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2020 MEXT Japanese Government Scholarship Qualifying Examinations

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

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{\...
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Solve: $\arcsin{\frac{3}{5}}=\arctan{x}$, $\arcsin{\frac{3}{5}}=2\arctan{x}$

Equations involving inverse trigonometric functions are solved for the unknown. The following relation is assumed proven: $$ \theta = \arcsin{\frac{a}{c}} = \arccos{\frac{\sqrt{c^2-a^2}}{c}}, $$ which is clear from the figure below. $\arcsin{\frac{3}{5}}=\arctan{x}$ $$\begin{align} \arctan{x} &= \arcsin{\frac{3}{5}} = \arctan{\frac{3}{\sqrt{5^2-3^2}}} = \arctan{\frac{3}{4}}\\ &\Rightarrow x = \frac{3}{4}.□ \end{align}$$ $\arcsin{\frac{3}{5}}=2\arctan{x}$ $$\begin{align} 2\arctan{x} &= \arcsin{\frac{3}{5}} = \arctan{\frac{3}{\sqrt{5^2-3^2}}} = \arctan{\frac{3}{4}}\\ \frac{3}{4} &= \tan{\left(2\arctan{x}\right)} = \frac{2\tan{\left(\arctan{x}\right)}}{1-\tan^2{\left(\arctan{x}\right)}} \\ &= \frac{2x}{1-x^2}\\ &\Rightarrow 0 = 3x^2 +8x -3 = (3x-1)(3x+3) \\ &\Rightarrow x = \frac{1}{3} ( x = -1 \hbox{ is extraneous}).□ \end{align}$$ References
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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...
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