Q1 For A $\sim$ K in the following sentences, choose the correct answer from among choices (0) $\sim$ (9) below. (1) Consider the quadratic function
$$y = a x ^ { 2 } + b x + c$$
whose graph is as shown in the figure at the right.
Then $a , b$ and $c$ satisfy the following expressions:
(i) $a \mathbf { A } 0 , b \mathbf { B } 0 , c \mathbf { C } 0$;
(ii) $a + b + c \mathbf { D } 0$;
(iii) $a - b + c \mathbf { E } 0$;
(iv) $4 a + 2 b + c \mathbf { F } 0$;
(v) $b ^ { 2 } - 4 a c \mathbf { G } 0$.
(2) Given the condition that $a , b$ and $c$ satisfy (i) and (ii) in (1), consider the case where the value of $a ^ { 2 } - 8 b - 8 c$ is minimized.
We see that $a = \mathbf { H }$. When we express $y = a x ^ { 2 } + b x + c$ in terms of $b$, we have
$$y = \mathbf { H } x ^ { 2 } + b x - b + \mathbf { I } \text {. }$$
Also, we see that the range of the values of $b$ is $\mathbf { J } < b < \mathbf { K }$. (0) 0
(1) 1
(2) 2
(3) 3
(4) 4
(5) - 2 (6) - 4 (7) $>$ (8) $=$ (9) $<$

Q2 Consider dice-X with the following property: when it is rolled, for numbers 1 through 5 the probabilities of that number's coming up are all the same, but the probability that 6 comes up is twice that of any other number.
(1) Denote by $p$ the probability that a particular number 1 through 5 comes up. The probability that number 6 comes up is $\mathbf { L }$ p. Since the probability of the whole event is $\mathbf { M }$, we have $p = \frac { \mathbf { N } } { \mathbf { O } }$.
(2) Dice-X is rolled twice in succession. Let us denote by $A$ the event that for both rolls the number that comes up is 1 through 5, and by $B$ the event that number 6 comes up at least once. Then the probability of event $A$, $P(A)$, and that of event $B$, $P(B)$, are
$$P ( A ) = \frac { \mathbf { PQ } } { \mathbf { RS } } , \quad P ( B ) = \frac { \mathbf { TU } } { \mathbf { VW } } .$$
Hence, we see that $\mathbf { X }$. (For $\mathbf { X }$, choose the correct answer from among choices (0) $\sim$ (4) below.) (0) $P ( A )$ is less than $P ( B )$ and the difference between them is not less than $\frac { 1 } { 36 }$.
(1) $P ( A )$ is less than $P ( B )$ and the difference between them is less than $\frac { 1 } { 36 }$.
(2) $P ( A )$ and $P ( B )$ are the same.
(3) $P ( A )$ is greater than $P ( B )$ and the difference between them is not less than $\frac { 1 } { 36 }$.
(4) $P ( A )$ is greater than $P ( B )$ and the difference between them is less than $\frac { 1 } { 36 }$.
(3) Next, dice-X is rolled three times in succession. Let us denote by $C$ the event that for all three rolls the number which comes up is 1 through 5, and by $D$ the event that the number 6 comes up at least once. When the probability $P(C)$ is compared with the probability $P(D)$, we see that $\mathbf { Y }$. (For $\mathbf { Y }$, choose the correct answer from among choices (0) $\sim$ (4) below.) (0) $P ( C )$ is less than $P ( D )$ and $P ( D )$ is not less than twice $P ( C )$.
(1) $P ( C )$ is less than $P ( D )$ and $P ( D )$ is less than twice $P ( C )$.
(2) $P ( C )$ and $P ( D )$ are the same.
(3) $P ( C )$ is greater than $P ( D )$ and $P ( C )$ is not less than twice $P ( D )$.
(4) $P ( C )$ is greater than $P ( D )$ and $P ( C )$ is less than twice $P ( D )$.

