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2021
eju-math__session1 12
2020
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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
2014 eju-math__session1

12 maths questions

QCourse1-I-Q1 Solving quadratics and applications Determining quadratic function from given conditions View
Q1 A quadratic function $y = ax^2 + bx + \frac{3}{a}$ satisfies the following two conditions:
(i) $y$ is maximized at $x = 3$,
(ii) the value of $y$ at $x = 1$ is 2.
We are to find the values of $a$ and $b$.
Using conditions (i) and (ii), we obtain the following relationships between $a$ and $b$:
$$\left\{ \begin{aligned} b & = \mathbf{AB}a \\ \mathbf{C} & = a + b + \frac{\mathbf{D}}{a}. \end{aligned} \right.$$
From these two equalities, we have the equation
$$\mathbf{E}a^2 + \mathbf{F}a - \mathbf{G} = 0$$
and hence
$$a = \mathbf{HI}, \quad b = \mathbf{J}.$$
Thus the maximum value of this function is $\mathbf{K}$.
QCourse1-I-Q2 Polynomial Division & Manipulation View
Q2 Consider
$$E = P^2 - 4Q^2 - 3P + 6Q$$
where $P$ and $Q$ are the integral expressions
$$P = 2x^2 - x + 2, \quad Q = x^2 - 2x + 1.$$
(1) By factorizing the right side of $E$, we obtain
$$E = (P - \mathbf{LL}Q)(P + \mathbf{M}Q - \mathbf{MN}).$$
(2) When we express $E$ in terms of $x$, we have
$$E = \mathbf{O}x(x - \mathbf{P})(\mathbf{Q})(\mathbf{Q} - \mathbf{Q}).$$
(3) If $x = -\frac{1 - \sqrt{5}}{3 - \sqrt{5}}$, then the value of $E$ is $\mathbf{S} + \mathbf{T}\sqrt{\mathbf{U}}$.
QCourse1-II-Q1 Binomial Distribution Find n or Threshold from Cumulative Probability Condition View
Q1 In a box, there are $n$ red balls and $(20 - n)$ white balls, where $0 < n < 20$. In each trial, a ball is taken out of the box, its color is examined, and it is returned to the box.
(1) Let $x$ be the probability that the ball taken out in one trial is red. Then, $x = \frac{n}{\mathbf{AB}}$.
(2) Let $p$ be the probability that in two trials a white ball is taken out at least once. Then $p$ can be expressed as $p = \mathbf{C} - x^{\mathbf{D}}$, where $x$ is the $x$ of (1).
(3) Let $q$ be the probability that in four trials a white ball is taken out at least twice. Then $q$ can be expressed as
$$q = \mathbf{E} - \mathbf{H}x^{\mathbf{G}} + \mathbf{H}x^{\mathbf{I}},$$
where $x$ is the $x$ of (1).
(4) For $p$ and $q$ of (2) and (3), we are to find the maximum value of $n$ such that $p < q$.
From the inequality $p < q$, we obtain the inequality
$$\mathbf{J}x^2 - \mathbf{K} \square \text{ } \square$$
When we solve this, we have
$$x < \frac{1}{\mathbf{L}}$$
Thus the maximum value of $n$ is $\square\mathbf{M}$.
QCourse1-II-Q2 Number Theory Linear Diophantine Equations View
Q2 Let $p$ be a prime number, and let $x$ and $y$ be positive integers. Then we are to find all triples of $p$, $x$ and $y$ which satisfy
$$\frac{p}{x} + \frac{7}{y} = p.$$
We can transform this equation into
$$(x - \mathbf{N})(py - \mathbf{OO}) = \mathbf{OP}.$$
From this, we obtain
$$x - \mathbf{N} = \mathbf{Q} \text{ or } \mathbf{R}, \quad (\text{note: have } \mathbf{Q} < \mathbf{R})$$
and hence
$$x = \mathbf{S} \text{ or } \mathbf{T}. \quad (\text{note: have } S < T)$$
First, if $x = S$, then
$$p = \mathbf{U}, \quad y = \mathbf{V}$$
or
$$p = \mathbf{W}, \quad y = \mathbf{X}. \quad (\text{note: have } \mathbf{U} < \mathbf{W})$$
Next, if $x = T$, then
$$p = \mathbf{Y}, \quad y = \mathbf{Z}.$$
QCourse1-III Curve Sketching Number of Solutions / Roots via Curve Analysis View
Consider a quadratic function in $x$
$$y = ax^2 + bx + c$$
such that the graph of function (1) passes through the two points $(-1, -1)$ and $(2, 2)$.
(1) When we express $b$ and $c$ in terms of $a$, we have
$$b = \mathbf{A} - a, \quad c = \mathbf{BC}a.$$
(2) Suppose that one of the points of intersection of the graph of function (1) and the $x$-axis is within the interval $0 < x \leqq 1$. Then the range of values of $a$ is [see figure].
