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Mathematics Extension 1

2020 HSC Maths Ext 1 Exam Paper Solutions

In this post, we reveal the solutions to the 2020 HSC Maths Extension 1 Exam paper.

Our Mathematics team have been hard at work putting together the 2020 HSC Maths Extension 1 Exam Paper solutions for you. Here they are!

 

2020 HSC Maths Ext 1 Exam Paper Solutions

Have a look through these worked solutions and see how you did!

Section 1: Multiple Choice

QuestionAnswerSolution
1 A  The quadratic factorises to \(x^2 -2x – 3 = (x-3)(x+1)\), so the solution is \(x>3 \cup x<-1\).
2C The domain and range of \(f(x) = 1+\sqrt{x}\) is \(D:x\geq 0, R: y\geq 1\), so conversely the domain and range of \(f^{-1}(x)\) is \(C:x\geq 1, R: y\geq 0\)
3D The standard antiderivative involved is \(\frac{1}{x^2+a^2} = \frac{1}{a}\tan^{-1}(\frac{x}{a})+c\), and since \(\frac{1}{4x^2+1} = \frac{1}{4} \frac{1}{x^2 + \frac{1}{4}}\), we have the antiderivative \(\frac{1}{4} \frac{1}{1/2}\tan^{-1}(\frac{x}{1/2})= \frac{1}{2} \tan^{-1}(2x)+c\).
4B First, we sum up all the vectors to get Maria’s final position, which is \(2i + 3j + 3i – 2j + 4i – 3j = 9i-2j\). The magnitude of this is \(\sqrt{9^2 + 2^2} = \sqrt{85}\).
5CSince the polynomial is monic, its leading coefficient must be +1, making it overall concave up. This eliminates A and B. Next, \(x^2 + x + 1\) has no solutions, hence the polynomial can only touch the x-axis at its double root (it cannot cross it). This leaves C as the correct answer.
6D For \(a-b\), we put \(a\) and \(b\) tip-to-tip and join the tails.
7AChecking the quadrants, we expect the derivative to be positive in the 2nd and 4th quadrants and negative in the 1st and 3rd quadrants. This eliminates D (positive all quadrants) and B (some negative in 2nd quadrant). Next, we note that as we approach the \(x\)-axis, \(dy/dx\) should become larger as the denominator tends to zero; rather than flattening out. Hence the solution is A.
8CFirst we choose 6 from 10, giving us \( ^{10} C_6 =\frac{10!}{4!6!}\). Next, we choose 4 from the 6, this time caring about order, giving \( ^6 P_4=\frac{6!}{2!}\). In total this is \(\frac{10!}{6!4!} \frac{6!}{2!}=\frac{10!}{4!2!}\).
9BWe can draw a diagram to show the situation:

hsc 2020 maths ext1 question 9 diagram

Hence, the answer should be \({4 \choose 8} + {4 \choose 8} – \left({6 \choose 7}-{4 \choose8}\right)={2 \choose 9}\).

10AFirst, we choose 6 from 10, giving us \( ^{10} C_6 =\frac{10!}{4!6!}\). Next, we choose 4 from the 6, this time caring about order, giving \( ^6 P_4=\frac{6!}{2!}\). In total this is \(\frac{10!}{6!4!} \frac{6!}{2!}=\frac{10!}{4!2!}\).

 

Section 2: Long response

Question 11a

(i)

\(2^3 + 3(2^2) – 13(2) + 6 = 8 + 12 – 26 + 6 = 0\).

 

(ii)

Since the quadratic polynomial is monic, the linear factor must be \(A(x) = x-2\). Therefore

\begin{equation*}
P(x) = (x-2)(x^2 + bx + c)
\end{equation*}

By comparing constants, we get \(-2c = 6\), so \(c = -3\). Comparing the coefficients of \(x^2\), we get \(b – 2 = 3\), so \(b = 5\). Therefore

\begin{equation*}
P(x) = (x-2)(x^2 + 5x – 3)
\end{equation*}

Question 11b

Two vectors are perpendicular if and only if their dot product is 0. Hence we require

\begin{equation*}
{ a \choose -1 } \cdot  { 2a \choose 3 }{ 2 } = 0
\end{equation*}

Therefore

\begin{equation*}
a(2a – 3) – 2 = 0 \Rightarrow 2a^2 – 3a – 2 = 0 \Rightarrow (a – 2)(2a + 1) = 0 \\
\end{equation*}

We obtain the two solutions, \(a = 2, -\frac{1}{2}\).

