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
$$\mathbf { I } < a \leqq \frac { \mathbf { J K } } { \mathbf { L M } } .$$

We have four cards of different sizes. We are to paint each card either red, black, blue or yellow. However, we may paint more than one card with the same color.
(1) There are a total of $\mathbf { N O P }$ ways of painting the cards.
(2) There are $\mathbf{QR}$ ways of painting them using all four colors.
(3) There are $\mathbf{ST}$ ways of painting two cards red, one card black, and one card blue.
(4) There are $\mathbf{UVW}$ ways of painting the four cards using three colors.
(5) There are $\mathbf { X Y }$ ways of painting them using two colors.

We are to find the positive number $a$ satisfying $a ^ { 3 } = 9 + \sqrt { 80 }$.
Let us consider the positive number $b$ which satisfies $b ^ { 3 } = 9 - \sqrt { 80 }$. Then
$$\left\{ \begin{aligned} a ^ { 3 } + b ^ { 3 } & = \mathbf { A B } \\ a b & = \mathbf { C } \end{aligned} \right.$$
holds.
First, using (2), (1) can be transformed into
$$( a + b ) ^ { 3 } - \mathbf { D } ( a + b ) = \mathbf { A B } .$$
Then, setting $x = a + b$, we have
$$x ^ { 3 } - \mathrm { D } x = \mathrm { AB } .$$
Transforming this equation, we obtain
$$x ^ { 3 } - 27 = \mathbf { D } ( x - \mathbf { E } ) \text {, }$$
which gives
$$( x - \mathbf { F } ) \left( x ^ { 2 } + \mathbf { G } x + \mathbf{H} \right) = 0 .$$
From that we have $x = \mathbf{I}$ and hence
$$a + b = \mathbf{I} \text {. }$$
Thus, from (2), (3) and $a > b$, we have
$$a = \frac { \mathbf { J } + \sqrt { \mathbf { K } } } { \mathbf{L} } .$$

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 $n$ be a two-digit natural number such that the remainder of $n ^ { 3 }$ divided by 66 is $n$. We are to find the number of such $n$ 's and to find the prime numbers among them.
From the conditions we have
$$n ^ { 3 } = \mathbf { A B } p + n \quad ( 0 < n \leqq \mathbf { C D } ) ,$$
where $p$ is the integer quotient of $n ^ { 3 }$ divided by 66 . This can be transformed into
$$n ( n - 1 ) ( n + 1 ) = \mathrm { AB } p$$
Since either $n - 1$ or $n$ has to be a multiple of $\mathbf { E }$ and either $n - 1 , n$ or $n + 1$ has to be a multiple of $\mathbf { F }$, and furthermore $\mathbf { E }$ and $\mathbf { F }$ are mutually prime, we know that $n ( n - 1 ) ( n + 1 )$ is a multiple of $\mathbf { G }$. (Write the answers in the order $1 < \square < \mathbf { E } < \mathbf { F } < \mathbf { G }$.) Hence one of $n - 1 , n$ and $n + 1$ must be a multiple of $\mathbf{HI}$.
So, since $n \leqq \mathrm { CD }$, the number of $n$ 's where $n - 1$ is a multiple of $\mathbf{HI}$ is $\mathbf{J}$, where $n$ is a multiple of $\mathbf{HI}$ is $\mathbf { K }$, and where $n + 1$ is a multiple of $\mathbf{HI}$ is $\mathbf { L }$.
Thus, the number of $n$ 's is $\mathbf { M N }$ and the prime numbers among them are $\mathbf { O P } , \mathbf { Q R }$, $\mathbf{ST}$, in ascending order.

