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2021
eju-math__session1 12
2020
eju-math__session2 12
2019
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2018
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2017
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2016
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2015
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2014
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2013
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2012
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2011
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2010
eju-math__session1 12
2020 eju-math__session2

12 maths questions

QCourse1-I-Q1 Completing the square and sketching Parabola through specified points with x-axis intersection properties View
Let $a$ be a positive constant. When we move the graph of the quadratic function $y = \frac{1}{4}x^2$ by parallel translation, the resulting parabola and the $x$-axis intersect at $(-2a, 0)$ and $(4a, 0)$. Let us consider the equation $y = f(x)$ of this parabola.
(1) The function $f(x)$ can be expressed as $$f(x) = \frac{\mathbf{A}}{\mathbf{B}}(x - \mathbf{C}a)(x + \mathbf{D}a)$$
(2) The range of values of $x$ such that the value of $y = f(x)$ is less than or equal to $10a^2$ can be obtained by solving the inequality $$x^2 - \mathbf{E}ax - \mathbf{FG}a^2 \leqq 0,$$ and it is $-\mathbf{H}a \leqq x \leqq \mathbf{I}a$.
(3) Suppose that the length of the segment between the intersections of the straight line $y = 10a$ and the graph of $y = f(x)$ is 10. Since $\mathbf{J}\sqrt{\mathbf{K}}a^2 + \mathbf{LM}a = 10$, we see that the value of $a$ is $\frac{\mathbf{N}}{\mathbf{O}}$.
QCourse1-I-Q2 Permutations & Arrangements Linear Arrangement with Constraints View
There is a staircase of 10 steps which we are to climb. We can go up one step at a time or two steps at a time, but we have to use each method at least once.
(1) Suppose we can go up two steps at a time twice or more in a row. (i) If we climb the staircase going up two steps at a time just 3 times, we will go up one step at a time just $\mathbf{P}$ times, and there are $\mathbf{QR}$ different ways of climbing the staircase. (ii) If we can go up two steps at a time twice or more in a row, there are altogether $\mathbf{ST}$ different ways of climbing the staircase.
(2) Suppose we cannot go up two steps at a time twice or more in a row. (i) If we climb the staircase going up two steps at a time just twice, we will go up one step at a time just $\mathbf{U}$ times, and there are $\mathbf{VW}$ different ways of climbing the staircase. (ii) If we cannot go up two steps at a time twice or more in a row, there are altogether $\mathbf{XY}$ different ways of climbing the staircase.
QCourse1-II-Q1 Indices and Surds Number-Theoretic Reasoning with Indices View
Let $m$ and $n$ be positive integers satisfying $0 < m - n\sqrt{2} < 1$. Denote the integral part of $(m + n\sqrt{2})^3$ by $a$ and the fractional part by $b$.
(1) We are to prove that $a$ is an odd number and $(m - n\sqrt{2})^3 = 1 - b$.
If $(m + n\sqrt{2})^3 = p + q\sqrt{2}$, where $p$ and $q$ are integers, then $$p = m^3 + \mathbf{A}mn^2, \quad q = \mathbf{B}m^2n + \mathbf{C}n^3.$$ Thus, we see that $(m - n\sqrt{2})^3 = p - q\sqrt{2}$.
Furthermore, the integral part of $(m - n\sqrt{2})^3$ is $\mathbf{D}$. When we denote its fractional part by $c$, the following two equations hold: $$\left\{\begin{array}{l} p + q\sqrt{2} = a + b \\ p - q\sqrt{2} = c \end{array}\right.$$
From these we obtain $$\mathbf{E} \quad p - a = b + c.$$
Here, since the left side is an integer and the range of values which the right side takes is $\mathbf{F} < b + c < \mathbf{G}$, we see that $$b + c = \mathbf{H}$$
Hence we see that $a = \mathbf{E}p - \mathbf{H}$, which shows that $a$ is an odd number and that $(m - n\sqrt{2})^3 = 1 - b$.