 

Question 11c

In the following graph, the original \(f(x)\) is shown in blue and \(\frac{1}{f(x)}\) is in red.

hsc 2020 maths ext1 question 11c diagram

 

Question 11d

Observe that

\begin{equation*}
A \sin(x + \alpha) = A(\sin x \cos \alpha + \cos x \sin \alpha)
\end{equation*}

Equating this with the expression \(\sqrt{3} \sin x + 3 \cos x\) gives

\begin{equation*}
\begin{cases}
A \cos \alpha = \sqrt{3} \quad &(1) \\ A \sin \alpha = 3 \quad &(2)
\end{cases}
\end{equation*}

By considering \((2)\) divided by \((1)\), we deduce \(\tan \alpha = \sqrt{3}\), therefore \(\alpha = \frac{\pi}{3}\). Squaring both equations and summing up, we obtain

\begin{equation*}
A^2(\cos^2\alpha + \sin^2\alpha) = 3 + 9 = 12
\end{equation*}

Therefore \(A = \sqrt{12} = 2\sqrt{3}\), and we get

\begin{equation*}
\sqrt{3} \sin x + 3 \cos x = 2\sqrt{3} \sin \left( x + \frac{\pi}{3} \right)
\end{equation*}

The equation to be solved can now be rewritten as

\begin{align*}
2\sqrt{3} \sin \left( x + \frac{\pi}{3} \right) &= \sqrt{3} \\
\sin \left( x + \frac{\pi}{3} \right) &= \frac{1}{2} \\
x + \frac{\pi}{3} &= \frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{6}, \ldots \\
\Rightarrow x &= -\frac{\pi}{6}, \frac{3\pi}{6} = \frac{\pi}{2}, \frac{11\pi}{6}
\end{align*}

The first solution is invalid (as \( \frac{-\pi}{6}\) is outside the domain of interest), so we have the solutions

\begin{equation*}
x = \frac{\pi}{2}, \frac{11\pi}{6}
\end{equation*}

 

Question 11e

This is a separable differential equation. To obtain \(x\) as a function of \(y\), we separate variables and integrate both sides:

\begin{align*}
\int \frac{dy}{e^{2y}} &= \int \,dx \\
\int e^{-2y} \,dy &= x \\
\Rightarrow x &= -\frac{1}{2}e^{-2y} + C
\end{align*}

Question 12a

When \(n = 1\),

\begin{align*}
\text{LHS} &= 1 \times 2 \\
&= 2 \,
\end{align*}

and

\begin{align*}
\text{RHS} &= 1^2 \left(1 + 1\right) \\
&= 1 \times 2 \\
&= 2 \,
\end{align*}

so the statement is true for \(n = 1\). Assume that the statement is true for \(n = k\), i.e.

\begin{equation*}
1 \times 2 + 2 \times 5 + 3 \times 8 + \dots + k \left(3k – 1\right) = k^2 \left(k + 1\right) \,.
\end{equation*}

When \(n = k + 1\),

\begin{align*}
\text{LHS} &= 1 \times 2 + 2 \times 5 + 3 \times 8 + \dots + k \left(3k – 1\right) + \left(k + 1\right) \left[3 \left(k + 1\right) – 1\right] \\
&= k^2 \left(k + 1\right) + \left(k + 1\right) \left[3 \left(k + 1\right) – 1\right] \quad \text{(applying the inductive assumption)} \\
&= k^2 \left(k + 1\right) + \left(k + 1\right) \left(3k + 2\right) \\
&= \left(k + 1\right) \left(k^2 + 3k + 2\right) \\
&= \left(k + 1\right) \left(k + 1\right) \left(k + 2\right) \\
&= \left(k + 1\right)^2 \left[\left(k + 1\right)+1\right] \,.
\end{align*}

Now, since the statement is true for \(n=1\) and \(k+1\) given \(k\), by the principle of mathematical induction, the statement is true for all integers \(n \ge 1\).