The triangle ABC satisfies
$$\mathrm { AB } = 4 , \quad \mathrm { AC } = 3 \quad \text { and } \quad \angle \mathrm { B } = 30 ^ { \circ } .$$
D is the point on side BC such that $\mathrm { AC } = \mathrm { AD }$. Let us consider the circumscribed circle O of triangle ACD.
(1) Since $\sin B = \frac { \mathbf { A } } { \mathbf { B } }$, we have $\sin C = \frac { \mathbf { C } } { \mathbf{D} }$.
Hence the radius of circle O is $\frac { \mathbf { E } } { \mathbf{F} }$.
(2) We have
$$\mathrm { BC } = \mathrm { G } \sqrt { \mathrm { H } } + \sqrt { \mathrm{H} }$$
and
$$\mathrm { BD } = \mathrm { J } \sqrt { \mathrm {~K} } - \sqrt { \mathrm { K } } .$$
Let us denote the intersection of side AB and circle O by E . Then
$$\mathrm { BE } = \frac { \mathbf { M } } { \mathbf{N} } .$$
Hence the relationships between the areas of triangles $\mathrm { BDE } , \mathrm { ADE }$ and ACD are
$$\begin{aligned} & \triangle \mathrm { BDE } : \triangle \mathrm { ADE } = \mathbf { O } : \mathbf { P } , \\ & \triangle \mathrm { BDE } : \triangle \mathrm { ACD } = \mathbf { Q } ( \mathbf{J} \sqrt { \mathbf { K } } - \sqrt { \mathrm { K } } ) : \mathbf { R S } \sqrt { \mathbf { T } } . \end{aligned}$$

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
$$\mathbf { I } < a \leqq \frac { \mathbf { J K } } { \mathbf { L M } } .$$

We have four cards of different sizes. We are to paint each card either red, black, blue or yellow. However, we may paint more than one card with the same color.
(1) There are a total of $\mathbf { N O P }$ ways of painting the cards.
(2) There are $\mathbf{QR}$ ways of painting them using all four colors.
(3) There are $\mathbf{ST}$ ways of painting two cards red, one card black, and one card blue.
(4) There are $\mathbf{UVW}$ ways of painting the four cards using three colors.
(5) There are $\mathbf{XY}$ ways of painting them using two colors.

The parallelepiped satisfies
$$\begin{aligned} & \mathrm { AB } = 2 , \quad \mathrm { AD } = 3 , \quad \mathrm { AE } = 1 \\ & \angle \mathrm { BAD } = 60 ^ { \circ } , \quad \angle \mathrm { BAE } = 90 ^ { \circ } , \quad \angle \mathrm { DAE } = 120 ^ { \circ } \end{aligned}$$
and M is the midpoint of edge GH. Let us take points P and Q on edges BF and DH, respectively, such that the four points A, P, M, and Q are on the same plane. We are to find the points P and Q which maximize the length of the line segment PQ.
(1) Setting $\vec { a } = \overrightarrow { \mathrm { AB } } , \vec { b } = \overrightarrow { \mathrm { AD } }$ and $\vec { c } = \overrightarrow { \mathrm { AE } }$, the inner products of these vectors are
$$\vec { a } \cdot \vec { b } = \mathbf { A } , \quad \vec { b } \cdot \vec { c } = - \frac { \mathbf { B } } { \mathbf{C} } , \quad \vec { c } \cdot \vec { a } = \mathbf { D } .$$
(2) Let $s$ and $t$ satisfy $0 \leqq s \leqq 1,0 \leqq t \leqq 1$, and set $\mathrm { BP } : \mathrm { PF } = s : ( 1 - s )$, $\mathrm { DQ } : \mathrm { QH } = t : ( 1 - t )$. Since the four points A, P, M and Q are on the same plane, there exist two real numbers $\alpha$ and $\beta$ such that
$$\overrightarrow { \mathrm { AM } } = \alpha \overrightarrow { \mathrm { AP } } + \beta \overrightarrow { \mathrm { AQ } } .$$
Hence $s$ and $t$ satisfy
$$s = \mathbf { E } ( \mathbf { F } - t ) .$$
Then $| \overrightarrow { \mathrm { PQ } } |$ may be expressed in terms of $t$ as
$$| \overrightarrow { \mathrm { PQ } } | ^ { 2 } = \mathbf { G } t ^ { 2 } - \mathbf { H I } t + \mathbf { J K } .$$
Hence the length of segment PQ is maximized when $\square$ L. Here, for $\square$ L choose the correct answer from among choices (0) $\sim$ (5) below. (0) $s = 0 , \quad t = 1$
(1) $s = 0 , \quad t = \frac { 1 } { 2 }$
(2) $s = \frac { 1 } { 2 } , t = \frac { 3 } { 4 }$
(3) $s = \frac { 2 } { 3 } , t = \frac { 2 } { 3 }$
(4) $s = 1 , \quad t = \frac { 1 } { 2 }$
(5) $s = 1 , \quad t = \frac { 2 } { 3 }$

