kyotsu-test

QCourse1-I-Q1 Completing the square and sketching Graph translation and resulting quadratic equation View

Let $a$ and $b$ be constants where $a > 0$. Translate the graph of the quadratic function
$$y = 4x^2 + 2ax + b$$
by $a$ in the $x$-direction and by $1 - 7a$ in the $y$-direction. If this graph passes through the point $(0, 4)$, we have
$$b = \mathbf{AB}\, a^2 + \mathbf{C}\, a + \mathbf{D},$$
and the quadratic function representing the graph resulting from these translations is
$$y = \mathbf{E}\, x^2 - \mathbf{F}\, ax + \mathbf{G}.$$
When the graph of quadratic function (1) is tangent to the $x$-axis, we have $a = \frac{\mathbf{H}}{\mathbf{I}}$, and the $x$-coordinate of the point of tangency is $x = \mathbf{J}$.

QCourse1-I-Q2 Factor & Remainder Theorem Factorization and Root Analysis View

Consider the polynomial
$$P = x^2 + 2(a-1)x - 8a - 8.$$
(1) Let $a$ be a rational number. If the value of $P$ is a rational number when $x = 1 - \sqrt{2}$, then $a =$ $\mathbf{K}$ and in this case the value of $P$ is $P =$ $\mathbf{LM}$.
(2) Let $x$ and $a$ be positive integers. We are to investigate $x$ and $a$ which are such that the value of $P$ is a prime number.
When we factorize $P$, we have
$$P = (x - \mathbf{N}\mathbf{N})(x + \mathbf{O}a + \mathbf{P}).$$
Hence $x$ must be $\mathbf{Q}$.
Furthermore, the smallest possible $a$ is $\mathbf{R}$, and in this case the value of $P$ is $P = \mathbf{ST}$.

QCourse1-II-Q1 Binomial Distribution Compute Exact Binomial Probability View

Let P be a point in a plane with a coordinate system that is initially located at the origin $(0,0)$ and moves in the plane according to the following rule:
One dice is thrown. When the number on the dice is a multiple of three, point P moves 1 unit in the positive direction of the $x$-axis, and when the number on the dice is not a multiple of three, point P moves 1 unit in the positive direction of the $y$-axis.
Assume that the dice is thrown four times.
(1) The probability that P reaches point $(3,1)$ is $\frac{\mathbf{A}}{\mathbf{BC}}$.
(2) Altogether, the number of the points which P can reach is $\mathbf{D}$, and the coordinates of these points can be expressed in terms of an integer $k$ as
$$(k,\, \mathbf{E} - k) \quad (\mathbf{F} \leq k \leq \mathbf{G}).$$
Let us denote the probability that P can reach a given point $(k, \mathbf{E} - k)$ by $p_k$. Then the maximum value of $p_k$ is $\frac{\mathbf{HI}}{\mathbf{HI}}$, and the minimum value of $p_k$ is $\frac{\mathbf{J}}{\mathbf{BC}}$.
(3) The probability that $P$ passes through point $(1,1)$ and reaches point $(2,2)$ is $\frac{\mathbf{KL}}{\mathbf{BC}}$.

QCourse1-II-Q2 Vectors Introduction & 2D Area Computation Using Vectors View

Let D, E and F be the three points which divide internally the three sides AB, BC and CA, respectively, of a triangle ABC in the ratio of $k:(1-k)$, where $0 < k \leq \frac{1}{2}$.
(1) When $k = \frac{1}{3}$, we are to find how many times greater the area of triangle ABC is than the area of triangle DEF. Since
$$\triangle\mathrm{ADF} = \triangle\mathrm{BED} = \triangle\mathrm{CFE} = \frac{\mathbf{M}}{\mathbf{N}} \triangle\mathrm{ABC},$$
it follows that
$$\triangle\mathrm{ABC} = \mathbf{O}\, \triangle\mathrm{DEF}.$$
(2) The area of the triangle DEF is half of the area of the triangle ABC when
$$k(1-k) = \frac{\mathbf{P}}{\mathbf{Q}},$$
that is, when
$$k = \frac{\mathbf{R} - \sqrt{\mathbf{S}}}{\mathbf{T}}.$$

