Solutions to 2023 A Level H3 Mathematics

Question 1

1(a) Prove that, for any real numbers $a_1,a_2,\dots,a_n$,

$a_1 + a_2 + \cdots + a_n \leq \sqrt{n}\sqrt{a_1^2 + a_2^2 + \cdots + a_n^2}$

1(b) Prove that, for any positive real numbers $x, y$ and $z$,

$\sqrt{\frac{x+y}{x+y+z}} + \sqrt{\frac{y+z}{x+y+z}} + \sqrt{\frac{z+x}{x+y+z}} \leq \sqrt{6}$

1(c) Hence solve the equation

$2\bigg(\sqrt{\frac{x+3}{x+6}}\bigg) + \sqrt{\frac{6}{x+6}} = \sqrt{6}$

Proof

For part (a), it suffices to prove that

$(a_1 + a_2 + \cdots + a_n)^2 \leq n(a_1^2 + a_2^2 + \cdots + a_n^2)$

By Cauchy-Schwarz inequality, we see that

$n(a_1^2 + a_2^2 + \cdots + a_n^2) = (1^2 + 1^2 + \cdots + 1^2)(a_1^2 + a_2^2 + \cdots + a_n^2) \geq (a_1 + a_2 + \cdots + a_n)^2$

Hence it completes the proof of part (a). For part (b), it suffices to prove that

$\sqrt{x + y} + \sqrt{y + z} + \sqrt{z + x} \leq \sqrt{6}\sqrt{x + y + z}$

Using part (a) by letting $a_1=\textstyle \color{red}\sqrt{\frac{x+y}{2}}$, $a_2=\color{blue}\sqrt{\frac{y+z}{2}}$, $a_3=\color{lime}\sqrt{\frac{z+x}{2}}$, hence we have

$\sqrt{\frac{x+y}{2}} + \sqrt{\frac{y+z}{2}} + \sqrt{\frac{z+x}{2}} \leq \sqrt{3}\sqrt{\frac{x+y}{2}+\frac{y+z}{2}+\frac{z+x}{2}} = \sqrt{3}\sqrt{x+y+z}$

Hence it completes the proof of part (b). For part (c), we have

$\sqrt{\frac{x+3}{x+6}} + \sqrt{\frac{x+3}{x+6}} + \sqrt{\frac{6}{x+6}} = \sqrt{6}$

Notice that the equality occurs when $b_i=ta_i$ for all index $i$ in Cauchy-Schwarz inequality, where $t$ is a constant. Now let $x=x, y=z=3$ in part (b), we have

$\sqrt{\frac{x+3}{x+6}} + \sqrt{\frac{x+3}{x+6}} + \sqrt{\frac{6}{x+6}} \leq \sqrt{6}$

with equality if and only if $1=ta_1=\textstyle t\color{red}{\sqrt{\frac{x+3}{2}}}$, $1=ta_2=t\color{blue}\sqrt{\frac{6}{2}}$, $1=ta_3=t\color{lime}\sqrt{\frac{3+x}{2}}$, which implies that $t=\textstyle\frac{1}{\sqrt{3}}$, hence we have $x=3$. $\Box$

Pretty standard question I must say.

Question 2

The curve $C$, defined for $x>0$, satisfies the differential equation $\frac{dy}{dx}=\frac{y}{x}-\frac{y^2}{x^3}$ and passes through the point $(a,b)$ where $a>0$ and $b\not=0$.
2(a) By using the substitution $u=\frac{x}{y}$, find the equation of the curve $C$.
2(b) State, with reasons, necessary and sufficient conditions on $a$ and $b$ for the curve $C$ to have two asymptotes. Find the equations of these asymptotes.

