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2021 HSC Maths Ext 1 Exam Paper Solutions

The 2021 HSC Mathematics Extension (3 Unit) exam paper solutions are here!

2021 HSC Maths Extension 1 Exam Paper Solutions

Have you seen the 2021 HSC Mathematics Extension (3 Unit) exam paper yet?

In this post, we will work our way through the 2021 HSC Maths Extension 1 (3 Unit) paper and give you the solutions, written by our Head of Mathematics Oak Ukrit and his team. (Doing practice papers? See the solutions for the 2020 HSC Maths Ext 1 Exam here.)

Read on to see how to answer all of the 2021 questions.

 

Section 1. Multiple Choice

\(\)
Question Answer Solution
1 C \begin{align*}
\overrightarrow{PQ} &= \overrightarrow{OQ} – \overrightarrow{OP} \\
&= \binom{2}{5} – \binom{-3}{1}\\
&= \binom{5}{4} \\
\end{align*}
2 B \begin{align*}
\text{cos}6x &= 1- 2\text{sin}^23x \\
\text{sin}^2 3x &= \frac{1-\text{cos}6x}{2} \\
∴ \int \text{sin}^2 3x \ dx &= \int \frac{1 – \text{cos}6x}{2} \ dx \\
\end{align*}
3 D \begin{align*}
\text{Use the Remainder Theorem:} \\
P(-2) &= -(-2)^3 -2(-2)^2-3(-2)+8 \\
&= 14 \\
\end{align*}
4 A \begin{align*}
\int y \ dy &= \int x \ dx \\
2 \int y \ dy &= 2 \int x \ dx \\
y^2 &= x^2 + C \\
\end{align*}
5 B \begin{align*}
\overrightarrow{OA} · \overrightarrow{OB} =|\overrightarrow{OA}||\overrightarrow{OB}| \text{cos} \theta &< 0 \\
∴ \text{cos} \theta &< 0 \\
∴ \theta & \text{ is obtuse}\\
\end{align*}
6 A \begin{align*}
r &= 1-P(X=0) \\
&= 1- \binom{10}{0} p^0(1-p)^{10} \\
&= 1 – 0.1^{10} \\
&= 0.9999999999 > 0.9 \\
\end{align*}
7 D \begin{align*}
\text{Let } f(x) = -a \text{sin}x + b \text{cos} x &= R \text{cos}(x+ \alpha). \text{ Then:} \\
-a \text{sin} x +b \text{cos} x &= R\text{cos}(x+ \alpha) \\
&= R \text{cos}(\alpha) \text{cos}(x) – R \text{sin}(\alpha)\text{sin}(x) \\
-a \text{sin} x + b \text{cos} x &= -R \text{sin} (\alpha) \text{sin} (x) + R \text{cos} (\alpha) \text{cos} (x) \\
\end{align*}Now, \(R \text{cos} \alpha = b >0\), \(R \text{sin} \alpha = a >0\). The signs of the trigonometric ratios indicate that our solution is in the first quadrant, i.e. \(0< \alpha < \frac{pi}{2}\).Therefore, \(f(x)\) is a \(\text{cos}(x)\) graph shifted left by \(\alpha\).Alternatively, consider the gradient and the sign of \(f(x)\) at \(x=0\).

  • \(f(0)=-a \text{sin}(0)+b \text{cos}(0) = b > 0\), so \(f(x)\) at \(x=0\) is positive.
  • \(f'(x)=-a \text{cos}(x) – b\text{sin}(x)\) and \(f'(0)=-a \text{cos}(0) – b\text{sin}(0)=-a<0\), so \(f'(x)\) at \(x=0\) is decreasing.

Therefore, the graph of \(f(x)\) at \(x=0\) should be positive and decreasing; this matches (D).

8 C In the domain \(– \frac{ \pi}{2} ≤ t ≤ \frac{ \pi}{2} \), \( -1 ≤ \text{sin}(t) ≤ 1 \) and \( 0 ≤ \text{cos}(t) ≤ 1 \).