(2) Let us find the values of $m$ and $n$ when $a = 197$.
Since $a = 197$, we see that $p = \mathbf{IJ}$, that is, $m^3 + \mathbf{A}mn^2 = \mathbf{IJ}$. The positive integers $m$ and $n$ satisfying this equation are $$m = \mathbf{K}, \quad n = \mathbf{L}.$$
QCourse1-II-Q2 Stationary points and optimisation Find absolute extrema on a closed interval or domain View
Let $a$ be a real number satisfying $a \geqq 0$. We are to express the maximum value $M$ of the function $f(x) = |x^2 - 2x|$ on the range $a \leqq x \leqq a + 1$ in terms of $a$. Furthermore, we are to find the minimum value of $M$ over the range $a \geqq 0$.
(1) The function $f(x)$ can be expressed without using the absolute value symbol as follows: when $x \leqq \mathbf{M}$ or $x \geqq \mathbf{N}$, then $f(x) = x^2 - 2x$; when $\mathbf{M} < x < \mathbf{N}$, then $f(x) = -x^2 + 2x$.
Hence, the maximum value of $f(x)$ on $a \leqq x \leqq a + 1$ is the following: when $0 \leqq a \leqq \mathbf{O}$, then $M = \mathbf{P}$; when $\mathbf{O} < a \leqq \frac{\mathbf{Q} + \sqrt{\mathbf{R}}}{\mathbf{S}}$, then $M = -a^2 + \frac{\mathbf{T}}{\mathbf{S}}a$; when $a > \frac{\mathbf{Q} + \sqrt{\mathbf{R}}}{\mathbf{S}}$, then $M = a^2 - \mathbf{U}$.
(2) The minimum value of $M$ over the range $a \geqq 0$ is $\frac{\sqrt{\mathbf{V}}}{\mathbf{W}}$.
QCourse1-III Proof Proof Involving Combinatorial or Number-Theoretic Structure View
Consider integers $a$ and $b$ satisfying the equation $$14a + 9b = 147. \tag{1}$$
(1) We are to find the positive integers $a$ and $b$ satisfying equation (1).
Since $$14a = \mathbf{A}(\mathbf{BC} - \mathbf{D}b) \text{ and } 9b = \mathbf{E}(\mathbf{FG} - \mathbf{H}a),$$ $a$ is a multiple of $\mathbf{A}$, and $b$ is a multiple of $\mathbf{E}$.
So, when we set $a = \mathbf{A}m$ and $b = \mathbf{E}n$, where $m$ and $n$ are integers, from (1) we have $$\mathbf{I}m + \mathbf{J}n = \mathbf{K}.$$
Since the positive integers $m$ and $n$ satisfying this are $$m = \mathbf{L} \text{ and } n = \mathbf{M},$$ we obtain $$a = \mathbf{N} \text{ and } b = \mathbf{O}.$$
(2) We are to find the solutions $a$ and $b$ of equation (1) satisfying $0 < a + b < 5$.
Since $14 \times \mathbf{N} + 9 \times \mathbf{O} = 147$, from this equality and (1) we have $$14(a - \mathbf{N}) = 9(\mathbf{O} - b).$$
Since 14 and 9 are relatively prime, $a$ and $b$ can be expressed in terms of an integer $k$ as $$a = \mathbf{P} + \mathbf{Q}k, \quad b = -\mathbf{RS}k + \mathbf{T}.$$
Since $0 < a + b < 5$, we have $k = \mathbf{U}$, and we obtain $$a = \mathbf{VW}, \quad b = -\mathbf{XY}.$$
QCourse1-IV Sine and Cosine Rules Multi-step composite figure problem View
Consider a triangle ABC and its circumscribed circle O, where the lengths of the three sides of the triangle are $$\mathrm{AB} = 2, \quad \mathrm{BC} = 3, \quad \mathrm{CA} = 4.$$ Below, the area of a triangle such as PQR is expressed as $\triangle\mathrm{PQR}$.