 

Question 12b

(i)

Recall that the expectation of a binomial random variable is given by \(E(X) = np\), \(n\) is the number of trials, and \(p = \) probability of success. In this experiment, let “heads” be considered a success, and \(n=100\). Then \(E(X) = np = 100 \times \frac{3}{5} = 60\).

 

(ii)

Recall that the variance of a binomial random variable is \(V(X) = np(1-p)\). In our experiment, we have

\begin{equation*}
V(X) = (100 \times \frac{3}{5}) \times \frac{2}{5} = 60 \times \frac{2}{5} = 24
\end{equation*}

Hence the standard deviation is \(\sigma = \sqrt{V(X)} = \sqrt{24} \approx 5\).

 

(iii)

Given the standard deviation is approximately 5 and the expected value is 60, the bounds 55 and 65 correspond to \(z\)-scores of \(-1\) and \(+1\) respectively.

For 100 coin tosses we can use a normal approximation of the binomial distribution; and a normal distribution has \(68 \%\) of values lying between \(z=-1\) and \(z=+1\).

Hence the approximate probability is 0.68.

 

Question 12c

There are exactly \( {8 \choose 3} = 56\) ways to choose a subset of 3 from a set of 8 courses. Now, with 400 students as pigeons and 56 topic combinations as pigeonholes, notice that

\begin{equation*}
\frac{400}{56} = 7.142\ldots > 7 \\
\end{equation*}

Hence, by the pigeonhole principle, there are at least 8 (the next integer up) students who passed exactly the same three topics.

Question 12d

Using one of the product-to-sum formulae, we have

\begin{align*}
\sin \left(3x\right) \cos \left(5x\right) &= \frac{1}{2} \left[\sin \left(3x + 5x\right) + \sin \left(3x – 5x\right)\right] \\
&= \frac{1}{2} \left[\sin \left(8x\right) + \sin \left(-2x\right)\right] \\
&= \frac{1}{2} \left[\sin \left(8x\right) – \sin \left(2x\right)\right] \,.
\end{align*}

Therefore,

\begin{align*}
\int_0^{\pi/2} \cos \left(5x\right) \sin \left(3x\right) \, dx &= \int_0^{\pi/2} \frac{1}{2} \left[\sin \left(8x\right) – \sin \left(2x\right)\right] \, dx \\
&= \frac{1}{2} \int_0^{\pi/2} \left[\sin \left(8x\right) – \sin \left(2x\right)\right] \, dx \\
&= \frac{1}{2} \left[-\frac{1}{8} \cos \left(8x\right) + \frac{1}{2} \cos \left(2x\right)\right]_0^{\pi/2} \\
&= \frac{1}{2} \left\{\left[-\frac{1}{8} \cos \left(4 \pi\right) + \frac{1}{2} \cos \left(\pi\right)\right] – \left[-\frac{1}{8} \cos 0 + \frac{1}{2} \cos \left(0\right)\right]\right\} \\
&= \frac{1}{2} \left\{ \left[-\frac{1}{8} \left(1\right) + \frac{1}{2} \left(-1\right)\right] – \left[-\frac{1}{8}\left(1\right) + \frac{1}{2} \left(1\right)\right]\right\} \\
&= -\frac{1}{2} \,.
\end{align*}

Question 12e

Separating variables and integrating both sides gives

\begin{equation*}
\int y \, dy = – \int x \, dx \,,
\end{equation*}

so

\begin{equation*}
\frac{y^2}{2} = – \frac{x^2}{2} + C \,,
\end{equation*}

where \(C\) is a constant. Substituting \(x = 1\) and \(y = 0\) into the general solution above gives

\begin{equation*}
\frac{0^2}{2} = – \frac{1^2}{2} + C \,,
\end{equation*}

so

\begin{equation*}
C = \frac{1}{2} \,.
\end{equation*}

Therefore, the curve required is

\begin{equation*}
\frac{y^2}{2} = – \frac{x^2}{2} + \frac{1}{2} \,,
\end{equation*}

which can be written as

\begin{equation*}
y^2 = -x^2 + 1
\end{equation*}

after multiplying both sides of the equation by \(2\).

