kyotsu-test

QCourse1-I-Q1 Polynomial Division & Manipulation View

Set $P = 10a^2 + 14ab - 21bc - 15ca$.
(1) Factorizing $P$, we obtain $$P = (\mathbf{A}a + \mathbf{B}b)(\mathbf{C}a - \mathbf{D}c).$$
(2) If $5a = \sqrt{6}$, $14b = \sqrt{2} + \sqrt{3} - \sqrt{6}$ and $15c = \sqrt{12} - \sqrt{8}$, then $$P = \frac{\mathbf{E}}{\mathbf{E} + \mathbf{F}} \frac{\mathbf{G}}{\mathbf{H}}$$ and hence the greatest integer less than $P$ is $\mathbf{I}$.

QCourse1-I-Q2 Combinations & Selection Combinatorial Probability View

There are two bags, A and B. Bag A contains four white balls and one red ball, and bag B contains two white balls and three red balls. Two balls are taken simultaneously out of bag A, then two balls are taken simultaneously out of bag B.
(1) The probability that two white balls are taken out of A, and one white ball and one red ball are taken out of B is $\frac{\mathbf{J}}{\mathbf{J}}$.
(2) The probability that the four balls taken out consist of three white balls and one red ball is $\frac{\mathbf{M}}{\mathbf{M}}$.
(3) The probability that the four balls taken out all have the same color is $\square$ PQ
(4) The probability that of the four balls taken out, two or fewer are white balls is $\frac{\mathbf{RS}}{\mathbf{TU}}$.

QCourse1-II-Q1 Solving quadratics and applications Determining quadratic function from given conditions View

Consider the two parabolas $$\begin{aligned} \ell : & & y = ax^2 + 2bx + c \\ m : & & y = (a+1)x^2 + 2(b+2)x + c + 3. \end{aligned}$$ Four points A, B, C and D are assumed to be in the relative positions shown in the figure. One of the two parabolas passes through the three points A, B and C, and the other one passes through the three points B, C and D.
(1) The parabola passing through the three points A, B and C is $\mathbf{A}$. Here, for $\mathbf{A}$ choose the correct answer from (0) or (1), just below. (0) parabola $\ell$ (1) parabola $m$
(2) Since both parabolas $\ell$ and $m$ pass through the two points B and C, the $x$-coordinates of B and C are the solutions of the quadratic equation $$x^2 + \mathbf{B}x + \mathbf{C} = 0.$$ Hence, the $x$-coordinate of point B is $\mathbf{DE}$, and the $x$-coordinate of point C is $\mathbf{FG}$.
(3) In particular, we are to find the values of $a$, $b$ and $c$ when $\mathrm{AB} = \mathrm{BC}$ and $\mathrm{CO} = \mathrm{OD}$.
Since the two points C and D are symmetric with respect to the $y$-axis, we have $b = \mathbf{H}$. On the other hand, since $\mathrm{AB} = \mathrm{BC}$, the straight line $x = \mathbf{IJ}$ is the axis of symmetry of $\mathbf{A}$. Hence we have $a = -\frac{\mathbf{K}}{\mathbf{L}}$. And we have $c = \frac{\mathbf{M}}{\mathbf{L}}$.

QCourse1-II-Q2 Number Theory Divisibility and Divisor Analysis View

We are to find the natural number $a$ such that $3a + 1$ is a divisor of $a^2 + 5$.
Set $b = 3a + 1$. Then we have $$a^2 + 5 = \frac{b^2 - \mathbf{OO}b + \mathbf{PQ}}{\mathbf{R}}.$$
On the other hand, since $b$ is a divisor of $a^2 + 5$, $a^2 + 5$ can be expressed as $$a^2 + 5 = bc$$ for some natural number $c$. From (1) and (2), we have $$b(\mathbf{S}c - b + \mathbf{IT}) = \mathbf{UV}.$$
This shows that $b$ must also be one of the divisors of UV. Of these, only $b = \mathbf{WX}$ is a number such that $a$ is a natural number. Hence, $a = \mathbf{YZ}$.

QCourse1-III Inequalities Solve Polynomial/Rational Inequality for Solution Set View

We have a triangle which has sides of the lengths 15, 19 and 23. We make it into an obtuse triangle by shortening each of its sides by $x$. What is the range of values that $x$ can take?
First, since $15 - x$, $19 - x$ and $23 - x$ can be the lengths of the sides of a triangle, it follows that $$x < \mathbf{AB}.$$
In addition, such a triangle is an obtuse triangle only when $x$ satisfies $$x^2 - \mathbf{CD}x + \mathbf{EF} < 0.$$
By solving this quadratic inequality, we have $$\mathbf{G} < x < \overline{\mathbf{HI}}.$$
Hence, the range of $x$ is $$\mathbf{J} < x < \mathbf{KL}.$$

