# 2021 HSC Maths Ext 2 Exam Paper Solutions

The 2021 HSC Maths Extension 2 Exam Paper solutions are here! Read on to see the full worked solutions written by our Head of Mathematics, Oak Ukrit and his team.

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## 2021 HSC Maths Ext 2 Exam Paper Solutions

Have you seen the 2021 HSC Mathematics Extension 2 Exam Paper yet?

In this post, we will work our way through the 2021 HSC Maths Extension 2 exam 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 2 Exam here.)

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

## Section 1. Multiple Choice

 Question Number Answer Solution 1. B Since $$\overrightarrow{AB} = \overrightarrow{CD}$$ (both right and flat along the table) and $$\overrightarrow{CQ} = \overrightarrow{DP}$$ (both diagonals across the cube upwards + rightwards + into the page) , we find that $$\overrightarrow{AB}+\overrightarrow{CQ} = \overrightarrow{CD}+\overrightarrow{DP} = \overrightarrow{CP}$$. 2. A Integrate by parts. \begin{align*} \textrm{Let } u &= x^5 \textrm { and } v’ = e^{7x}\\ u’ &= 5x^4 \textrm { and } v = \frac{1}{7}e^{7x}\\ \int x^5 e^{7x} \,dx &= vu – \int u’v \,dx\\ &= \frac{1}{7}e^{7x}x^5 – \int 5x^4\times \frac{1}{7}e^{7x} \,dx\\ &= \frac{1}{7}x^5 e^{7x} – \frac{5}{7}\int x^4 e^{7x} \,dx. \end{align*} 3. B The direction vector needs to be parallel to \begin{align*} \overrightarrow{BA} = \begin{pmatrix} 4 \\ 2 \\5 \end{pmatrix} – \begin{pmatrix} -2 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 6 \\ 0 \\4 \end{pmatrix}. \end{align*} Out of the 4 options, it is clear that the only direction vector parallel to $$\overrightarrow{BA}$$ is $$(3, 0, 2)$$, hence the solution is \begin{align*} \underset{\text{~}}{r} = \begin{pmatrix} 4 \\ 2 \\5 \end{pmatrix}+\lambda \begin{pmatrix} 3 \\ 0 \\2 \end{pmatrix} \end{align*} 4. C The contrapositive of a statement “if A then B” is “if not B then not A”. We split this statement into parts at the “if” and “then”: \begin{align*} \textrm{For all integers n, } | \textrm{ if } | \textrm{ n is a multiple of 6, } | \textrm{ then } | \textrm{ n is a multiple of 2}. \end{align*} We leave the “for all integers n” at the front, since it is context. Next, we negate the “B” part, i.e. “n is not a multiple of 2”, and also negate the “A” part, i.e. “n is not a multiple of 6”. Finally, we structure as “If Not B then Not A”, and add our context back, to get \begin{align*} \textrm{For all integers n, } | \textrm{ if } | \textrm{ n is not a multiple of 2, } | \textrm{ then } | \textrm{ n is not a multiple of 6}. \end{align*} Check: Remember, the contrapositive must be logically equivalent to the original. If you were asked to prove the original statement, you could equivalently prove “for all integers n, if n is not a multiple of 2, then n is not a multiple of 6”, which is (C). Remark: Observe that (A) is actually the negation! 5. D We can solve this by considering each of the answers: A. The function $$x^3$$ is increasing on its domain. Therefore $$ae^{-b}$$. C. The function $$\ln(x)$$ is increasing on its domain. Therefore $$a 0$$ is a small increment in time. At time $$t+\Delta t$$, the particle has moved slightly further along the curve. Since the speed is increasing, the arrow representing $$\mathbf{v}(t+\Delta t)$$ should be slightly longer than $$\mathbf{v}(t)$$.   Now we consider the difference $$\mathbf{a}(t) \Delta t \approx \mathbf{v}(t+\Delta t) – \mathbf{v}(t)$$:   The vector $$\mathbf{a(t)}\Delta t$$ is $$\mathbf{a}(t)$$ rescaled by $$\Delta t$$, but the key thing is that they should approximately have the same direction when $$\Delta t$$ is sufficiently small. Consequently we see that (D) is the best match. 9. B This is a classic logic puzzle :). It is best resolved by thinking about the contrapositive, which is always logically equivalent to the original statement: “If a card does not have WIN on one side, then the other side does not have RED”. Sam needs to check both the direct implication and the contrapositive — hence they should flip RED to see if WIN is on the other side, and then flip TRY AGAIN to make sure RED is not on the other side. 10. C If $$z = u+iv$$ and $$w = x+iy$$, then we identify $$z$$ and $$w$$ with the vectors $$z = (u,v)$$ and $$w = (x,y)$$ respectively. The projection of $$z$$ onto $$w$$ is given by \begin{align*} \text{Proj}_w(z) = \frac{z \cdot w}{|w|^2}w. \end{align*} Now observe that $$z \cdot w = ux + vy$$, and $$\text{Re}(zw) = ux – vy$$. Unfortunately this has a minus sign! However, we can easily fix this by considering $$\bar{w}$$ instead: \begin{align*} \text{Re}(z\bar{w}) = ux + vy = z \cdot w. \end{align*} Finally, notice that \begin{align*} \text{Re}\left(\frac{z}{w}\right) = \text{Re}\left(\frac{z\bar{w}}{|w|^2}\right) = \frac{1}{|w|^2}\text{Re}(z\bar{w}) = \frac{z \cdot w}{|w|^2}. \end{align*} This shows that (C) is correct.

