertexes be
$$\begin{array} { l l } \mathrm { A } ( 2 t , 0 ) , \quad \mathrm { B } ( 0,2 t ) , & \mathrm { C } ( - 2 t , 0 ) , \quad \mathrm { D } ( 0 , - 2 t ) , \\ \mathrm { P } \left( 4 - t ^ { 2 } , 4 - t ^ { 2 } \right) , & \mathrm { Q } \left( - 4 + t ^ { 2 } , 4 - t ^ { 2 } \right) , \\ \mathrm { R } \left( - 4 + t ^ { 2 } , - 4 + t ^ { 2 } \right) , & \mathrm { S } \left( 4 - t ^ { 2 } , - 4 + t ^ { 2 } \right) , \end{array}$$
where $0 < t < 2$. Denote the areas of the two squares ABCD and PQRS by $S _ { 1 }$ and $S _ { 2 }$, respectively.
Then we have
$$S _ { 1 } = \mathbf { M } t ^ { 2 } \text { and } S _ { 2 } = \mathbf { N } \left( t ^ { 2 } - \mathbf { O } \right) ^ { 2 } .$$
(1) $S _ { 1 } + S _ { 2 }$ is minimized at $t = \sqrt { \mathbf { P } }$, and the minimum value is $\mathbf { Q } \mathbf { R }$.
(2) For $\mathbf { W }$ and $\mathbf { X }$ below, choose the correct answer from among (0) $\sim$ (9), and for the other $\square$, enter the correct numbers.
We are to find the range of $t$ such that $S _ { 1 } < S _ { 2 }$. If $S _ { 1 } < S _ { 2 }$, then $t$ satisfies the inequality
$$t ^ { 4 } - \mathbf { ST } t ^ { 2 } + \mathbf { UV } > 0 .$$
From the above inequality, a condition on $t ^ { 2 }$ is $\mathbf { W }$. Hence, $S _ { 1 } < S _ { 2 }$ if and only if $t$ satisfies $\mathbf { X }$.
(0) $t ^ { 2 } < 4$ or $6 < t ^ { 2 }$ (1) $4 < t ^ { 2 } < 6$ (2) $t ^ { 2 } < 2$ or $8 < t ^ { 2 }$ (3) $2 < t ^ { 2 } < 8$ (4) $t ^ { 2 } \neq 4$ (5) $0 < t < 2$ (6) $0 < t < \sqrt { 2 }$ (7) $\sqrt { 2 } < t < 2$ (8) $2 < t < \sqrt { 6 }$ (9) $t \neq 2$

Let $n$ be a positive integer, and $x$ and $y$ be non-negative integers. We are to examine the solutions of the following equation in $x$ and $y$
$$x ^ { 2 } - y ^ { 2 } = n . \tag{1}$$
First of all, by transforming (1), we obtain
$$( x + y ) ( x - y ) = n . \tag{2}$$
(1) When we find the solutions $( x , y )$ of (1) in the cases where $n = 8$ and $n = 9$, we have that if $n = 8$, then $( x , y ) = ( \mathbf { A } , \mathbf { B } )$, and if $n = 9$, then $( x , y ) = ( \mathbf { C } , \mathbf { D } ) , ( \mathbf { E } , \mathbf { F } )$. Note that you should write the solutions in the order such that $\mathbf{C} \leq \mathbf{E}$.
(2) For each of $\mathbf { G }$ $\sim$ $\mathbf { R }$ in the following sentences, choose the correct answer from among (0) $\sim$ (9) given below.
The following is a proof that (3) given below is the necessary and sufficient condition for (1) to have a solution.
Proof: First, suppose that $( x , y )$ satisfies (1). If $x$ and $y$ are both even or both odd, then both $x + y$ and $x - y$ are $\mathbf { G }$. Hence, by (2) we see that $n$ is a multiple of $\mathbf { H }$.
Next, if one of $x$ and $y$ is even and the other is odd, then both $x + y$ and $x - y$ are $\mathbf{I}$, and hence $n$ is $\mathbf{J}$.
Thus we see that
$$\text{``} n \text{ is a multiple of } \mathbf{H} \text{, or } n \text{ is } \mathbf{J} \text{''} \quad \ldots\ldots (3)$$
is a necessary condition for (1) to have a solution.
Conversely, suppose that $n$ satisfies the condition (3). If $n$ is a multiple of $\mathbf{H}$, then $n$ can be represented as $n = \mathbf{H} \cdot k$, where $k$ is a positive integer. So, if for example we take $x + y = \mathbf { K } \cdot k$ and $x - y = 2$, then $( x , y ) = ( k + \mathbf { L } , k - \mathbf { M } )$, which shows that (1) has a solution.
On the other hand, if $n$ is $\mathbf{J}$, then $n$ can be represented as $n = \mathbf { N } \ell + \mathbf { O }$, where $\ell$ is a non-negative integer. So, if for example we take $x + y = \mathbf { P } \ell + \mathbf { Q }$ and $x - y = 1$, then $( x , y ) = ( \ell + \mathbf { R } , \ell )$, which shows that (1) has a solution.
From the above, we see that the necessary and sufficient condition for (1) to have a solution is (3).
(0) 0 (1) 1 (2) 2 (3) 3 (4) 4 (5) 5 (6) 6 (7) even (8) odd (9) prime