We note that \(y\) is always less than the centre 2; so we want \(y = 2 – \text{cos} (t)\), which is (C).

9 A \begin{align*}
\frac{dy}{dx} &= \frac{ \text{cos}x}{ \sqrt{1 – \text{sin}^x}} \\
&= \frac{ \text{cos}x}{|\text{cos} x|} \\
&=
\begin{cases}
1 & \text{when } \text{cos}x > 0\\
-1 & \text{when } \text{cos}x < 0\\
\end{cases}
\end{align*}As the domain is \(x∈ \bf{R}\), the solution must be (A).Alternatively, we note that the inner function \(\text{sin} (x) \) is periodic and has a domain of \(x∈ \bf{R}\), eliminating (C) and (D). Next, neither the range of \(\text{sin}\) nor \(\text{sin}^{-1}\) have any discontinuities, so option (A) fits best.
10 C  First note that \(\frac{3543}{15} = 236.2\). To minimise the votes of the winning candidate, we could have that each candidate receives 236 votes each, with 3 votes left to distribute. For a winner, they must have 2 of these 3 remaining votes, so \(236 + 2 = 238\).

Need more help understanding these questions?

 

Long response questions

Question 11a

\((\underset{\text{~}}i+6\underset{\text{~}}j)+(2\underset{\text{~}}i-7\underset{\text{~}}j)=3\underset{\text{~}}i-\underset{\text{~}}j\)

 

Question 11b

\begin{align*}
(2a-b)^4 &= (2a)^4-4(2a)^3b+6(2a)^2b^2-4(2a)b^3+b^4 \\
&= 16a^4-32a^3b+24a^2b^2-8ab^3+b^4 \\
\end{align*}

 

Question 11c

Let \(u=x+1\). \(\frac{du}{dx}=1\), \(du=dx\).\begin{align*}
\int x \sqrt{x+1} \ dx &= \int (u-1) \sqrt{u} \ du \\
&= \int u^{\frac{3}{2}} – u^ \frac{1}{2} \ du \\
&= \frac{2}{5}u^{\frac{5}{2}} – \frac{2}{3} u^{\frac{3}{2}} + C \\
&= \frac{2}{5} (x+1)^2 \sqrt{x+1} – \frac{2}{3}(x+1) \sqrt{x+1} + C \\
\end{align*}

 

Question 11d

We choose the 5 men from the 10 first, then choose 3 women from the 8 to obtain

\(\binom{10}{5} \times \binom{8}{3} = 14112\)

 

Question 11e

We have that \( \frac{dr}{dt}= 0.2 \ \text{mm/s}\), so we hope to obtain the change of the volume through the chain rule:\(\frac{dV}{dt} = \frac{dr}{dt} \times \frac{dV}{dr} = 0.2 \times \frac{dV}{dr}\)

It remains to find \(\frac{dV}{dr}\) by noting that \(V= \frac{4}{3} \pi r^3\) and so \( \frac{dV}{dr} = 4 \pi r^2 = \frac{36}{25} \pi\) at \(r=0.6 \ \text{mm}\).

Substituting into the change rule above, we arrive at:

\( \frac{dV}{dt} ≈ 0.9 \text{mm}^3 \text{/s}\)

 

Question 11f

This is a standard integral:

\begin{align*}
\int_0^{\sqrt{3}} \frac{1}{\sqrt{4-x^2}} \ \mathrm{d} x &= \left[ \sin^{-1}\frac{x}{2}\right]_0^{\sqrt{3}}\\
&= \sin^{-1}\frac{\sqrt{3}}{2} – \sin^{-1}0 \\
&= \frac{\pi}{3}.
\end{align*}

 

