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!

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

Question | Answer | Solution |

1 | A | The quadratic factorises to \(x^2 -2x – 3 = (x-3)(x+1)\), so the solution is \(x>3 \cup x<-1\). |

2 | C | 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 \(C:x\geq 1, R: y\geq 0\) |

3 | D | 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 \(\frac{1}{4} \frac{1}{1/2}\tan^{-1}(\frac{x}{1/2})= \frac{1}{2} \tan^{-1}(2x)+c\). |

4 | B | 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 \(\sqrt{9^2 + 2^2} = \sqrt{85}\). |

5 | C | Since 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. |

6 | D | For \(a-b\), we put \(a\) and \(b\) tip-to-tip and join the tails. |

7 | A | Checking 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. |

8 | C | First 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 \(\frac{10!}{6!4!} \frac{6!}{2!}=\frac{10!}{4!2!}\). |

9 | B | We can draw a diagram to show the situation:
Hence, the answer should be \({4 \choose 8} + {4 \choose 8} – \left({6 \choose 7}-{4 \choose8}\right)={2 \choose 9}\). |

10 | A | First, 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 \(\frac{10!}{6!4!} \frac{6!}{2!}=\frac{10!}{4!2!}\). |

**(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*}

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, \(a = 2, -\frac{1}{2}\).

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

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*}

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*}

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 \(n \ge 1\).

**(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.

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.

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*} |

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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**(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*} |

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*} |

**(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.

**(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*} |

**(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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