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2020 MEXT Japanese Government Scholarship Undergraduate Students Natural Sciences Qualifying Examination Mathematics (B): Problem 1(5)

This problem appears at the Qualifying Examinations for Applicants for Japanese Government (MEXT) Scholarships 2020 . There are two mathematics exams: one for biology-related natural sciences (Mathematics A), and another for physics- and engineering-related natural sciences (Mathematics B). This problem is from the 2020 Mathematics (B) questionnaire . The official answer key is here . Problem 1(5) The angle

$\theta\left( 0\lt \theta \lt \frac{\pi}{2}\right)$ between the two lines

$y=(2-\sqrt{3})x$ and

$y=(\sqrt{3}-2)x$ on the

$xy$ -plane is

$\fbox{ A }$ . Solution For convenience, let us name the lines

$l_1$ and

$l_2$ .

$\begin{align} l_1: y = (2-\sqrt{3})x\\ l_2: y = (\sqrt{3}-2)x \end{align}$ Let us also call the angle that

$l_1$ makes with the positive

$x$ -axis

$\theta_1$ , and the angle that

$l_2$ makes with the positive

$x$ -axis

$\theta_2$ . The problem seeks to find the angle

$\theta$ which is the smaller of

$|\theta_1-\theta_2|$ and $\pi-|\t...

Find the limit of the sequence. $a_1=1, a_{n+1} = \frac{3a_n+4}{2a_n+3}$

We take the limits of sequences defined recursively by first showing the existence of the limits then, actually computing them. Existence is demonstrated by showing that the sequence is both (1) monotonic and (2) bounded. Problem Find the limit of the following sequence.

$\begin{align} $a_1&=1\\ a_{n+1} &= \frac{3a_n+4}{2a_n+3}$ \end{align}$ Solution Monotonicity : We use mathematical induction to show that for all

$n\in\mathbb{N}$ ,

$a_{n+1} - a_n \geq 0$ and therefore,

$\{a_n\}$ is monotonic increasing. For the base case at

$n=1$ , it is easy to see that

$a_{n+1}-a_n = \frac{7}{5} - 1 = \frac{2}{5} >0.$ For the inductive case at

$n=k$ , we assume that

$\begin{align} a_{k+1} - a_{k} & = \frac{3a_k+4}{2a_k+3} - \frac{3a_{k-1}+4}{2a_{k-1}+3}\\ & = \frac{a_k - a_{k-1}}{(2a_k+3)(2a_{k-1}+3)} \geq 0. \end{align}$ For $n=k...

2020 MEXT Japanese Government Scholarship Undergraduate Students Natural Sciences Qualifying Examination Mathematics (B): Problem 1(4)

$f(x)$ by

$(x-1)^2$ gives the remainder of

$x+1$ , and that by

$x^2$ gives the remainder

$2x+3$ . Thus, the remainder of the division of

$f(x)$ by

$x^2(x-1)$ is

$\fbox{ A }x^2 + \fbox{ B }x + \fbox{ C } .$ Solution We need to find the remainder when

$f(x)$ is divided by

$x^2(x-1)$ . Because

$x^2(x-1)$ is of order

$n=3$ , the remainder will be of at most the order

$n=2$ , which means that it is of the form

$Ax^2 + Bx + C$ . The problem is now to find the coefficients

$A,B$ and

$C$ such that ...

2020 MEXT Japanese Government Scholarship Undergraduate Students Natural Sciences Qualifying Examination Mathematics (B): Problem 1(3)

$a$ and

$b$ be real numbers with

$b \geq 0$ . When the equation

$x^4 + ax^2 + b = 0$ has exactly two real solutions, the minimum value of

$a + 2b$ is

$\fbox{ A }$ , and the maximum value of

$\lceil a - b \rceil$ is

$\fbox{ B }$ . Here,

$\lceil r \rceil$ denotes the smallest integer that is larger than or equal to the real number

$r$ . Solution The most important hint is the phrase " has exactly two real solutions ". This means that there are no complex solutions, and at least one or both of the real solutions have a multiplicit...

2020 MEXT Japanese Government Scholarship Undergraduate Students Natural Sciences Qualifying Examination Mathematics (B): Problem 1(2)

$f(x) = 1 + \frac{1}{x-1}(x\neq 1)$ . The solution of the equation

$f\left(f\left(x\right)\right) = f\left(x\right)$ is

$x = \fbox{A }, \fbox{B }$ . Solution The right-hand side of the equation is

$f(x)$ , which is already given in the problem to be

$f\left(x\right) = 1 + \frac{1}{x-1}.$ The left-hand side of the equation is $$\begin{align} f\left(f\left(x\right)\right) &= 1 + \frac{1}{f(x) - 1}\\ &= 1 + \frac{1}{\left(1+\frac{1}{x-1}\right) - 1}\\ &= 1 + \frac{1}{\frac{1}{x-1}}\\ &= 1 + {x-1}\\ &= x. \end{align}...

2020 MEXT Japanese Government Scholarship Undergraduate Students Natural Sciences Qualifying Examination Mathematics (B): Problem 1(1)

$\log_2{3}\cdot\log_3{4}\cdot\log_4{5}\cdots\cdots\log_{2019}{2020}$ is

$\fbox { A }$ . Solution The key idea is the following relationship that holds for positive numbers

$p, q,$ and

$b$ .

$\log_p{q} = \frac{\log_b{q}}{\log_b{p}},$ where

$p\neq1$ and

$b\neq1$ . For convenience, let

$L \equiv \log_2{3}\cdot\log_3{4}\cdot\log_4{5}\cdots\cdots\log_{2019}{2020}.$ Selecting

$b=e$ , we obtain $$\begin{align} L &=\frac{\cancel{\log{3}}}{\log{2}}\cdot \frac{\cancel{\...

Prove: $\arctan{\frac{1}{2}}+\arctan{\frac{1}{3}} = \frac{\pi}{4}$ , $\arcsin{\frac{16}{65}}+\arcsin{\frac{5}{13}} = \arcsin{\frac{3}{5}}$

We prove identities involving inverse trigonometric functions. Sum formulas from trigonometry are exploited.

$\arctan{\frac{1}{2}}+\arctan{\frac{1}{3}} = \frac{\pi}{4}$ Suppose that

$\arctan{\frac{1}{2}}+\arctan{\frac{1}{3}} \neq \frac{\pi}{4}.$ Then,

$\begin{align} \tan{\left(\arctan{\frac{1}{2}}+\arctan{\frac{1}{3}}\right)} &\neq \tan{\frac{\pi}{4}}\\ \frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\cdot\frac{1}{3}} &\neq 1 \\ \frac{5}{6-5} &\neq 1 \\ 1 &\neq 1, \end{align}$ which is a contradiction. Thus,

$\arctan{\frac{1}{2}}+\arctan{\frac{1}{3}} = \frac{\pi}{4}.□$

$\arcsin{\frac{16}{65}}+\arcsin{\frac{5}{13}} = \arcsin{\frac{3}{5}}$ Suppose that

$\arcsin{\frac{16}{65}}+\arcsin{\frac{5}{13}} \neq \arcsin{\frac{3}{5}}.$ Then, $$\begin{align} \sin{\left(\arcsin{\frac{16}{65}}+\arcsin{\fra...