Algebra Solutions

Part of the Oxford MAT Livestream.

Revision Questions

We can rearrange the first equation for $x$ and substitute into the second equation; $x=1-4y$ so $2(1-4y)-y=3$ which is a linear equation for $y$ with solution $y=-1/9$. Then from the equation $x=1-4y$ we have $x=13/9$.
We can rearrange the second equation for $y=2-x$ and substitute into the first equation to get $x^2+2x+x(2-x)+(2-x)^2=5$ which rearranges to $x^2-1=0$ so $x=1$ or $x=-1$. We can then use $y=2-x$ to find the corresponding values of $y$. The solution is that $(x,y)$ is $(1,1)$ or $(-1,3)$.
We can rearrange the first equation for $y=2-x^2$ and substitute into the second equation to get $x^4-4x^2+x+2=0$. The first two terms are the difference of two squares, so this is $x^2(x+2)(x-2)+(x+2)=0$. So $x=2$ is a solution, or $x^3-2x^2+1=0$. This has a root at $x=1$ and two other roots when $x^2-x-1=0$ which we can find with the quadratic formula.
Substituting these back into the equation $y=2-x^2$ we have four solutions for $(x,y)$; \[(-2,-2)\quad\text{or}\quad (1,1)\quad\text{or}\quad \left(\frac{1+\sqrt{5}}{2},\frac{1-\sqrt{5}}{2}\right)\quad\text{or}\quad \left(\frac{1-\sqrt{5}}{2},\frac{1+\sqrt{5}}{2}\right).\]
We can use the binomial theorem here to get $(2x)^3+3(2x)^2(3y)+3(2x)(3y)^2+(3y)^3$ which is $8x^3+36x^2y+54xy^2+27y^3$.
The $x^2$ term is $\displaystyle \binom{4}{2}(3x)^2(-1)^2$, so the coefficient is 54.
$(x+2)^3=x^3+6x^2+12x+8$, and so the sum of the coefficients is $1+6+12+8=27$. Hopefully you spotted that this is $3^3$, and that the sum of the coefficients of any polynomial is just the value of that polynomial at $x=1$. Then the sum of the coefficients of $(x+2)^{300}$ is $3^{300}$.
Thinking about the shape of $y=x^2+4x+3$, we should look for any points where $y=0$ because if there are two roots then the function will be negative in between those roots. We have $x^2+4x+3=0$ when $x=-1$ or $x=-3$, so $x^2+4x+3>0$ if $x>-1$ or if $x<-3$.
$a^2$ could be as small as zero, because $a$ could be zero. It definitely can't be negative, so that's a lower bound on $a^2$. On the other hand, $a^2$ could be almost as large as $4$, but not equal to 4 or any larger. So the most that we can say is that $0\leq a^2<4$.
In the first case, there's nothing we can say about the relationship between $ac$ and $bd$; those could be any two numbers.
In the second case, we can say something! We have $c<d$ and then because $a>0$ we can multiply each side by $a$ to get $ac<ad$. Separately, we can start with $a<b$ and multiply by $d$ to get $ad<bd$ since $d>0$ (because $d>c$ and $c>0$). Combining these two results, we have $ac<bd$.
There are $2^5=32$ possible sequences of heads/tails that I could get when I flip five coins. Exactly three of the coins are heads (and two are tails) in some of these sequences; if I think about choosing different combinations of coins to be the ones that come up heads then I can see that the number of sequences with exactly three heads is precisely $\binom{5}{3}=10$. So the probability is $\frac{10}{32}=\frac{5}{16}$.
We could consider what happens as we deal out the cards one at a time. It doesn't matter if the first card is even or odd, but then the probability the second card is the opposite parity is $\frac{3}{5}$. If that continues the pattern, then we've got one odd number and one even number dealt out and four cards to go. The probability that the next one continues the pattern is $\frac{2}{4}$, then if we continue to think about the probability that each card fits the pattern we have the probabilities $\frac{2}{3}$ then $\frac{1}{2}$ then $\frac{1}{1}$. Multiplying all these probabilities together, because we want all these events to happen one after the other, we get a probability of $\frac{1}{10}$.
Alternatively, consider trying to count how many of the $6!=720$ possible shuffles have the alternating property. There are some that start with an even number and some that start with an odd number. Consider the ones that start with an even number; we must have the $2$ and $4$ and $6$ cards as the first and third and fifth cards dealt, and there are $3!=6$ ways to do this. Similarly, there are $6$ permutations for the odd cards, so $6\times 6 =36$ shuffles match this pattern. Separately, there are $36$ shuffles that start with an odd number and then alternate, for a total of $72$ out of a possible $720$. That's a probability of $\frac{72}{720}=\frac{1}{10}$.
There are $\binom{10}{3}$ ways to choose the finalists. If contestant 1 is chosen, there are $\binom{8}{2}$ ways to choose two more finalists who are not contestant 2. The probability is
$$\frac{\binom{8}{2}}{\binom{10}{3}}=\frac{7}{30}$$
There are $\binom{20}{4}$ ways for the teacher to select four students. How many of those possibilities have exactly three girls and one boy? There are $\binom{10}{3}$ ways to choose the girls for such a group, and in each of those cases there are $\binom{10}{1}$ ways to choose the boy for the group. The probability is therefore
$$\frac{\binom{10}{3}\binom{10}{1}}{\binom{20}{4}}=\frac{80}{323}$$
The probability of getting all ten questions right is $\left(\frac{1}{5}\right)^{10}$. The probability of getting exactly 9 right is $\binom{10}{9}\left(\frac{1}{5}\right)^9\left(\frac{4}{5}\right)$. The probability of getting exactly 8 right is $\binom{10}{8}\left(\frac{1}{5}\right)^8\left(\frac{4}{5}\right)^2$. Add those together!