Proof

Using the substitution $u=\textstyle\frac{x}{y} \iff uy=x$, which implies that $y\textstyle\frac{du}{dx}+u\frac{dy}{dx}=1$, we have

$\frac{1}{u}\bigg(1 - y\frac{du}{dx}\bigg) = u^{-1} - \frac{u^{-2}}{x} \iff 1 - y\frac{du}{dx} = 1 - \frac{u^{-1}}{x} \iff \frac{du}{dx} = \frac{u}{x}\frac{u^{-1}}{x} = \frac{1}{x^2}$

Hence we have

$\int 1\,du = \int \frac{1}{x^2}\,dx \Longrightarrow u = -x^{-1} + c$

Hence we obtain

$y = \frac{x}{u} = \frac{x}{-x^{-1}+c} = \frac{x^2}{-1+cx}$

Since the curve passes through the point $(a,b)$, hence we have $c=u+x^{-1}=\textstyle\frac{a}{b}+\frac{1}{a}=\frac{a^2+b}{ab}$. Therefore the equation of the curve $C$ is

$y = \frac{x^2}{-1+(\frac{a^2+b}{ab})x} = \frac{abx^2}{-ab+(a^2+b)x}$

For part (b), by long polynomial division, we have

$y = \frac{ab}{a^2+b}x + \bigg(\frac{ab}{a^2+b}\bigg)^2 + \frac{(\frac{ab}{a^2+b})^{2}}{-1+(\frac{a^2+b}{ab})x}$

Hence the two asymptotes of the curve $C$ are

$y = \frac{ab}{a^2+b}x + \bigg(\frac{ab}{a^2+b}\bigg)^2, \quad x = \frac{ab}{a^2+b}$

Since $x>0$ and $a>0$, hence we require $x=\textstyle\frac{ab}{a^2+b}>0\Longrightarrow \frac{b}{a^2+b}>0$, hence we need $b>0$ or $a^2+b<0\iff b<-a^2<0$. The converse is also true. $\Box$

Another easy question to score, it doesn’t even feel like it was H3 Math. Probably the only trick here is the necessary and sufficient condition, which is if and only if, although I had knew that when I saw the phrase, yet I still didn’t prove the other direction.

Question 3

Let $n$ be a positive integer.
3(a) Write down the binomial expansion of $(1+x)^n$.
3(b) Prove that

$3^n = \sum_{j=0}^n \bigg(\binom{n}{j}\sum_{i=0}^j \binom{j}{i}\bigg)$

3(ci) Let $m=p_1^{m_1}p_2^{m_2}\cdots p_k^{m_k}$, where $p_1,p_2,\dots,p_k$ are distinct prime numbers and $m_1,m_2,\dots,m_k$ are positive integers. Determine the number of positive divisors of $m$ that have no square factors other than $1$, justifying your answer.
3(cii) The function $\mu$ is defined on the positive integers as follows.
$\mu(1)=1$ and for $m>1$ with $m=p_1^{m_1}p_2^{m_2}\cdots p_k^{m_k}$, as in part (i),

$\mu(n) = \left\{ \begin{array}{cc} 0 & \text{if}\;m_i>1\;\text{for some}\;i=1,2,\dots,k \\ (-1)^k & \text{otherwise.} \end{array} \right.$

Write down the values of $\mu(2), \mu(3), \mu(4), \mu(6)$ and $\mu(12)$.
3(ciii) Prove that, for $m>1$,

$\sum_{d\mid m} \mu(d) = 0$

where $d\mid m$ denotes that $d$ is a positive divisor of $m$.

Proof

For part (a), use MF26. For part (b), we use part (a) and obtain

$3^n = (1+2)^n = \sum_{j=0}^n \binom{n}{j}2^j = \sum_{j=0}^n \bigg(\binom{n}{j}\sum_{i=0}^j \binom{j}{i}\bigg)$

For part (ci), we have two ways to think of it. For an integer $m=p_1^{m_1}p_2^{m_2}\cdots p_k^{m_k}$, the number of positive divisors of $m$ such that they have no square factors other than $1$, i.e. the number of square-free divisors of $m$, is the total number of ways of choosing the distinct prime numbers of $m$ such that there are no repeats of the prime numbers, hence we have

$\binom{k}{0} + \binom{k}{1} + \cdots + \binom{k}{k} = \sum_{j=0}^k \binom{k}{j} = (1+1)^k = 2^k$

Another way to look at it is that the indices $m_i$ can only be from $\{0,1\}$, hence the number of such divisors is $2^k$ by Multiplicative Principle.