(1) We see that $\cos\angle\mathrm{ABC} = \frac{\mathbf{AB}}{\mathbf{C}}$.
(2) Let us take a point D on the circumference of circle O such that it is on the opposite side of the circle from point B with respect to AC and $$\frac{\triangle\mathrm{ABD}}{\triangle\mathrm{BCD}} = \frac{8}{15}.$$ We are to find the lengths of line segments AD and CD.
First, since $$\angle\mathrm{BAD} = \mathbf{DEF}^\circ - \angle\mathrm{BCD},$$ we have $\sin\angle\mathrm{BAD} = \sin\angle\mathrm{BCD}$. Hence from (1) we have $$\frac{\mathrm{AD}}{\mathrm{CD}} = \frac{\mathbf{G}}{\mathbf{H}},$$ so we set $\mathrm{AD} = \mathbf{G}k$ and $\mathrm{CD} = \mathbf{H}k$, where $k$ is a positive number. Furthermore, since $$\angle\mathrm{ADC} = \mathbf{IJK}^\circ - \angle\mathrm{ABC},$$ we have $\cos\angle\mathrm{ADC} = \frac{\mathbf{L}}{\mathbf{L}}$. Hence, we obtain $k = \frac{\mathbf{N}}{\sqrt{\mathbf{OP}}}$, and then $$\mathrm{AD} = \frac{\mathbf{QR}\sqrt{\mathbf{OP}}}{\mathbf{OP}},$$ $$\mathrm{CD} = \frac{\mathbf{ST}\sqrt{\mathbf{OP}}}{\mathbf{OP}}.$$
(3) When we denote the point of intersection of the straight line DA and the straight line CB by E, we have $$\frac{\triangle\mathrm{ABE}}{\triangle\mathrm{CDE}} = \frac{\mathbf{UV}}{\mathbf{WXY}}.$$
QCourse2-I-Q1 Completing the square and sketching Parabola through specified points with x-axis intersection properties View
Let $a$ be a positive constant. When we move the graph of the quadratic function $y = \frac{1}{4}x^2$ by parallel translation, the resulting parabola and the $x$-axis intersect at $(-2a, 0)$ and $(4a, 0)$. Let us consider the equation $y = f(x)$ of this parabola.
(1) The function $f(x)$ can be expressed as $$f(x) = \frac{\mathbf{A}}{\mathbf{B}}(x - \mathbf{C}a)(x + \mathbf{D}a)$$
(2) The range of values of $x$ such that the value of $y = f(x)$ is less than or equal to $10a^2$ can be obtained by solving the inequality $$x^2 - \mathbf{E}ax - \mathbf{FG}a^2 \leqq 0,$$ and it is $-\mathbf{H}a \leqq x \leqq \mathbf{I}a$.
(3) Suppose that the length of the segment between the intersections of the straight line $y = 10a$ and the graph of $y = f(x)$ is 10. Since $\mathbf{J}\sqrt{\mathbf{K}}a^2 + \mathbf{LM}a = 10$, we see that the value of $a$ is $\frac{\mathbf{N}}{\mathbf{O}}$.
QCourse2-I-Q2 Permutations & Arrangements Linear Arrangement with Constraints View
There is a staircase of 10 steps which we are to climb. We can go up one step at a time or two steps at a time, but we have to use each method at least once.
(1) Suppose we can go up two steps at a time twice or more in a row. (i) If we climb the staircase going up two steps at a time just 3 times, we will go up one step at a time just $\mathbf{P}$ times, and there are $\mathbf{QR}$ different ways of climbing the staircase. (ii) If we can go up two steps at a time twice or more in a row, there are altogether $\mathbf{ST}$ different ways of climbing the staircase.