Question 11g

We factorise first:\begin{align*}
2\sin^3x + 2\sin^2x-\sin x-1 &=0 \\
2\sin^2x(\sin x+1) – (\sin x + 1) &= 0 \\
(\sin x+1)(2 \sin^2 x-1) &= 0
\end{align*}to obtain that\begin{equation*}
\sin x = -1 \qquad \mathrm{or} \qquad \sin x = \pm \frac{1}{\sqrt{2}}.
\end{equation*}The first solution gives \(x=\frac{3\pi}{2}\) only, whereas the second solution consists of the four quadrant solutions corresponding to the related angle of \(\frac{\pi}{4}\): \(x=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4}\) and \(\frac{7\pi}{4}\). In total, we have five solutions for \(x\):\begin{equation*}
x = \frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{3\pi}{2} \ \mathrm{and} \ \frac{7\pi}{4}.
\end{equation*}

 

Question 11h

We express the quantity \(\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\delta}+\frac{1}{\gamma}\) in terms of the various sums and products of roots:\begin{align*}
\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\delta}+\frac{1}{\gamma} &= \frac{\alpha \beta \gamma + \alpha \beta \delta + \alpha \gamma \delta + \beta \gamma \delta}{\alpha \beta \gamma \delta} \\
&= \frac{-d/a}{e/a} \\
&= \frac{-(-3/1)}{6/1} \\
&= \frac{1}{2}.
\end{align*}where the original quartic is of the form \(ax^4+bx^3+cx^2+dx+e\) with \(a=1,b=c=0,d=-3,e=6\).

 

Question 12a

Various solutions will likely be permitted – as long as the curve generally conforms to the direction field, and goes through the point P. Two alternatives are presented below.

2021 HSC Maths Ext 1 Exam Paper Solutions Q12a

 

Question 12b

(i)

We solve the differential equation by separating variables and integrating both sides.

\begin{align*}
\frac{dT}{T-25} &= k \,dt \\
\int \frac{1}{T-25} \,dT &= k \int \,dt \\
\ln|T-25| &= kt + C.
\end{align*}

Initially (\(t=0\)), the water is at \(5 ° \textrm{C}\), so \(T-25 < 0\) and we flip the terms in the absolute value. Substituting this point yields:

\begin{align*}
%\ln(25-T) = kt+C \overset{(t=0, T=5)}{\Longrightarrow} C = \ln 20.
\ln(25-T) &= kt+C \\
\ln(25-5) &= k\times 0+C \\
C &= \ln 20.
\end{align*}

Now we can find \(k\).

\begin{align*}
\ln(25-T) &= kt + \ln 20 \\
kt &= \ln(25-T) – \ln 20 = \ln\left(\frac{25-T}{20}\right). \qquad (*)
\end{align*}

At \(t=8\), we are given that \(T=10\). Hence

\begin{equation*}
8k = \ln\left(\frac{15}{20}\right) = \ln\left(\frac{3}{4}\right) ⇒ k = \frac{1}{8}\ln\left(\frac{3}{4}\right).
\end{equation*}

We keep writing \(k\) until the last step of the calculation in order to avoid propagating round-off errors.

Now we determine the solution to the differential equation.

\begin{align*}
kt &= \ln\left(\frac{25-T}{20}\right) \\
e^{kt} &= \frac{25-T}{20} \\
20e^{kt} &= 25-T \\
⇒ T(t) &= 25 – 20e^{kt}.
\end{align*}

Substituting \(k\) in and solving for \(T=20\), we have:

\begin{align*}
20 &= 25 – 20e^{\frac{1}{8}\ln(\frac{3}{4})t}\\
\frac{-5}{-20} &= e^{\frac{1}{8}\ln(\frac{3}{4})t}\\
\ln\left(\frac{1}{4}\right) &= \frac{1}{8}\ln\left(\frac{3}{4}\right)t\\
t &= \frac{\ln(\frac{1}{4})}{\frac{1}{8}\ln(\frac{3}{4})}\\
&= 38.55 \approx 39 \ \textrm{min.}
\end{align*}

