kyotsu-test

QCourse1-I-Q1 Completing the square and sketching Symmetry of parabolas (with respect to origin, line, or point) View

Let $a \neq 0$. Let $G$ be a curve which is symmetric with respect to the origin $(0,0)$ to the graph of the quadratic function in $x$
$$y = ax^2 - 4x - 4a. \tag{1}$$
(1) The coordinates of the vertex of the graph of (1) are
$$\left( \frac{\mathbf{A}}{a}, -\frac{\mathbf{B}}{a} - 4a \right).$$
(2) Among the following choices, the quadratic function whose graph is $G$ is $\square$ C. (0) $y = ax^2 + 4x + 4a$
(1) $y = ax^2 + 4x - 4a$
(2) $y = ax^2 - 4x + 4a$
(3) $y = -ax^2 + 4x + 4a$
(4) $y = -ax^2 - 4x + 4a$
(5) $y = -ax^2 - 4x - 4a$
(3) The curve $G$ intersects the graph of the quadratic function (1) at the two points
$$(\mathrm{DE},\ \mathrm{F}) \text{ and } (\mathrm{G},\ \mathrm{HI}).$$
(4) Let $a = 2$. Then over the interval $\mathrm{DE} \leqq x \leqq \square$, the maximum and the minimum values of the quadratic function whose graph is $G$ are $\square$ JK and $\square$ LM, respectively.

QCourse1-I-Q2 Modulus function Solving equations involving modulus View

Consider the following equation in $x$
$$|ax - 11| = 4x - 10, \tag{1}$$
where $a$ is a constant.
(1) Equation (1) can be rewritten without using the absolute value symbol as
$$\begin{aligned} & \text{when } ax \geqq 11, \quad \text{then } (a - \mathbf{N})x = \mathbf{O}; \\ & \text{when } ax < 11, \quad \text{then } (a + \mathbf{P})x = \mathbf{QR}. \end{aligned}$$
(2) When $a = \sqrt{7}$, the solution of equation (1) is
$$x = \frac{\mathrm{S}}{\mathrm{T}} - \sqrt{\mathrm{U}}.$$
(3) Let $a$ be a positive integer. When equation (1) has a positive integral solution, we have $a = \mathbf{W}$, and that solution $x = \mathbf{X}$.

QCourse1-II-Q1 Discrete Probability Distributions Probability Computation for Compound or Multi-Stage Random Experiments View

There are two boxes, A and B.
In box A, there are three cards on which the number 0 is written, two cards on which the number 2 is written, and one card on which the number 3 is written.
In box B, there are two cards on which the number 1 is written, and three cards on which the number 2 is written.
Take two cards together from box A and one card from box B. Denote the product of the numbers on the three cards by $X$.
The total number of values which $X$ can take is A. The maximum value which $X$ can take is $\mathbf{BC}$. The minimum value which $X$ can take is D.
The probability that $X = \mathrm{BC}$ is $\frac{\mathbf{E}}{\mathrm{F}}$, and the probability that $X = \square\mathrm{D}$ is $\frac{\mathbf{H}}{\mathbf{I}}$.

QCourse1-II-Q2 Sine and Cosine Rules Multi-step composite figure problem View

Consider a triangle ABC where $\mathrm{AB} = 8$, $\mathrm{AC} = 5$ and $\angle\mathrm{BAC} = 60^\circ$. Take point D and point E on sides AB and AC, respectively, such that the segment DE divides triangle ABC into two parts with the equal areas. Set $x = \mathrm{AD}$.
(1) When we represent AE in terms of $x$, we have $\mathrm{AE} = \frac{\mathbf{JK}}{x}$.
(2) When E moves along side AC, the range of the values which $x$ can take is
$$\mathbf{L} \leq x \leqq \mathbf{M}.$$
(3) Since $\mathrm{DE}^2 = \left(x - \frac{\mathbf{NO}}{x}\right)^2 + \mathbf{PQ}$, the length of segment DE is minimized at $x = \mathbf{R}\sqrt{\mathbf{S}}$, and its length there is $\mathbf{T}$.$\mathbf{U}$.

