Q1 The quadratic function $f ( x ) = 2 x ^ { 2 } + a x - 1$ in $x$ satisfies
$$f ( - 1 ) \geqq - 3 , \quad f ( 2 ) \geqq 3 .$$
Let us consider the minimum value $m$ of $f ( x )$.
(1) $\quad m$ can be expressed in terms of $a$ as
$$m = -\frac{\mathbf{A}}{\mathbf{B}} a^2 - \mathbf{C}$$
(2) The range of the values of $a$ such that $f ( x )$ satisfies condition (1) is
$$\mathbf { D E } \leqq a \leqq \mathbf { F } .$$
(3) The value of $m$ is maximized when the axis of symmetry of the graph of $y = f ( x )$ is the straight line $x = \mathbf { G }$, and then the value of $m$ is $\mathbf { H I }$.
(4) The value of $m$ is minimized when the axis of symmetry of the graph of $y = f ( x )$ is the straight line $x = \mathbf { J K }$, and then the value of $m$ is $\mathbf { L M }$.

Q2 A triangle ABC is drawn on a plane, and a ball is placed on vertex A. A dice is rolled, and the ball is moved according to the following rules:
(i) when the ball is on A, if the number on the dice is 1 the ball is moved to B, otherwise it stays on A;
(ii) when the ball is on B, if the number on the dice is less than or equal to 4 the ball is moved to C, otherwise it stays on B.
If the ball is moved to C, the trials are stopped.
We are to find the probability that the ball is moved to C within 4 rolls of the dice.
(1) The probability that the ball is moved to C on the second roll of the dice is $\frac { 1 } { \mathbf{N} }$
(2) The probability that the ball is moved to C on the third roll of the dice is $\frac{\mathbf{O}}{\mathbf{PQ}}$
(3) The probability that the ball is moved to C on the fourth roll of the dice is $\frac{\mathbf{RS}}{\mathbf{TUV}}$
Therefore, find the probability that the ball is moved to C within 4 rolls of the dice.

Q1 Let $a$ and $b$ be rational numbers and let $p$ be a real number. Consider the quadratic equation
$$x ^ { 2 } + a x + b = 0 \tag{1}$$
which has a solution $x = \frac { \sqrt { 5 } + 3 } { \sqrt { 5 } + 2 }$, and consider the inequality
$$x + 1 < 2 x + p + 3 . \tag{2}$$
(1) First, we are to find the values of $a$ and $b$.
When we rationalize the denominator of $x = \frac { \sqrt { 5 } + 3 } { \sqrt { 5 } + 2 }$, we have
$$x = \sqrt { \mathbf { A } } - \mathbf { B }$$
Since this is a solution of equation (1), by substituting this in (1) we have
$$- a + b + \mathbf { C } + ( a - \mathbf { D } ) \sqrt { \mathbf { E } } = 0 .$$
Hence we see that
$$a = \mathbf { F } \text { and } b = \mathbf { G H } .$$
(2) Next, we are to find the smallest integer $p$ such that both solutions of equation (1) satisfy inequality (2).
When we solve inequality (2), we have
$$x > - p - 1 .$$
Since both solutions of equation (1) satisfy this, we see that
$$p > \sqrt { \mathbf { J } } - \mathbf { K } .$$
Hence the smallest integer $p$ is $\mathbf { L }$.

Q2 Consider the quadratic function
$$f ( x ) = \frac { 3 } { 4 } x ^ { 2 } - 3 x + 4$$
Let $a$ and $b$ be real numbers satisfying $0 < a < b$ and $2 < b$. We are to find the values of $a$ and $b$ such that the range of the values of the function $y = f ( x )$ on $a \leqq x \leqq b$ is $a \leqq y \leqq b$.
Since the equation of the axis of symmetry of the graph of $y = f ( x )$ is $x = \mathbf { M }$, we divide the problem into two cases as follows:
(i) $\mathbf{M} \leqq a$;
(ii) $0 < a < \mathbf{M}$.
In the case of (i), since the values of $f ( x )$ increase with $x$ on $a \leqq x \leqq b$, the equations $f ( a ) = a$ and $f ( b ) = b$ have to be satisfied. By solving these, we obtain $a = \frac { \mathbf { N } } { \mathbf { O } }$ and $b = \mathbf { P }$. However, this $a$ does not satisfy (i).
In the case of (ii), since the minimum value of $f ( x )$ on $a \leqq x \leqq b$ is $\mathbf { Q }$, we have
$$a = \mathbf { R } .$$
This satisfies (ii). Then since $f ( a ) = \frac { \mathbf { S } } { \mathbf { T } } < b$, we have $f ( b ) = b$. Hence, we obtain
$$b = \mathbf { U } .$$