Q1 Let $a = \sqrt { 5 } + \sqrt { 3 }$ and $b = \sqrt { 5 } - \sqrt { 3 }$. We are to find the integers $x$ satisfying
$$2 \left| x - \frac { a } { b } \right| + x < 10$$
(1) We see that $\frac { a } { b } = \mathbf { A } + \sqrt { \mathbf { BC } }$. Hence the largest integer less than $\frac { a } { b }$ is $\mathbf{D}$.
(2) For $\mathbf { F }$ and $\mathbf { H }$ in the following sentence, choose the correct answer from among choices (0) $\sim$ (7) below, and for $\mathbf { E }$ and $\mathbf { G }$, enter the correct numbers.
When $x$ is an integer, the left side of the inequality can be expressed without using the absolute value symbol as follows:
$$\left\{ \begin{array} { l } \text { if } x \leqq \mathbf { E } , \text { then } 2 \left| x - \frac { a } { b } \right| + x = \mathbf { F } , \\ \text { if } x \geqq \mathbf { G } , \text { then } 2 \left| x - \frac { a } { b } \right| + x = \mathbf { H } . \end{array} \right.$$
(0) $x - 6 - 2 \sqrt { 10 }$
(1) $x + 8 + 2 \sqrt { 15 }$
(2) $- x + 8 + 2 \sqrt { 15 }$
(3) $- x + 6 + 2 \sqrt { 10 }$
(4) $3 x - 6 - 2 \sqrt { 10 }$
(5) $3 x - 8 - 2 \sqrt { 15 }$ (6) $- 3 x + 8 + 2 \sqrt { 15 }$ (7) $- 3 x + 6 + 2 \sqrt { 10 }$
(3) Thus, the integers $x$ satisfying inequality $2 \left| x - \frac { a } { b } \right| + x < 10$ are those greater than or equal to $\mathbf { I }$ and less than or equal to $\mathbf { J }$.

Q2 Let $a$ be a real number. For the two quadratic functions in $x$
$$\begin{aligned} & f ( x ) = x ^ { 2 } + 2 a x + a ^ { 2 } - a , \\ & g ( x ) = 4 - x ^ { 2 } , \end{aligned}$$
answer the following questions.
(1) The range of the values of $a$ such that the equation $f ( x ) = g ( x )$ has two different solutions is
$$- \mathbf { K } < a < \mathbf { L } .$$
(2) In the case of (1), the parabolas $y = f ( x )$ and $y = g ( x )$ intersect at two points. We are to find the range of the values of $a$ such that both of the $y$ coordinates of these points of intersection are positive.
First, let $h ( x ) = f ( x ) - g ( x )$. Since the solutions of the equation $f ( x ) = g ( x )$ are the $x$ coordinates of the points of intersection of parabolas $y = f ( x )$ and $y = g ( x )$, the solutions of $h ( x ) = 0$ have to be between $- \mathbf { M }$ and $\mathbf { N }$. Accordingly, we have
$$\begin{aligned} & h \left( - \mathbf { M } \right) = a ^ { 2 } - \mathbf { O } a + \mathbf { P } > 0 , \quad \cdots \cdots \cdots (1)\\ & h ( \mathbf { N } ) = a ^ { 2 } + \mathbf { Q } a + \mathbf { R } > 0 . \quad \cdots \cdots \cdots (2) \end{aligned}$$
Also, from the position of the axis of the parabola $y = h ( x )$ we have that
$$- \mathbf { S } < a < \mathbf { T } . \quad \cdots \cdots \cdots (3)$$
Therefore, from (1), (2), (3) and (4) we obtain
$$- \mathbf { U } < a < \mathbf{V}.$$

Where $m$ and $n$ are positive integers, consider the rational number
$$r = \frac { m } { 3 } + \frac { n } { 7 } .$$
We are to find $m$ and $n$ such that among all $r$'s satisfying $r < \sqrt { 2 }$, $r$ is closest to $\sqrt { 2 }$.
It is sufficient that among all $m$'s and $n$'s which satisfy the inequality
$$\mathbf { A } m + \mathbf { B } n < \mathbf { CD } \sqrt { 2 }$$
we find the $m$ and $n$ such that $\mathrm { A } m + \mathrm { B } n$ is closest to $\mathrm { CD } \sqrt { 2 }$.
Squaring both sides of (1), we have
$$(\mathrm { A } m + \mathrm { B } n ) ^ { 2 } < \mathbf { EFG } .$$
Here, the greatest square number which is smaller than EFG is $\mathbf{HIJ} = \mathbf{KL}^{2}$. So, let us consider the equation
$$\mathrm { A } m + \mathrm { B } n = \mathrm { KL } .$$
Transforming this equation, we have
$$n = \frac { \mathbf { MN } - \mathbf { O } m } { \mathbf { P } } .$$
Since $n$ is an integer, $\mathbf { MN } - \mathbf { O } m$ is a multiple of $\mathbf { Q }$. Thus, we obtain
$$m = \mathbf { R } , \quad n = \mathbf { S } .$$