QCourse1-III Straight Lines & Coordinate Geometry Area Computation in Coordinate Geometry View

Consider a figure made by cutting two corners from a rectangle, as in the diagram to the right. The lengths of the sides are
$$\begin{array}{lll} \mathrm{AB} = 11, & \mathrm{BC} = 4, & \mathrm{CD} = 2\sqrt{13} \\ \mathrm{DE} = 5, & \mathrm{EF} = 2\sqrt{5}, & \mathrm{FA} = 6 \end{array}$$
We are to find the area of this figure.
First, extend the sides of the figure as in the diagram and denote the sides forming the right angles by $x, y, u$ and $v$. Then
$$u = \mathbf{A} - y, \quad v = x + \mathbf{B}.$$
Substituting these expressions in the equation $u^2 + v^2 = \mathbf{CD}$ and also using the equation $x^2 + y^2 = \mathbf{EF}$, we obtain
$$x = \mathbf{G}.\, y - \mathbf{H}.$$
Then, since
$$\mathbf{I}y^2 - \mathbf{J} = \mathbf{J}y - \mathbf{K} = 0,$$
we obtain $y = \mathbf{L}$.
From this we have $x = \mathbf{M}$, and further $u = \mathbf{N}$ and $v = \mathbf{O}$. Finally we conclude that the area of this figure is $\mathbf{PQ}$.

QCourse1-IV Inequalities Optimization Subject to an Algebraic Constraint View

Let $x$ and $y$ be real numbers which satisfy
$$3x^2 + 2xy + 3y^2 = 32. \tag{1}$$
Then we are to find the ranges of the values of $x + y$ and $xy$.
First, we set
$$a = x + y. \tag{2}$$
By eliminating $y$ from (1) and (2), we obtain the quadratic equation in $x$
$$\mathbf{A}x^2 - \mathbf{B}ax + \mathbf{C}a^2 - 32 = 0.$$
Since $x$ is a real number, we have
$$\mathbf{DE} \leqq a \leqq \mathbf{F}.$$
Next, we set
$$b = xy. \tag{4}$$
From (1), (2) and (4) we obtain
$$b = \frac{\mathbf{G}}{\mathbf{H}}a^2 - \mathbf{I}.$$
Hence from (3) and (5) we have
$$\mathbf{JK} \leqq b \leqq \mathbf{L}.$$

QCourse2-I-Q1 Completing the square and sketching Symmetry of parabolas (with respect to origin, line, or point) View

Let $a \neq 0$. Let $G$ be a curve which is symmetric with respect to the origin $(0,0)$ to the graph of the quadratic function in $x$
$$y = ax^2 - 4x - 4a. \tag{1}$$
(1) The coordinates of the vertex of the graph of (1) are
$$\left( \frac{\mathbf{A}}{a}, -\frac{\mathbf{B}}{a} - 4a \right).$$
(2) Among the following choices, the quadratic function whose graph is $G$ is $\square$ C. (0) $y = ax^2 + 4x + 4a$
(1) $y = ax^2 + 4x - 4a$
(2) $y = ax^2 - 4x + 4a$
(3) $y = -ax^2 + 4x + 4a$
(4) $y = -ax^2 - 4x + 4a$
(5) $y = -ax^2 - 4x - 4a$
(3) The curve $G$ intersects the graph of the quadratic function (1) at the two points
$$(\mathrm{DE},\ \mathrm{F}) \text{ and } (\mathrm{G},\ \mathrm{HI}).$$
(4) Let $a = 2$. Then over the interval $\mathrm{DE} \leqq x \leqq \square\mathrm{G}$, the maximum and the minimum values of the quadratic function whose graph is $G$ are JK and LM, respectively.

