University of Tokyo science mathematics 2015: problems and solutions

These are this site's independently written solutions to the 2015 science-track mathematics paper, not official University of Tokyo answers. Difficulty comments and score-planning suggestions are the original author's retrospective judgments, not admissions thresholds. For an approach to reviewing past papers, see the textbook review.

Problem 1: the region swept by a family of parabolas

For every positive real $a$, consider

$$C_a:\quad y=ax^2+\frac{1-4a^2}{4a}.$$

Describe the region swept out as $a$ ranges over all positive real numbers.

Strategy and difficulty

This is a standard region problem. You can either fix $x$ and find the range of $y$, or ask whether a proposed point gives an admissible positive solution for $a$. The first approach keeps the cases particularly transparent; see swept regions.

Solution

Fix $x$ and write

$$y=f_x(a)=(x^2-1)a+\frac1{4a},\qquad a>0.$$

Its derivative is $f_x'(a)=x^2-1-1/(4a^2)$.

If $|x|<1$, the function is strictly decreasing, tends to $+\infty$ as $a\to0+$, and tends to $-\infty$ as $a\to\infty$. Continuity therefore gives every real $y$.

If $|x|=1$, it is $1/(4a)$, whose range is $(0,\infty)$.

If $|x|>1$, its derivative changes from negative to positive at

$$a=\frac1{2\sqrt{x^2-1}},$$

where the attained minimum is $\sqrt{x^2-1}$. Both endpoint limits are $+\infty$, so its range is $[\sqrt{x^2-1},\infty)$.

Consequently, the swept region is

$$\boxed{\begin{cases} y\in\mathbb R&|x|<1,\\ y>0&|x|=1,\\ y\ge\sqrt{x^2-1}&|x|>1. \end{cases}}$$

The curved boundary is included, while $(\pm1,0)$ and the nonpositive parts of $x=\pm1$ are excluded. The source's strict inequality in the third case is a typo: the minimum is attained.

Problem 2: random strings of unequal-length blocks

Roll a fair die repeatedly, appending a block at each roll: outcomes 1, 2, or 3 append AA, outcome 4 appends B, outcome 5 appends C, and outcome 6 appends D. For example, rolls $2,5,6,3,4$ produce AACDAAB.

1. After $n\ge1$ rolls, find the probability that the $n$th character from the left is A.
2. For $n\ge2$, find the probability that the $(n-1)$st character is A and the $n$th is B.

Strategy and difficulty

This is a standard probability-recurrence problem with an important modeling choice: condition on the first block, not just the last roll. After $n$ rolls there are at least $n$ characters, so later appends cannot alter the character positions in question.

Part 1

Let the requested probability be $p_n$. For a target position $n+2$, the first block has length one with probability $1/2$, leaving target position $n+1$ in the remaining string, or length two with probability $1/2$, leaving position $n$. Thus

$$p_{n+2}=\frac12p_{n+1}+\frac12p_n.$$

The initial values are $p_1=1/2$ and $p_2=3/4$. For the latter, either the first block is AA, or it is a one-character block followed by AA.

The recurrence yields

$$p_{n+2}-p_{n+1}=-\frac12(p_{n+1}-p_n),$$ $$p_{n+2}+\frac12p_{n+1}=p_{n+1}+\frac12p_n.$$

Hence

$$p_{n+1}-p_n=\left(-\frac12\right)^{n+1},\qquad p_{n+1}+\frac12p_n=1.$$

Eliminating $p_{n+1}$ gives

$$\boxed{p_n=\frac23\left\{1-\left(-\frac12\right)^{n+1}\right\}.}$$

Part 2

Let $q_n$ be the probability of AB at positions $n-1,n$. The same first-block split gives $q_{n+2}=(q_{n+1}+q_n)/2$ for $n\ge2$.

We have $q_2=0$: a string cannot begin AB because A arrives only in a double block. Also $q_3=1/12$, obtained by first appending AA and then B. Therefore

$$q_{n+1}-q_n=\frac1{12}\left(-\frac12\right)^{n-2}, \qquad q_{n+1}+\frac12q_n=\frac1{12}.$$

Solving,

$$\boxed{q_n=\frac1{18}\left\{1-\left(-\frac12\right)^{n-2}\right\}\quad(n\ge2).}$$

The recurrence guide explains these two geometric combinations.