 

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Question 13a

(i)

\begin{equation*}
\dfrac{d}{d \theta} \left(\sin^3 \theta\right) = 3 \sin^2 \theta \cos \theta
\end{equation*}

(ii)

Let \(x = \tan \theta\). Then \(\tfrac{dx}{d \theta} = \sec^2 \theta\), so

\begin{equation*}
dx = \sec^2 \theta \, d\theta \,.
\end{equation*}

When \(x = 0\), we have \(\tan \theta = 0\) so \(\theta = 0\); when \(x = 1\), we have \(\tan \theta = 1\) so \(\theta = \tfrac{\pi}{4}\). Therefore,

\begin{align*}
\int_0^1 \frac{x^2}{\left(1 + x^2\right)^{5/2}} \, dx &= \int_0^{\pi/4} \frac{\tan^2 \theta}{\left(1 + \tan^2 \theta\right)^{5/2}} \left(\sec^2 \theta\right) \, d\theta \\
&= \int_0^{\pi/4} \frac{\tan^2 \theta \sec^2 \theta}{\left(\sec^2 \theta\right)^{5/2}} \, d\theta \\
&= \int_0^{\pi/4} \frac{\tan^2 \theta}{\sec^3 \theta} \, d\theta \\
&= \int_0^{\pi/4} \left(\frac{\sin^2 \theta}{\cos^2 \theta}\right) \cos^3 \theta \, d\theta \\
&= \int_0^{\pi/4} \sin^2 \theta \cos \theta \, d\theta \\
&= \left[\frac{1}{3} \sin^3 \theta\right]_0^{\pi/4} \quad \text{by (a)(i)} \\
&= \frac{1}{3} \left(\frac{\sqrt{2}}{2}\right)^3 \\
&= \frac{\sqrt{2}}{12} \,.
\end{align*}

 

Question 13b

Let us first find the point of intersection between the two graphs. Of course, we can simply make the educated guess \(x = \frac{\pi}{6}\) (which is correct), but we include a more systematic method too. We solve \(\cos (2x) = \sin x\), which is equivalent to

\begin{equation*}
1 – 2\sin^2 x = \sin x \Rightarrow 2 \sin^2 x + \sin x – 1 = 0 \\
\end{equation*}

Notice that we obtain quadratic in \(\sin x\), which may be factorised to give

\begin{equation*}
(2 \sin x – 1)(\sin x + 1) = 0
\end{equation*}

The second factor gives \(\sin x = -1 \Rightarrow x = \pi\), which is outside the domain. Hence the point of intersection must be given by \(2 \sin x = 1 \Rightarrow x = \frac{\pi}{6}\).

Now we can compute the required volume. Observe that it is obtained by subtracting the volume generated by the curve \(y = \sin x\) from the volume generated by the \(y = \cos (2x)\) curve. Hence,

\begin{align*}
V &= \pi \int_0^{\pi/6} \cos^2 (2x) \,dx – \pi \int_0^{\pi/6} \sin^2 x \,dx \\
&= \frac{\pi}{2} \int_0^{\pi/6} (1 + \cos(4x)) \,dx – \frac{\pi}{2} \int_0^{\pi/6} (1 – \cos(2x)) \,dx \\
&= \frac{\pi}{2} \int_0^{\pi/6} \cos(4x) + \cos(2x) \,dx \\
&= \frac{\pi}{2} \left[ \frac{1}{4} \sin(4x) + \frac{1}{2} \sin(2x) \right]^{\pi/6}_0 \\
&= \frac{\pi}{2} \left( \frac{1}{4} \sin \frac{2\pi}{3} + \frac{1}{2} \sin \frac{\pi}{3} \right) \\
&= \frac{\pi}{2} \cdot \frac{3}{4} \sin \frac{\pi}{3} \\
&= \frac{3\sqrt{3}\pi}{16} \text{ units}^3
\end{align*}

 

Question 13c

(i)

We calculate the derivative of \(f\):

\begin{align*}
f'(x) &= \sec^2 \left( \cos^{-1}(x) \right) \frac{d}{dx}(\cos^{-1}(x)) \\
(*) \quad &= \frac{1}{x^2} \cdot \left( \frac{-1}{\sqrt{1 – x^2}} \right) \\
&= \frac{-1}{x^2 \sqrt{1-x^2}}
\end{align*}

In the step marked (\(*\)), note carefully(!) that \(\sec^2 \theta = (\sec \theta)^2 = \dfrac{1}{(\cos \theta)^2}\) by definition, and hence, for \(\theta = \cos^{-1}(x)\), we find that

\begin{equation*}
\sec^2 \left( \cos^{-1}(x) \right) = \frac{1}{x^2}
\end{equation*}

Now we turn to \(g\), and use the quotient rule carefully:

\begin{align*}
g'(x) &= \frac{ \frac{-x}{\sqrt{1-x^2}}x – \sqrt{1-x^2} }{x^2} \\
&= \frac{1}{x^2 \sqrt{1-x^2}} ( -x^2 – (1-x^2)) \\
&= \frac{-1}{x^2 \sqrt{1-x^2}}
\end{align*}

This proves that \(f'(x) = g'(x)\).