A side note, we can also use the following theorem: If $f$ is multiplicative, we have

$\sum_{d\mid n} \mu(d)f(d) = \prod_{p\mid n} (1-f(p))$

Using the fact that the Mobius function $\mu$ is multiplicative, hence by letting $f(d)=\mu(d)$, we have

$\sum_{d\mid m} \mu^2(d) = \prod_{p\mid m} (1-\mu(p)) = \prod_{p\mid m} 2 = \prod_{i=1}^k 2 = 2^k$

For part (cii), it is obvious that $\mu(2)=\mu(3)=-1$, $\mu(4)=\mu(12)=0$ and $\mu(6)=1$. For part (ciii), notice that $\mu(d)=0$ for all divisors $d$ of $m$ with square factors, hence we only need to care about those that are square-free, i.e. we have

$\begin{align*}\sum_{d\mid m} \mu(d) &= \sum_{\substack{d\mid m\\ d\;\text{is square-free}}} \mu(d) = \mu(1) + \sum_{r=1}^{k} \sum_{1\leq j_1<j_2<\cdots<j_r\leq k} \mu(p_{j_1}p_{j_2}\cdots p_{j_r})\\ &= \mu(1) + \bigg[\mu(p_1) + \cdots + \mu(p_k)\bigg] + \bigg[\mu(p_1p_2) + \cdots + \mu(p_{k-1}p_k)\bigg] + \cdots + \mu(p_1p_2\cdots p_k)\\ &= 1 + \binom{k}{1} (-1) + \binom{k}{2} (-1)^2 + \cdots + \binom{k}{k} (-1)^k = (1-1)^k = 0\quad\Box\end{align*}$

When I saw this question, I am over the moon because I know Mobius function and I have proven such results before. Moreover, part (a), (b) and (cii) are just free marks to score.

Question 4

Let $n$ stones be placed in fixed positions on a line. Each stone is painted using one of four colours (red, white, yellow or blue) in such a way that no two adjacent stones are the same colour. Let $r_n$ be the number of ways of painting the stones such that the first and last stones are both red. Let $s_n$ be the number of ways of painting the stones so that the first stone is red but the last stone is not red.
4(a) Explain why $r_1=1, r_2=0, s_1=0, s_2=3$.
4(b) Find a formula for $r_n+s_n$ and explain why $r_{n+1}=s_n$.
4(c) Using Mathematical induction, or otherwise, prove that for all $n\geq1$,

$r_n = \frac{3^{n-1}+3(-1)^{n-1}}{4}.$

4(d) Now let $n$ stones, where $n>1$, be placed on a circle with numbered positions. Find the number of ways of painting these stones, using at most four distinct colours, in such a way that no two adjacent stones are the same colour.

Proof

For part (a), $r_1=1$ since if there is only $1$ stone, there is only one way to put the (red) stone such that the first and last stones are both red. $r_2=0$ since the two adjacent stones must not have the same colour, hence there is no way. $s_1=0$ since there is no way for the first stone to be red and the last stone to be not red (superposition state lol). $s_2=3$ since the first must be red and the second stone must be from the other $3$ colours so that the two adjacent stones are not the same colour.

For part (b), notice that $r_n=s_{n-1}\iff r_{n+1}=s_n$ since in order for the last stone to be red, we must have the case where the $(n-1)$th stones before the $n$th last (red) stone to be first stone red and last stone not red, i.e. the case $s_{n-1}$, so that there are no two adjacent stones of the same colour.

For there to be first stone red and last stone not red, we have two cases:

Case (A): the case with $r_{n-1}$, with red in the $(n-1)$th stone, hence the $n$th stone can be from the $3$ other colours, hence there is $3r_{n-1}$ ways.
Case (B): the case with $s_{n-1}$, with not red in the $(n-1)$th stone, hence the $n$th stone can be from the $2$ remaining colours so that there are no two adjacent stones of the same colour, hence there is $2s_{n-1}$ ways.

Taking all together, we have $s_n=3r_{n-1}+2s_{n-1}$ by Addition Principle, hence we obtain $v_n=r_n+s_n=s_{n-1}+3r_{n-1}+2s_{n-1}=3(r_{n-1}+s_{n-1})=3v_{n-1}=\cdots=3^{n-1}v_1=3^{n-1}$.