For a quadrilateral ABCD inscribed in a circle of radius 1, let $\mathrm { AB } : \mathrm { AD } = 1 : 2$ and $\angle \mathrm { BAD } = 120 ^ { \circ }$. Also, when the point of intersection of diagonals BD and AC is denoted by E, let $\mathrm { BE } : \mathrm { ED } = 3 : 4$.
We are to find the area of quadrilateral $\mathrm{ABCD}$.
In order to find the area of quadrilateral ABCD, we are to find the area of triangle ABD, denoted by $\triangle \mathrm { ABD }$, and the area of triangle BCD, denoted by $\triangle \mathrm { BCD }$.
First, let us find $\triangle \mathrm { ABD }$. Since
$$\mathrm { BD } = \sqrt { \mathbf { A } } , \quad \mathrm { AB } = \frac { \sqrt { \mathbf { BC } } } { \mathbf { D } } ,$$
we have
$$\triangle \mathrm { ABD } = \frac { \mathbf { E } \sqrt { \mathbf { F } } } { \mathbf { GH } } .$$
Next, let us find $\triangle \mathrm { BCD }$. Since
$$\triangle \mathrm { ABC } : \triangle \mathrm { ACD } = \mathbf{I} : \mathbf { J } ,$$
we see that $\mathrm { BC } : \mathrm { CD } = \mathbf { K } : \mathbf { L }$. (Give the answers using the simplest integer ratios.)
Hence we have $\mathrm { BC } = \frac { \mathbf{M} \sqrt { \mathbf { N } } } { \mathbf{O} }$ and
$$\triangle \mathrm { BCD } = \frac { \mathbf { PQ } \sqrt { \mathbf { R } } } { \mathbf { ST } }$$
Thus, from (1) and (2) we obtain the result that the area of quadrilateral ABCD is $\frac{\mathbf{U}\sqrt{\mathbf{V}}}{\mathbf{WX}}$.

(Course 2) Q1 For A $\sim$ K in the following sentences, choose the correct answer from among choices (0) $\sim$ (9) below. (1) Consider the quadratic function
$$y = a x ^ { 2 } + b x + c$$
whose graph is as shown in the figure at the right.
Then $a , b$ and $c$ satisfy the following expressions:
(i) $a \mathbf { A } 0 , b \mathbf { B } 0 , c \mathbf { C } 0$;
(ii) $a + b + c \mathbf { D } 0$;
(iii) $a - b + c \mathbf { E } 0$;
(iv) $4 a + 2 b + c \mathbf { F } 0$;
(v) $b ^ { 2 } - 4 a c \mathbf { G } 0$.
(2) Given the condition that $a , b$ and $c$ satisfy (i) and (ii) in (1), consider the case where the value of $a ^ { 2 } - 8 b - 8 c$ is minimized.
We see that $a = \mathbf { H }$. When we express $y = a x ^ { 2 } + b x + c$ in terms of $b$, we have
$$y = \mathbf { H } x ^ { 2 } + b x - b + \mathbf { I } \text {. }$$
Also, we see that the range of the values of $b$ is $\mathbf { J } < b < \mathbf { K }$. (0) 0
(1) 1
(2) 2
(3) 3
(4) 4
(5) - 2 (6) - 4 (7) $>$ (8) $=$ (9) $<$