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## Section 2. Long Response Questions

Question 11a

We have

 \begin{align*} zw &= 2e^{i\frac{\pi}{2}} \times 6e^{i\frac{\pi}{6}} \\ &=(2 \times 6) e^{i\frac{\pi}{2} + i\frac{\pi}{6}} \\&= 12 e^{i\frac{2\pi}{3}}. \end{align*}

Question 11b

It is easiest to calculate directly:

 \begin{align*} \sum_{n=1}^5 i^n &= i + i^2 + i^3 + i^4 + i^5 \\&= i – 1 – i + 1 + i \\&= i \end{align*}

Question 11c

We have:

 \begin{align*} a \cdot b &= -6 + 0 + 8 = 2\\ |a| &= \sqrt{2^2 + 0^2 + 4^2} = 2\sqrt{5}\\ |b| &= \sqrt{(-3)^2+1^2+2^2} = \sqrt{14} \end{align*}

Thus

 \begin{align*} \cos\theta = \frac{a\cdot b}{|a||b|} = \frac{2}{2\sqrt{5} \sqrt{14}} = \frac{1}{\sqrt{70}} \end{align*}

and $$\theta = \cos^{-1}(\frac{1}{\sqrt{70}}) \approx 83.1^\circ$$.

Question 11d

i) Let $$z=x+iy$$, and $$z^2 = -i$$ so that $$z$$ represents the square roots of $$-i$$.

Expanding, $$z^2 = (x^2-y^2) + 2xyi = -i$$.

Equating real and imaginary parts, $$x^2 = y^2$$ and $$2xy = -1$$.

The first equation gives $$x=\pm y$$. If $$x=y$$, then $$2xy=-1$$ has no real solutions.

Thus $$x=-y$$, and the second equation yields

 \begin{align*} -2x^2 = -1 \Rightarrow x^2 = \frac{1}{2} \Rightarrow x = \pm \frac{1}{\sqrt{2}}. \end{align*}

ii) We use the quadratic formula, with $$a=1$$, $$b=2$$, $$c=1+i$$:

 \begin{align*} z &= \frac{-2 \pm \sqrt{4 – 4(1+i)}}{2}\\ &= -1 \pm \sqrt{-i} \end{align*}

Now, from (i):

 \begin{align*} z&=-1 \pm \sqrt{-i} \\ &= -1 \pm \frac{1}{\sqrt{2}}(1-i) \end{align*}

Thus the two solutions are $$z = (-1+\tfrac{1}{\sqrt{2}}) – \tfrac{1}{\sqrt{2}}i$$ and $$z = -(1+\tfrac{1}{\sqrt{2}}) + \tfrac{1}{\sqrt{2}}i$$.