For part (c), notice that $r_n+r_{n+1}=3(r_{n-1}+r_n)\iff r_{n+1}-2r_n-3r_{n-1}=0$ using $s_{n}=r_{n+1}$, hence we have the characteristic equation $m^2-2m-3=0\iff (m+1)(m-3)=0$, which implies that

$r_n = A(-1)^n + B(3)^n, \quad n\geq1$

Since $1=r_1=-A+3B$ and $0=r_2=A+9B$, hence we have $B=\textstyle\frac{1}{12}$ and $A=-9B=-\textstyle\frac{3}{4}$. Therefore we have

$r_n = \frac{3^n}{12} - \frac{3(-1)^n}{4} = \frac{3^{n-1}+3(-1)^{n-1}}{4} \quad\Box$

For part (d), since we require no two adjacent stones with the same colour in a circle, hence we need the case of $s_n$ so that the first stone and last stone are not of the same colour. Since the position is numbered, hence there will be no over-counting, and since we can start with any of the $4$ colours in the first position, hence the number of way is

$4s_n = 4r_{n+1} = 3^n + 3(-1)^n$

Pretty interesting question.

Question 5

5(a) Let $p$ be a prime number greater than $2$. Write down the possible remainders of $p$ when divided by $4$.
Fermat’s Little Theorem states that if $p$ is prime and $a$ is an integer which is not divisible by $p$, then $a^{p-1}\equiv 1\pmod{p}$.
5(b) Use Fermat’s Little Theorem to prove that if $p$ is a prime number greater than $2$, and there exists an integer $z$ such that $z^2\equiv -1\pmod{p}$, then $p$ is not congruent to $3$ (modulo $4$).
5(c) Write down the possible remainders of $w^2$ when divided by $8$ where $w$ is an integer.
In the remainder of this question $x$ and $y$ are integers. You will prove, by contradiction, that the equation $y^2=x^3+7$ has no solution in integers.
5(di) Show that if $y^2=x^3+7$ then $x$ must be odd.
5(dii) Show that if $y^2=x^3+7$ then $y^2+1$ must be divisible by a prime $p$ with $p\equiv 3\pmod{4}$.
5(diii) Hence deduce that $y^2=x^3+7$ has no solution in integers.

Proof

For part (a), consider the prime $p$ of the forms: $4k, 4k+1, 4k+2, 4k+3$. For $p=4k$ and $p=4k+2$, it is an even number, hence not an odd prime. Hence we have $p\equiv 4k+1, 4k+3\equiv 1,3\pmod{4}$, therefore the possible remainders are $\{1,3\}$.

For part (b), consider $z^4\equiv (-1)(-1)\equiv 1\pmod{p}$ and by Fermat’s Little Theorem, we have $z^{p-1}\equiv 1\pmod{p}$ since $p(k)+z(-z)=1$ implies that $\gcd(p,z)=1$. Suppose that $p\equiv 3\pmod{4}$, i.e. $p=4k+3$ for some integer $k$, hence we have

$z^{p-1} \equiv z^{4k+2} \equiv (z^4)^k z^2 \equiv z^2 \equiv 1 \pmod{p}$

which contradicts with $z^2\equiv -1\pmod{p}$, hence we must have $p$ not congruent to $3$ (modulo $4$).

For part (c), consider $w$ of the forms $8k,8k+1,\dots,8k+7$, hence we have $w^2\equiv (8k)^2,\dots,(8k+7)^2\equiv 0,1,4\pmod{8}$.

For part (di), suppose that $x$ is even, hence $x=2k$ for some integer $k$. We have $y^2=(2k)^3+7=8k^2+7\equiv -1\pmod{8}$, which is not any remainders in part (c), hence $x$ must be odd.

Question 6

Let $f(x)=e^{mx}\sin(mx)$ where $m$ is a positive integer.
6(a) Find $\int f(x)\,dx$.
6(b) Find $\int f(x)f(x-\frac{\pi}{2})\,dx$ in terms of $m$, describing carefully all the possible cases that arise.

Proof

Using Integration by parts twice, we have

$\begin{align*}\int f(x)\,dx &= \int e^{mx}\sin(mx)\,dx = \frac{e^{mx}\sin(mx)}{m} - \int e^{mx}\cos(mx)\,dx\\ &= \frac{e^{mx}(\sin(mx)-\cos(mx))}{m} - \int e^{mx}\sin(mx)\,dx \end{align*}$

Hence we obtain

$\int f(x)\,dx = \int e^{mx}\sin(mx)\,dx = \frac{e^{mx}(\sin(mx)-\cos(mx))}{2m} + c$