Problem 3: a power curve and a logarithm meet once

Let $a>0$ and let $p$ be a positive rational number. Suppose $y=ax^p$ and $y=\log x$, for $x>0$, have exactly one common point $Q$.

1. Express $a$ and the $x$-coordinate of $Q$ in terms of $p$.
2. Find the volume formed by rotating the region bounded by these two curves and the $x$-axis about the $x$-axis.
3. Find $p$ when that volume is $2\pi$.

You may use $x^p/\log x\to\infty$ as $x\to\infty$ without proof.

Part 1: separate the parameter

Rather than trying to solve $ax^p=\log x$ explicitly, write

$$a=F(x)=\frac{\log x}{x^p},\qquad F'(x)=\frac{1-p\log x}{x^{p+1}}.$$

The function increases up to $x=e^{1/p}$ and decreases afterward. It tends to $-\infty$ at zero, is zero at one, has maximum $1/(ep)$, and tends to zero from above at infinity.

For positive $a$, a unique intersection occurs precisely at the maximum. Therefore

$$\boxed{a=\frac1{ep},\qquad Q=\left(e^{1/p},\frac1p\right).}$$

The positive condition on $a$ matters: nonpositive horizontal levels have different intersection behavior.

Part 2: disks followed by washers

On $0\le x\le1$ the lower boundary is the axis, while on $1\le x\le e^{1/p}$ it is $\log x$. The power curve provides the upper boundary throughout.

Put $E=e^{1/p}$. The volume is

$$V=\pi a^2\int_0^Ex^{2p}\,dx-\pi\int_1^E(\log x)^2\,dx.$$

The first term is $\pi E/(p^2(2p+1))$. Integration by parts twice gives

$$\int(\log x)^2\,dx=x\{(\log x)^2-2\log x+2\}+C.$$

Thus

$$\begin{aligned} \frac V\pi &=\frac{E}{p^2(2p+1)}-E\left(\frac1{p^2}-\frac2p+2\right)+2\\ &=\frac{2(1-2p)}{1+2p}e^{1/p}+2. \end{aligned}$$

Therefore

$$\boxed{V=\pi\left\{\frac{2(1-2p)}{1+2p}e^{1/p}+2\right\}.}$$

Part 3

Since the exponential and denominator are positive, $V=2\pi$ exactly when $1-2p=0$. Hence

$$\boxed{p=\frac12.}$$

The initial parameter separation is the main obstacle; after it, the volume calculation is a standard application of integration by parts.

Problem 4: a nonlinear recurrence hides Fibonacci numbers

Define

$$p_1=1,\qquad p_2=2,\qquad p_{n+2}=\frac{p_{n+1}^2+1}{p_n}\quad(n\ge1).$$

1. Show that $(p_{n+1}^2+p_n^2+1)/(p_{n+1}p_n)$ is independent of $n$.
2. Express $p_{n+1}+p_{n-1}$ using $p_n$, for $n\ge2$.
3. If $q_1=q_2=1$ and $q_{n+2}=q_{n+1}+q_n$, prove $p_n=q_{2n-1}$.

Part 1: find the invariant

Positivity follows inductively, so all divisions are valid. Substituting the recurrence and cancelling gives

$$\frac{p_{n+2}^2+p_{n+1}^2+1}{p_{n+2}p_{n+1}} =\frac{p_{n+1}^2+p_n^2+1}{p_{n+1}p_n}.$$

The common value is its value at $n=1$, namely $(4+1+1)/2=3$.

Part 2: turn it into a linear recurrence

The invariant says $p_{n+1}^2+p_n^2+1=3p_{n+1}p_n$. Replace $p_{n+1}^2+1$ by $p_{n+2}p_n$ and divide by $p_n>0$:

$$p_{n+2}+p_n=3p_{n+1}.$$

Shifting the index,

$$\boxed{p_{n+1}+p_{n-1}=3p_n\quad(n\ge2).}$$

This avoids a potentially zero factor $p_{n+1}-p_{n-1}$ that would arise from subtracting squared identities.

Part 3: compare recurrences and initial values

The odd-index Fibonacci subsequence satisfies

$$\begin{aligned} q_{2k+3}&=2q_{2k+1}+q_{2k}\\ &=2q_{2k+1}+(q_{2k+1}-q_{2k-1})\\ &=3q_{2k+1}-q_{2k-1}. \end{aligned}$$

This is the same second-order recurrence as for $p_n$. The first two values match: $p_1=q_1=1$ and $p_2=q_3=2$. Two-step induction therefore proves

$$\boxed{p_n=q_{2n-1}\quad(n\ge1).}$$

The induction guide explains why two initial cases are needed.

