Q1 Consider the equation $$(x-1)^2 = |3x-5|.$$ (1) Among all solutions of equation (1), the solutions satisfying $x \geqq \frac{5}{3}$ are $x = \mathbf{A}$ and $x = \mathbf{B}$, where $\mathbf{A} < \mathbf{B}$. (2) Equation (1) has a total of $\mathbf{C}$ solutions. When the minimum one is denoted by $\alpha$, the integer $m$ satisfying $m-1 < \alpha \leqq m$ is $\mathbf{DE}$.
Q2 Consider the following three conditions (a), (b) and (c) on two real numbers $x$ and $y$: (a) $x+y=5$ and $xy=3$, (b) $x+y=5$ and $x^2+y^2=19$, (c) $x^2+y^2=19$ and $xy=3$. (1) Using the equality $x^2+y^2=(x+y)^2 - \square\mathbf{F}\, xy$, we see that $$\text{condition (b) gives } xy = \mathbf{G},$$ $$\text{condition (c) gives } x+y = \mathbf{H} \text{ or } x+y = \mathbf{IJ}.$$ (2) For each of the following $\mathbf{K} \sim \mathbf{M}$, choose the most appropriate answer from among the choices (0)$\sim$(3) below. (i) (a) is $\mathbf{K}$ for (b). (ii) (b) is $\mathbf{L}$ for (c). (iii) (c) is $\mathbf{M}$ for (a). (0) a necessary and sufficient condition (1) a sufficient condition but not a necessary condition (2) a necessary condition but not a sufficient condition (3) neither a necessary condition nor a sufficient condition
Q1 Using the five numerals $0,1,2,3,4$, we are to make four-digit integers. (Note that "0123", etc. are not allowed.) (1) The total possible number of integers where the digits are all different numerals is $\mathbf{AB}$. Among them, the total number of integers that do not use 0 is $\mathbf{CD}$. (2) If we are allowed to use the same numeral repeatedly, then the total possible number of four-digit integers is $\mathbf{EFG}$. Among them (i) the total number of integers that use both 1 and 3 twice is $\mathbf{H}$, (ii) the total number of integers that use both 0 and 4 twice is $\mathbf{I}$, (iii) the total number of integers that use both of two numerals twice is $\mathbf{JK}$.
Q2 As shown in the figure, a triangle $ABD$ is inscribed in a semi-circle with the diameter $BC$, where $$\mathrm{AB}=3, \quad \mathrm{BD}=5, \quad \tan\angle\mathrm{ABD}=\frac{3}{4}.$$ We are to find the lengths of the three sides $BC$, $CD$ and $DA$ of the quadrangle $ABCD$ and the area $S$ of the quadrangle $ABCD$. First, since $\cos\angle\mathrm{ABD}=\frac{\mathbf{L}}{\mathbf{M}}$, we have $\mathrm{DA}=\sqrt{\mathbf{NO}}$. Also, since $\sin\angle\mathrm{ABD}=\frac{\mathbf{P}}{\mathbf{Q}}$, we have $\mathrm{BC}=\frac{\square\mathbf{R}\sqrt{\mathbf{ST}}}{\mathbf{U}}$ and thus $\mathrm{CD}=\frac{\mathbf{V}}{\mathbf{W}}$. From these we obtain $$S = \frac{\mathbf{XXY}}{\mathbf{XY}}.$$
Consider the following quadratic equations in $x$ $$x^2+2x-15=0 \tag{1}$$ $$2x^2+3x+a^2+12a=0 \tag{2}$$ Let us denote the two solutions of (1) by $\alpha$ and $\beta$ ($\alpha < \beta$). We are to find the range of values which $a$ in (2) can take, in order that (2) has two real solutions $\gamma$ and $\delta$ and they satisfy $$\alpha < \gamma < \beta < \delta.$$ (1) $\alpha = \mathbf{AB}$ and $\beta = \mathbf{C}$. (2) When we set $b = a^2+12a$, from the condition $\alpha < \gamma$ we have $$b > \mathbf{DEF}$$ and from the condition $\gamma < \beta < \delta$ we have $$b < \mathbf{GHI}.$$ Hence the range of the values which $a$ can take is $$\mathbf{JK} < a < \mathbf{LM}, \quad \mathbf{NO} < a < \mathbf{PQ},$$ where $\mathrm{JK} < \mathrm{NO}$.
