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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...

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{\...