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

Let $m$ be a real number. On a plane with the coordinate system, in which the origin is denoted by O, consider the parabola $y = x^2$ and the two points on it,
$$\mathrm{A}(a,\, ma+1), \quad \mathrm{B}(b,\, mb+1) \quad (a < 0 < b)$$
(1) The $x$-coordinates $a$ and $b$ of the two points A and B can be expressed in terms of $m$ as
$$a = \frac{m - \sqrt{D}}{\mathbf{A}}, \quad b = \frac{m + \sqrt{D}}{\mathbf{B}},$$
where the expression $D$ is
$$D = m^2 + \mathbf{C}.$$
(2) Let the coordinates of the point of intersection of the segment AB and the $y$-axis be denoted by $(0, c)$. Then $c = \mathbf{D}$.
(3) Further, when the area $S$ of the triangle OAB with the three vertices O, A and B is expressed in terms of $a$ and $b$, we have
$$S = \frac{1}{2}\mathbf{E},$$
where $E$ is the appropriate choice from among (0) $\sim$ (5). (0) $a + b$
(1) $a - b$
(2) $b - a$
(3) $a^2 + b^2$
(4) $a^2 - b^2$
(5) $b^2 - a^2$
Also, when $S$ is represented in terms of $m$, we have
$$S = \frac{\mathbf{F}}{\mathbf{G}} \sqrt{m^2 + \mathbf{H}}.$$
Hence the value of $S$ is minimalized when $m = \mathbf{I}$, and its minimum value is $S = \mathbf{J}$.

QCourse1-IV Discriminant and conditions for roots Intersection/tangency conditions between two curves View

Let $a$ be a real number. Consider the quadratic expressions in $x$
$$\begin{aligned} & A = x^2 + ax + 1 \\ & B = x^2 + (a+3)x + 4 \end{aligned}$$
(1) The range of values taken by $a$ such that there exists a real number $x$ satisfying $A + B = 0$ is
$$a \leq -\sqrt{\mathbf{AB}} - \frac{\mathbf{C}}{\mathbf{D}} \text{ or } \sqrt{\mathbf{AB}} - \frac{\mathbf{C}}{\mathbf{D}} \leq a.$$
(2) The range of values taken by $a$ such that there exists a real number $x$ satisfying $AB = 0$ is
$$a \leq \mathbf{EF} \text{ or } \mathbf{G} \leq a.$$
(3) There exists a real number $x$ satisfying $A^2 + B^2 = 0$ only when $a = \mathbf{H}$. In this case $x = \mathbf{IJ}$.

QCourse2-I-Q1 Completing the square and sketching Graph translation and resulting quadratic equation View

Let $a$ and $b$ be constants where $a > 0$. Translate the graph of the quadratic function
$$y = 4x^2 + 2ax + b$$
by $a$ in the $x$-direction and by $1 - 7a$ in the $y$-direction. If this graph passes through the point $(0, 4)$, we have
$$b = \mathbf{AB}\, a^2 + \mathbf{C}\, a + \mathbf{D}$$
and the quadratic function representing the graph resulting from these translations is
$$y = \mathbf{E}\, x^2 - \mathbf{F}\, ax + \mathbf{G}.$$
When the graph of quadratic function (1) is tangent to the $x$-axis, we have $a = \frac{\mathbf{H}}{\mathbf{I}}$, and the $x$-coordinate of the point of tangency is $x = \mathbf{J}$.

QCourse2-I-Q2 Factor & Remainder Theorem Factorization and Root Analysis View

Consider the polynomial
$$P = x^2 + 2(a-1)x - 8a - 8.$$
(1) Let $a$ be a rational number. If the value of $P$ is a rational number when $x = 1 - \sqrt{2}$, then $a =$ $\mathbf{K}$ and in this case the value of $P$ is $P =$ $\mathbf{LM}$.
(2) Let $x$ and $a$ be positive integers. We are to investigate $x$ and $a$ which are such that the value of $P$ is a prime number.
When we factorize $P$, we have
$$P = (x - \mathbf{N})(x + \mathbf{O}a + \mathbf{P}).$$
Hence $x$ must be $\mathbf{Q}$.
Furthermore, the smallest possible $a$ is $\mathbf{R}$, and in this case the value of $P$ is $P = \mathbf{ST}$.

