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In this article, we reveal 2024 HSC Maths Extension 2 Exam Solutions, complete with full explanations written by the Matrix Maths Team.
Have you seen the 2024 HSC Mathematics Extension 2 Exam Paper yet? For reference, you can find it, here.
Doing past papers? Find all the solutions to the 2017 – 2021 HSC Maths Ext 2 Exams, here.
Question | Answer | Solution |
1 | D | Non-zero vectors are perpendicular if their dot product is zero. Of the four choices, \(3\mathbf i – 2 \mathbf j + \mathbf k\) is the only valid option: \begin{align*} |
2 | C | We read this phrase left-to-right. The “\(\forall\)” symbol reads “for all,” meaning the answer must be (C) or (D). The symbol “\(\in\)” reads “is in,” so the statement \(\theta \in \left(\frac{\pi}{2}, \pi\right)\) says “theta is in \(\left(\frac{\pi}{2}, \pi\right)\)“, which is the third quadrant. The answer must then be (C). |
3 | A | Let \(P\) be the polygon in question and rewrite the statement as $$\text{Statement} : \qquad \text{\(P\) is a square} \Rightarrow \text{\(P\) is a rectangle}.$$ The converse statement reverses the direction of implication, and would be The converse statement would then read the same as (A). |
4 | B | If a polynomial with real coefficients has non-real root \(\omega\), then the conjugate \(\overline \omega\) must also be a root. It follows that not only are \(3\) and \(2 + i\) roots of \(f(x)\), but so is \(\overline{2 + i} = 2 – i\). Since \(f(x)\) is of degree three, it has at most these three distinct roots. Because \(f(x)\) is monic, this gives \begin{align*} |
5 | D | We construct a graph of the motion of this particle in the first \(10\) seconds. We’ll label the central point of motion \(c\), and label \(8\) metres to the right and left as \(c + 8\) and \(c – 8\), respectively. Since the particle starts at the central point of motion and moves left, we draw a sinusoidal curve starting at \(c\) and moving in the negative direction. From the graph, we can see that after \(7.5\) seconds, the particle ends up stationary, \(8\) metres right of the central point of motion. The answer is thus (D). |
6 | A | The total mass of the system is \(9+5=14\). Summing forces we get: \begin{align*} |
7 | C | The equation \(|z – 1 + i| = 2\) describes the locus of a circle in the complex plane, centred at \(1 – i\), with radius \(2\). Graphing this gives The point labelled \(A\) is the point on the circle farthest from \(0\), with distance \(2 + \sqrt 2\) units. |
8 | A | We can see that \(e^{\overline z} = e^{x – i y}\). Looking at (B), (C) and (D), $$\begin{matrix} None of these are equal to what we got for \(e^{\overline z}\) before, so it must be (A), \begin{align*} |
9 | B | The roots of \(z^4 = -9\) make the vertices of a square, given on the Argand diagram below Taking the product of the two roots with positive principal arguments gives |
10 | D | We know that \(|a|=|b|=|c|=1\) and \(a\cdot b=b\cdot c=0\). In particular this means that \begin{align*} Thus the angle \(\theta\) is determined by \begin{align*} If we are minimising \(\theta\) then we want to maximise \(\cos\theta\). Since \(|a\cdot c|=\big||a||c|\cos(\theta)\big|=|\cos(\theta)|\leq 1\), we are essentially trying to find the maximum of the function \( \frac{1+x}{\sqrt{3+2x}}\)over \(-1\leq x \leq 1\). By sketching or by differentiating, it can be seen that this function is increasing over \(-1\leq x \leq 1\), and so the maximum occurs at \(x=1\). |
(a).
Let \(u = x\) and \(\frac{dv}{dx} = e^x\) in integration by parts to obtain
$$\begin{matrix}
u = x & \dfrac{dv}{dx} = e^x \\
\dfrac{du}{dx} = 1 & v = e^x
\end{matrix}$$
This gives
\begin{align*}
\int x e^x \,dx
&= \int u \frac{dv}{dx} \,dx \\
&= uv – \int v \frac{du}{dx} \,dx \\
&= (x)(e^x) – \int (e^x)(1) \,dx + C_0\tag{where \(C \in \mathbb R\)} \\
&= x e^x – e^x + C
\end{align*}
(b).
(i).
\begin{align*}
z + \overline w
&= (2 + 3 i) + (1 + 5 i) \\
&= 3 + 8 i
\end{align*}
(ii).
\begin{align*}
z^2
&= (2 + 3 i)^2 \\
&= (2)^2 + 2(2)(3i) + (3i)^2 \\
&= 4 + 12 i – 9 \\
&= -5 + 12 i
\end{align*}
(c).
The dot product formula states
\(\mathbf u \cdot \mathbf v = |\mathbf u| |\mathbf v| \cos \theta\)where \(\theta\) is the angle between \(\mathbf u\) and \(\mathbf v\).