Problem 5: the first even binomial coefficient

Find the smallest positive integer $m\le2015$ for which $\binom{2015}{m}$ is even.

Solution

Track powers of two instead of computing huge binomial coefficients. Let $v_2(k)$ denote the exponent of two dividing a positive integer $k$. Consecutive coefficients satisfy

$$\binom{2015}{m}=\binom{2015}{m-1}\frac{2016-m}{m}.$$

For $1\le m<32$, write $m=2^su$ with $u$ odd and $s\le4$. Since $2016=2^5\cdot63$,

$$2016-m=2^s(2^{5-s}\cdot63-u),$$

and the bracket is odd. Therefore $v_2(2016-m)=v_2(m)$, so each step preserves the coefficient's two-adic valuation. Starting from $\binom{2015}{0}=1$, every coefficient through $m=31$ is odd.

At $m=32$,

$$2016-32=1984=2^6\cdot31,\qquad v_2(32)=5.$$

The valuation increases by one for the first time, giving

$$\boxed{m=32.}$$

The original solution checks the sixteen even values through 32 individually; the argument above groups those checks by their exact power of two. Either approach needs to establish that every earlier coefficient is odd. Looking only at whether a ratio is an integer would be insufficient, since the intermediate ratios can be nonintegral even when their two-adic valuation is zero.

Problem 6: concentrating integrals

For a positive integer $n$, define

$$g(x)=\begin{cases}(1+\cos\pi x)/2&|x|\le1,\\0&|x|>1,\end{cases}$$ $$h(x)=\begin{cases}-(\pi/2)\sin\pi x&|x|\le1,\\0&|x|>1.\end{cases}$$

1. Let $f$ be continuous. If $p\le f(x)\le q$ for $|x|\le1/n$, prove

$$p\le n\int_{-1}^{1}g(nx)f(x)\,dx\le q.$$

2. Find $\lim_{n\to\infty}n^2\int_{-1}^{1}h(nx)\log(1+e^{x+1})\,dx$.

Part 1: normalize the averaging kernel

Substitute $t=nx$. Since $g(t)=0$ outside $[-1,1]$,

$$n\int_{-1}^{1}g(nx)f(x)\,dx =\int_{-1}^{1}g(t)f(t/n)\,dt.$$

On this interval $g(t)\ge0$, and

$$\int_{-1}^{1}g(t)\,dt =\left[\frac t2+\frac{\sin\pi t}{2\pi}\right]_{-1}^{1}=1.$$

Multiplying $p\le f(t/n)\le q$ by the nonnegative weight and integrating proves the bounds. Nonnegativity and total weight one are both essential.

Part 2: use the derivative relationship

We have $g'=h$, including at the endpoints, and $(g(nx))'=nh(nx)$. Integration by parts gives

$$\begin{aligned} n^2\int_{-1}^{1}h(nx)\log(1+e^{x+1})\,dx &=n[g(nx)\log(1+e^{x+1})]_{-1}^{1}\\ &\quad-n\int_{-1}^{1}g(nx)\frac{e^{x+1}}{1+e^{x+1}}\,dx\\ &=-n\int_{-1}^{1}g(nx)F(x)\,dx, \end{aligned}$$

where $F(x)=e^{x+1}/(1+e^{x+1})$. The boundary term vanishes because $g(\pm n)=0$ for every integer $n\ge1$.

Since $F'(x)=e^{x+1}/(1+e^{x+1})^2>0$, Part 1 gives

$$-F(1/n)\le-n\int_{-1}^{1}g(nx)F(x)\,dx\le-F(-1/n).$$

Both bounds tend to $-F(0)$. The squeeze theorem therefore gives

$$\boxed{-\frac e{1+e}.}$$

Reviewing this paper

The author's assessment was that Problem 5's unusual parity calculation could consume disproportionate time, while the remaining problems mostly reward familiar region, recurrence, and calculus methods. The original suggested practice targets of roughly 80–90 points for the strongest medical-track candidates and 60–70 for other science-track candidates; these are historical strategy opinions, not official requirements or predictions of admission.

A useful review records the modeling choice in each problem: fixing a coordinate, conditioning on the first block, separating a parameter, finding an invariant, tracking a prime valuation, or recognizing an averaging kernel. These decisions are more transferable than memorizing the final numerical answers.