A more efficient method: To find the time at which \(T=20\), we may return to the earlier step \((*)\):

\begin{equation*}
kt = \ln\left(\frac{25-20}{20}\right) = \ln\left(\frac{5}{20}\right) = \ln\left(\frac{1}{4}\right).
\end{equation*}

Therefore
\begin{equation*}
t = \frac{1}{k}\ln\left(\frac{1}{4}\right) = 8 \frac{\ln(\tfrac{1}{4})}{\ln(\frac{3}{4})} =38.5507\ldots \approx 39 \ \text{min}
\end{equation*}

to the nearest minute.

 

(ii)

We plot the function \(T(t)\) below, showing the horizontal asymptote.

 

Question 12c

If \(n=1\), then the right hand side is\begin{equation*}
\frac{1}{4} – \frac{1}{2\times 2\times 3} = \frac{2}{3\times 4} = \frac{1}{6}
\end{equation*}while the left hand side is simply \(\frac{1}{1\times 2\times 3} = \frac{1}{6}\). Hence the base case is true.Now assume the statement holds for some integer \(n\ge 1\). We need to prove that\begin{equation*}
\frac{1}{1\times 2\times 3} + \frac{1}{2\times 3\times 4} + \ldots + \frac{1}{n(n+1)(n+2)} + \frac{1}{(n+1)(n+2)(n+3)} = \frac{1}{4} – \frac{1}{2(n+2)(n+3)} \quad (\star)
\end{equation*}Starting with the left hand side, we obtain\begin{multline*}
\frac{1}{1\times 2\times 3} + \frac{1}{2\times 3\times 4} + \ldots + \frac{1}{n(n+1)(n+2)} + \frac{1}{(n+1)(n+2)(n+3)} \\
= \frac{1}{4} – \frac{1}{2(n+1)(n+2)} + \frac{1}{(n+1)(n+2)(n+3)}
\end{multline*}by the induction assumption. Now we calculate

\begin{align*}
– \frac{1}{2(n+1)(n+2)} + \frac{1}{(n+1)(n+2)(n+3)} &= \frac{-(n+3) + 2}{2(n+1)(n+2)(n+3)} \\
&= \frac{-(n+1)}{2(n+1)(n+2)(n+3)} \\
&= \frac{-1}{2(n+2)(n+3)}.
\end{align*}

Therefore equation (\(\star\)) holds, which completes the induction step.

 

Question 12d

(i)

We may first consider the basic parabola defined by \(y = (1-\frac{x}{2})^2\) for \(x\) in the domain \((-\infty, 2]\), which looks like this:

In Page - 2021 HSC Maths Ext 1 Exam Paper Solutions

Then \(y = 4-(1-\frac{x}{2})^2\) is obtained by reflecting the above graph about the \(x\)-axis, and then shifting the vertex of the parabola up by 4 units:

In Page - 2021 HSC Maths Ext 1 Exam Paper Solutions

(ii)

We consider the function \(f(x) = 4-(1-\frac{x}{2})^2\) with domain \((-\infty,2]\) and range \((-\infty, 4]\). Thus the domain of the inverse function \(f^{-1}(x)\) is \((-\infty, 4]\). Now we can determine \(f^{-1}\) by first writing \(x = 4-(1-\frac{y}{2})^2\), and then solve for \(y\):

\begin{align*}
\left(1-\frac{y}{2}\right)^2 &= 4-x \\
1-\frac{y}{2} &= \pm \sqrt{4-x} \\
\frac{y}{2} &= 1 \mp \sqrt{4-x} \\
⇒ y &= 2 \mp 2\sqrt{4-x}.
\end{align*}

Finally we need to determine the correct choice of sign. Since \(f\) passes through the point \((x,y) = (-2,0)\), it follows that \(f^{-1}\) passes through \((0,-2)\). Plugging in \(x=0\) into \(2 \mp 2\sqrt{4-x}\), we find that \(–\) is correct.