QCourse2-I-Q2 Inequalities Absolute Value Inequality View

Consider the following equation in $x$
$$|ax - 11| = 4x - 10, \tag{1}$$
where $a$ is a constant.
(1) Equation (1) can be rewritten without using the absolute value symbol as
$$\begin{aligned} & \text{when } ax \geqq 11, \quad \text{then } (a - \mathbf{N})x = \mathbf{O}; \\ & \text{when } ax < 11, \quad \text{then } (a + \mathbf{P})x = \mathbf{QR}. \end{aligned}$$
(2) When $a = \sqrt{7}$, the solution of equation (1) is
$$x = \frac{\mathrm{S}\left(\frac{\mathrm{T}}{\mathrm{V}} - \sqrt{\mathrm{U}}\right)}{\mathrm{V}}.$$
(3) Let $a$ be a positive integer. When equation (1) has a positive integral solution, we have $a = \mathbf{W}$, and that solution $x = \mathbf{X}$.

QCourse2-II Vectors 3D & Lines Section Division and Coordinate Computation View

Suppose that a triangle ABC which is inscribed in a circle O of radius 2 satisfies
$$3\overrightarrow{\mathrm{OA}} + 4\overrightarrow{\mathrm{OB}} + 2\overrightarrow{\mathrm{OC}} = \vec{0}.$$
Let D denote the point of intersection of the straight line AO and the segment BC. We are to find the lengths of the segments AD and BD.
(1) When we set $\overrightarrow{\mathrm{OD}} = k\overrightarrow{\mathrm{OA}}$ where $k$ is a real number, we have
$$\overrightarrow{\mathrm{OD}} = -\frac{\mathbf{A}}{\mathbf{B}}k\overrightarrow{\mathrm{OB}} - \frac{\mathbf{C}}{\mathbf{D}}k\overrightarrow{\mathrm{OC}}.$$
As the three points B, C and D are located on a straight line, we obtain $k = \frac{\mathbf{EF}}{\mathbf{GH}}$. From this we derive that $\mathrm{OD} = \mathbf{H}$ and finally obtain
$$\mathrm{AD} = 1.$$
(2) From (1) we see that $\mathrm{BD} = \frac{\mathbf{J}}{\mathbf{K}}\mathrm{BC}$. So in order to find the length of the segment BD, we should find the length of the segment BC.
First we note that
$$\mathrm{BC}^2 = \mathbf{L} - \mathbf{M}\overrightarrow{\mathrm{OB}} \cdot \overrightarrow{\mathrm{OC}}$$
where $\overrightarrow{\mathrm{OB}} \cdot \overrightarrow{\mathrm{OC}}$ represents the inner product of $\overrightarrow{\mathrm{OB}}$ and $\overrightarrow{\mathrm{OC}}$. Since we know from (1) that $|4\overrightarrow{\mathrm{OB}} + 2\overrightarrow{\mathrm{OC}}|^2 = \mathbf{NO}$, we have
$$\overrightarrow{\mathrm{OB}} \cdot \overrightarrow{\mathrm{OC}} = \frac{\mathbf{PQR}}{\mathbf{S}}.$$
Hence we obtain $\mathrm{BC} = \frac{\square\sqrt{\mathbf{U}}}{\square\mathbf{V}}$ and finally from that
$$\mathrm{BD} = \frac{\sqrt{\mathrm{W}}}{\mathrm{X}}.$$

QCourse2-III Laws of Logarithms Optimize a Logarithmic Expression View

Let $x$ and $y$ be positive numbers which satisfy
$$\left(\log_2 x\right)^2 + \left(\log_2 y\right)^2 = \log_2 \frac{8x^2}{y^2}. \tag{1}$$
We are to find the maximum value of $xy^2$ and the values of $x$ and $y$ at that point.
(1) The right side of (1) can be transformed into
$$\log_2 \frac{8x^2}{y^2} = \mathbf{A}\log_2 x - \mathbf{B}\log_2 y + \mathbf{C}.$$
So, setting $X = \log_2 x$ and $Y = \log_2 y$, we can express (1) in terms of $X$ and $Y$ as
$$(X - \mathbf{D})^2 + (Y + \mathbf{E})^2 = \mathbf{F}.$$
(2) Set $k = \log_2 xy^2$. Using $X$ and $Y$ above, this equality can be transformed into
$$X + \mathbf{GG}\,Y - k = 0.$$
If we graph (2) and (3) on a plane with coordinates $(X, Y)$, the graph of (2) is a circle, and the graph of (3) is a straight line. When $k$ is maximized, the graph of (3) is tangent to the graph of (2). Hence, when $k = \mathbf{H}$, $xy^2$ takes the maximum value IJ and in this case $x = \mathbf{K}$ and $y = \mathbf{L}$.