For any $x$ and $y$ satisfying $x > 0$ and $y > 0$, let $m$ be the smallest value among $\frac { y } { x } , x$ and $\frac { 8 } { y }$.
Also, let $A$ be the set of points $( x , y )$ where $m = \frac { y } { x }$, and let $B$ be the set of points $( x , y )$ where $m = \frac { 8 } { y }$.
(1) For $\mathbf { M } \sim$ S in the following sentence, choose the correct answer from among choices (0) $\sim$ (7) below. $A$ and $B$ can be expressed as follows:
$$\begin{aligned} & A = \{ ( x , y ) \mid \mathbf { M } \leqq \mathbf { N } , \quad \mathbf { O } \leqq 8\mathbf { P } \} \\ & B = \{ ( x , y ) \mid 8 \mathbf { Q } \leqq \mathbf { R } , \quad 8 \leqq \mathbf { S } \} . \end{aligned}$$
(0) $x$
(1) $y$
(2) $x + y$
(3) $x - y$
(4) $x ^ { 2 }$
(5) $x y$ (6) $y ^ { 2 }$ (7) $x ^ { 2 } + y ^ { 2 }$
(2) For $\mathbf { T }$ and $\mathbf { U }$ in the following sentence, choose the correct answer from among choices (0) $\sim$ (8).
When sets $A$ and $B$ are indicated on the $xy$-plane, $A$ is the shaded portion of graph $\mathbf{T}$ and $B$ is the shaded portion of graph $\mathbf{U}$. Note that the $x$ and $y$ axes are not included in the shaded portions.
(3) We are to find the maximum value of $m$ when a point $\mathrm { P } ( x , y )$ moves within $A \cup B$.
When $\mathrm { P } ( x , y ) \in A$, since $y = m x$, we need to find the point P which maximizes the slope of the straight line passing through the origin O and P.
Also, when $\mathrm { P } ( x , y ) \in B$, since $m = \frac { 8 } { y }$, we need to find the point P at which the $y$ coordinate of P is minimized.
From the above, at $( x , y ) = ( \mathbf { V } , \mathbf { W } ) , m$ takes the maximum value $\mathbf { X }$.

We are to find the maximum and the minimum values of the function
$$f ( x ) = 4 \sin ^ { 3 } x + 4 \cos ^ { 3 } x - 8 \sin 2 x - 7$$
where $0 \leqq x \leqq \pi$.
Set $t = \sin x + \cos x$. Since
$$\sin x + \cos x = \sqrt { \mathbf { A } } \sin \left( x + \frac { \mathbf { B } } { \mathbf { C } } \pi \right) , \quad ( \text { note: have } \mathbf { B } < \mathbf { C } )$$
the range of values which $t$ takes is $- \mathbf { D } \leqq t \leqq \sqrt { \mathbf{E} }$. Next, since
$$\sin 2 x = t ^ { 2 } - \mathbf { F }$$
and
$$4 \sin ^ { 3 } x + 4 \cos ^ { 3 } x = - \mathbf { G } t ^ { 3 } + \mathbf { H } t ,$$
we have
$$f ( x ) = - \mathbf { G } t ^ { 3 } - \mathbf { I } t ^ { 2 } + \mathbf { H } t + \mathbf { J } . \tag{1}$$
When we set the right side of (1) as $g ( t )$ and differentiate with respect to $t$, we have
$$g ^ { \prime } ( t ) = - \mathbf { K } ( \mathbf { L } t - \mathbf { M } ) \left( t + \mathbf { N } \right) .$$
Hence at $t = \frac { \mathbf { O } } { \mathbf { P } } , g ( t ) ( = f ( x ) )$ takes the maximum value $\frac { \mathbf { Q R } } { \mathbf{S} }$, and at $t = \sqrt { \mathbf { U } }$, it takes the minimum value $\mathbf { V } \sqrt { \mathbf { W } } - \mathbf { X Y }$.