This can be solved uniquely for \(\theta\) as
\(\theta = \cos^{-1} \left(\frac{\mathbf u \cdot \mathbf v}{|\mathbf u| |\mathbf v|}\right),\)since the angle between any two vectors is at least \(0\) and at most \(\pi\) radians. This gives
\begin{align*}
\theta
&= \cos^{-1} \left(\frac{\mathbf u \cdot \mathbf v}{|\mathbf u| |\mathbf v|}\right) \\
&= \cos^{-1} \left(\frac{(1)(4) + (2)(-4) + (-2)(7)}{\sqrt{1^2 + 2^2 + (-2)^2} \sqrt{4^2 + (-4)^2 + 7^2}}\right) \\
&= \text{$2.3$ rad (1 d.p.).}
\end{align*}
(d).
Using the substitution \(t = \tan\left(\frac{\theta}{2}\right)\) gives \(t\left(\frac{\pi}{2}\right) = 1\), \(t(0) = 0\),
\(\sin \theta = \frac{2 t}{1 + t^2}\)
and, by solving \(t = \tan\left(\frac{\theta}{2}\right)\) for \(\theta\),
It follows that
\begin{align*}
\int_0^{\frac{\pi}{2}} \frac{1}{\sin(\theta) + 1} \,d\theta
&= \int_0^1 \frac{1}{\frac{2 t}{1 + t^2} + 1} \times \frac{2}{1 + t^2} \,dt \\
&= \int_0^1 \frac{2}{2 t + (1 + t^2)} \,dt \\
&= 2 \int_0^1 \frac{1}{(t + 1)^2} \,dt \\
&= 2 \left[-\frac{1}{t + 1}\right]_0^1 \\
&= -2 \left(\frac{1}{2} – 1\right) \\
&= 1
\end{align*}
(e).
(i).
If we sketch \(\sqrt 3 + i\):
We can see that \(\sqrt 3 + i\) exists in the first quadrant of the complex plane, with angle \(\tan^{-1}\left(\frac{1}{\sqrt 3}\right) = \frac{\pi}{6}\) made with the horizontal axis, at a distance of \(2\) units from the origin, giving
\(\sqrt 3 + i = 2\,e^\frac{i \pi}{6}.\)
(ii).
Using the result in the previous part,
\begin{align*}
(\sqrt 3 + i)^7
&= \left(2 \, e^{\frac{i \pi}{6}}\right)^7 \\
&= 2^7 \, e^{\frac{7 i \pi}{6}} \\
&= 128 \, e^{- \frac{5 i \pi}{6}}.
\end{align*} A quick sketch places \(e^{-5 i \pi/6}\) in the following position:
It follows that:
\begin{align*}
e^{- \frac{5 i \pi}{6}}
&= – \frac{\sqrt{3}}{2} – \frac{1}{2} i,
\end{align*}
and that:
\begin{align*}
(\sqrt{3} + i)^7
&= 128 \, e^{- \frac{5 i \pi}{6}}, \\
&= 128 \, \left(- \frac{\sqrt{3}}{2} – \frac{1}{2} i\right), \\
&= – 64 \sqrt{3} – 64 i.
\end{align*}
(f).
The region defined by \(|z| < 3\) is the circle of radius \(3\), centred at origin, not including its boundary:
The region defined by \(0 \leq \arg(z – i) \leq \frac{\pi}{2}\) is the set of all complex \(z\) for which the complex number from \(i\) to \(z\) has its principal argument in the set \([0, \frac{\pi}{2}]\). That gives the following region:
Overlapping these two regions and taking only the intersection gives the following graph:
(a).
(i).
It can be seen that
\begin{align*}
\frac{\mathbf a \cdot \mathbf b}{\mathbf b \cdot \mathbf b} \mathbf b
&= \frac{(1)(2) + (2)(0) + (3)(-4)}{2^2 + 0^2 + (-4)^2} \begin{pmatrix}2\\0\\-4\end{pmatrix} \\
&= – \frac{1}{2} \begin{pmatrix}2\\0\\-4\end{pmatrix} \\
&= \begin{pmatrix}-1\\0\\2\end{pmatrix}
\end{align*}
(ii).
From the previous part,
\(\frac{\mathbf a \cdot \mathbf b}{\mathbf b \cdot \mathbf b} \mathbf b = \begin{pmatrix}-1\\0\\2\end{pmatrix},\)
and so
Taking the dot product of this vector with \(\mathbf b\) gives
\begin{align*}
\left(\mathbf a – \frac{\mathbf a \cdot \mathbf b}{\mathbf b \cdot \mathbf b} \mathbf b\right) \cdot \mathbf b
&= \begin{pmatrix}2\\2\\1\end{pmatrix} \cdot \begin{pmatrix}2\\0\\-4\end{pmatrix} \\
&= (2)(2) + (2)(0) + (1)(-4) \\
&= 4 + 0 – 4 \\
&= 0
\end{align*}
Since the dot product of \(\mathbf a – \frac{\mathbf a \cdot \mathbf b}{\mathbf b \cdot \mathbf b} \mathbf b\) and \(\mathbf b\) is zero, and these vectors are non-zero, they are perpendicular.