(3) When the value of $a$ varies within interval (2), the range of values of $a + bc$ is
$$\frac{\mathbf{GH}}{\square\mathbf{I}} \leqq a + bc \leqq \square.$$
QCourse1-IV Sine and Cosine Rules Circumradius or incircle radius computation View
A triangle ABC satisfies
$$\mathrm{AB} = 7, \quad \mathrm{BC} = 8, \quad \mathrm{CA} = 6.$$
We denote the center and the radius of the circumscribed circle of this triangle ABC by O and $r$, respectively. We draw two straight lines which are tangent to this circumscribed circle at the points B and C, and denote the point of intersection of these straight lines by D.
We see that
$$\cos\angle\mathrm{BAC} = \frac{\mathbf{A}}{\mathbf{B}}, \quad \sin\angle\mathrm{BAC} = \frac{\sqrt{\mathbf{CD}}}{\mathbf{E}},$$
$$r = \frac{\mathbf{FG}\sqrt{\mathbf{HI}}}{\mathbf{JK}}, \quad \mathrm{BD} = \mathbf{LM}.$$
Furthermore, if P is a point on the circumscribed circle, the shortest possible length of the segment DP is $\frac{\mathbf{NO}\sqrt{\mathbf{PQ}}}{\mathbf{R}}$.
QCourse2-I-Q1 Solving quadratics and applications Determining quadratic function from given conditions View
Q1 A quadratic function $y = ax^2 + bx + \frac{3}{a}$ satisfies the following two conditions:
(i) $y$ is maximized at $x = 3$,
(ii) the value of $y$ at $x = 1$ is 2.
We are to find the values of $a$ and $b$.
Using conditions (i) and (ii), we obtain the following relationships between $a$ and $b$:
$$\left\{ \begin{aligned} b & = \mathbf{AB}a \\ \mathbf{C} & = a + b + \frac{\mathbf{D}}{a}. \end{aligned} \right.$$
From these two equalities, we have the equation
$$\mathbf{E}a^2 + \mathbf{F}a - \mathbf{G} = 0$$
and hence
$$a = \mathbf{HI}, \quad b = \mathbf{J}.$$
Thus the maximum value of this function is $\mathbf{K}$.
QCourse2-I-Q2 Polynomial Division & Manipulation View
Q2 Consider
$$E = P^2 - 4Q^2 - 3P + 6Q$$
where $P$ and $Q$ are the integral expressions
$$P = 2x^2 - x + 2, \quad Q = x^2 - 2x + 1.$$
(1) By factorizing the right side of $E$, we obtain
$$E = (P - \mathbf{LL}Q)(P + \mathbf{M}Q - \mathbf{MN}).$$
(2) When we express $E$ in terms of $x$, we have
$$E = \mathbf{O}x(x - \mathbf{P})(\mathbf{Q})(\mathbf{Q} - \mathbf{Q}).$$
(3) If $x = -\frac{1 - \sqrt{5}}{3 - \sqrt{5}}$, then the value of $E$ is $\mathbf{S} + \mathbf{T}\sqrt{\mathbf{U}}$.
QCourse2-II Vectors 3D & Lines Vector Algebra and Triple Product Computation View
In a regular tetrahedron OABC, each side of which has the length 1, let L denote the point which divides segment OA internally in the ratio $3:1$, let M denote the midpoint of side BC, and let P denote the point which divides segment LM internally in the ratio $t:(1-t)$, where $0 < t < 1$.
(1) When we set $\overrightarrow{\mathrm{OA}} = \vec{a}$, $\overrightarrow{\mathrm{OB}} = \vec{b}$, $\overrightarrow{\mathrm{OC}} = \vec{c}$ and express $\overrightarrow{\mathrm{OP}}$ in terms of $\vec{a}, \vec{b}, \vec{c}$, we have
$$\overrightarrow{\mathrm{OP}} = \frac{\mathbf{A}}{\mathbf{B}}(\mathbf{C} - t)\vec{a} + \frac{\mathbf{D}}{\mathbf{E}}t(\vec{b} + \vec{c}).$$
Since $\vec{a} \cdot (\vec{b} + \vec{c}) = \mathbf{F}$ and $|\vec{b} + \vec{c}|^2 = \mathbf{G}$, we have
$$|\overrightarrow{\mathrm{OP}}| = \frac{1}{\mathbf{H}}\sqrt{\mathbf{I}t^2 - \mathbf{JJ}t + \mathbf{K}},$$
where $\vec{a} \cdot (\vec{b} + \vec{c})$ means the inner product of the vectors $\vec{a}$ and $(\vec{b} + \vec{c})$.
(2) The value of $t$ at which $|\overrightarrow{\mathrm{OP}}|$ is minimized is
$$t = \frac{\mathbf{L}}{\mathbf{L}}$$
and the minimum value of $|\overrightarrow{\mathrm{OP}}|$ is $\frac{\sqrt{\mathbf{N}}}{\mathbf{O}}$.