 

(ii)

Define the function \(h(x) = f(x) – g(x)\). Then by part (i), we deduce

\begin{equation*}
h'(x) = f'(x) – g'(x) = 0
\end{equation*}

Therefore \(h(x) = c\) where \(c\) is a constant! But what is the value of this constant? We need to plug in a value that lies within the domain of \(h\), which in turn depends on the domains of \(f\) and \(g\).

Now the inverse cosine function is defined only for \(x \in [-1, 1]\), and \(g(x)\) is defined for \(x \in [-1, 1]\) excluding \(x = 0\). Thus we can plug in \(x = 1\) to discover

\begin{equation*}
h(1) = f(1) – g(1) = \tan(\cos^{-1}(1)) – 0 = \tan(0) = 0
\end{equation*}

Hence \(h(x) = 0 \Rightarrow f(x) = g(x)\) for all \(x \in [-1, 1], x \ne 0\).

Remark: This was a rather tricky question, requiring careful observation and understanding of the relationship between functions.

 

Question 14a

(i)

We have the identity

\begin{align*}
(1+x)^{2n} = (1+x)^n (1+x)^n
\end{align*}

Note that \({2n \choose n}\) is the coefficient of the \(x^n\) term in the expansion of \((1+x)^{2n}\); and from the RHS we notice that there are n ways of constructing the \(x^n\) term as:

\begin{align*}
{2n \choose n}x^n = {n \choose 0}x^0 \cdot {n \choose n} x^n +{n \choose 1}x^1 \cdot {n \choose n-1} x^{n-1} + \cdot\cdot\cdot
+{n \choose n}x^n \cdot {n \choose 0} x^0
\end{align*}

Now, since \({n \choose k}= {n \choose n-k}\), we can equate coefficients to obtain:

\begin{align*}
{2n \choose n} x^n&= {n \choose 0} \cdot {n \choose 0} x^n +{n \choose 1} \cdot {n \choose 1} x^n + \cdot\cdot\cdot
+{n \choose n}\cdot {n \choose n} x^n \\
{2n \choose n} &= {n \choose 0}^2 +{n \choose 1}^2 + \cdot\cdot\cdot
+{n \choose n}^2
\end{align*}

as required.

(ii)

Assume \(2k\) people are chosen for the group, with \(k\) men and \(k\) women. There are \({n \choose k}\) ways of choosing the men, multiplied by \({n \choose k}\) ways of choosing the women.

Since \(k\) can be from \(0\) to \(n\), we have a total of

\begin{align*}
{n \choose 0}^2 +\binom{n}{1}^2 + \cdot\cdot\cdot
+{n \choose n}^2
\end{align*}

total ways of choosing an even-sized group; and from part i) this is equal to \({2n \choose n}\) as required.

 

(iii)

In the group of \(k\) men, there are \(k\) ways of choosing a leader; and similarly in the group of \(k\) women there are \(k\) ways of choosing a leader. Hence, for a group of size \(2k\), there are

\begin{align*}
k^2 {n \choose k}^2
\end{align*}

ways of choosing groups with a leader. Again since \(k\) can range from 1 to \(n\) (as there cannot be a leader in an empty group), the total number of ways is

\begin{align*}
1^2 {n \choose 1}^2 +2^2{n \choose 2}^2 + \cdot\cdot\cdot
+n^2 {n \choose n}^2
\end{align*}

 

as required.