(b).
Equate the integrand to the following expression
\(\frac{3x^2 + 2x + 1}{(x – 1)(x^2 + 1)} \equiv \frac{a}{x – 1} + \frac{bx + c}{x^2 + 1}\)
for some real constants \(a\), \(b\) and \(c\). Multiplying through by \((x – 1)(x^2 + 1)\) gives
Letting \(x = 1\) shows that
\begin{align*}
3(1)^2 + 2(1) + 1 &= a((1)^2 + 1) \\
2a &= 6 \\
a &= 3
\end{align*}
Expanding the equivalence instead gives
\begin{align*}
3x^2 + 2x + 1
&\equiv a x^2 + a + bx^2 – bx + cx – c \\
&\equiv (a + b)x^2 + (c – b)x + (a – c)
\end{align*}
Equating the \(x^2\) coefficients gives \(a + b = 3\), and \(b = 0\). Equating the constants gives \(a – c = 1\), and \(c = 2\). Applying this to the integral gives
\begin{align*}
\int \frac{3x^2 + 2x + 1}{(x – 1)(x^2 + 1)} \,dx
&= \int \frac{3}{x – 1} + \frac{2}{x^2 + 1} \,dx \\
&= 3 \int \frac{1}{x – 1} \,dx + 2 \int \frac{1}{x^2 + 1} \,dx \\
&= 3 \ln |x – 1| + 2 \tan^{-1}(x) + C
\end{align*}
for some real \(C\).
(c).
(i).
The left hand side is a real number, and has no imaginary component. The right hand side must also have no imaginary component.
\begin{align*}
\Im(z + 8 + 12 i) &= 0 \\
\Im(z) + 12 &= 0 \\
b &= -12 \tag{since \(\Im(z)=b\)}
\end{align*}
(ii).
Using the fact that \(b = -12\),
\begin{align*}
|z| &= z + 8 + 12i \\
|a – 12i| &= a – 12i + 8 + 12i \\
\sqrt{a^2 + 144} &= a + 8 \\
a^2 + 144 &= a^2 + 16a + 64 \\
16a &= 80 \\
a &= 5
\end{align*}
Hence, \(z = 5 – 12 i\).
(d).
First suppose that \(n\in\mathbb{Z}\) is even. Then
\begin{align*}
(n+1)^{41}-79n^{40}&=(\text{odd})^{41}-79\times(\text{even})^{40}\\
&=(\text{odd})-(\text{even})\\
&=(\text{odd})
\end{align*}
Otherwise, if \(n\) is odd then
\begin{align*}
(n+1)^{41}-79n^{40}&=(\text{even})^{41}-79\times(\text{odd})^{40}\\
&=(\text{even})-(\text{odd})\\
&=(\text{odd})
\end{align*}
Thus \((n+1)^{41}-79n^{40}\) is always odd and cannot be equal to \(2\).
(e).
(i).
One vector equation for the line \(\ell\) is
\begin{align*}
\ell(\lambda)&=\begin{pmatrix} 3 \\ 5 \\ -4 \end{pmatrix}+\lambda\left[\begin{pmatrix} 7 \\ 0 \\ 2 \end{pmatrix}-\begin{pmatrix} 3 \\ 5 \\ -4 \end{pmatrix}\right]\\ &=\begin{pmatrix} 3 \\ 5 \\ -4 \end{pmatrix}+\lambda\begin{pmatrix} 4 \\ -5 \\ 6 \end{pmatrix}
\end{align*}
(ii).
By the previous part we try to solve the following three equations for \(\lambda\):
\begin{align*}
3+4\lambda&=10\tag{1}\\
5-5\lambda&=5\tag{2}\\
-4+6\lambda&=-2\tag{3}
\end{align*}
However, solving eqution (1) gives \(\lambda=\frac{7}{4}\) while, solving equation (2) gives \(\lambda=0\). Thus the point does not lie on the line.
(a).
(i).
We simply calculate that
\((ii).
From the previous part, \(|AB|^{2}=6p^{2}-24p+125\). since this is a quadratic, the minimum can be found with the axis of symmetry formula,
\begin{align*}
p&=-\frac{b}{2a}\\
&=-\frac{24}{2\times 6}\\
&=2
\end{align*}
Thus the minimum distance is given by
\begin{align*}
|AB|&=\sqrt{6(2)^{2}-24(2)+125}\\
&=\sqrt{101}
\end{align*}
(b).
We will use the formula that relates \(x\) and \(v\) for a particle in SHM:
\begin{align*}
v^{2} &= n^{2}[A^{2} – (x – x_{0})^{2}].