(3) When $|\overrightarrow{\mathrm{OP}}|$ is minimized as in (2), we have $\cos\angle\mathrm{AOP} = \frac{\square\mathbf{P}\sqrt{\mathbf{Q}}}{\square\mathbf{R}}$.
QCourse2-III Quadratic trigonometric equations View
Consider the following two equations in $x$
$$\sin 2x + a\cos x = 0 \tag{1}$$ $$\cos 2x + a\sin x = -2 \tag{2}$$
over the interval $-\frac{\pi}{2} < x < \frac{\pi}{2}$, where $a > 0$.
Let $a = \sqrt{2}$. Then the value of $x$ which satisfies (1) is
$$x = \frac{\mathbf{AB}}{\mathbf{A}}$$
However, at this $x$ the value of the left side of (2) is $\mathbf{DE}$, and so equation (2) does not hold. Hence, when $a = \sqrt{2}$, (1) and (2) have no common solution.
Now, let us find a value of $a$ such that (1) and (2) have a common solution, and also the common solution $x$.
First, from (1) we have
$$\sin x = \frac{\mathbf{FG}}{\mathbf{H}}a, \quad \cos 2x = \mathbf{I} - \frac{a^2}{\mathbf{J}}.$$
When we substitute these into (2), we obtain
$$a^2 = \mathbf{K}.$$
Thus $a = \sqrt{\mathbf{K}}$, and the common solution is
$$x = \frac{\mathbf{LM}}{\mathbf{N}}$$
QCourse2-IV-Q1 Applied differentiation Tangent line computation and geometric consequences View
Q1 Let $a > 0$. Consider two curves
$$C_1: y = e^{6x}$$ $$C_2: y = ax^2.$$
We are to find the condition on $a$ such that there exist two straight lines, each of which is tangent to both $C_1$ and $C_2$.
The equation of the tangent to $C_1$ at a point $(t, e^{6t})$ is
$$y = \mathbf{A}e^{6t}x - e^{6t}(\mathbf{B}t - \mathbf{C})$$
This is tangent also to $C_2$ under the condition that the quadratic equation
$$ax^2 = \mathbf{A}e^{6t}x - e^{6t}(\mathbf{B}t - \mathbf{C})$$
has just one solution. Hence, the equation
$$\mathbf{D}e^{12t} - ae^{6t}(\mathbf{E}t - \mathbf{F}) = 0$$
must hold for $a$ and $t$. From this equation we obtain
$$a = \frac{\mathbf{D}}{\mathbf{E}t - \mathbf{F}}e^{6t}$$
Let $f(t)$ denote the right side of this equation. The condition under which there exist two straight lines each of which is tangent to both $C_1$ and $C_2$, is that the straight line $s = a$ intersects the graph of $s = f(t)$ at two points.
Now, the derivative of $f(t)$ is
$$f'(t) = \frac{108e^{6t}(\mathbf{G}t - \mathbf{H})}{(\mathbf{E}t - \mathbf{F})^2}.$$
Hence the condition on $a$ that we are seeking is
$$a > \square e^{\square}.$$
Note that $\lim_{t \to \infty} \frac{e^t}{t} = \infty$.
QCourse2-IV-Q2 Volumes of Revolution Revolution about a Non-Standard Line with Parameter View
Q2 For $\mathbf{K} \sim \mathbf{ZZ}$ in the following statements, choose the appropriate answer from among (0) $\sim$ (9) at the bottom of this page.
Let $a$ and $t$ be positive real numbers. Let $D$ denote the region of a plane bounded by the graph of the quadratic function in $x$
$$y = \frac{1}{t^2}\left(x - at^2\right)^2$$
the $x$-axis, and the $y$-axis. Let $V_1$ denote the volume of the solid obtained by rotating $D$ once about the $x$-axis, and $V_2$ denote the volume of the solid obtained by rotating $D$ once about the $y$-axis. Now, let us show that for a certain value of $a$, $V_1 = V_2$, independent of the value of $t$.
First, the value of $V_1$ is
$$\begin{aligned} V_1 &= \pi \int_{\mathbf{K}}^{\mathbf{L}} \frac{1}{t^{\mathbf{M}}}\left(x - at^2\right)^{\mathbf{N}} dx \\ &= \frac{\pi}{\mathbf{O}} a^{\mathbf{P}} t^{\mathbf{Q}} \end{aligned}$$
Next, the value of $V_2$ is
$$\begin{aligned} V_2 &= \pi \int_{\mathbf{R}}^{\mathbf{S}} (\cdots) \\ &= \frac{\pi}{\mathbf{T}} a^{\mathbf{W}} t^{\mathbf{W}} \end{aligned}$$
Hence, when $a = \frac{\mathbf{Y}}{\mathbf{Y}}$, then $V_1 = V_2$, independent of the value of $t$.
Options: (0) 0
(1) 1
(2) 2
(3) 3
(4) 4
(5) 5 (6) 6 (7) $t$ (8) $at^2$ (9) $a^2t^2$