QCourse1-IV Sine and Cosine Rules Multi-step composite figure problem View

In the figure to the right, let $$\mathrm{OA} = 6, \quad \mathrm{OB} = 3, \quad \angle\mathrm{AOB} = 120^\circ,$$ and let the point Q denote the point of intersection of the bisector of $\angle\mathrm{XAB}$ and the bisector of $\angle\mathrm{ABY}$. Let P denote the point of intersection of segment AB and segment OQ. We are to find the length of segment PQ.
(1) First of all, we see that $\mathrm{AB} = \square\sqrt{\mathbf{A}}$ and that the area of triangle OAB is $\mathbf{BC}\sqrt{\mathbf{D}}$.
(2) For $\mathbf{F}$ and $\mathbf{G}$ in the following, choose the correct answer from among choices (0) $\sim$ (4), just below. (0) AB (1) AP (2) AQ (3) BP (4) BQ
Since AQ is the bisector of the exterior angle of $\angle\mathrm{A}$ of triangle OAP and BQ is the bisector of the exterior angle of $\angle\mathrm{B}$ of triangle OBP, we have $$\begin{aligned} \mathrm{OQ} : \mathrm{PQ} &= \mathrm{OA} : \mathbf{F} \\ &= \mathrm{OB} : \mathbf{G} \end{aligned}$$
(3) Thus we see $\mathrm{AP} = \mathbf{H}\sqrt{\mathbf{I}}$. Since $\angle\mathrm{AOP} = \mathbf{JK}^\circ$, we have $\mathrm{OP} = \square\mathbf{L}$. Hence we have $\mathrm{PQ} = \mathbf{M} + \mathbf{N}\sqrt{\mathbf{O}}$.

QCourse2-I-Q1 Completing the square and sketching Two quadratic functions: intersection, tangency, or equality conditions View

Consider the two parabolas $$\begin{aligned} \ell : & & y = ax^2 + 2bx + c \\ m : & & y = (a+1)x^2 + 2(b+2)x + c + 3. \end{aligned}$$ Four points A, B, C and D are assumed to be in the relative positions shown in the figure. One of the two parabolas passes through the three points A, B and C, and the other one passes through the three points B, C and D.
(1) The parabola passing through the three points A, B and C is $\mathbf{A}$. Here, for $\mathbf{A}$ choose the correct answer from (0) or (1), just below. (0) parabola $\ell$ (1) parabola $m$
(2) Since both parabolas $\ell$ and $m$ pass through the two points B and C, the $x$-coordinates of B and C are the solutions of the quadratic equation $$x^2 + \mathbf{B}x + \mathbf{C} = 0.$$ Hence, the $x$-coordinate of point B is $\mathbf{DE}$, and the $x$-coordinate of point C is $\mathbf{FG}$.
(3) In particular, we are to find the values of $a$, $b$ and $c$ when $\mathrm{AB} = \mathrm{BC}$ and $\mathrm{CO} = \mathrm{OD}$.
Since the two points C and D are symmetric with respect to the $y$-axis, we have $b = \mathbf{H}$. On the other hand, since $\mathrm{AB} = \mathrm{BC}$, the straight line $x = \mathbf{IJ}$ is the axis of symmetry of $\mathbf{A}$. Hence we have $a = -\frac{\mathbf{K}}{\mathbf{L}}$. And we have $c = \frac{\mathbf{M}}{\mathbf{L}}$.

QCourse2-I-Q2 Combinations & Selection Combinatorial Probability View

There are two bags, A and B. Bag A contains four white balls and one red ball, and bag B contains two white balls and three red balls. Two balls are taken simultaneously out of bag A, then two balls are taken simultaneously out of bag B.
(1) The probability that two white balls are taken out of A, and one white ball and one red ball are taken out of B is $\frac{\mathbf{O}}{\mathbf{PQ}}$.
(2) The probability that the four balls taken out consist of three white balls and one red ball is $\frac{\mathbf{R}}{\mathbf{S}}$.
(3) The probability that the four balls taken out all have the same color is $\square$ T UV
(4) The probability that of the four balls taken out, two or fewer are white balls is $\frac{\mathbf{WX}}{\mathbf{W}}$.