QCourse2-II Arithmetic Sequences and Series Summation of Derived Sequence from AP View

Consider a sequence $\{a_n\}$ $(n = 1, 2, 3, \cdots)$ where the sum of the first $n$ terms is
$$\sum_{k=1}^{n} a_k = n^2 + 3n$$
(1) Then $a_n = \mathbf{A}\, n + \mathbf{B}$.
(2) For the sequence $\{b_n\}$ $(n = 1, 2, 3, \cdots)$, where $b_n = n^2 - 5n - 6$, the number of terms satisfying $b_n < 0$ is $\mathbf{C}$, and the sum of such terms is $-\mathbf{DE}$.
(3) It follows that for the sequences $\{a_n\}$ and $\{b_n\}$ in (1) and (2),
$$\sum_{k=1}^{n} \frac{k^2 b_k}{a_k} = \frac{1}{\mathbf{F}}\, n(n + \mathbf{G})\left(n^2 - \mathbf{H}\, n - \mathbf{I}\right).$$

QCourse2-III Circles Inscribed/Circumscribed Circle Computations View

Let $a$, $b$ and $c$ be positive real numbers. Consider a triangle ABC whose vertices are the three points $\mathrm{A}(a, 0)$, $\mathrm{B}(3, b)$ and $\mathrm{C}(0, c)$ on a plane with the coordinate system. Assume that the circumscribed circle of the triangle ABC passes through the origin $\mathrm{O}(0,0)$ and that $\angle\mathrm{BAC} = 60^\circ$.
(1) Since $\angle\mathrm{AOB} = \mathbf{AB}^\circ$, we obtain $b = \sqrt{\mathbf{C}}$.
(2) The equation of the circumscribed circle is
$$\left(x - \frac{a}{\mathbf{D}}\right)^2 + \left(y - \frac{c}{\mathbf{E}}\right)^2 = \frac{a^2 + c^2}{\mathbf{F}},$$
and $c$ can be expressed in terms of $a$ as $c = \sqrt{\mathbf{G}}\,(\mathbf{H} - a)$.
(3) Let D denote the point of intersection of the segment OB and the segment AC.
Set $\alpha = \angle\mathrm{OAC}$ and $\beta = \angle\mathrm{ADB}$. When $a = 2\sqrt{3}$, it follows that
$$\tan\alpha = \mathbf{I} - \sqrt{\mathbf{J}}, \quad \tan\beta = \mathbf{K}.$$

QCourse2-IV-Q1 Stationary points and optimisation Find critical points and classify extrema of a given function View

Let $a$ be a positive real number. We are to investigate local extrema of the function
$$f(x) = x^2 - 5 + 4a\log(2x + a + 8) \quad \left(-\frac{a}{2} - 4 < x < -2\right).$$
(1) When we differentiate the function $f(x)$ with respect to $x$, we obtain
$$f'(x) = \frac{\mathbf{A}(\mathbf{B}\, x + a)(x + \mathbf{C})}{\mathbf{D}\, x + a + \mathbf{E}}.$$
(2) Since a condition of $a$ is that $a > 0$ and the domain of $f(x)$ is $-\frac{a}{2} - 4 < x < -2$, the range of values of $a$ such that $f(x)$ has both a local maximum and a local minimum is
$$\mathbf{F} < a < \mathbf{G}.$$
In such a case, the sum of the local maximum and the local minimum is
$$\frac{a^2}{\mathbf{H}} + \mathbf{I} + \mathbf{IJ}\, a\log\mathbf{K}\, a.$$

QCourse2-IV-Q2 Integration by Parts Definite Integral Evaluation by Parts View

For a positive integer $n$ and a real number $a$, consider the function
$$f_n(a) = \int_0^{\pi} (\cos x + a\sin 2nx)^2\, dx$$
(1) When we transform $f_n(a)$ into
$$f_n(a) = \int_0^{\pi} \left\{\frac{1 + \cos \mathbf{L}\, x}{2} + a^2 \frac{1 - \cos \mathbf{M}\, nx}{2} + a(\sin(2n+1)x + \sin(2n-1)x)\right\} dx$$
and calculate the definite integral on the right side, we obtain
$$f_n(a) = \frac{\pi}{\mathbf{N}}\, a^2 + \frac{\mathbf{O}\, n}{\mathbf{P}\, n^2 - \mathbf{Q}}\, a + \frac{\pi}{\mathbf{R}}.$$
(2) Let $a_n$ denote the value of $a$ at which $f_n(a)$ is minimalized, and set $S_N = \sum_{n=1}^{N} \frac{a_n}{n}$.
Then
$$\begin{aligned} S_N &= -\frac{\mathbf{S}}{\pi} \sum_{n=1}^{N} \left(\frac{1}{2n - \mathbf{T}} - \frac{1}{2n + \mathbf{U}}\right) \\ &= -\frac{\mathbf{S}}{\pi} \left(\mathbf{V} - \frac{1}{\mathbf{W}N + \mathbf{U}}\right) \end{aligned}$$
Hence we obtain
$$\sum_{n=1}^{\infty} \frac{a_n}{n} = \lim_{N \to \infty} S_N = -\frac{\mathbf{Y}}{\pi}.$$

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2012 eju-math__session1