(Course 2) Q2 Consider dice-X with the following property: when it is rolled, for numbers 1 through 5 the probabilities of that number's coming up are all the same, but the probability that 6 comes up is twice that of any other number.
(1) Denote by $p$ the probability that a particular number 1 through 5 comes up. The probability that number 6 comes up is $\mathbf { L }$ p. Since the probability of the whole event is $\mathbf { M }$, we have $p = \frac { \mathbf { N } } { \mathbf { O } }$.
(2) Dice-X is rolled twice in succession. Let us denote by $A$ the event that for both rolls the number that comes up is 1 through 5, and by $B$ the event that number 6 comes up at least once. Then the probability of event $A$, $P(A)$, and that of event $B$, $P(B)$, are
$$P ( A ) = \frac { \mathbf { PQ } } { \mathbf { RS } } , \quad P ( B ) = \frac { \mathbf { TU } } { \mathbf { VW } }$$
Hence, we see that $\mathbf { X }$. (For $\mathbf { X }$, choose the correct answer from among choices (0) $\sim$ (4) below.) (0) $P ( A )$ is less than $P ( B )$ and the difference between them is not less than $\frac { 1 } { 36 }$.
(1) $P ( A )$ is less than $P ( B )$ and the difference between them is less than $\frac { 1 } { 36 }$.
(2) $P ( A )$ and $P ( B )$ are the same.
(3) $P ( A )$ is greater than $P ( B )$ and the difference between them is not less than $\frac { 1 } { 36 }$.
(4) $P ( A )$ is greater than $P ( B )$ and the difference between them is less than $\frac { 1 } { 36 }$.
(3) Next, dice-X is rolled three times in succession. Let us denote by $C$ the event that for all three rolls the number which comes up is 1 through 5, and by $D$ the event that the number 6 comes up at least once. When the probability $P(C)$ is compared with the probability $P(D)$, we see that $\mathbf { Y }$. (For $\mathbf { Y }$, choose the correct answer from among choices (0) $\sim$ (4) below.) (0) $P ( C )$ is less than $P ( D )$ and $P ( D )$ is not less than twice $P ( C )$.
(1) $P ( C )$ is less than $P ( D )$ and $P ( D )$ is less than twice $P ( C )$.
(2) $P ( C )$ and $P ( D )$ are the same.
(3) $P ( C )$ is greater than $P ( D )$ and $P ( C )$ is not less than twice $P ( D )$.
(4) $P ( C )$ is greater than $P ( D )$ and $P ( C )$ is less than twice $P ( D )$.

(Course 2) Q1 For $\mathbf { A } , \mathbf { B } , \mathbf { D } , \mathbf { E }$ and $\mathbf { G }$ in the following sentences, choose the correct answer from among choices (0) $\sim$ (9) below, and for the other $\square$, enter the correct number.
Given a sphere of radius 2 with the center at point O, we have a tetrahedron ABCD whose four vertices are on the sphere. Let $\mathrm { AB } = \mathrm { BC } = \mathrm { CA } = 2$ and side BD be a diameter of the sphere.
Set $\overrightarrow { \mathrm { OA } } = \vec { a } , \overrightarrow { \mathrm { OB } } = \vec { b }$ and $\overrightarrow { \mathrm { OC } } = \vec { c }$.
(1) Let M and N denote the midpoints of segments DA and BC, respectively. Then we have
$$\overrightarrow { \mathrm { DA } } = \mathbf { A } , \quad \overrightarrow { \mathrm { MN } } = \frac { \mathbf { B } } { \mathbf { C } } + \mathbf { D }$$
(2) When the midpoint of segment MN is denoted by P and the center of gravity of triangle BCD is denoted by G, we see that
$$\overrightarrow { \mathrm { OP } } = \frac { \mathbf { E } } { \mathbf { F } } , \quad \overrightarrow { \mathrm { OG } } = \frac { \mathbf { G } } { \mathbf { F } } , \quad | \overrightarrow { \mathrm { PG } } | = \frac { \sqrt { \mathbf { H } } } { \mathbf { I } }$$
Also, since $\overrightarrow { \mathrm { AG } } = \frac { \mathbf { K } } { \mathbf { L } } \overrightarrow { \mathrm { AP } }$, we see that the three points A, P and G are on a straight line. (0) $\vec { a }$
(1) $\vec { b }$
(2) $\vec { c }$
(3) $\vec { a } - \vec { b }$
(4) $\vec { b } - \vec { c }$
(5) $\vec { c } - \vec { a }$ (6) $\vec { a } + \vec { b }$ (7) $\vec { b } + \vec { c }$ (8) $\vec { c } + \vec { a }$ (9) $\vec { a } + \vec { b } + \vec { c }$

(Course 2) Q2 Let $\alpha , \beta$ and $\gamma$ be three complex numbers representing three different points A, B and C on a complex plane. Also, $\alpha , \beta$ and $\gamma$ satisfy
$$\begin{aligned} & ( \gamma - \alpha ) ^ { 2 } + ( \gamma - \alpha ) ( \beta - \alpha ) + ( \beta - \alpha ) ^ { 2 } = 0 \quad \cdots (1)\\ & | \beta - 2 \alpha + \gamma | = 3 \quad \cdots (2) \end{aligned}$$
We are to find the area of the triangle ABC.
Since from (1)
$$\frac { \gamma - \alpha } { \beta - \alpha } = \frac { - \mathbf { M } \pm \sqrt { \mathbf { N } } i } { \mathbf { O } } ,$$
we have
$$\left| \frac { \gamma - \alpha } { \beta - \alpha } \right| = \mathbf { P } , \quad \arg \frac { \gamma - \alpha } { \beta - \alpha } = \pm \frac { \mathbf { Q } } { \mathbf { R } } \pi ,$$
where $- \pi < \arg \frac { \gamma - \alpha } { \beta - \alpha } < \pi$. Also, since
$$\beta - 2 \alpha + \gamma = ( \beta - \alpha ) \cdot \frac { \mathbf { S } \pm \sqrt { \mathbf { T } } } { \mathbf { U } } ,$$
we have from (2) that
$$| \beta - \alpha | = \mathbf { V } .$$