Consider a rhombus ABCD with sides of length $a$, where $a$ is a constant. Let $r$ be the radius of the circle O inscribed in the rhombus ABCD, and $\mathrm{K}, \mathrm{L}$, $\mathrm{M}, \mathrm{N}$ be the points of tangency of the circle O and the rhombus. Let $S$ denote the area of the part of the rhombus outside circle O.
We are to find the range of the values of $r$, and the maximum value of $S$.
(1) For each of $\mathbf{A}$ $\sim$ $\mathbf{C}$ below, choose the correct answer from among (0) $\sim$ (9).
Let $\angle \mathrm{ABO} = \theta$. We have $\mathrm{OB} = \mathbf{A}$, and hence $\mathrm{OK} = \mathbf{B}$. Hence, since $( \cos \theta - \sin \theta ) ^ { 2 } \geqq 0$, the range of the values taken by $r$ is
$$0 < r \leqq \mathbf { C } .$$
(0) $a$ (1) $\frac { a } { 2 }$ (2) $\frac { a } { 3 }$ (3) $a \sin \theta$ (4) $a \cos \theta$ (5) $a \tan \theta$ (6) $a \sin ^ { 2 } \theta$ (7) $a \cos ^ { 2 } \theta$ (8) $a \sin \theta \cos \theta$ (9) $a \tan ^ { 2 } \theta$
(2) For each of $\mathbf { D } \sim$ $\mathbf{F}$ below, choose the correct answer from among (0) $\sim$ (9).
When the area $S$ is expressed in terms of $r$, we have
$$S = \mathbf { D } .$$
Here, we observe that when $r = \mathbf { E }$, the value of $\mathbf{D}$ is maximized, and this value for $r$ satisfies (1). Thus, at $r = \mathbf { E }$, $S$ takes the maximum value $\mathbf { F }$.
(0) $2 a r - \pi r ^ { 2 }$ (1) $a r - \frac { \pi } { 2 } r ^ { 2 }$ (2) $\frac { a } { 2 } r - \pi r ^ { 2 }$ (3) $\frac { a r - \pi r ^ { 2 } } { 2 }$ (4) $\frac { 2 a } { \pi }$ (5) $\frac { a } { \pi }$ (6) $\frac { a } { 2 \pi }$ (7) $\frac { 4 a ^ { 2 } } { \pi }$ (8) $\frac { a ^ { 2 } } { \pi }$ (9) $\frac { a ^ { 2 } } { 4 \pi }$

Consider the quadratic function
$$y = 3 x ^ { 2 } - 6 .$$
(1) Suppose that the graph obtained by a parallel translation of the graph of $y = 3 x ^ { 2 } - 6$ passes through the two points $( 1,5 )$ and $( 4,14 )$. The quadratic function of this graph is
$$y = \mathbf { A } x ^ { 2 } - \mathbf { BC } x + \mathbf { D E } .$$
This graph is the parallel translation of the graph of $y = 3 x ^ { 2 } - 6$ by $\mathbf{F}$ in the $x$-direction and by $\mathbf { G }$ in the $y$-direction.
(2) The quadratic function having the graph which is symmetric to the graph of $y = 3 x ^ { 2 } - 6$ with respect to the straight line $y = c$ is
$$y = - \mathbf { H } x ^ { 2 } + \mathbf { I } c + \mathbf { J } .$$
When the graphs of the two quadratic functions (1) and (2) have just one common point, it follows that $c = \mathbf { K }$, and the coordinates of the common point are ( $\mathbf { L } , \mathbf { M }$ ).