Hence the inverse function is

\begin{equation*}
f^{-1}(x) = 2 – 2\sqrt{4-x}
\end{equation*}

with domain \((-\infty, 4]\).

 

(iii)

We sketch \(f^{-1}\) (red) along with \(f\).

2021 HSC Maths Ext 1 Exam Paper Solutions

 

Question 13a

The volume of the sculpture is calculated as the difference in volume between two paraboloids. The larger paraboloid, formed by the curve \(y=x^2\) for \(0 \leq y \leq 2\), has volume\begin{equation*}
V_1 = \pi \int_0^2 x^2 \,dy = \pi \int_0^2 y \,dy = \pi \left[ \frac{y^2}{2} \right]^2_0 = 2\pi.
\end{equation*}On the other hand, the smaller paraboloid is formed by the curve \(y=x^2+1\) for \(1 \leq y \leq 2\). We notice that \(x^2 = y-1\) in this case, hence the volume is given by\begin{equation*}
V_2 = \pi \int_1^2 x^2 \,dy = \pi \int_1^2 y-1 \,dy = \pi \left[ \frac{y^2}{2}-y \right]^2_1 = \frac{\pi}{2}.
\end{equation*}Therefore
\begin{equation*}
V_{\text{sculpture}} = V_1 – V_2 = 2\pi – \frac{\pi}{2} = \frac{3\pi}{2} \text{ units}^3.
\end{equation*}

 

Question 13b

We are given the following initial data:

  • Initial velocity \(V=12 \text{ms}^{-1}\);
  • Launch angle \(\theta = 30 ° \); and
  • Initial height \(h=1\).

Hence we find that \(Vt \cos\theta = 6\sqrt{3}\, t\) and \(Vt\sin\theta = 6t\). The trajectory of the ball is therefore parametrised by
\begin{align*}
x(t) &= 6\sqrt{3}\,t \\
y(t) &= -5t^2 + 6t + 1.
\end{align*}
We can find the maximum height already by differentiating \(y(t)\).
\begin{equation*}
\dot{y}(t) = -10t + 6 = 0 ⇒ t = \frac{3}{5} \text{s}.
\end{equation*}
This is a maximum, since \(\ddot{y}(t) = -10 < 0\). Hence
\begin{equation*}
y_{\max} = -5 \left( \frac{3}{5} \right)^2 + 6\left( \frac{3}{5} \right) + 1 = \frac{-9 + 18 + 5}{5} = \frac{14}{5},
\end{equation*}
and we see that \(y_{\max} = 2.8 < 3\). Hence the ball does not hit the ceiling.

Now if we set \(x(t) = 10\), then \(6\sqrt{3} t = 10\), which gives
\begin{equation*}
t = \frac{5}{3\sqrt{3}} \text{s}.
\end{equation*}
This is the time at which the ball hits the far wall. However, we check on our calculator that
\begin{equation*}
y\left( \frac{5}{3\sqrt{3}} \right) = -5\left( \frac{5}{3\sqrt{3}} \right)^2 + 6\left( \frac{5}{3\sqrt{3}} \right) + 1 \approx 2.144 > 0
\end{equation*}
(where we have given the result correct to 3 decimal places). This shows the vertical displacement is positive, so we can deduce that the ball has not yet hit the ground.