For part (b), notice that $\sin(mx-\textstyle\frac{m\pi}{2})=(-1)^{\frac{m}{2}}\sin(mx)$ for even $m$ and $\sin(mx-\textstyle\frac{m\pi}{2})=(-1)^{\frac{m+1}{2}}\cos(mx)$ for odd $m$, hence we have

$\begin{align*}\int f(x)f\bigg(x-\frac{\pi}{2}\bigg)\,dx &= \int e^{mx}\sin(mx)e^{mx-\frac{m\pi}{2}}\sin\bigg(mx-\frac{m\pi}{2}\bigg)\,dx\\ &= e^{-\frac{m\pi}{2}}\int e^{2mx}\sin(mx)\sin\bigg(mx-\frac{m\pi}{2}\bigg)\,dx \end{align*}$

For even $m$, we have

$\begin{align*}\int f(x)f\bigg(x-\frac{\pi}{2}\bigg)\,dx &= e^{-\frac{m\pi}{2}}\int e^{2mx}\sin(mx)(-1)^{\frac{m}{2}}\sin(mx)\,dx = e^{-\frac{m\pi}{2}}(-1)^{\frac{m}{2}}\int \sin^2(mx)e^{2mx}\,dx\\ &= e^{-\frac{m\pi}{2}}(-1)^{\frac{m}{2}}\bigg(\sin^2(mx) \frac{e^{2mx}}{2m}\bigg) - e^{-\frac{m\pi}{2}}(-1)^{\frac{m}{2}}\int 2\sin(mx)\cos(mx)m\frac{e^{2mx}}{2m}\,dx\\ &= \frac{e^{-\frac{m\pi}{2}}(-1)^{\frac{m}{2}}e^{2mx}\sin^2(mx)}{2m} - \frac{e^{-\frac{m\pi}{2}}(-1)^{\frac{m}{2}}}{2}\int \sin(2mx)e^{2mx}\,dx\\ &= \frac{e^{-\frac{m\pi}{2}}(-1)^{\frac{m}{2}}e^{2mx}\sin^2(mx)}{2m} - \frac{e^{-\frac{m\pi}{2}}(-1)^{\frac{m}{2}}}{2}\frac{e^{2mx}(\sin(2mx)-\cos(2mx))}{4m} + c\\ &= \frac{e^{-\frac{m\pi}{2}}(-1)^{\frac{m}{2}}e^{2mx}(4\sin^2(mx)-\sin(2mx)+\cos(2mx))}{8m} + c \end{align*}$

using part (a). For odd $m$, we have

$\begin{align*}\int f(x)f\bigg(x-\frac{\pi}{2}\bigg)\,dx &= e^{-\frac{m\pi}{2}}\int e^{2mx}\sin(mx)(-1)^{\frac{m+1}{2}}\cos(mx)\,dx = \frac{e^{-\frac{m\pi}{2}}(-1)^{\frac{m+1}{2}}}{2}\int \sin(2mx)e^{2mx}\,dx\\ &= \frac{e^{-\frac{m\pi}{2}}(-1)^{\frac{m+1}{2}}e^{2mx}(\sin(2mx)-\cos(2mx))}{8m} + c \end{align*}$

Question 7

7(a) In the diagram below there are $5$ vertical lines and $4$ horizontal lines, with the same spacing. How many squares are there in the diagram?
7(b) Now suppose that the diagram has $m$ vertical lines and $n$ horizontal lines with $m\geq n>1$. Show that the number of squares in this diagram is given by

$\frac{1}{6}n(n-1)(3m-n-1).$

7(c) Hence find all pairs $(m,n)$ for which the number of squares in the diagram is $100$.

Proof

For part (a), we have total number of squares to be $3\times4+2\times3+1\times2=20$.