\end{align*}
Plugging in \(x = 0, v = \pm 4, n = 2, x_{0} = -1\) gives:
\begin{align*}
16 &= 4[A^{2} – 1].
\end{align*}
which rearranges to give \(A=\sqrt{5}\). Thus the total distance travelled in one full period (which is four times the amplitude) is \(4\sqrt{5}\).
(c).
(i).
We start with \(a=-kv^{2}\). Using the fact that \(a=v\frac{dv}{dx}\) we get
\(\(
(ii).
Use the conditions \(x=15,v=10\) we obtain
\begin{align*}
10&=40e^{-15k}\\
k&=\frac{\ln(4)}{15}
\end{align*}
(iii).
Again, start with \(a = -kv^{2}\) but this time use the fact that \(a = \frac{dv}{dt}\). This gives:
\begin{align*}
\frac{dv}{dt} &= -kv^{2}, \\
\int\frac{1}{v^{2}} \, dv &= \int -k \, dt, \\
-\frac{1}{v} &= -kt + C.
\end{align*}
The initial conditions \(x = 0, v = 40\) imply that \(C = -\frac{1}{40}\). Then plugging in \(v = 30\), we get:
\((d).
Taking \(x=b,y=c\) in the given inequality shows that
\(Using all three of these inequalities one gets
\begin{align*}
\sqrt{bc}+b\sqrt{ac}+c\sqrt{ab}&\leq\frac{a(b+c)+b(a+c)+c(a+b)}{2}\\
&=ab+bc+ac\\
&=\frac{abc}{c} + \frac{abc}{a} + \frac{abc}{b}\\
&=abc\left(\frac{1}{c} + \frac{1}{a} + \frac{1}{b}\right)\\
&=abc \tag{given \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}=1\)}
\end{align*}
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(a).
Suppose that \(a\) is an odd integer. Then either \(a=4k+1\) or \(a=4k+3\), for some integer \(k\) (i.e. it has a remainder of either \(1\) or \(3\) when divided by \(4\)). If \(a=4k+1\) then
\begin{align*}
a^{2} – 1 &= (4k + 1)^{2} – 1, \\
&= 16k^{2} + 8k, \\
&= 8(2k^{2} + k).
\end{align*}
This is divisible by \(8\). Otherwise, if \(a = 4k + 3\), then:
\begin{align*}
a^{2} – 1 &= (4k + 3)^{2} – 1, \\
&= 16k^{2} + 24k + 8, \\
&= 8(2k^{2} + 3k + 1).
\end{align*}
which is also divisible by \(8\). Hence we are done.
(b).
We prove
\(\binom{2n}{n}<2^{2n-2}\)
for all integers \(n\geq 5\) by induction.
Base case: When \(n=5\):
\begin{align*}
\text{LHS}&=\binom{10}{5}\\
&=252.\\
\text{RHS}&=2^{2\times 5 -2}\\
&=256
\end{align*}
Hence \(LHS < RHS\). So the statement is true for \(n=5\).
Induction step: Suppose the statement is true for some integer \(k\geq5\). That is, suppose:
\begin{align*}
\binom{2k}{k}<2^{2k-2}
\end{align*}
We prove the statement for \(n=k+1\). That is, we prove:
\begin{align*}
\binom{2k+2}{k+1}&<2^{2k}\\
\end{align*}
First note that
\begin{align*}
0&\leq 1\\
2k+1&\leq 2k+2\\
2k+1&\leq 2(k+1)\\
\frac{2k+1}{k+1}&\leq 2 \tag{result 1.}
\end{align*}
Division by \(k+1\) is permitted here as \(k+1>0\) since \(k\geq 5\). Finally,
\begin{align*}
\text{LHS}&=\binom{2k+2}{k+1}\\
&=\frac{(2k+2)!}{(k+1)!(k+1)!}\\
&=2\frac{(2k+1)}{(k+1)}\times\binom{2k}{k}\\
&<2\frac{(2k+1)}{(k+1)}\times2^{2k-2}\tag{induction hypothesis}\\
&<2\times 2 \times2^{2k-2}\tag{using result 1.}\\
&=2^{2k}\\
&=\text{RHS}
\end{align*}
It follows by mathematical induction that the statement is true for all integers \(n\geq5\).
(c).
Since \(\arg(\frac{z}{w})=-\frac{\pi}{2}\) it follows that
\begin{align*}
\frac{z}{w}&=re^{-\frac{i\pi}{2}}\\
\frac{z}{w}&=-ir \tag{for some \(r>0\)}\\
z&=-irw
\end{align*}
But then
\begin{align*}
\frac{z-w}{z+w}&=\frac{w(1+ri)}{w(1-ri)}\\
&=\frac{(1+ri)^{2}}{1+r^{2}}\\
&=\frac{(1-r^{2})+(2r)i}{1+r^{2}}
\end{align*}
Hence
\begin{align*}
\bigg|\frac{z-w}{z+w}\bigg|&=\frac{\sqrt{(1-r^{2})^{2}+(2r)^{2}}}{1+r^{2}}\\
&=\frac{\sqrt{(1+r^{2})^{2}}}{1+r^{2}}\\
&=1
\end{align*}
(d).