 

Question 13c

We apply the area bounded between two curves formula, noting that we have an even function and so we can simplify things by only finding the area from \(x=0\) to \(x=2\) and doubling our answer.
\begin{align*}
A &= \int_{-2}^{2} 2 – |x| – \left(1-\frac{8}{4+x^2}\right) \ \mathrm{d} x\\
&= 2 \int_0^2 1-x+\frac{8}{4+x^2} \; \mathrm{d}x\\
&= 2 \left[x-\frac{1}{2} x^2 + 4\tan^{-1} \frac{x}{2}\right]_0^2 \\
&= 2\pi \ \mathrm{units}^2.
\end{align*}

 

Question 13d

(i)

If \(A=B-d\) and \(C=B+d\) then
\begin{align*}
\frac{\sin A+\sin C}{\cos A + \cos C} &= \frac{\sin(B-d)+\sin(B+d)}{\cos(B-d)+\cos(B+d)} \\
&= \frac{2\sin B \cos d}{2\cos B \cos d} \\
&= \tan B
\end{align*}
where we make the use of the sum to product formulae in the first step.

 

(ii)

If we let \(B=\frac{11\theta}{14}\) and \(d=\frac{\theta}{14}\) in (i) then we get:\begin{equation*}
\frac{\sin \frac{5\theta}{7} + \sin \frac{6\theta}{7}}{\cos \frac{5\theta}{7} + \cos \frac{6\theta}{7}} = \tan \frac{11\theta}{14}
\end{equation*}and so we simplify our problem to solving the equation\begin{equation*}
\tan \frac{11\theta}{14} = \sqrt{3} \qquad \mathrm{where} \qquad 0\leq \theta \leq 2\pi ⇒0\leq \frac{11\theta}{14}\leq \frac{11\pi}{7}.
\end{equation*}Looking at first and third quadrant solutions, we obtain\begin{align*}
\frac{11\theta}{14} &= \frac{\pi}{3}, \frac{4\pi}{3} \\
∴\theta &= \frac{14\pi}{33}, \frac{56\pi}{33}
\end{align*}

 

Question 14a

Consider the following diagram, where the 175km/h vector is placed tip-to-tail with the 42 km/h wind vector. By alternate angles, the angle between the wind vector and the destination vector is \(63^\circ\).

2021 HSC Maths Ext 1 Exam Paper Solutions

Using the sine law,

\begin{align*}
\frac{\sin \theta}{42} &= \frac{\sin 63^\circ}{175} \\
\theta &= \sin^{-1}\left(\frac{42 \sin 63^\circ}{175}\right) \\
&\approx 12^\circ \; \text{(nearest degree)}
\end{align*}

Hence, the bearing is \(063^\circ T + \theta = 075^\circ T\) to the nearest degree.

 

Question 14b

This is a classic example of a logistic equation. We first separate variables\begin{equation*}
\frac{C}{P(C-P)} \,dP = 0.1\,dt
\end{equation*}and then integrate both sides, using the given fact. Note that we use \(K\) for our integration constant, since \(C\) is already taken!\begin{align*}
\int \frac{C}{P(C-P)} \,dP &= \int 0.1 \,dt = 0.1t + K \\
⇒ 0.1t + K &= \int \frac{1}{P} + \frac{1}{C-P} \,dP \\
&= \ln P – \ln(C-P) = \ln\left(\frac{P}{C-P}\right).
\end{align*}(We do not need absolute values here: we must have \(C \ge P(t)\) for all \(t\ge 0\), since \(C\) is the carrying capacity, i.e. the largest population the habitat can sustain).At \(t=0\) (the year 1980), we are given that the deer population was \(150\, 000\). Setting \(t=0\) and \(P=150\, 000\) in the above result, we find\begin{equation*}
K = \ln\left( \frac{150\, 000}{C – 150\, 000} \right).
\end{equation*}In the year 2000, which corresponds to \(t=20\), we are given that \(P = 600\, 000\). Therefore

\begin{equation*}
2 + K = \ln\left( \frac{600\, 000}{C-600\, 000} \right).
\end{equation*}