Question 11e

We start by realising the denominator, by multiplying top and bottom by $$\bar{w}$$:

 \begin{align*} \frac{\bar{z}}{w} &= \frac{\bar{z}\bar{w}}{w\bar{w}}\\ &= \frac{\bar{z}\bar{w}}{|w|^2} \\ &= \frac{(5-i)(2+4i)}{20} \\&= \frac{14+18i}{20} \\&= \frac{7}{10} + \frac{9}{10}i. \end{align*}

Question 11f

Note that the quadratic $$x^2+x+1$$ is irreducible (cannot be factorized) over $$ℝ$$, hence our partial fraction decomposition has the form

 \begin{equation*} \frac{3x^2-5}{(x-2)(x^2+x+1)} = \frac{A}{x-2} + \frac{Bx+C}{x^2+x+1}. \end{equation*}

Multiplying both sides by the denominator, we get

 \begin{equation*} A(x^2+x+1) + (Bx+C)(x-2) \equiv 3x^2-5. \end{equation*}

Setting $$x=2$$ to zero out the second term, we find that

 \begin{align*} A(2^2 + 2 + 1) + (Bx + C)(2-2) &= 3(2^2) + 5\\ 7A &= 7\\ A&=1. \end{align*}

Matching coefficients of $$x^2$$, we find $$A+B=3 \Rightarrow B=2$$. Finally, we match constants to obtain $$A-2C = -5 \Rightarrow C=3$$. Hence

 \begin{align*} \frac{3x^2-5}{(x-2)(x^2+x+1)} = \frac{1}{x-2} + \frac{2x+3}{x^2+x+1}. \end{align*}

Question 12a

Notice that $$x^2 + 2x + 2 = (x+1)^2 + 1$$, so the quadratic cannot be factored over $$ℝ$$, i.e. we cannot use partial fractions.

Instead, we notice that the derivative of the denominator is $$2x+2$$, which leaves 1 in the numerator when separated out:

 \begin{align*} \int \frac{2x+3}{x^2+2x+2} \,dx &= \int \frac{2x+2}{x^2+2x+2} \,dx + \int \frac{1}{x^2+2x+2} \,dx. \end{align*}

The first integral is a logarithm with its derivative on the numerator; the second one is an inverse tan.

 \begin{align*} &= \ln(x^2+2x+2) + \int \frac{1}{(x+1)^2+1} \,dx \\ &= \ln(x^2+2x+2) + \tan^{-1}(x+1) + C. \end{align*}

Question 12b

i) We obtain the converse by swapping the “if” and “then” statements: “if $$n$$ is even, then $$n^2$$ is even”.

ii) Assume that $$n$$ is even, so that $$n=2k$$ for some integer $$k$$. Then $$n^2 = (2k)^2 = 4k^2 = 2(2k^2)$$, hence $$n^2$$ is even.

Question 12c

Since the lines are perpendicular, their direction vectors are perpendicular, and so their dot products are 0:

 \begin{align*} \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} p \\ 3 \\ -1 \end{pmatrix} &=0 \\ p – 2&=0 \\p&=2 \end{align*}

Now we are given that the lines intersect.
Looking at the equation for the $$y$$-coordinate, we see that $$3\mu – 2 = 1 \Rightarrow \mu = 1$$.
Now the $$x$$-coordinate equation gives $$6 = \lambda – 2 \Rightarrow \lambda = 8$$.
Finally, the $$z$$-coordinate equation gives

 \begin{align*} 3+ 16 = q-1 \Rightarrow q = 20. \end{align*}

Question 12d

Firstly, we check the base case on our calculator: $$\sqrt{9!} = 602.3952\ldots$$ and $$2^9 = 512 < \sqrt{9!}$$.
Now assume the statement is true for some $$k\ge 9$$. We need to show that $$2^{k+1} < \sqrt{(k+1)!}$$. Starting with the left hand side of this inequality, we have

 \begin{align*} 2^{k+1} = 2^k \cdot 2 < 2\sqrt{k!} \end{align*}

by assumption.

Now if $$k \ge 9$$, then $$\sqrt{k+1} \ge \sqrt{10} > 2$$. Hence

 \begin{align*} 2^{k+1} < 2\sqrt{k!} < \sqrt{k+1} \sqrt{k!} = \sqrt{(k+1)!} \end{align*}

which completes the induction.

Question 12e

i) Since the diagonals of the square base bisect at $$H$$ we have that $$\overrightarrow{HA} =-\overrightarrow{HC}$$ and $$\overrightarrow{HB}=-\overrightarrow{HD}$$. We then have that

 \begin{align*} \overrightarrow{HA}+\overrightarrow{HB}+\overrightarrow{HC}+\overrightarrow{HD} &= (\overrightarrow{HA}+\overrightarrow{HC})+(\overrightarrow{HB}+\overrightarrow{HD}) \\ &= (-\overrightarrow{HC}+\overrightarrow{HC}) + (-\overrightarrow{HD}+\overrightarrow{HD}) \\ &= \mathbf{0}. \end{align*}

ii)  Starting from the given equation, we want to turn $$\overrightarrow{GA}, \overrightarrow{GB}, \overrightarrow{GC}, \overrightarrow{GD}$$ into $$\overrightarrow{GH}$$.