When integration by parts is used on indefinite integrals, technically another implicit constant is added to the right hand side. So a correct argument would actually yield
\begin{align*}\int\frac{1}{x}dx=1+\int\frac{1}{x}dx+C\end{align*}
for some constant \(C\) (note that this does not prove then that \(0=1\), rather that \(C=-1\)).
(e).
(i).
One simply calculates that
\(
\begin{align*}
\overrightarrow{OR} &= \overrightarrow{OP} + \overrightarrow{PR}, \\
&= \frac{3}{5} \overrightarrow{OA} + k \overrightarrow{PQ}, \\
&= \frac{3}{5} a + k (\overrightarrow{OQ} – \overrightarrow{OP}), \\
&= \frac{3}{5} a + k \left(3 \overrightarrow{OB} – \frac{3}{5} \overrightarrow{OA}\right), \\
&= \frac{3}{5}(1 – k) a + 3k b.
\end{align*}
\)
(ii).
By what was given we have that
\(Since \(a\) and \(b\) are not parallel, we equate the coefficients to give us two equations
\begin{align*}
\frac{3}{5}(1-k)&=1-h\tag{1}\\
3k&=h \tag{2}
\end{align*}
Solving these simultaneously gives \(k=\frac{1}{6}\) (and \(h=\frac{1}{2}\)).
(iii).
In part (ii) we found \(h=\frac{1}{2}\) implying \(\overrightarrow{OR} = \left(\frac{1}{2}\right)a + \left(\frac{1}{2}\right)b
\). Write
That is,
\(
\begin{align*}
\frac{\lambda}{2}a + \frac{\lambda}{2}b = \overrightarrow{OT} = (1 – \mu)a + 3\mu b.
\end{align*}
\)
Since \(a\) and \(b\) are not parallel, we equate coefficients to give us two equations
\begin{align*}
\frac{\lambda}{2}&=1-\mu \tag{1}\\
\frac{\lambda}{2}&=3\mu \tag{2}
\end{align*}
Solving for \(\mu\) we find that \(\mu=\frac{1}{4}\) and hence
\(
(a).
(i).
The line between \(A\) and \(B\) has equation
\(\ell(\lambda)=(1-\lambda)a+\lambda b\)which gives the point \(M\) for the value \(\lambda=1/2\).
(ii).
The line between \(M\) and \(C\) has the equation
\(\ell(\lambda)=(1-\lambda)\left(\frac{a+b}{2}\right)+\lambda c\)which gives the point \(G\) for the value \(\lambda=\frac{1}{3}\). Given the values of \(M,C\) and \(G\) correspond to \(\lambda\) values \(0,1\) and \(\frac{1}{3}\) respectively, we know \(G\) must lie between \(M\) and \(C\), since \(0<\frac{1}{3}<1\).
(iii).
By the triangle inequality, \(|x+w|\leq|x|+|w|\), with equality for non-zero complex numbers occurring if and only if \(\arg(x)=\arg(w)\). Since \(|x|=|w|\) is given, having \(\arg(x)=\arg(w)\) would imply \(x=w\), contradicting their uniqueness. We conclude \(|x+w|<|x|+|w|\). It follows that
\begin{align*}
|\alpha|
&= \frac{1}{3} |x + w + z| \\
&\leq \frac{1}{3} (|x + w| + |z|) \tag{triangle inequality} \\
&< \frac{1}{3} (|x| + |w| + |z|) \tag{as we’ve shown \(|x + w| < |x| + |w|\)} \\
&= \frac{1}{3} (1 + 1 + 1) \tag{since \(|x| = |y| = |z| = 1\)} \\
&= 1
\end{align*}
That is, we have shown \(|\alpha| < 1\). Assume for a contradiction that \(\alpha=\frac{1}{3}(x+w+z)\) is a cube root of \(xwz\). That is, that \(\alpha^{3}=xwz\). Taking the modulus of both sides gives us
\begin{align*}
|\alpha^{3}|&=|xwz|\\
|\alpha|^3&=|x||w||z|\\
|\alpha|&=1 \tag{since \(|x|=|w|=|z|=1\)}
\end{align*}
This contradicts what we found earlier, in that \(|\alpha| < 1\). There is thus no way that \(\frac{1}{3} (x + w + z)\) could be a cube root of \(x w z\).
(b).