Consider four natural numbers $a , b , c$ and $d$ satisfying $1 < a < b < c < d$. Suppose that two sets using these numbers, $A = \{ a , b , c , d \}$ and $B = \left\{ a ^ { 2 } , b ^ { 2 } , c ^ { 2 } , d ^ { 2 } \right\}$, satisfy the following two conditions:
(i) Just two elements belong to the intersection $A \cap B$, and the sum of these two elements is greater than or equal to 15, and less than or equal to 25.
(ii) The sum of all the elements belonging to the union $A \cup B$, is less than or equal to 300.
We are to find the values of $a , b , c$ and $d$.
First, set $A \cap B = \{ x , y \}$, where $x < y$. Since $x \in B$ and $y \in B$, it follows from (i) that $y = \mathbf { A B }$ and that $x$ is either $\mathbf{C}$ or $\mathbf { D }$. (Write the answers in the order $\mathbf { C } < \mathbf { D }$.) Here, when we consider (ii), we see that $x = \mathbf { E }$. Hence $A$ includes the elements $\mathbf { F }$, $\mathbf{F}$ and $\mathbf{F}$.
Furthermore, when we denote the remaining element of $A$ by $z$, from (ii) we see that $z$ satisfies
$$z ^ { 2 } + z \leqq \mathbf { G H } .$$
Hence we have $z = \mathbf { I }$. From the above we obtain
$$a = \mathbf { J } , \quad b = \mathbf { K } , \quad c = \mathbf { L } \text { and } d = \mathbf { M N } .$$

Let the lengths of the three sides of the triangle ABC be $\mathrm { AB } = 6 , \mathrm { BC } = 8$ and $\mathrm { CA } = 4$. Let $\mathrm { O } ^ { \prime }$ be the center of the circle which passes through the two points B and C and is tangent to the straight line AB. Let O be the center of the circle circumscribed about triangle ABC. We are to find the length of the line segment $\mathrm { OO } ^ { \prime }$.
(1) First, we have $\cos \angle \mathrm { ABC } = \frac { \mathbf { A } } { \mathbf { B } }$ and $\sin \angle \mathrm { ABC } = \frac { \sqrt { \mathbf { C D } } } { \mathbf { E } }$.
(2) The radius of the circle circumscribed about triangle ABC is $\frac { \mathbf { F G } \sqrt { \mathbf { H I } } } { \mathbf { J K } }$.
(3) When the intersection point of the straight line $\mathrm { OO } ^ { \prime }$ and the side BC is denoted by D, we have
$$\mathrm { OD } = \frac { \mathbf{N} \sqrt { \mathbf { L M } } } { \mathbf { O P } } \text { and } \mathrm { O } ^ { \prime } \mathrm { D } = \frac { \mathbf { Q R } \sqrt { \mathbf { S T } } } { \mathbf { U V } } .$$
Thus we have $\mathrm { OO } ^ { \prime } = \frac { \mathbf { W } \sqrt { \mathbf { X Y } } } { \mathbf { Z } }$.

Q1 The quadratic function $f ( x ) = 2 x ^ { 2 } + a x - 1$ in $x$ satisfies
$$f ( - 1 ) \geqq - 3 , \quad f ( 2 ) \geqq 3 .$$
Let us consider the minimum value $m$ of $f ( x )$.
(1) $m$ can be expressed in terms of $a$ as
$$m = - \frac { \mathbf { A } } { \mathbf{B} } a ^ { 2 } - \mathbf { C }$$
(2) The range of the values of $a$ such that $f ( x )$ satisfies condition (1) is
$$\mathbf { D E } \leqq a \leqq \mathbf { F } .$$
(3) The value of $m$ is maximized when the axis of symmetry of the graph of $y = f ( x )$ is the straight line $x = \mathbf { G }$, and then the value of $m$ is $\mathbf { H I }$.
(4) The value of $m$ is minimized when the axis of symmetry of the graph of $y = f ( x )$ is the straight line $x = \mathbf { J K }$, and then the value of $m$ is $\mathbf { L M }$.

