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 $C=3$.■