We can now eliminate \(K\) and start solving for \(C\):

\begin{align*}
K = \ln\left( \frac{150\, 000}{C – 150\, 000} \right) &= \ln\left( \frac{600\, 000}{C-600\, 000} \right) – 2 \\
\ln\left( \frac{600\, 000}{C-600\, 000} \right) – \ln\left( \frac{150\, 000}{C – 150\, 000} \right) &= 2 \\
\ln\left(\frac{600\,000}{C – 600\, 000} \cdot \frac{C – 150\, 000}{150\, 000} \right) &= 2 \\
\ln\left( 4\frac{C-150\,000}{C-600\,000}\right) = 2.
\end{align*}

Taking the exponential of both sides, we continue to solve:

\begin{align*}
4\frac{C-150\,000}{C-600\,000} &= e^2 \\
\frac{C-150\,000}{C-600\,000} &=\frac{1}{4}e^2 \\
C – 150\,000 &= \frac{1}{4} e^2(C – 600\,000) \\
\left(\frac{1}{4}e^2 – 1\right)C &= 150\, 000(e^2 – 1) \\
⇒ C &= 150\,000 \frac{e^2 -1}{\frac{1}{4}e^2 – 1} \approx 1\, 131\, 121.
\end{align*}

The carrying capacity is therefore \(C \approx 1\, 130\, 000\), to 3 significant figures.

 

Question 14c

(i)

The easiest way to do this is algebraically. Let \(v= x \underset{\text{~}}{i} + y\underset{\text{~}}{j}\) then \(\underset{\text{~}}{v}\cdot\underset{\text{~}}{v} = x^2+y^2.\) We now check the right hand side, \(|\underset{\text{~}}{v}|^2 = \left(\sqrt{x^2+y^2}\right)^2=x^2+y^2\) and so we conclude that \(\underset{\text{~}}{v}\cdot\underset{~}{v} = |\underset{\text{~}}{v}|^2\).

(ii)

Note that\begin{equation*}
\overrightarrow{AC} = \overrightarrow{AB}+\overrightarrow{BC} = \underset{\text{~}}{a}+\underset{\text{~}}{b} \qquad \mathrm{and} \qquad \overrightarrow{BD} = \overrightarrow{AD}-\overrightarrow{AB} = k \underset{\text{~}}{b}-\underset{\text{~}}{a}
\end{equation*}Using the fact that \(|\overrightarrow{AC}| = |\overrightarrow{BD}|\), or rather (in order to use (i)), \(|\overrightarrow{AC}|^2 = |\overrightarrow{BD}|^2\) to obtain\begin{equation*}
\overrightarrow{AC}\cdot\overrightarrow{AC} = \overrightarrow{BD}\cdot\overrightarrow{BD}
\end{equation*}Substituting for our expressions for \(\overrightarrow{AC}\) and \(\overrightarrow{BD}\) and rearranging, we get\begin{equation*}
(\underset{\text{~}}{a}+\underset{\text{~}}{b})\cdot(\underset{\text{~}}{a}+\underset{\text{~}}{b}) = (k\underset{\text{~}}{b}-\underset{\text{~}}{a})\cdot(k\underset{\text{~}}{b}-\underset{\text{~}}{a})
\end{equation*}Now we expand both sides and make use of (i) to obtain the required result:\begin{align*}
\underset{\text{~}}{a}\cdot \underset{\text{~}}{a}+2\underset{\text{~}}{a}\cdot \underset{\text{~}}{b}+\underset{\text{~}}{b}\cdot\underset{\text{~}}{b}&=k^2\underset{\text{~}}{b}\cdot\underset{\text{~}}{b}-2k\underset{\text{~}}{a}\cdot\underset{\text{~}}{b}+\underset{\text{~}}{a}\cdot\underset{\text{~}}{a} \\
2\underset{\text{~}}{a}\cdot\underset{\text{~}}{b}+|\underset{\text{~}}{b}|^2&=k^2|\underset{\text{~}}{b}|^2-2k\underset{\text{~}}{a}\cdot \underset{\text{~}}{b}\\
2(1+k)\underset{\text{~}}{a}\cdot\underset{\text{~}}{b} &= (k^2-1)|\underset{\text{~}}{b}|^2 \\
2(1+k)\underset{\text{~}}{a}\cdot\underset{\text{~}}{b} &= -(1+k)(1-k)|\underset{\text{~}}{b}|^2 \\
2\underset{\text{~}}{a}\cdot\underset{\text{~}}{b} &= -(1-k)|\underset{\text{~}}{b}|^2 \quad \text{since } 1+k \neq 0\\
2\underset{\text{~}}{a}\cdot\underset{\text{~}}{b}+(1-k)|\underset{\text{~}}{b}|^2 &= 0.
\end{align*}