Q2 A triangle ABC is drawn on a plane, and a ball is placed on vertex A. A dice is rolled, and the ball is moved according to the following rules:
(i) when the ball is on A, if the number on the dice is 1 the ball is moved to B, otherwise it stays on A;
(ii) when the ball is on B, if the number on the dice is less than or equal to 4 the ball is moved to C, otherwise it stays on B.
If the ball is moved to C, the trials are stopped.
We are to find the probability that the ball is moved to C within 4 rolls of the dice.
(1) The probability that the ball is moved to C on the second roll of the dice is $\frac{1}{\mathbf{N}}$
(2) The probability that the ball is moved to C on the third roll of the dice is $\frac{\mathbf{O}}{\mathbf{PQ}}$
(3) The probability that the ball is moved to C on the fourth roll of the dice is $\frac{\mathbf{RS}}{\mathbf{TUV}}$
Therefore, find the probability that the ball is moved to C within 4 rolls of the dice.

Q1 We are to find the general term $a _ { n }$ of the sequence $\left\{ a _ { n } \right\}$ which is determined by the recurrence formula
$$a _ { 1 } = 18 , \quad a _ { n + 1 } - 12 a _ { n } + 3 ^ { n + 2 } = 0 \quad ( n = 1,2,3 , \cdots ) .$$
When we define a sequence $\left\{ b _ { n } \right\}$ by
$$b _ { n } = \frac { a _ { n } } { \mathbf{A}^n } \quad ( n = 1,2,3 , \cdots ) ,$$
$\left\{ b _ { n } \right\}$ satisfies
$$b _ { 1 } = \mathbf { B } , \quad b _ { n + 1 } - \mathbf { C } \; b _ { n } + \mathbf { D } = 0 \quad ( n = 1,2,3 , \cdots ) .$$
This recurrence formula can be transformed into
$$b _ { n + 1 } - \mathbf { E } = \mathbf{F} ( b_n - \mathbf{E} )$$
Next, when we define a sequence $\left\{ c _ { n } \right\}$ by
$$c _ { n } = b _ { n } - \mathbf { E } \quad ( n = 1,2,3 , \cdots ) ,$$
$\left\{ c _ { n } \right\}$ is a geometric progression such that the first term is $\mathbf{G}$ and the common ratio is $\mathbf{H}$.
Hence we have
$$a _ { n } = \mathbf { I } ^ { n } \left( \mathbf { J } \cdot \mathbf { K } ^ { n - 1 } + \mathbf { L } \right) \quad ( n = 1,2,3 , \cdots ) .$$

Q2 As shown in the figure, on an $xy$-plane whose origin is O, let us consider an isosceles triangle ABC satisfying $\mathrm { AB } = \mathrm { AC }$. Furthermore, suppose that side AB passes through $\mathrm { P } ( - 1,5 )$ and side AC passes through $\mathrm{Q}(3, 3)$.
Let us consider the radius of the inscribed circle of the triangle ABC.
Denote the straight line passing through the two points A and B by $\ell _ { 1 }$ and the straight line passing through the two points A and C by $\ell _ { 2 }$. When we denote the slope of $\ell _ { 1 }$ by $a$, the equations of $\ell _ { 1 }$ and $\ell _ { 2 }$ are
$$\begin{aligned} & \ell _ { 1 } : y = a x + a + \mathbf { M } , \\ & \ell _ { 2 } : y = - a x + \mathbf { N } a + \mathbf { O } . \end{aligned}$$
Denote the center and the radius of the inscribed circle by I and $r$, respectively. Then the coordinates of I are $\left( \mathbf { P } - \frac { \mathbf { Q } } { a } , r \right)$.
Hence $r$ can be expressed in terms of $a$ as
$$r = \frac { \mathbf { R } } { \mathbf { T } + \sqrt { \mathbf { S } } }$$
In particular, when $r = \frac { 5 } { 2 }$, the coordinates of vertex A are $\left( \frac { \mathbf { V } } { \mathbf{U} } , \frac { \mathbf { X Y } } { \mathbf { W } } \right)$.