Consider that for each of A, B, C and D:

 \begin{align*} \overrightarrow{GA} &= \overrightarrow{GH}+\overrightarrow{HA}\\ \overrightarrow{GB} &= \overrightarrow{GH}+\overrightarrow{HB}\\ \overrightarrow{GC} &= \overrightarrow{GH}+\overrightarrow{HC}\\ \overrightarrow{GD} &= \overrightarrow{GH}+\overrightarrow{HD} \end{align*}

Performing these substitutions, we have:

 \begin{align*} \overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}+\overrightarrow{GD}+\overrightarrow{GS} &= (\overrightarrow{GH}+\overrightarrow{HA})+(\overrightarrow{GH}+\overrightarrow{HB})+(\overrightarrow{GH}+\overrightarrow{HC})+(\overrightarrow{GH}+\overrightarrow{HD})+\overrightarrow{GS}\\ &= (\overrightarrow{HA}+\overrightarrow{HB}+\overrightarrow{HC}+\overrightarrow{HD}) + 4\overrightarrow{GH} +\overrightarrow{GS} \\ &=4\overrightarrow{GH} + \overrightarrow{GS}. \end{align*}

Now, from the question, $$G$$ is such that $$\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}+\overrightarrow{GD}+\overrightarrow{GS}=\mathbf{0}$$, so we obtain

 \begin{align*} 4\overrightarrow{GH} + \overrightarrow{GS} = \mathbf{0}. \end{align*}

iii) Using vector addition, we can write $$\overrightarrow{GS} = \overrightarrow{GH}+\overrightarrow{HS}$$.

Substituting into (ii) gives

 \begin{align*} 4\overrightarrow{GH} + \overrightarrow{GS} &= \mathbf{0} \\ 4\overrightarrow{GH} + \overrightarrow{GH}+\overrightarrow{HS} &= \mathbf{0} \\ 5\overrightarrow{GH} &= -\overrightarrow{HS} \\ \overrightarrow{GH} &= -\frac{1}{5} \overrightarrow{HS} \\ \overrightarrow{HG} &= \frac{1}{5}\overrightarrow{HS} \end{align*}

so we conclude that $$\lambda = \frac{1}{5}$$.

Question 13a

Each of the 4th roots has the same magnitude, and are rotated by $$\frac{2\pi}{4} = 90\text{˚}$$. Since the magnitude of $$a+ib$$ as indicated is greater than 1, our resultant magnitude should also be greater than 1 (i.e. outside the unit circle), but smaller than the original. Additionally, we use the lines provided to draw a vector with a quarter of the modulus; then we rotate it 4 times by $$90 \text{˚}$$, giving us our solution as indicated in pink below:

Question 13b

The most straightforward substitution is $$u = \sqrt{x^2 – 9}$$. This gives:

 \begin{equation*} du = \frac{x}{\sqrt{x^2 – 9}} \,dx \Rightarrow u\,du = x\,dx. \end{equation*}

When $$x=\sqrt{10}$$ and $$\sqrt{13}$$, we have $$u=1$$ and 2 respectively. Therefore

 \begin{align*} \int_{\sqrt{10}}^{\sqrt{13}} x^3 \sqrt{x^2 – 9} \,dx &= \int_{\sqrt{10}}^{\sqrt{13}} x^2 \sqrt{x^2 – 9} \,(x\,dx) \\ &= \int_1^2 (u^2+9)u^2 \,du \\ &= \int_1^2 u^4 + 9u^2 \,du \\ &= \left[ \frac{u^5}{5} + 3u^3 \right]^2_1 \\ &= \left(\frac{32}{5}+24\right) – \left(\frac{1}{5}+3\right) \\ &= \frac{136}{5}. \end{align*}

i) We use integration by parts with $$u = (\ln x)^n$$ and $$dv = 1$$. Thus $$du = n(\ln x)^{n-1} \frac{1}{x}$$ and $$v=x$$. Using these parts, we can derive the reduction formula from the question:

