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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(2)

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

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 ...
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Compute $\sin(\sin(...\sin(x)))$

Problem Find the limit $\sin(\sin(...\sin(x)))$ for any real number $x$. Solution The problem can be precisely formulated as follows: Find the limit $L$ such that $$ L \equiv\lim_{n\rightarrow\infty}{f_n(x)}, $$ where $x\in\mathbb{R}$, $n\in\mathbb{N}$, and $$\begin{align} f_1(x) &= \sin(x)\\ f_n(x) &= \sin(f_{n-1}(x)). \end{align}$$ We observe that, if the limit $L$ exists, $$\begin{align} \lim_{n\rightarrow\infty} f_{n-1}(x) &= L\\ \lim_{n\rightarrow\infty} f_n(x) &= L, \end{align}$$ and thus, $$ L = \sin{L}, $$ which we show to be $L=0$ in the following. First, we show that the limit exists. We accept the Axiom of Continuity of the Real Line, which says that any bounded monotonic sequence converges. Thus, to show that $L$ exists, we need only show that (1) $f_n(x)$ is a bounded sequence of $n$, and that (2) $f_n(x)$ is a monotonic sequence of $n$. $f_n(x)$ is clearly bounded because $f_n(...
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