We are to find the range of the values of $k$ such that the inequality
$$\frac { \log 3 x } { 4 x + 1 } \leqq \log \left( \frac { 2 k x } { 4 x + 1 } \right) \tag{1}$$
holds for all positive real numbers $x$, where $\log$ is the natural logarithm.
(1) For $\mathbf{A}$ and $\mathbf{B}$ in the following sentences, choose the correct answer from among (0) $\sim$ (8) below.
By transforming inequality (1) we obtain
$$\log k \geqq \mathbf { A } . \tag{2}$$
Here, when the right side of (2) is denoted by $g ( x )$ and this $g ( x )$ is differentiated with respect to $x$, we have
$$g ^ { \prime } ( x ) = \mathbf { B } .$$
(0) $\frac { \log 3 x } { 4 x + 1 } - \log ( 4 x + 1 ) - \log 2 x$
(1) $\frac { \log 3 x } { 4 x + 1 } - \log ( 4 x + 1 ) + \log 2 x$
(2) $\frac { \log 3 x } { 4 x + 1 } + \log ( 4 x + 1 ) + \log 2 x$
(3) $\frac { \log 3 x } { 4 x + 1 } + \log ( 4 x + 1 ) - \log 2 x$
(4) $\frac { 4 \log 3 x } { ( 4 x + 1 ) ^ { 2 } }$
(5) $\frac { 3 x + 2 + \log 3 x } { ( 4 x + 1 ) ^ { 2 } }$ (6) $- \frac { 4 \log 3 x } { ( 4 x + 1 ) ^ { 2 } }$ (7) $\frac { 3 x - 2 - \log 2 x } { ( 4 x + 1 ) ^ { 2 } }$ (8) $- \frac { 3 \log 2 x } { ( 4 x + 1 ) ^ { 2 } }$
(2) In the following sentences, for $\mathbf { E } , \mathbf { F }$ and $\mathbf { G }$, choose the correct answer from among (0) $\sim$ (3) below. For the other blanks, enter the correct number.
Over the interval $0 < x < \frac { \mathbf { C } } { \mathbf{D} }$, $g ( x )$ is $\mathbf { E }$ and over the interval $\frac { \mathbf{C} } { \mathbf{D} } < x$, $g ( x )$ is $\mathbf { F }$. Hence at $x = \frac { \mathbf { C } } { \mathbf{D} }$, $g ( x )$ is $\mathbf { G }$.
From the above, the range of the value of $k$ such that inequality (1) holds for all positive real numbers $x$ is
$$k \geqq \frac { \mathbf { H } } { \mathbf { I } }$$
(0) increasing
(1) decreasing
(2) maximized
(3) minimized

Consider the following two curves
$$x ^ { 2 } + y ^ { 2 } = 1 , \tag{1}$$ $$4 x y = 1 , \tag{2}$$
where $x > 0 , y > 0$. We are to find the area $S$ of the region bounded by curve (1) and curve (2).
(1) First, let P and Q be the intersection points of curves (1) and (2), and let us denote the $x$-coordinates of P and Q by $p$ and $q$ $(p < q)$, respectively.
From (1), the coordinates $( x , y )$ of the intersection points of curves (1) and (2) can be expressed as $x = \cos \theta , y = \sin \theta \left( 0 < \theta < \frac { \pi } { 2 } \right)$. Then from (2) we have
$$\sin \mathbf { A } \theta = \frac { \mathbf { B } } { \mathbf{C} } .$$
From this we know that
$$\theta = \frac { \mathbf { D } } { \mathbf { E F } } \pi \quad \text { or } \quad \frac { \mathbf { G } } { \mathbf { H I } } \pi$$
(Write the answers in the order such that $\frac { \mathbf{D} } { \mathbf{EF} } < \frac { \mathbf{G} } { \mathbf{HI} }$.) Hence we have
$$p = \cos \frac { \mathbf { J } } { \mathbf { KL } } \pi , \quad q = \cos \frac { \mathbf { M } } { \mathbf { N } } \pi .$$
(2) Now we can find the value of $S$. Since
$$S = \int _ { p } ^ { q } \left( \sqrt { 1 - x ^ { 2 } } - \frac { 1 } { 4 x } \right) d x$$
we have to find the values of
$$I = \int _ { p } ^ { q } \sqrt { 1 - x ^ { 2 } } \, d x , \quad J = \int _ { p } ^ { q } \frac { 1 } { x } \, d x$$
For $I$, when we set $x = \cos \theta$ and calculate by substituting it for $x$ in the integral, we have
$$I = \frac { \mathbf { P } } { \mathbf { Q } }$$
For $J$, we have
$$J = \log \left( \mathbf { R } ^ { \mathbf{S} } + \sqrt { \mathbf { S } } \right) ,$$
where $\log$ is the natural logarithm. From these, we obtain
$$S = \frac { \mathbf{P} } { \mathbf{Q} } \pi - \frac { \mathbf { T } } { \mathbf{U} } \log ( \mathbf { R } + \sqrt { \mathbf { S } } ) .$$