 \begin{align*} I_n = \int_1^e (\ln x)^n \,dx &= \bigg[(\ln x)^n x \bigg]^e_1 – \int_1^e n(\ln x)^{n-1} \,dx \\ &= e – nI_{n-1} \qquad \text{for all } n\ge 1. \end{align*}

ii) We find $$V_B$$ first, which is found by subtracting the volume of the solid of revolution under $$y=\ln x$$ from $$x=1$$ to $$x=e$$ from the volume of the cylinder with radius $$1$$ and height $$e-1$$:

 \begin{align*} V_B &= \pi (e-1) – \pi \int_1^e (\ln x)^2 \,dx \\ &= \pi (e-1) – \pi I_2 \\ &= \pi (e-1) – \pi(e – 2I_1) \\ &= 2\pi I_1 – \pi. \end{align*}

using part (i). Using the reduction formula again and noting that $$I_0 = e-1$$, we have

 \begin{equation*} 2\pi I_1 – \pi = 2\pi(e – I_0) – \pi = 2\pi(e – (e-1)) – \pi = \pi \end{equation*}

and therefore $$V_B = \pi$$. Now $$V_A$$ can be found by subtracting the volume of half a solid sphere of radius 1 from the volume of a cylinder of height 1 and base radius 1. Hence

 \begin{equation*} V_A = \pi – \frac{1}{2}\times\frac{4}{3}\pi = \frac{\pi}{3} = \frac{1}{3}V_B. \end{equation*}

Question 13c

i) Noting that $$\ddot x = -4(x-3)$$, we identify that $$n=2$$ and the centre of the motion is $$x_0=3$$.

Using the relationship

 \begin{equation*} v^2 = n^2\left(A^2-(x-x_0)^2\right) \end{equation*}

and the fact that when $$x=0$$, $$v^2=64$$ we may solve for the amplitude of the motion through

 \begin{align*} 64 &= 4(A^2-9) \\ 16 &= A^2-9 \\ A^2 &= 25 \\ A &= 5 \qquad (A>0). \end{align*}

We have that

 \begin{equation*} x_0 – A \leq x \leq x_0 + A \Longrightarrow -2 \leq x \leq 8 \end{equation*}

and so the particle oscillates between the values $$x=-2$$ and $$x=8$$.

ii) We first model the motion of the particle, then solve for $$x=0$$.

Since the particle starts moving back towards the origin (cosine) rather than towards the endpoint (sine), we can let for the displacement-time equation be

 \begin{equation*} x = x_0 + A \cos(nt+\alpha) \end{equation*}

where $$x_0=3,A=5,n=2$$ and $$0<\alpha<\frac{\pi}{2}$$. So we have

 \begin{equation*} x = 3 + 5 \cos (4t + \alpha) \end{equation*}

We now substitute the condition that $$x(0)=5.5$$ to find $$\alpha$$:

 \begin{align*} 5.5 &= 3 + 5\cos\alpha \\ 5 \cos \alpha &= 2.5 \\ \cos \alpha &= \frac{1}{2} \\ \alpha &= \frac{\pi}{3} \end{align*}

Finally, solving for the first $$t>0$$ for which $$x=0$$:

 \begin{align*} 3 + 5 \cos\left(4t+ \frac{\pi}{3}\right) &= 0 \\ \cos\left(4t + \frac{\pi}{3}\right) &= -\frac{3}{5} \\ 4t + \frac{\pi}{3} &= \cos^{-1}\left(-\frac{3}{5}\right) \\ t &= \frac{1}{4} \left( \cos^{-1}\left( -\frac{3}{5}\right) – \frac{\pi}{3}\right) \\ &\approx 0.29 \text{ correct to 2 decimal places.} \end{align*}

Question 14a

We use the $$t$$-substitution $$t=\tan\left(\frac{x}{2}\right)$$ where $$\mathrm{d}x = \frac{2}{1+t^2} \ \mathrm{d}t$$. When $$x=0,t=0$$ and when $$x=\frac{\pi}{2}, t=1$$. Evaluating the integral we get:

 \begin{align*} \int_0^{\pi/2} \frac{1}{3+5\cos x} \ \mathrm{d}x &= \int_0^1 \frac{1}{3+5\frac{1-t^2}{1+t^2}} \cdot \frac{2}{1+t^2} \ \mathrm{d}t \\ &= \int_0^1 \frac{2}{3(1+t^2) + 5(1-t^2)} \ \mathrm{d}t \\ &= \int_0^1 \frac{1}{4-t^2} \ \mathrm{d}t \\ &= \int_0^1 \frac{1}{(2-t)(2+t)} \ \mathrm{d}t \\ &= \frac{1}{4} \int_0^1 \frac{1}{2-t} + \frac{1}{2+t} \ \mathrm{d}t \\ &= \frac{1}{4} \left[\ln \left| \frac{2+t}{2-t}\right| \right]_0^1 \\ &= \frac{\ln{3}}{4}. \end{align*}