We use integration by parts with \(u=x^{n+\frac{1}{2}}, dv=(a-x)^{\frac{1}{2}}dx\). This gives \(du=(n+\frac{1}{2})x^{n-\frac{1}{2}}dx, \\ v=-\frac{2}{3}(a-x)^{\frac{3}{2}.}\)
Hence
\begin{align*}
I_{n}&=\bigg[x^{n+\frac{1}{2}}\times-\frac{2}{3}(a-x)^{\frac{3}{2}}\bigg]_{0}^{a}-\int_{0}^{a}\left(n+\frac{1}{2}\right)x^{n-\frac{1}{2}}\times-\frac{2}{3}(a-x)^{\frac{3}{2}}dx\\
&=\frac{2n+1}{3}\int_{0}^{a}x^{n-\frac{1}{2}}(a-x)^{\frac{1}{2}}(a-x)dx\\
&=\frac{2n+1}{3}\bigg[a\int_{0}^{a}x^{n-\frac{1}{2}}(a-x)^{\frac{1}{2}}dx-\int_{0}^{a}x^{n+\frac{1}{2}}(a-x)^{\frac{1}{2}}dx\bigg]\\
&=\frac{2n+1}{3}(aI_{n-1}-I_{n})
\end{align*}
That is, we have shown that \(I_{n}=\frac{2n+1}{3}(aI_{n-1}-I_{n})\). Rearranging, we have
\begin{align*}
3I_{n}&=a(2n+1)I_{n-1}-(2n+1)I_{n}\\
(2n+4)I_{n}&=a(2n+1)I_{n-1}
\end{align*} as required.
(c).
(i).
Using \(\frac{d}{dx}\left(\frac{1}{2}v^{2}\right) =a\), we have
\begin{align*}
\frac{d}{dx}\left(\frac{1}{2}v^{2}\right) &= \frac{27g}{x^{3}} – g, \\
\frac{1}{2}v^{2} &= \int\left(\frac{27g}{x^{3}} – g\right) \, dx, \\
v^2 &= 2g\bigg(-\frac{27}{2x^{2}} – x\bigg) + C.
\end{align*}
The conditions \(x = 6, v = 0\) imply:
\begin{align*}
0 &= 2g\bigg(-\frac{27}{2(6)^{2}} – 6\bigg) + C, \\
C &= \frac{51g}{4}.
\end{align*}
Hence:
\begin{align*}
v^2 &= 2g\bigg(-\frac{27}{2x^{2}} – x\bigg) + \frac{51g}{4}, \\
v^{2} &= g\bigg(\frac{51}{4} – 2x – \frac{27}{x^{2}}\bigg).
\end{align*}
Note that by construction \(x=6\) is a root of this cubic, so simple long division gives
\((x-6)(8x^{2}-3x-18)=0\)Using the quadratic formula to solve the quadratic factor one finds the other positive root to be
\(\frac{3+\sqrt{585}}{16}\approx1.7\,(1 \text{ d.p.})\)which is where the object comes to rest.
Let \(x=2\sin^{2}\theta\) so that \(vdx=4\sin\theta\cos\theta \, d\theta\). Hence
\begin{align*}
I&=\int\frac{2x^{2}}{\sqrt{x(2-x)}}\,dx\\
&=\int\frac{8\sin^{4}\theta}{\sqrt{2\sin^{2}\theta(2-2\sin^{2}\theta)}}\times4\sin\theta\cos\theta \,d\theta\\
&=16\int\sin^{4}\theta \,d\theta
\end{align*}
By using the double angle identities or else by another method, it can be shown that
\begin{align*}
\sin^{4}\theta=\frac{3}{8}-\frac{1}{2}\cos(2\theta)+\frac{1}{8}\cos(4\theta)
\end{align*}
Hence
\begin{align*}
I&=\int6-8\cos(2\theta)+2\cos(4\theta)d\theta
\\&=6\theta-4\sin(2\theta)+\frac{1}{2}\sin(4\theta)+C
\end{align*}
Substituting back in for \(x\) is a bit of a pain. Note that \(\theta=\sin^{-1}(\sqrt{\frac{x}{2}})\).
Hence
\begin{align*}
\sin(2\theta)&=2\sin\theta\cos\theta\\
&=2\left(\sqrt{\frac{x}{2}}\sqrt{1-\frac{x}{2}}\right)
\end{align*}
and
\begin{align*}
\sin(4\theta)&=2\sin(2\theta)\cos(2\theta)\\
&=4\sin\theta\cos\theta(1-2\sin^{2}\theta)\\
&=4\sqrt{\frac{x}{2}}\sqrt{1-\frac{x}{2}}(1-x)
\end{align*}
(a).
We seek a point \(P(a,b)\) on the curve \(y=\cos(kx)\) such that the vector \(\overrightarrow{OP}\) is perpendicular to the tangent to the curve at \(P\). A vector tangent to the curve at \(P\) will point in the same direction as the tangent. We use the gradient of the tangent
\begin{align*}
\frac{d}{dx}\cos(kx)=-k\sin(kx)
\end{align*}
at \(P\) to obtain the vector \(\underset{\sim}{i}-k\sin(ka)\underset{\sim}{j}\), as shown in the diagram below.