MAT Questions

MAT 2014 Q1A

Let $u=x^2$. Note that $u\geq 0$. We'd like to know whether $u^2<8u+9$.
Rearranging and factorising the quadratic, this is equivalent to the question of whether $(u-9)(u+1)<0$. That happens for $-1<u<9$.
But remember that $u\geq 0$. So the range we're interested in is $0\leq u <9$.
That corresponds to $0\leq x^2 < 9$, which happens for $-3<x<3$.
The answer is (a).

MAT 2014 Q1G

First consider the expansion of $(1+(xy+y^2))^n$.
The term that we're interested in is $(xy+y^2)^4$, because that includes something proportional to $x^3y^5$.
The coefficient of $(xy+y^2)^4$ in the expansion is $\displaystyle \binom{n}{4}$.
Now we should think about expanding $(xy+y^2)^4$. That expansion has a term $\displaystyle \binom{4}{3}(xy)^3(y^2)^1$.
Putting this together, we expand the original expression to get $\displaystyle \binom{n}{4}$ times an expression which itself has a term $4 x^4y^5$. So the coefficient is $\displaystyle 4\binom{n}{4}$.
The answer is (d).

MAT 2015 Q2

(i) I decided to multiply out the brackets by first multiplying each term in the second bracket by $a$ and then by $(-b)$. This gives me \begin{align*}a^{n+1}+ & a^nb+a^{n-1}b^2+\dots + a^2b^{n-1}+ab^n\\ - & \left(a^n b + a^{n-1}b^2 + a^{n-2}b^3+\dots + ab^n+b^{n+1}\right).\end{align*}

I've tried to line up the terms there to make it clear what's going to happen next; almost all the terms cancel, to leave just $a^{n+1}-b^{n+1}$.

This is a powerful and general result, which we can use later in the question. We might choose particular values for $a$ and $b$ and $n$ in what comes next.

(ii) First, let's check the fact about 3. $2\times 2 -1=3$ so it's true. That gives us an idea for the general expression for "one less than a square number"; we're looking for prime numbers of the form $x^2-1$, where $x$ is a whole number.

That expression factorises as $(x-1)(x+1)$. You might recognise that as the difference of two squares, or you might have applied the result from the previous part of the question with $a=x$ and $b=1$ and $n=2$.