Question 14b

i) Consider the diagram below:

Resolving the gravity force of $$5g$$ down the slope, we have that gravity contributes to a force of $$5g \cos(30^\circ)$$ down the slope. We have the component of gravity equal to $$5g \cos(30^\circ)=\frac{5\sqrt{3}}{2} g$$ and two opposing forces $$2v$$ and $$2v^2$$. Hence the resultant force down the slope is given by

 \begin{equation*} \frac{5\sqrt{3}}{2}g – 2v -2v^2. \end{equation*}

ii) For the object to slide down the slope at constant speed, we have that the acceleration is zero. Using Newton’s second law, we have that

 \begin{equation*} 5a = \frac{5\sqrt{3}}{2}g – 2v -2v^2. \end{equation*}

Setting $$a=0$$ and $$g=10$$ gives the quadratic:

 \begin{equation*} 2v^2 + 2v – 25\sqrt{3}=0 \end{equation*}

 \begin{equation*} v = \frac{-2 + \sqrt{4+200\sqrt{3}}}{4} \approx 4.2 \mathrm{ms}^{-1} \end{equation*}

where we take the positive square root as $$v$$ is directed down the slope.

Question 14c

i) Using De Moivre’s Theorem, we have that

 \begin{equation*} (\cos\theta+i\sin\theta)^5 = \cos5\theta + i \sin5\theta \end{equation*}

However by the binomial theorem, we have that

 \begin{align*} (\cos\theta+i\sin\theta)^5 &= \cos^5\theta+5\cos^4\theta (i\sin\theta) – 10\cos^3\theta \sin^2\theta -10\cos^2\theta (i\sin^3\theta) + 5 \cos\theta \sin^4\theta + i\sin^5\theta \\ &= (\cos^5\theta- 10\cos^3\theta \sin^2\theta+5 \cos\theta \sin^4\theta) + i(\ldots\text{imaginary part}\ldots). \end{align*}

Equating the real parts of both these expressions for $$(\cos\theta+i\sin\theta)^5$$ we obtain

 \begin{equation*} \cos5\theta = \cos^5\theta- 10\cos^3\theta \sin^2\theta+5 \cos\theta \sin^4\theta \end{equation*}

To express this as a polynomial in $$\cos \theta$$ only, we use the identity $$\sin^2\theta = 1- \cos^2\theta$$ to obtain

 \begin{align*} \cos5\theta &= \cos^5\theta- 10\cos^3\theta \sin^2\theta+5 \cos\theta \sin^4\theta \\ &= \cos^5\theta- 10\cos^3\theta (1-\cos^2\theta)+5 \cos\theta (1-\cos^2\theta)^2 \\ &= \cos^5\theta -10\cos^3\theta+10\cos^5\theta +5\cos\theta(1-2\cos^2\theta+\cos^4\theta) \\ &= 16\cos^5\theta – 20 \cos^3\theta +5 \cos\theta \end{align*}

as required.

ii) Note that

 $$\text{Re}(e^{i\frac{\pi}{10}}) = \text{R}e(\cos\frac{\pi}{10}+i\sin\frac{\pi}{10})=\cos\frac{\pi}{10}$$.

Let $$\theta = \frac{\pi}{10}$$; then, using (i):

 \begin{align*} 16\cos^5{\theta} – 20\cos^3{\theta} + 5\cos{\theta} &= \cos\left(\frac{5\pi}{10}\right)=0. \end{align*}

If we now let $$x=\cos{\theta}$$, we can write:

 \begin{align*} 16x^5 – 20x^3 + 5x &= 0 \\ 16x^4 – 20x^2 + 5 &= 0 \end{align*}

where we note that $$x\neq0$$, making it valid to divide both sides by $$x$$. Treating this as a quadratic equation in $$x^2$$ we obtain

 \begin{align*} x^2 &= \frac{-(-20)\pm\sqrt{(20)^2 – 4\times 16\times 5}}{2\times16} \\ &= \frac{20\pm\sqrt{80}}{32} \\ &= \frac{5\pm\sqrt{5}}{8} \\ x &= \pm \sqrt{\frac{5\pm\sqrt{5}}{8}}. \end{align*}

Now, $$\cos\frac{\pi}{10}>0$$ as it is in the first quadrant; so we have

 \begin{align*} x &= \sqrt{\frac{5\pm\sqrt{5}}{8}} = 0.951 \textrm{ or } 0.5877. \end{align*}

where 0.951 is taking $$+\sqrt{5}$$ and 0.5877 is taking $$-\sqrt{5}$$.