We have four white cards, three red cards and three black cards. A different number is written on each of the ten cards.
(1) Choose two of the ten cards and put one in box A, and one in box B. There are $\mathbf{NO}$ ways of putting two cards in the two boxes.
(2) There are $\mathbf { P Q }$ ways of choosing two cards of the same color, and $\mathbf { R S }$ ways of choosing two cards of different colors.
Next, put the ten cards in a box and take out one card and without returning it to the box, take out second card.
(3) The probability that the two cards taken out have the same color is $\square\mathbf{T UV}$
(4) The probability that the color of the first card taken out is white or red, and the color of the second card taken out is red or black is $\frac { \mathbf { W X } } { \mathbf { Y } \mathbf { Y } }$.

A triangle OAB and a triangle OAC share one side OA. Further, they satisfy the following two conditions:
(i) $\overrightarrow { \mathrm { OC } } = x \overrightarrow { \mathrm { OA } } + \frac { 1 } { 2 } \overrightarrow { \mathrm { OB } }$;
(ii) the center of gravity G of the triangle OAC is on the segment AB.
We are to find the value of $x$, and express $\overrightarrow { \mathrm { OG } }$ in terms of $\overrightarrow { \mathrm { OA } }$ and $\overrightarrow { \mathrm { OB } }$.
Let us denote the point of intersection of the segment OC and the segment AB by D. Then we have
$$\overrightarrow { \mathrm { OD } } = \frac { x } { \mathbf { A } } \overrightarrow { \mathrm { OA } } + \frac { \mathbf { B } } { \mathbf { A } } \overrightarrow { \mathrm { OB } } .$$
Since D is on the segment AB, we have $x = \frac { \mathbf { D } } { \mathbf { E } }$. Hence we obtain
$$\overrightarrow { \mathrm { OG } } = \frac { \mathbf { F } } { \mathbf { G } } \overrightarrow { \mathrm { OA } } + \frac { \mathbf { H } } { \mathbf { I } } \overrightarrow { \mathrm { OB } } .$$
In particular, when $\mathrm { OA } = 1 , \mathrm { OB } = 2$ and $\angle \mathrm { AOB } = 60 ^ { \circ }$, we have $\mathrm { OG } = \frac { \sqrt { \mathbf { J K } } } { \mathbf { L } }$.

Let $z$ be a complex number satisfying $| z | = 2$. In the complex number plane with the origin O, let A and B be the points representing $1 + z$ and $1 - \frac { 1 } { 2 } z$, respectively.
First of all, we can express the complex number $z$ as
$$z = \mathbf { M } ( \cos \theta + i \sin \theta ) \quad ( - \pi \leqq \theta < \pi ) .$$
(1) If $z$ is not a real number, then the area $S$ of the triangle OAB is $S = \mathbf { N }$. For $\mathbf{N}$, choose the correct answer from among (0) $\sim$ (8) below.
Hence, when $\theta = \pm \frac { \mathbf { O } } { \mathbf { P } } \pi$, $S$ is maximized.
(0) $\frac { 1 } { 2 } \left| \sin \left( \theta + \frac { 1 } { 3 } \pi \right) \right|$ (1) $\frac { 1 } { 2 } | \sin \theta |$ (2) $\frac { 1 } { 2 } \left| \sin \left( \theta - \frac { 1 } { 3 } \pi \right) \right|$ (3) $\left| \sin \left( \theta + \frac { 1 } { 3 } \pi \right) \right|$ (4) $| \sin \theta |$ (5) $\left| \sin \left( \theta - \frac { 1 } { 3 } \pi \right) \right|$ (6) $\frac { 3 } { 2 } \left| \sin \left( \theta + \frac { 1 } { 3 } \pi \right) \right|$ (7) $\frac { 3 } { 2 } | \sin \theta |$ (8) $\frac { 3 } { 2 } \left| \sin \left( \theta - \frac { 1 } { 3 } \pi \right) \right|$
(2) When the triangle OAB is an isosceles triangle where $\mathrm { OA } = \mathrm { OB }$, we see that
$$| 1 + z | = \left| 1 - \frac { 1 } { 2 } z \right| = \sqrt { \mathbf { Q } }$$
and
$$\arg ( 1 + z ) = \pm \frac { \mathbf { R } } { \mathbf { S } } \pi , \quad \arg \left( 1 - \frac { 1 } { 2 } z \right) = \mp \frac { \mathbf { T } } { \mathbf{U} } \pi ,$$
where the right-hand sides of the equations are of opposite signs, and where $- \pi \leqq \arg ( 1 + z ) < \pi$ and $- \pi \leqq \arg \left( 1 - \frac { 1 } { 2 } z \right) < \pi$.