(2) Suppose we cannot go up two steps at a time twice or more in a row. (i) If we climb the staircase going up two steps at a time just twice, we will go up one step at a time just $\mathbf{U}$ times, and there are $\mathbf{VW}$ different ways of climbing the staircase. (ii) If we cannot go up two steps at a time twice or more in a row, there are altogether $\mathbf{XY}$ different ways of climbing the staircase.
QCourse2-II-Q1 Arithmetic Sequences and Series Optimization Involving an Arithmetic Sequence View
Let $\{a_n\}$ be a sequence such that the sum $S_n$ of the terms from the first term to the $n$-th term is $$S_n = \frac{n^2 - 17n}{4},$$ and let $\{b_n\}$ be the sequence defined by $$b_n = a_n \cdot a_{n+5} \quad (n = 1, 2, 3, \cdots)$$
(1) For $\mathbf{A}$ $\sim$ $\mathbf{C}$ in the following sentences, choose the correct answer from among choices (0) $\sim$ (9) below.
Let us find the sum $T_n$ of the terms of sequence $\{b_n\}$ from the first term to the $n$-th term.
Since $a_n = \mathbf{A}$, we have $b_n = \mathbf{B}$. Hence we obtain $$T_n = \mathbf{C}.$$
(0) $\frac{n-7}{2}$
(1) $\frac{n-9}{2}$
(2) $\frac{n-11}{2}$
(3) $\frac{n^2 - 12n + 27}{4}$
(4) $\frac{n^2 - 13n + 36}{4}$
(5) $\frac{n^2 - 14n + 45}{4}$ (6) $\frac{n(n^2 - 17n + 83)}{12}$ (7) $\frac{n(n^2 - 17n + 89)}{12}$ (8) $\frac{n(n^2 - 18n + 83)}{12}$ (9) $\frac{n(n^2 - 18n + 89)}{12}$
(2) Next, let us find the minimum value of $T_n$.
When $n \leqq \mathbf{D}$ or $\mathbf{EF} \leqq n$, we see that $b_n > 0$. On the other hand, when $\mathbf{G} \leqq n \leqq \mathbf{H}$, we see that $b_n < 0$.
Hence $T_n$ is minimized at $n = \mathbf{I}$, $n = \mathbf{J}$ and $n = \mathbf{K}$, and its minimum value is $\mathbf{L}$. (Answer in the order such that $\mathbf{I} < \mathbf{J} < \mathbf{K}$.)
QCourse2-II-Q2 Complex Numbers Arithmetic Trigonometric/Polar Form and De Moivre's Theorem View
Answer the following questions.
(1) When we express the complex number $8 + 8\sqrt{3}i$ in polar form, we have $$\mathbf{MN}\left(\cos\frac{\pi}{\mathbf{O}} + i\sin\frac{\pi}{\mathbf{P}}\right).$$
(2) Consider the complex numbers $z$ that satisfy $z^4 = 8 + 8\sqrt{3}i$ in the range $0 \leqq \arg z < 2\pi$.
We see that $|z| = \mathbf{Q}$. There are 4 such complex numbers $z$. When these are denoted by $z_1, z_2, z_3, z_4$ in the ascending order of their arguments, we have $$\arg\frac{z_1 z_2 z_3}{z_4} = \frac{\pi}{\mathbf{R}}.$$
(3) Consider the complex numbers $w$ that satisfy $w^8 - 16w^4 + 256 = 0$ in the range $0 \leqq \arg w < 2\pi$. There are 8 such complex numbers $w$. Let us denote them by $w_1, w_2, w_3, w_4, w_5, w_6, w_7, w_8$ in the ascending order of their arguments. Then four of these coincide with numbers $z_1, z_2, z_3, z_4$ in (2). That is, $$w_{\mathbf{S}} = z_1, \quad w_{\mathbf{T}} = z_2, \quad w_{\mathbf{U}} = z_3, \quad w_{\mathbf{V}} = z_4.$$ Also, we have that $w_1 w_8 = \mathbf{W}$ and $w_3 w_4 = \mathbf{XY}$.
