Skip to main content

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 $$ f(x) = Q(x)\cdot x^2(x-1) + (Ax^2 + Bx + C) $$ where $Q(x)$ is the quotient and $(Ax^2 + Bx + C)$ is the remainder.

We first write down the given conditions.

  1. Division of $f(x)$ by $(x-1)^2$: $$ f(x) = q_1(x)\cdot(x-1)^2 + (x+1), $$ where $q_1(x)$ is the quotient and $x+1$ is the remainder.

    From this, we know that $$ （★）：f(1) = q_2(x)\cancelto{0}{(1-1)} + (1+1) = 2. $$

  2. Division of $f(x)$ by $x^2$: $$ （★★）:f(x) = q_2(x)\cdot x^2 + (2x+3), $$ where $q_2(x)$ is the quotient and $2x+3$ is the remainder.

    We can also write this using $Q(x)$ and grouping $Ax^2$ with it so that we get $$ 　（★★★）:f(x) = \left(Q(x)\cdot (x-1) + A\right)x^2 + (Bx + C), $$ which we can compare with （★★）to conclude that for all values of $x$, $$ 2x + 3 = Bx + C \Leftrightarrow B = 2, C =3. $$ Furthermore, we can set $x=1$ in （★★★） and use the values of $B = 2$ and $C =3$ to obtain $$\begin{align} f(1) = \left(Q(x)\cdot \cancelto{0}{(1-1)} + A\right) \cdot 1^2 + (2\cdot 1 + 3) = A + 5, \end{align}$$ which we can now compare with （★）to obtain $$ \begin{align} 2 = f(1) &= A + 5\\ A &= -3. \end{align}$$

Therefore, the coefficients are $A=-3, B=2$ and $C=3$.■

Labels:

Comments

Post a Comment

Popular posts from this blog

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(...
Post a Comment
Read more

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...
Post a Comment
Read more

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...
Post a Comment
Read more