Is it possible that a prime number factorises as $(x-1)(x+1)$? If that happens, then those factors need to be trivial, and one factor or the other is equal to 1. $x-1=1$ gives $x=2$ which corresponds to the prime 3 we already had. The other possibility $x+1=1$ gives $x=0$, but $0^2-1$ isn't prime. So there are no other prime numbers that are one less than a square number.

(iii) This is similar to the previous part. A number is one more than a cube number if we can write it as $x^3+1$ for some whole number $x$. But $x^3+1$ factorises. You might notice that you can use the first part of the question with $a=x$ and $b=-1$ and $n=3$ to write $x^3+1=(x+1)(x^2-x+1)$. Or you might spot that $-1$ is a root of the equation $x^3+1=0$, so $(x+1)$ is a factor of the polynomial $x^3+1$.

Either way, we're must now consider whether $(x+1)(x^2-x+1)$ could be prime. If $x+1=1$ then we have $x=0$. If $x^2-x+1=1$ then $x^2-x=0$, so $x=0$ or $x=1$. The number $0^3+1$ is not a prime, but the number $1^3+1=2$ is a prime.

So 2 is the only prime number that's one more than a cube number.

(iv) No. The expression is close to $k^3$ but slightly larger. To be precise, it's between $k^3$ and $(k+1)^3$, because $2k^2+2k+1>0$ and $2k^2+2k+1<3k^2+3k+1$. That last inequality is true because $k^2+k>0$.

Extension

No, because that's $\left(3^p\right)^q-\left(2^p\right)^q$, which is a multiple of $3^p-2^p$ by the same logic as part (iv) of the question above, using the expression you worked out for part (i).
Yes, $k=2$ gives the value $27$, which is $3^3$.

MAT 2011 Q3

(i) The gradient of the line is $m$. The gradient of the cubic is $3x^2-1$, which is $3b^2-1$ at $x=b$. The line and the curve tough at $x=b$, so they have the same gradient at that point.

(ii) I haven't used the fact that the line goes through the point $B$ yet. The value of the cubic is $b^3-b$, and the value of the line is $m(b-a)$. This rearranges to
$$
a=\frac{mb-b^3+b}{m}.
$$
Next I'll substitute to get rid of the $m$, using $m=3b^2-1$, and then the numerator simplifies a bit to give the expression in the question.

(iii) Note that the cubic on the left-hand side has zero $x^2$ coefficient. If we multiply out the right-hand side, we get $(x^2-2bx+b^2)(x-c)=x^3-2bx^2-cx^2+b^2 x +2bcx-b^2c$. Comparing the coefficient of $x^2$, we conclude that $-c-2b=0$ so $c=-2b$.

(iv) The expression we just found is really large if $b$ is very negative, or if $b$ is close to $\pm 1/\sqrt{3}$. If $b$ is very negative, then the approximate value of $a$ is $2b/3$. But we know that $a<b$. We're also told that $b<0$ so $b$ is close to $-1/\sqrt{3}$.

(v) The area is
$$
\int_b^c m(x-a)-x^3+x\,\mathrm{d}x
$$
where I've written it as the area underneath the line minus the area under the cubic (or you can think of this as an expression for the difference in values between the curves). I'll use the fact from part (iv), the expansion we worked out in part (iv), and the fact that $c=-2b$ to write this as
\begin{equation*}
- \int_b^{-2b} x^3-3b^2 x +2b^3 \,\mathrm{d}x=-\left[\frac{x^4}{4}-3b^2\frac{x^2}{2}+2b^3 x\right]_b^{-2b}=\frac{27}{4}b^4
\end{equation*}
If we remember that $b$ is negative, we can make this expression as large as possible by making $b$ as negative as possible. The corresponding value of $a$ (using the equation in part (ii)) is $-1$. The corresponding value of $R$ is $27/4$.

Extension

The cubic on the left is the difference between the cubic and the line, so it has roots when those curves meet. The root at $b$ is a repeated root because the line and the cubic touch there. There are no other roots, and the leading coefficient is 1.
This is the same cubic, but transformed by a stretch factor $k^{1/2}$ parallel to the $x$-axis and by a stretch factor of $k^{3/2}$ parallel to the $y$-axis. The maximum area for $R$ becomes $27k^2/4$.