 

(iv)

If we choose the leaders first, we have \(n\) ways of selecting a male leader and \(n\) ways of selecting a female leader. We now have \(2(n-1)\) people left to choose from, and we have to choose up to \(n-1\) males and up to \(n-1\) females. This is exactly the same scenario as part (ii) with fewer members; so this is equal to

\begin{align*}
n^2 \textrm{(Pick the leaders)} \times {2(n-1) \choose n-1} = n^2 {2n-2 \choose n-1}
\end{align*}

 

Question 14b

(i)

Expanding \(\sin(3\theta)\), we have:

\begin{align*}
\sin (3\theta) &= \sin (2\theta) \cos \theta + \cos(2\theta) \sin \theta\\
&=2 \sin \theta \cos^2 \theta + (1-2\sin^2 \theta)\sin \theta\\
&= 2 \sin \theta (1-\sin^2\theta) + \sin \theta – 2 \sin^3 \theta\\
&= 2 \sin\theta – 2 \sin^3 \theta + \sin \theta – 2 \sin^3 \theta \\
&= 3 \sin^3 \theta – 4 \sin^3 \theta
\end{align*}

Substituting this into the entire expression above, we have:

\begin{align*}
\sin^3 \theta – \frac{3}{4}\sin\theta + \frac{\sin(3\theta)}{4} &= \sin^3 \theta – \frac{3}{4}\sin \theta + \frac{3 \sin^3 \theta – 4 \sin^3\theta}{4}\\
&= \sin^3 \theta-\sin^3 \theta+\frac{3}{4}\sin\theta-\frac{3}{4}\sin\theta\\
&=0 \\
\end{align*}

as required.

 

(ii)

Let \(x = 4 \sin \theta\). Then the equation \(x^3 – 12x + 8 = 0\) becomes

\begin{align*}
\left(4 \sin \theta\right)^3 – 12 \left(4 \sin \theta\right) + 8 &= 0 \\
64 \sin^3 \theta – 48 \sin \theta + 8 &= 0 \\
4 \sin^3 \theta – 3 \sin \theta + \frac{1}{2} &= 0 \quad \text{(dividing the whole equation by $16$)} \\
3 \sin \theta – 4 \sin^3 \theta &= \frac{1}{2} \,.
\end{align*}

 

Rearranging the equation from part (b)(i) gives

\begin{align*}
\frac{\sin \left(3 \theta\right)}{4} &= \frac{3}{4} \sin \theta – \sin^3 \theta \\
\sin \left(3 \theta\right) &= 3 \sin \theta – 4 \sin^3 \theta
\end{align*}

Finally, substituting the RHS from above, we have

\begin{align*}
\sin \left(3 \theta\right) &= \frac{1}{2}
\end{align*}

as required.

 

(iii)

We note that if \(\theta = \frac{\pi}{18}\), then \(\sin(3\theta) = \frac{1}{2}\), so \(\sin \theta\) where \(\theta = \frac{\pi}{18}\) is a solution to the equation in part (ii).

Similarly,

\begin{align*}
\sin(3\cdot\frac{5\pi}{18}) &=
\sin{\frac{15\pi}{18}} \\
&= \sin{\frac{5\pi}{6}} \\
&=\frac{1}{2}
\end{align*}

and

\begin{align*}
\sin(3 \cdot \frac{25\pi}{18}) &= \sin{\frac{25\pi}{6}}\\
&=\sin({2\pi +\frac{25\pi}{6}}) \\
&=\frac{1}{2}
\end{align*}

so

\begin{align*}
\alpha &= 4\sin \frac{5\pi}{18}\\
\beta &= 4\sin\frac{5\pi}{18} \\
\gamma&=4\sin{\frac{25\pi}{18}}
\end{align*}

are also roots of the polynomial \(x^3 – 12x + 8\) from part (ii).

Reframing it as a polynomial question, we need to find

\begin{align*}
\frac{\alpha^2 + \beta^2 + \gamma^2}{4^2} &= \frac{1}{16}((\alpha + \beta + \gamma)^2 – 2(\alpha\beta + \beta\gamma + \alpha\gamma))\\
&= \frac{1}{16}(\frac{-b}{a}^2 – 2 \frac{c}{a})\\
&= \frac{1}{16}(0 – 2 \times \frac{-12}{1})\\
&= \frac{24}{16}\\
&=\frac{3}{2}
\end{align*}

 

as required.

 

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Written by Matrix Maths Team

The Matrix Maths Team are tutors and teachers with a passion for Mathematics and a dedication to seeing Matrix Students achieving their academic goals.

 

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