Shoulders, Giants, and Cats🐈

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-|\theta_1-\t

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(4) The division of a polynomial function $f(x)$ 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 $$ f(x) = Q(x)\

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(3) Let $a$ 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 multiplicity greater

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(2) Let $f(x) = 1 + \frac{1}{x-1}(x\neq 1)$. 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}

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(1) The largest one among natural numbers that are less than $$ \log_2{3}\cdot\log_3{4}\cdot\log_4{5}\cdots\cdots\log_{2019}{2020} $$ 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{\