 

Question 14d

With \(p = \frac{3}{500}\), we can calculate the following quantities for \(\hat p\),\begin{align*}
\mu &= p = \frac{3}{500} \\
\sigma^2 &= \frac{p(1-p)}{n} = \frac{1491}{500^2 n}
\end{align*}The aim is to find the probability \(P(\hat p \geq \frac{4}{500})\), which we will change into z-scores,\begin{align*}
P(\hat p \geq \frac{4}{500}) &= P \left(Z \geq \frac{\frac{4}{500}-\mu}{\sigma}\right)\\
&= P \left(Z \geq \frac{\frac{4}{500} – \frac{3}{500}}{\frac{1}{500} \sqrt{\frac{1491}{n}}} \right) \\
&= P \left(Z \geq \frac{\sqrt{n}}{\sqrt{1491}} \right)
\end{align*}From the empirical rule, 95% of the data is between z-scores -2 and 2.So, 2.5% is to the right of 2.Hence, we need\begin{align*}
\frac{\sqrt{n}}{\sqrt{1491}} &\geq 2 \\
n &\geq 4 \times 1491 = 5964.
\end{align*}

Hence, the sample size required is 6000 to the nearest thousand.

 

Question 14e

We have that \(g(x)=x^3+4x-2\) and \(g'(x)=3x^2+4\). In order to find the gradient of the tangent to \(f(x)=x g^{-1}(x)\) at \(x=3\) we need to find \(f'(3)\). Differentiating \(f(x)\) using the product rule, we get\begin{align*}
f'(x) = g^{-1}(x) + x \frac{d}{dx}\left(g^{-1}(x)\right)
\end{align*}It becomes necessary to differentiate \(g^{-1}(x)\). To find what this is, consider the function \(y=g^{-1}(x)\): we have \(x=g(y)\), and we want to find \(\frac{dy}{dx}\).Then\begin{align*}
\frac{dx}{dy} &= g'(y) \\
\frac{dy}{dx} &= \frac{1}{g'(y)}
\end{align*}However, note here that \(y\) in this context is not the same as \(g(x)\), because in this context, \(y = g^{-1}(x)\). We should instead try to express \(\frac{dy}{dx} = \frac{d}{dx}(g^{-1}(x))\) in terms of \(x\) only, by substituting \(y = g^{-1}(x)\):\begin{align*}
\frac{dy}{dx} = \frac{d}{dx} (g^{-1}(x))&= \frac{1}{g'(y)} \\
&= \frac{1}{g’\left(g^{-1}(x)\right)}
\end{align*}So the derivative of \(f(x)\) is given by

\begin{equation*}
f'(x) = g^{-1}(x) + \frac{x}{g’\left(g^{-1}(x)\right)}
\end{equation*}

Since \(g(x)\) passes through the point \((1,3)\), we have \(g(1)=3\) so \(g^{-1}(3)=1\). Now substituting \(x=3\) we have

\begin{align*}
f'(3) &= g^{-1}(3) + \frac{3}{g’\left(g^{-1}(3)\right)} \\
&= 1 + \frac{3}{g'(1)} \\
&= 1 + \frac{3}{3\times1^2+4} \\
&= \frac{10}{7}
\end{align*}

We conclude that the gradient of the tangent to \(f(x)\) at \(x=3\) is \(\frac{10}{7}\).

 

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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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