Let $x$, $y$ and $z$ satisfy the following two equations: $$x+y-z=0 \tag{1}$$ $$2x-y+1=0 \tag{2}$$ We are to find the values of $a$, $b$ and $c$ such that the equation $$ax^2+by^2+cz^2=1 \tag{3}$$ holds for all $x$, $y$ and $z$ satisfying (1) and (2). First, given (1) and (2), we may express $y$ and $z$ in terms of $x$ as $$y = \mathbf{A}x+\mathbf{B}, \quad z = \mathbf{C}x+\mathbf{D}. \tag{4}$$ This shows that the values of both $y$ and $z$ depend on the value of $x$. Next, when (4) is substituted into (3) and the left side is arranged in descending order of powers of $x$, we obtain $$(a+\mathbf{E}b+\mathbf{F}c)x^2+(\mathbf{G}b+\mathbf{H}c)x+b+c=1.$$ Since this equation holds for any $x$, it holds also when $x=0$, $x=1$ and $x=-1$ are substituted into it, from which we obtain $$\left\{\begin{aligned} b+c &= 1 \\ a+9b+\mathbf{IJ}c &= 1 \\ a+b+\mathbf{K}c &= 1 \end{aligned}\right.$$ When we regard these as simultaneous equations and solve them for $a$, $b$ and $c$, we have $$a = \mathbf{L}, \quad b = \mathbf{M}, \quad c = \mathbf{NO}.$$
Q1 Consider the equation $$(x-1)^2 = |3x-5|.$$ (1) Among all solutions of equation (1), the solutions satisfying $x \geqq \frac{5}{3}$ are $x = \mathbf{A}$ and $x = \mathbf{B}$, where $\mathbf{A} < \mathbf{B}$. (2) Equation (1) has a total of $\mathbf{C}$ solutions. When the minimum one is denoted by $\alpha$, the integer $m$ satisfying $m-1 < \alpha \leqq m$ is $\mathbf{DE}$.
Q2 Consider the following three conditions (a), (b) and (c) on two real numbers $x$ and $y$: (a) $x+y=5$ and $xy=3$, (b) $x+y=5$ and $x^2+y^2=19$, (c) $x^2+y^2=19$ and $xy=3$. (1) Using the equality $x^2+y^2=(x+y)^2-\square\mathbf{F}\,xy$, we see that condition (b) gives $xy=\mathbf{G}$, condition (c) gives $x+y=\mathbf{H}$ or $x+y=\mathbf{IJ}$. (2) For each of the following $\mathbf{K}\sim\mathbf{M}$, choose the most appropriate answer from among the choices (0)$\sim$(3) below. (i) (a) is $\mathbf{K}$ for (b). (ii) (b) is $\mathbf{L}$ for (c). (iii) (c) is $\mathbf{M}$ for (a). (0) a necessary and sufficient condition (1) a sufficient condition but not a necessary condition (2) a necessary condition but not a sufficient condition (3) neither a necessary condition nor a sufficient condition
Consider two straight lines $$y=1, \quad y=-1$$ and the point $\mathrm{A}(0,3)$ in the $xy$-plane. Take a point P on the straight line $y=1$ and a point Q on the straight line $y=-1$ such that $$\angle\mathrm{PAQ}=90^\circ.$$ Let the two points P and Q move preserving the above conditions. We are to find the minimum value of the length of the line segment PQ. First, denote the coordinates of P by $(\alpha,1)$ and the coordinates of Q by $(\beta,-1)$. Then the condition $\angle\mathrm{PAQ}=90^\circ$ is reduced to the conditions $\alpha\neq 0$, $\beta\neq 0$ and $$\alpha\beta = \mathbf{AB}.$$ Since we know that $\alpha$ and $\beta$ have opposite signs, let us assume that $\alpha<0<\beta$. Then we have $$\begin{aligned}
\mathrm{PQ}^2 &= (\beta-\alpha)^2+\mathbf{C} \\
&= \alpha^2+\beta^2+\mathbf{DE} \\
&\geqq 2|\alpha\beta|+\mathbf{DE} = \mathbf{FG}.
\end{aligned}$$ So we have $$\mathrm{PQ}\geqq\mathbf{H}.$$ Hence, when $$\alpha=\mathbf{IJ}\sqrt{\mathbf{K}} \text{ and } \beta=\mathbf{L}\sqrt{\mathbf{M}},$$ PQ takes the minimum value $\mathbf{H}$.