Let
$$a _ { n } = \int _ { 0 } ^ { 1 } x ^ { 2n } \sqrt { 1 - x ^ { 2 } } \, d x \quad ( n = 0,1,2 , \cdots )$$
We are to find the value of the limit $\lim _ { n \rightarrow \infty } \frac { a _ { n } } { a _ { n - 1 } }$.
(1) First, let us find $a _ { 0 }$ and $a _ { 1 }$. Since the area of a circle with the radius 1 is $\pi$, we see that
$$a _ { 0 } = \int _ { 0 } ^ { 1 } \sqrt { 1 - x ^ { 2 } } \, d x = \frac { \pi } { \mathbf { A } }$$
Next, the partial integral method applied to $a _ { 1 }$ gives
$$\begin{aligned} a _ { 1 } & = \int _ { 0 } ^ { 1 } x ^ { 2 } \sqrt { 1 - x ^ { 2 } } \, d x \\ & = - \frac { \mathbf { B } } { \mathbf { C } } \left[ x \left( 1 - x ^ { 2 } \right) ^ { \frac { \mathbf { D } } { \mathbf { E } } } \right] _ { 0 } ^ { 1 } + \frac { \mathbf { F } } { \mathbf { G } } \int _ { 0 } ^ { 1 } \left( 1 - x ^ { 2 } \right) ^ { \frac { \mathbf { H } } { \mathbf{I} } } d x \\ & = \frac { \mathbf { J } } { \mathbf { K } } \left\{ \int _ { 0 } ^ { 1 } \sqrt { 1 - x ^ { 2 } } \, d x - \int _ { 0 } ^ { 1 } x ^ { \mathbf { L } } \sqrt { 1 - x ^ { 2 } } \, d x \right\} \end{aligned}$$
Thus we have
$$a _ { 1 } = \frac { \pi } { \mathbf { M N } } .$$
(2) For $\mathbf { O } \sim \mathbf { U }$ in the following sentences, choose the correct answer from among choices (0) $\sim$ (9) below.
When the partial integral method is applied to $a _ { n }$ in the same way as for $a _ { 1 }$, we get
$$a _ { n } = \frac { \mathbf { O } } { \mathbf { P } } \left\{ \int _ { 0 } ^ { 1 } x ^ { \mathbf { Q } } \sqrt { 1 - x ^ { 2 } } \, d x - \int _ { 0 } ^ { 1 } x ^ { \mathbf { R } } \sqrt { 1 - x ^ { 2 } } \, d x \right\} \quad ( n = 1,2,3 , \cdots )$$
Hence we have
$$( \mathbf { S } ) a _ { n } = ( \mathbf { T } ) a _ { n - 1 } ,$$
and so
$$\lim _ { n \rightarrow \infty } \frac { a _ { n } } { a _ { n - 1 } } = \mathbf { U }$$
(0) 0
(1) 1
(2) 2
(3) 3
(4) 4
(5) $2 n - 2$ (6) $2 n - 1$ (7) $2 n$ (8) $2 n + 1$ (9) $2 n + 2$