QCourse2-III Tangents, normals and gradients Find tangent line with a specified slope or from an external point View
Consider the function $f(x) = x^3 - 4x + 4$. Let the straight line $\ell$ be the tangent to the graph of $y = f(x)$ at the point $\mathrm{A}(-1, 7)$, and the straight line $m$ be the tangent to the graph of $y = f(x)$ that passes through the point $\mathrm{B}(0, -12)$. Also, let C be the point of intersection of $\ell$ and $m$. Let us denote the angle formed by $\ell$ and $m$ at C by $\theta$ $\left(0 < \theta < \frac{\pi}{2}\right)$. We are to find $\tan\theta$.
(1) The derivative $f'(x)$ of $f(x)$ is $$f'(x) = \mathbf{A}x^{\mathbf{B}} - \mathbf{C}.$$ Hence, the slope of $\ell$ is $\mathbf{DE}$, and the equation of $\ell$ is $$y = \mathbf{DE}x + \mathbf{F}.$$
(2) Let us denote by $a$ the $x$-coordinate of the tangent point of the graph of $y = f(x)$ and line $m$. Then the equation of $m$ can be expressed in terms of $a$ as $$y = (\mathbf{G}a^{\mathbf{H}} - \mathbf{I})x - \mathbf{J}a^{\mathbf{K}} + \mathbf{K}.$$ Since line $m$ passes through point $\mathrm{B}(0, -12)$, we see that $a = \mathbf{M}$, and the equation of $m$ is $$y = \mathbf{N}x - \mathbf{OP}.$$ Hence, the coordinates of point C, the intersection of $\ell$ and $m$, are $(\mathbf{Q}, \mathbf{R})$.
(3) Let us denote by $\alpha$ the angle between the positive direction of the $x$-axis and line $\ell$, and by $\beta$ the angle between the positive direction of the $x$-axis and line $m$. Then we see that $$\tan\alpha = \mathbf{ST}, \quad \tan\beta = \mathbf{U},$$ and hence $$\tan\theta = \frac{\mathbf{V}}{\mathbf{W}}.$$
QCourse2-IV Indefinite & Definite Integrals Definite Integral Evaluation (Computational) View
Consider the function $$f(x) = \sin x + \frac{\sin 2x}{2} + \frac{\sin 3x}{3}$$ on the interval $0 \leqq x \leqq \pi$. We are to show that $f(x) > 0$ on $0 < x < \pi$, and to find the area $S$ of the region bounded by the graph of $y = f(x)$ and the $x$-axis.
(1) For $\mathbf{K}$, $\mathbf{N}$, $\mathbf{Q}$, $\mathbf{R}$ in the following sentences, choose the correct answer from the following two choices: (0) increasing, (1) decreasing, and for the other blanks, enter the correct number.
When we differentiate $f(x)$, we have $$f'(x) = (\mathbf{A}\cos^2 x - \mathbf{B})(\mathbf{C}\cos x + \mathbf{D}).$$ Hence, over the range $0 \leqq x \leqq \pi$, there are three $x$'s at which $f'(x) = 0$, and when they are arranged in ascending order, they are given accordingly.
Next, looking at whether $f(x)$ is increasing or decreasing, the behaviour is described accordingly.
Also, we have $$f(0) = 0, \quad f(\pi) = 0, \quad f\left(\frac{\mathbf{L}}{\mathbf{M}}\pi\right) = \frac{\sqrt{\mathbf{S}}}{\mathbf{T}} > 0.$$ Hence we see that $f(x) > 0$ on $0 < x < \pi$.
(2) The area $S$ of the region bounded by the graph of $y = f(x)$ and the $x$-axis is $$S = \frac{\mathbf{UV}}{\mathbf{W}}.$$