For part (b), we have the total number of squares in a diagram with $m$ vertical lines and $n$ horizontal lines is

$\begin{align*}\sum_{i=1}^{n-1} (n-i)(m-i) &= \sum_{i=1}^{n-1} nm - (n+m) \sum_{i=1}^{n-1} i + \sum_{i=1}^{n-1} i^2\\ &= nm(n-1) - (n+m)\frac{(n-1)(1+n-1)}{2} + \frac{1}{6}(n-1)(n)(2n-1)\\ &= \frac{1}{6}n(n-1)(6m-3(n+m)+2n-1)\\ &= \frac{1}{6}n(n-1)(3m-n-1)\end{align*}$

For part (c), we require

$\frac{1}{6}n(n-1)(3m-n-1) = 100 \iff n(n-1)(3m-n-1) = 600 = 2^3\cdot3\cdot5^2$

Hence we find all the consecutive divisors of $600$, which are $1,2,3,4,5,6,24,25$, we need $m=\textstyle\frac{200}{n(n-1)}+\frac{n+1}{3}$ to be an integer, hence we see that the pairs $(m,n)$ are $(101,2),(12,5),(9,6)$, we reject $(9,25)$ since $m<n$.

Question 8

8(a) Consider the list of $2400$ fractions

$\frac{1}{2400}, \frac{2}{2400}, \frac{3}{2400}, \cdots, \frac{2400}{2400}$

How many of these fractions are expressed in their simplest terms?
A number of the form $\frac{1}{N}$, where $N$ is an integer greater than $1$, is called a unit fraction.
8(bi) Show that if $N, a$ and $b$ are integers such that

$\frac{1}{N} = \frac{1}{a} + \frac{1}{b}$

then

$(a-N)(b-N) = N^2.$

8(bii) Hence prove that if $N$ is prime, then there is exactly one way to express $\frac{1}{N}$ as a sum of two distinct unit fractions.
Let $f(x)=Ax^2-Bx+1$ where $A$ and $B$ are integers.
8(ci) Show that if $r$ is a rational root of $f(x)=0$, then $\lvert r \rvert$ may be expressed as a unit fraction.
8(cii) Find $A$ and $B$ such that $f(x)=0$ has two distinct positive rational roots $r_1, r_2$ such that $r_1+r_2=\frac{7}{13}$.

Proof

We let $\lvert A_i\rvert$ be the number of divisors of $2400$ that are divisible by $i$, since $2400=2^5\cdot 3\cdot 5^2$, we require the set

$\begin{align*}\lvert A_2'\cap A_3' \cap A_5'\rvert &= \lvert S\rvert - \lvert A_2\cup A_3 \cup A_5\rvert\\ &= 2400 - \lvert A_2\rvert - \lvert A_3\rvert - \lvert A_5\rvert + \lvert A_2\cap A_3\rvert + \lvert A_2\cap A_5\rvert + \lvert A_3\cap A_5\rvert - \lvert A_2\cap A_3\cap A_5\rvert\\ &= 2400 - \frac{2400}{2} - \frac{2400}{3} - \frac{2400}{5} + \frac{2400}{6} + \frac{2400}{10} + \frac{2400}{15} - \frac{2400}{30} = 640\end{align*}$

For part (bi), we have

$N = \frac{ab}{a+b} \Longrightarrow (a-N)(b-N) = ab - (a+b)N + N^2 = N^2$

For part (bii), since $N$ is prime, $(a-N)$ and $(b-N)$ are integers. WLOG, assume that $a\leq b$, hence we have

$\left\{ \begin{array}{ll}a-N = 1 \\b-N = N^2 \end{array}\right. \quad \text{or} \quad \left\{\begin{array}{ll}a-N = N \\b-N = N\end{array}\right.$

by Fundamental Theorem of Arithmetic. Hence there is only one way to express $\textstyle\frac{1}{N}$ as a sum of two distinct unit fractions:

$\frac{1}{N} = \frac{1}{a} + \frac{1}{b} = \frac{1}{N+1} + \frac{1}{N^2+N}$

For part (cii), since $r$ may be expressed as a unit fraction using part (ci), hence we have

$r_1 + r_2 = \frac{1}{q_1} + \frac{1}{q_2} = \frac{7}{13} \iff \frac{1}{7q_1} + \frac{1}{7q_2} = \frac{1}{13}$

Since $13$ is prime, hence using (bii) we have $7q_1=13+1=14\iff q_1=2$ and $7q_2=13^2+13=182\iff q_2=26$, hence we obtain

$f(x) = A(x-r_1)(x-r_2) = A\bigg(x-\frac{1}{2}\bigg)\bigg(x-\frac{1}{26}\bigg) = 0 \iff Ax^2 - \frac{7A}{13}x + \frac{A}{52} = 0$

which implies that $A=52$ and $B=\textstyle\frac{7A}{13}=28$. $\Box$