Since \(\overrightarrow{OP}\) and the tangent are perpendicular, the dot product of the two vectors is 0.
Hence,
\begin{align*}
\begin{pmatrix}1\\-k\sin(ka)\end{pmatrix} \cdot \begin{pmatrix}a\\b\end{pmatrix} &= 0, \\
a – kb\sin(ka) &= 0, \\
kb\sin(ka) &= a.
\end{align*}
Since \(P(a, b)\) lies on \(y = \cos(kx)\), we know \(b = \cos(ka)\):
\begin{align*}
k\cos(ka)\sin(ka) &= a, \\
k\sin(2ka) &= 2a \tag{double angle formula}, \\
k(2ka) &> 2a \tag{since \(x > \sin(x)\) for all \(x > 0\)}, \\
k^2 &> 1, \\
k &> 1 \tag{since \(k > 0\)}.
\end{align*}
(b).
(i).
Note that \(\gamma\) is a ninth root of unity. In particular, this means that \(\gamma^{9} = 1\) and also that \(\bar{\gamma} = \gamma^{8}\). Clearly, \(\gamma + \bar{\gamma} = 2\Re(\gamma)\) is real. Also, one has that:
\(Hence, \(\gamma + \bar{\gamma}\) is a root of \(z^3 – 3z + 1 = 0\).
(ii).
We prove:
\(\cos\left(\frac{2^n\pi}{9}\right)\cos\left(\frac{2^{n+1}\pi}{9}\right)\cos\left(\frac{2^{n+2}\pi}{9}\right) = -\frac{1}{8}\)for all integers \(n\geq 1\) by induction.
Base case: When \(n = 1\):
\begin{align*}
\gamma^3 &= e^{\frac{2\pi i}{3}}, \\
\gamma^3 &= e^{i\left(\frac{2\pi}{3} + 2k\pi\right)} \tag{for \(k \in \mathbb{Z}\)}, \\
\gamma &= e^{\frac{i}{3}\left(\frac{2\pi}{3} + 2k\pi\right)}, \\
\gamma &= e^{\left(\frac{2\pi}{9}\right)i}, \, e^{\left(\frac{8\pi}{9}\right)i}, \, e^{\left(\frac{14\pi}{9}\right)i}, \\
\gamma &= e^{\left(\frac{2\pi}{9}\right)i}, \, e^{\left(\frac{8\pi}{9}\right)i}, \, e^{-\left(\frac{4\pi}{9}\right)i}.
\end{align*}
Recall from part (i) that \(\gamma + \bar{\gamma} = 2\Re(\gamma)\) is a root of \(z^3 – 3z + 1 = 0\). Hence, the three roots of \(z^3 – 3z + 1 = 0\) are:
$$
2\Re\left(e^{\frac{2\pi i}{9}}\right), \, 2\Re\left(e^{\frac{8\pi i}{9}}\right), \text{ and } 2\Re\left(e^{-\frac{4\pi i}{9}}\right).
$$
The product of roots gives:
\begin{align*}
2\Re\left(e^{\frac{2\pi i}{9}}\right) \cdot 2\Re\left(e^{\frac{8\pi i}{9}}\right) \cdot 2\Re\left(e^{-\frac{4\pi i}{9}}\right) &= -1, \\
2\cos\left(\frac{2\pi}{9}\right) \cdot 2\cos\left(-\frac{4\pi}{9}\right) \cdot 2\cos\left(\frac{8\pi}{9}\right) &= -1, \\
8\cos\left(\frac{2\pi}{9}\right)\cos\left(\frac{4\pi}{9}\right)\cos\left(\frac{8\pi}{9}\right) &= -1, \\
\cos\left(\frac{2\pi}{9}\right)\cos\left(\frac{2^2\pi}{9}\right)\cos\left(\frac{2^3\pi}{9}\right) &= -\frac{1}{8}.
\end{align*}
So, the statement is true for \(n = 1\).
Induction step: Suppose the statement is true for some integer \(k \geq 1\). That is, suppose:
\begin{align*}
\cos\left(\frac{2^k\pi}{9}\right)\cos\left(\frac{2^{k+1}\pi}{9}\right)\cos\left(\frac{2^{k+2}\pi}{9}\right) = -\frac{1}{8}.
\end{align*}
We prove the statement for \(n = k + 1\). That is, we prove:
\begin{align*}
\cos\left(\frac{2^{k+1}\pi}{9}\right)\cos\left(\frac{2^{k+2}\pi}{9}\right)\cos\left(\frac{2^{k+3}\pi}{9}\right) = -\frac{1}{8}.