(Course 2) On a coordinate plane, consider a circle $C$ with the radius of 1 centered at the origin O. We denote by P and Q the points of intersection of $C$ and the radii which are rotated at angles of $\theta$ and $3\theta$ respectively from the positive section of the $x$ axis, where $0 \leqq \theta \leqq \pi$.
Also, we denote by A the point at which the straight line which is perpendicular to the $x$ axis and passes through point P intersects the $x$ axis, and we denote by B the point at which the straight line which is perpendicular to the $x$ axis and passes through point Q intersects the $x$ axis. Furthermore, we denote the length of line segment AB by $\ell$.
(1) When $\theta = \frac { \pi } { 3 }$, we see that $\ell = \frac { \mathbf { A } } { \mathbf { B } }$.
(2) We are to find the maximum value of $\ell$. When we set $\cos \theta = t$ and express $\ell$ in terms of $t$, we have
$$\ell = \left| \mathbf { C } t ^ { \mathbf { D } } - \mathbf { E } t \right| .$$
Next, when we set $g ( t ) = \mathrm { C } t ^ { \mathrm { D } } - \mathrm { E } t$, we have
$$g ^ { \prime } ( t ) = \mathbf { F } \left( \mathbf { G } t ^ { \mathbf { H } } - 1 \right) .$$
Hence, when
$$\cos \theta = \pm \frac { \sqrt { \mathbf { J } } } { \mathbf { J } }$$
$\ell$ is maximized and its value is $\frac { \mathbf { K } \sqrt { \mathbf { L } } } { \mathbf { M } }$.
(3) For $\mathbf { N } \sim \mathbf{S}$ in the following sentence, choose the correct answer from among choices (0) $\sim$ (9) below.
There are two pairs of points P and Q at which $\ell$ is maximized, and their coordinates are
$$\mathrm { P } \left( \frac { \sqrt { \mathbf{I} } } { \mathbf{J} } , \mathbf{N} \right) \text{ and } \mathrm { Q } \left( \mathbf{O} , \mathbf{P} \right)$$
and
$$\mathrm { P } \left( - \frac { \sqrt { \mathbf{I} } } { \mathbf{J} } , \mathbf{Q} \right) \text{ and } \mathrm { Q } \left( \mathbf{R} , \mathbf{S} \right)$$
(0) $\frac { \sqrt { 6 } } { 3 }$
(1) $\frac { \sqrt { 6 } } { 2 }$
(2) $\frac { 4 \sqrt { 3 } } { 9 }$
(3) $- \frac { 4 \sqrt { 3 } } { 9 }$
(4) $\frac { 5 \sqrt { 3 } } { 9 }$
(5) $- \frac { 5 \sqrt { 3 } } { 9 }$ (6) $\frac { \sqrt { 6 } } { 9 }$ (7) $- \frac { \sqrt { 6 } } { 9 }$ (8) $\frac { 2 \sqrt { 6 } } { 9 }$ (9) $- \frac { 2 \sqrt { 6 } } { 9 }$

(Course 2) Answer the following questions, where log is the natural logarithm.
(1) Let $f ( x ) = x - 1 - \log x$. We are to find the minimum value of $f ( x )$.
First, we have
$$f ^ { \prime } ( x ) = \mathbf { A } - \frac { \mathbf { B } } { x } .$$
Examining the increases and decreases of the value of $f ( x )$, we see that at $x = \mathbf { C }$ the function is minimized and its value is $\mathbf { D }$. From this, we derive the inequality $x - 1 \geqq \log x$.
(2) For $\mathbf { G }$ in the following sentences, choose the correct answer from among choices (0) $\sim$ (3) below. For the other $\square$, enter the correct number.
Let $k$ be a positive real number and $n$ be a positive integer. We denote by $S$ the area of the figure bounded by the three straight lines $y = \frac { x } { n }$, $x = k$ and $y = 0$, and by $T$ the area of the figure bounded by the curve $y = \log x$, the straight line $x = k$ and the $x$ axis.