QCourse2-II-Q1 Vectors Introduction & 2D Angle or Cosine Between Vectors View

Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{a}| = 1$, $|\vec{b}| = 2$ and the angle formed by $\vec{a}$ and $\vec{b}$ is $60^\circ$. Set $\vec{u} = x\vec{a} + \vec{b}$ and $\vec{v} = x\vec{a} - \vec{b}$ for a real number $x$. When $x > 1$, we are to find the value of $x$ such that the angle formed by $\vec{u}$ and $\vec{v}$ is $30^\circ$. In the following, $\vec{u} \cdot \vec{v}$ denotes the inner product of $\vec{u}$ and $\vec{v}$, and $\vec{a} \cdot \vec{b}$ denotes the inner product of $\vec{a}$ and $\vec{b}$.
First of all, since the angle formed by $\vec{u}$ and $\vec{v}$ is $30^\circ$, we obtain $$(\vec{u} \cdot \vec{v})^2 = \frac{\mathbf{A}}{\mathbf{B}}|\vec{u}|^2|\vec{v}|^2.$$
When we express this equation in terms of $x$, noting $\vec{a} \cdot \vec{b} = \mathbf{C}$, we have $$x^4 - \mathbf{DE}x^2 + \mathbf{FG} = 0.$$
By transforming this, we also have $$\left(x^2 - \mathbf{H}\right)^2 = (\mathbf{I}x)^2.$$
When this is solved for $x$, we obtain $$x = \mathbf{J} + \sqrt{\mathbf{KL}},$$ noting $x > 1$.

QCourse2-II-Q2 Complex Numbers Argand & Loci Locus Identification from Modulus/Argument Equation View

In a complex number plane, consider the complex numbers $z$ such that $z^3$ is a real number.
(1) Let $C$ be the figure formed by the set of complex numbers $z = x + iy$ satisfying the above condition. Since the arguments of the complex numbers $z$ satisfy $$\arg z = \frac{\pi}{\mathbf{M}}k \quad (k : \text{integer}),$$ figure $C$ consists of three straight lines represented in terms of $x$ and $y$ by the equations $$y = \mathbf{N}, \quad y = \sqrt{\mathbf{O}}\,x, \quad y = -\sqrt{\mathbf{O}}\,x.$$
(2) Suppose that on $C$ there exists only one complex number $z$ satisfying $|z - 1 - i| = r$. Then the value of $r$ is $$r = \frac{\sqrt{\mathbf{Q}} - \square\mathbf{R}}{\square}$$ and the value of $z$ is $$z = \frac{\mathbf{T} + \sqrt{\mathbf{U}}}{\square\mathbf{V}}(1 + \sqrt{\mathbf{W}}\,i).$$

QCourse2-III Stationary points and optimisation Find absolute extrema on a closed interval or domain View

We are to find the range of the values of a real number $t$ such that the maximum value of the cubic function $$f(x) = \frac{1}{3}x^3 - \frac{t+2}{2}x^2 + 2tx + \frac{2}{3}$$ over the interval $x \leqq 4$ is greater than 6.
First of all, since the derivative of $f(x)$ is $$f'(x) = (x - \mathbf{A})(x - t),$$ we consider the problem by dividing the range of the values of $t$ as follows:
(i) When $t > \mathbf{A}$, $f(x)$ has a local maximum at $x = \mathbf{A}$ and a local minimum at $x = t$. Since $f(4) = \mathbf{B}$, we only have to find the range of the values of $t$ satisfying $f(\mathbf{A}) > 6$.
(ii) When $t = \mathbf{A}$, the maximum value of $f(x)$ over the interval $x \leqq 4$ is $f(\mathbf{C}) = \mathbf{D}$, and hence the condition is not satisfied.
(iii) When $t < \mathbf{A}$, $f(x)$ has a local maximum at $x = t$ and a local minimum at $x = \mathbf{A}$. Since $f(4) = \mathbf{B}$, we only have to find the range of the values of $t$ satisfying $f(t) > 6$.
Here, we note $$f(t) - 6 = -\frac{1}{6}(t + \mathbf{E})(t - \mathbf{EF})^2.$$
From the above, the range of the values of $t$ is to be determined.

QCourse2-IV Integration by Substitution Substitution to Transform Integral Form (Show Transformed Expression) View

Consider the function $$f(x) = \frac{\sin x}{3 - 2\cos x} \quad (0 \leqq x \leqq \pi)$$
(1) The derivative of $f(x)$ is $$f'(x) = \frac{\mathbf{A}\cos x - \mathbf{B}}{(\mathbf{C} - \mathbf{D}\cos x)^2}.$$ Let $\alpha$ be the value of $x$ at which $f(x)$ has a local extremum. Then we have $$\cos\alpha = \frac{\mathbf{E}}{\mathbf{F}}.$$
(2) The portion of the plane bounded by the graph of the function $y = f(x)$ and the $x$-axis is divided into two parts by the straight line $x = \alpha$. Let $S_1$ be the area of the part located on the left side of the line. Then we have $$S_1 = \int_{\frac{\mathbf{G}}{\mathbf{H}}}^{\mathbf{I}} \frac{dt}{\mathbf{J} - \mathbf{K}t} = \frac{\mathbf{L}}{\mathbf{L}}\log\frac{\mathbf{L}}{\mathbf{L}}.$$ Let $S_2$ be the area of the part located on the right side. We have $$S_2 = \frac{\mathbf{P}}{2}\log\mathbf{Q}.$$

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2015 eju-math__session1