\end{align*}
LHS:
\begin{align*}
\text{LHS} &= \cos\left(\frac{2^{k+1}\pi}{9}\right)\cos\left(\frac{2^{k+2}\pi}{9}\right)\cos\left(\frac{2^{k+3}\pi}{9}\right), \\
&= \cos\left(\frac{2^{k+1}\pi}{9}\right)\cos\left(\frac{2^{k+2}\pi}{9}\right)\cos\left(\frac{2^{k+3}\pi}{9} – 2^k\pi\right), \\
&= \cos\left(\frac{2^{k+1}\pi}{9}\right)\cos\left(\frac{2^{k+2}\pi}{9}\right)\cos\left(\frac{8 \cdot 2^k\pi – 9 \cdot 2^k\pi}{9}\right), \\
&= \cos\left(\frac{2^{k+1}\pi}{9}\right)\cos\left(\frac{2^{k+2}\pi}{9}\right)\cos\left(-\frac{2^k\pi}{9}\right), \\
&= \cos\left(\frac{2^{k+1}\pi}{9}\right)\cos\left(\frac{2^{k+2}\pi}{9}\right)\cos\left(\frac{2^k\pi}{9}\right), \\
&= -\frac{1}{8} \tag{by induction hypothesis}, \\
&= \text{RHS}.
\end{align*}
It follows by mathematical induction that the statement is true for all integers \(n \geq 1\).
(c).
For particle \(A\) we have that \(a=g-kv\). Hence
\begin{align*}
\frac{dv}{dt}&=g-kv\\
\int\frac{dv}{g-kv}&=\int dt\\
-\frac{1}{k}\ln|g-kv|&=t+C.
\end{align*}
The conditions \(t=0,v=v_{0}\) imply that \(vC=-\frac{1}{k}\ln|g-kv_{0}|\) and so, after rearranging, one has that
\begin{align*}
v &= \frac{g}{k} – \bigg(\frac{g – kv_{0}}{k}\bigg)e^{-kt}.
\end{align*}
Hence:
\begin{align*}
x &= \int \frac{g}{k} – \bigg(\frac{g – kv_{0}}{k}\bigg)e^{-kt} \, dt, \\
&= \frac{g}{k}t + \bigg(\frac{g – kv_{0}}{k^{2}}\bigg)e^{-kt} + C.
\end{align*}
We will choose the starting position of each particle to be its own origin (and the direction of motion of each particle to be positive). So, we take \(x=0,t=0\) which gives \(C=-\frac{g-kv_{0}}{k^{2}}\).
Hence
\(x=\frac{g}{k}t+\bigg(\frac{g-kv_{0}}{k^{2}}\bigg)(e^{-kt}-1).\)For particle \(B\) we have \(a=-g-kv\).
Hence
\begin{align*}
\int\frac{dv}{g+kv}&=\int-1dt\\
\frac{1}{k}\ln|g+kv|&=-t+C.
\end{align*}
The conditions \(t=0,v=v_{0}\) imply that \(C=\frac{1}{k}\ln|g+kv_{0}|\) and hence after rearranging one gets
\begin{align*}
v &= -\frac{g}{k} + \bigg(\frac{g + kv_{0}}{k}\bigg)e^{-kt}.
\end{align*}
Hence:
\begin{align*}
x &= \int -\frac{g}{k} + \bigg(\frac{g + kv_{0}}{k}\bigg)e^{-kt} \, dt, \\
&= -\frac{g}{k}t – \bigg(\frac{g + kv_{0}}{k^{2}}\bigg)e^{-kt} + C.
\end{align*}
Taking \(t=0,x=0\) gives \(C=\frac{g+kv_{0}}{k^{2}}\) and so
\begin{align*} x=-\frac{g}{k}t-\bigg(\frac{g+kv_{0}}{k^{2}}\bigg)(e^{-kt}-1).
\end{align*}
Since we have chosen each particles starting point as its own origin the two particles will intersect when \(x_{A}+x_{B}=d\).
Hence
\begin{align*}
\frac{g}{k}t+\bigg(\frac{g-kv_{0}}{k^{2}}\bigg)(e^{-kt}-1)-\frac{g}{k}t-\bigg(\frac{g+kv_{0}}{k^{2}}\bigg)(e^{-kt}-1)=d.
\end{align*}
Since the linear terms cancel this equation can be solved directly to give
\begin{align*}
t=-\frac{1}{k}\ln\bigg(1-\frac{kd}{2v_{0}}\bigg).
\end{align*}
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Written by Oak Ukrit
Oak is the Head of Mathematics at Matrix Education and has been teaching for over 12 years and has been helping students at Matrix since 2016. He has 1st class honours in Aeronautical Engineering from UNSW where he taught for over 4 years while he was undertaking a PhD. When not plane spotting he enjoys landscape photography.© Matrix Education and www.matrix.edu.au, 2023. Unauthorised use and/or duplication of this material without express and written permission from this site’s author and/or owner is strictly prohibited. Excerpts and links may be used, provided that full and clear credit is given to Matrix Education and www.matrix.edu.au with appropriate and specific direction to the original content.