Consider a triangle OAB. We denote by M the point dividing the side OA internally in the ratio 3:1 and denote by N the point dividing the side OB internally in the ratio 1:2. Also we denote by P the point of intersection of the line segment AN and the line segment BM. (1) When the vectors $\overrightarrow{\mathrm{OA}}$ and $\overrightarrow{\mathrm{OB}}$ are denoted by $\vec{a}$ and $\vec{b}$ respectively, we are to express the vector $\overrightarrow{\mathrm{OP}}$ in terms of $\vec{a}$ and $\vec{b}$. When we set $$\begin{array}{ll}
\mathrm{AP}:\mathrm{PN} = s:(1-s) & (0we have $$\overrightarrow{\mathrm{OP}} = (\mathbf{A} - t)\vec{b} = \frac{\mathbf{D}}{\mathbf{E}}t\vec{a}+\frac{\mathbf{B}}{\mathbf{E}}s\vec{b},$$ from which we obtain $$s=\frac{\mathbf{G}}{\mathbf{H}}, \quad t=\frac{\square}{\mathbf{I}}.$$ Hence, $\overrightarrow{\mathrm{OP}}$ can be expressed in terms of $\vec{a}$ and $\vec{b}$ as $$\overrightarrow{\mathrm{OP}} = \frac{\mathbf{K}}{\mathbf{L}}\vec{a}+\frac{\mathbf{M}}{\mathbf{N}}\vec{b}.$$ (2) We are to look at the relation between the length of the line segment OP and the size of $\angle\mathrm{AOB}$, when $\mathrm{OA}=6$ and $\mathrm{OB}=9$. Let us denote the length of OP by $\ell$. When we express $\ell^2$ in terms of $\vec{a}\cdot\vec{b}$, we have $$\ell^2 = \frac{\mathbf{Q}}{\mathbf{PQ}}\vec{a}\cdot\vec{b}+\mathbf{RS}$$ where $\vec{a}\cdot\vec{b}$ represents the inner product of $\vec{a}$ and $\vec{b}$. Hence, for instance, if $\ell=4$, then we have $$\cos\angle\mathrm{AOB}=\frac{\mathbf{TU}}{\mathbf{V}}$$ When the size of $\angle\mathrm{AOB}$ varies, the range of the values which $\ell$ can take is $$\mathbf{W}<\ell<\mathbf{X}.$$
Q1 Let $f(x)=\log(4x-\log x)$, where $\log$ is the natural logarithm. We are to find a local extremum of $f(x)$ by using $f''(x)$. For $\mathbf{K}$ and $\mathbf{L}$, choose the most appropriate answer from among the choices (0)$\sim$(6) below. First of all, we have $$\begin{aligned}
f'(x) &= \frac{\mathbf{A}-\frac{\mathbf{B}}{x}}{4x-\log x} \\
f''(x) &= \frac{\frac{1}{x^{\mathbf{C}}}(4x-\log x)-\left(\mathbf{A}-\frac{\square}{x}\right)^{\mathbf{D}}}{(4x-\log x)^2}
\end{aligned}$$ which give $$\begin{aligned}
f'\left(\frac{\mathbf{E}}{\mathbf{F}}\right) &= 0 \\
f''\left(\frac{\mathbf{E}}{\mathbf{F}}\right) &= \frac{\mathbf{GH}}{\mathbf{I}+\log\mathbf{J}}.
\end{aligned}$$ Since $$f''\left(\frac{\mathbf{E}}{\mathbf{F}}\right) \mathbf{K} \, 0,$$ $f(x)$ has a $\mathbf{L}$ at $x=\frac{\mathbf{E}}{\mathbf{F}}$, and this value is $\log(\mathbf{M}+\log\mathbf{N})$. (0) $=$ (1) $>$ (2) $\geqq$ (3) $<$ (4) $\leqq$ (5) local maximum (6) local minimum
Q2 Suppose that the curve $y=2\cos 2x$ and the curve $y=4\cos x+k$ have a common tangent at $x=a\left(0(1) We set $f(x)=2\cos 2x$ and $g(x)=4\cos x+k$. Since we have assumed that $y=f(x)$ and $y=g(x)$ have a common tangent at $x=a$, we see that $$f'(a)=g'(a), \quad f(a)=g(a).$$ Since $f'(a)=g'(a)$ and $0Hence the coordinates of the tangent point are $\left(\frac{\pi}{\mathbf{O}},-\mathbf{Q}\right)$, and the equation of the common tangent line is $$y=-\mathbf{R}\sqrt{\mathbf{S}}\left(x-\frac{\pi}{\mathbf{T}}\right)-\mathbf{U}.$$ (2) We are to find the area $S$ of the region bounded by these two curves over the range $-\frac{\pi}{2}\leqq x\leqq\frac{\pi}{2}$. Since both of these curves are symmetric with respect to the $y$-axis, by putting $b=\mathbf{V}$ and $c=\frac{\pi}{\mathbf{O}}$ we have $$S=\mathbf{W}\int_{b}^{c}(2\cos 2x-4\cos x-k)\,dx$$ By calculating this, we obtain $$S=\mathbf{X}\cdot\pi-\mathbf{Y}\cdot\mathbf{Z}.$$