2020 HSC Maths Ext 1 Exam Paper Solutions

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

Section 2: Long response

Question 11a

(i)

(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-3} \choose 2} = 0
\end{equation*}

Therefore

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.

Question 11d

Observe that

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

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.

When \(n = k + 1\),

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

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

Therefore,

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 \(2\).

Want to learn to solve challenging problems like these?

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,

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

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,

Question 13c

(i)

We calculate the derivative of \(f\):

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:

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

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:

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

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

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

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

Question 14b

(i)

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

Substituting this into the entire expression above, we have:

as required.

(ii)

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

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

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

as required.

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