Consider the function $y = \frac { 2 ^ { x ^ { 2 } } } { 5 ^ { 3 x } }$, where $x \geqq 0$.
(1) We are to find the $x$ at which $y$ is minimized.
When we differentiate $y$, we have
$$\frac { d y } { d x } = \frac { 2 ^ { x ^ { 2 } } } { 5 ^ { 3 x } } \left( 2 x \log _ { e } \mathbf { A } - \mathbf { B } \log _ { e } \mathbf { C } \right) .$$
Hence, when we express the value of $x$ at which $y$ is minimized using the common logarithm, we have
$$x = \frac { \mathbf { D } } { \mathbf { F } \left( 1 - \log _ { 10 } \mathbf { E } \right) } .$$
(2) We are to find the smallest positive integer $x$ satisfying $\frac { 2 ^ { x ^ { 2 } } } { 5 ^ { 3 x } } > 1000$.
From the inequality $y > 1000$, we obtain
$$x ^ { \mathbf { H } } \log _ { 10 } \mathbf { I } - \mathbf { J } x \log _ { 10 } \mathbf { L } - \mathbf { K } > 0 .$$
When we solve the inequality using $0.3$ as an approximate value for $\log _ { 10 } 2 = 0.301 \cdots$, the smallest positive integer $x$ satisfying $y > 1000$ is $\mathbf{Q}$.

Consider the function $f ( x ) = x \sin ^ { 2 } x$ on the interval $0 \leqq x \leqq \pi$. Let $\ell$ be the tangent to the curve $y = f ( x )$ that passes through the origin, where $\ell$ is not the $x$-axis. We are to find the area $S$ of the region bounded by the curve $y = f ( x )$ and the tangent $\ell$.
(1) For each of $\mathbf{A}$ $\sim$ $\mathbf{D}$ in the following sentences, choose the correct answer from among (0) $\sim$ (9) below.
When we denote the point of tangency of the curve $y = f ( x )$ and the tangent $\ell$ by $( t , f ( t ) )$, we have the equality $\mathbf{A}$, since $\ell$ passes through the origin. Further, since
$$f ^ { \prime } ( t ) = \mathbf { B } + 2 t \, \mathbf { C }$$
the $x$-coordinate of the point of tangency is $t = \mathbf { D }$.
(0) $f ( t ) = t f ^ { \prime } ( t )$ (1) $f ^ { \prime } ( t ) = t f ( t )$ (2) $\sin t$ (3) $\sin ^ { 2 } t$ (4) $\cos ^ { 2 } t$ (5) $\sin t \cos t$ (6) $\frac { \pi } { 2 }$ (7) $\frac { \pi } { 3 }$ (8) $\frac { \pi } { 4 }$ (9) $\frac { \pi } { 6 }$
(2) For each of $\mathbf { E }$ $\sim$ $\mathbf { G }$ in the following sentences, choose the correct answer from among (0) $\sim$ (9) below.
The antiderivative of the function $f ( x )$ is
$$\int f ( x ) d x = \mathbf { E } \left( 2 x ^ { 2 } - 2 x \mathbf { F } - \mathbf { G } \right) + C ,$$
where $C$ is the integral constant.
(0) $\frac { 1 } { 8 }$ (1) $\frac { 1 } { 4 }$ (2) $\frac { 1 } { 2 }$ (3) 2 (4) 4 (5) 8 (6) $\sin x$ (7) $\cos x$ (8) $\sin 2 x$ (9) $\cos 2 x$
(3) Thus, the area $S$ of the region bounded by the curve $y = f ( x )$ and the tangent $\ell$ is
$$S = \frac { \mathbf { H } } { \mathbf { I J } } \pi ^ { \mathbf { K } } - \frac { \mathbf { L } } { \mathbf{M} } .$$