QCourse2-IV-Q1 Differentiating Transcendental Functions Determine parameters from function or curve conditions View

Let $a$ be a constant. Assume that the function
$$f(x) = 2\sin^3 x + a\sin 2x + \frac{9}{2}\cos 2x - 9\cos x - 2ax + 6$$
takes a local extremum at $x = \frac{\pi}{3}$. We consider about the maximum and minimum values of $f(x)$ over the interval $0 \leqq x \leqq \frac{\pi}{2}$.
(1) Since $f(x)$ takes a local extremum at $x = \frac{\pi}{3}$, it follows that $a = \frac{\mathbf{A}}{\mathbf{B}}$.
Hence the derivative $f'(x)$ of $f(x)$ can be expressed as
$$f'(x) = \mathbf{C}\cos x(\mathbf{D}\cos x - 1)(\sin x - \mathbf{E}).$$
(2) It can be seen from the result of (1) that $f(x)$ over $0 \leqq x \leqq \frac{\pi}{2}$ takes the maximum value at $x = \mathbf{F}$ and the minimum value at $x = \mathbf{G}$, where $\mathbf{F}$ and $\mathbf{G}$ are the appropriate expressions from among (0) $\sim$ (4) below. (0) $0$
(1) $\frac{\pi}{6}$
(2) $\frac{\pi}{4}$
(3) $\frac{\pi}{3}$
(4) $\frac{\pi}{2}$

QCourse2-IV-Q2 Integration by Parts Reduction Formula or Recurrence via Integration by Parts View

A sequence $\{a_n\}$ is defined as
$$a_n = \int_0^{\frac{1}{4}} x^n e^{-x}\, dx \quad (n = 1, 2, 3, \cdots).$$
Then
$$a_1 = -\frac{\mathbf{H}}{\mathbf{I}} e^{\frac{\mathbf{JK}}{\mathbf{L}}} + 1.$$
Also $a_{n+1}$ can be expressed in terms of $a_n$ as
$$a_{n+1} = -\left(\frac{\mathbf{M}}{\mathbf{N}}\right)^{n+1} e^{\frac{\mathrm{JK}}{\mathrm{L}}} + (n + \mathbf{O})a_n \quad (n = 1, 2, 3, \cdots).$$
When this is transformed into
$$na_n = a_{n+1} - a_n + \left(\frac{\mathrm{M}}{\mathrm{N}}\right)^{n+1} e^{\frac{\mathrm{JK}}{\mathrm{L}}},$$
we have
$$\sum_{k=1}^{n} ka_k = a_{n+1} - a_1 + \frac{\mathbf{P}}{\mathbf{PQ}} e^{\frac{\mathrm{JK}}{\mathrm{L}}} \left\{1 - \left(\frac{\mathbf{S}}{\mathbf{I}}\right)^n\right\}.$$
Since, for $0 \leqq x$, the range of values of $e^{-x}$ is $0 < e^{-x} \leqq \mathbf{U}$, it follows that
$$0 < a_n < \int_0^{\frac{1}{4}} \square\, x^n\, dx = \frac{1}{\square^{n+1}(n+1)}.$$
Thus, since
$$\lim_{n \to \infty} a_n = \mathbf{W},$$
we obtain
$$\lim_{n \to \infty} \sum_{k=1}^{n} ka_k = \frac{\mathbf{X}}{\mathbf{Y}} e^{\frac{\mathrm{JK}}{\mathrm{L}}} - 1.$$

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2012 eju-math__session2