For the inner sign, we note that $$0<\frac{\pi}{10}<\frac{\pi}{4}$$, and so

 \begin{equation*} 0.707\approx \cos \frac{\pi}{4} < \cos \frac{\pi}{10} < \cos 0 = 1 \end{equation*}

and hence we must choose the inner $$+$$ sign to obtain $$x \approx 0.951$$ since $$0.5877$$ is too small. We can now conclude that

 \begin{equation*} \cos \frac{\pi}{10} = \sqrt{ \frac{5+\sqrt{5}}{8} }. \end{equation*}

Question 15a

i) We are given $$\sqrt{xy} \le \frac{x+y}{2}$$.

We begin by substituting $$x=ab$$ and $$y=c$$ to get the LHS:

 \begin{align*} \sqrt{xy} &\le \frac{x+y}{2}\\ \sqrt{(ab)c} &\le \frac{ab+c}{2} \end{align*}

Next, we substitute $$x=a^2$$ and $$y=b^2$$ to eliminate the ab term on our RHS:

 \begin{align*} \sqrt{a^2b^2} &\le \frac{a^2+b^2}{2}\\ ab &\le \frac{a^2+b^2}{2}. \end{align*}

Substituting, we have:

 \begin{align*} \sqrt{(ab)c} &\le \frac{ab+c}{2}\\ &\le \frac{\frac{a^2+b^2}{2}+c}{2}\\ &= \frac{a^2+b^2+2c}{4} \end{align*}

as required.

ii) The key observation is that the left hand side of the inequality in (i) is unchanged if the labels $$a,b,c$$ are swapped with each other. Hence we obtain 2 other inequalities for a total of 3:

 \begin{align*} \sqrt{abc} &\le \frac{a^2+b^2+2c}{4} \\ \sqrt{abc} &\le \frac{b^2+c^2+2a}{4} \\ \sqrt{abc} &\le \frac{a^2+c^2+2b}{4}. \end{align*}

Adding up the inequalities just obtained, we find

 \begin{align*} 3\sqrt{abc} &\le \frac{2(a^2+b^2+c^2)+2(a+b+c)}{4} \\ \sqrt{abc} &\le \frac{(a^2+b^2+c^2)+(a+b+c)}{6} \end{align*}

as required.

Question 15b

i) The question asks us to prove that every odd triangular number is a hexagonal number.

Let $$n=2k+1$$ be an odd integer, where $$k$$ is an integer $$\ge 0$$. Then

 \begin{align*} t_n &= \frac{n(n+1)}{2} \\ &= \frac{(2k+1)(2k+2)}{2} \\ &= (k+1)(2k+1). \end{align*}

Now note that the $$m$$-th hexagonal number is given by $$h_m = 2m^2 – m = m(2m-1)$$. If we set $$m = k+1$$ to match the brackets, then $$2m-1= 2(k+1)-1 = 2k+1$$, so we can write:

 \begin{align*} t_{2k+1} &= (k+1)(2k+1) \\ &= m(2m-1)\\ &= 2m^2 – m \\ &= h_m. \end{align*}

Thus the $$(2k+1)$$-th triangular number is the $$(k+1)$$-th hexagonal number; or put another way, the $$(2k-1)$$-th triangular number is the $$k$$-th hexagonal number.

ii) We note that in the previous step, we’ve mapped each hexagonal number one-to-one with each triangular number; i.e. $$h_1 = t_3$$, $$h_2 = t_5$$, $$h_m = t_{2m-1}$$.

Suppose $$t_{2k}$$, an even triangular number, was a hexagonal number $$h_n$$.

We know that triangular numbers are increasing, so $$t_{2k-1}<t_{2k}<t_{2k+1}$$.

Since each of the surrounding odd triangular numbers is a hexagonal number, then $$h_{k}<t_{2k}<h_{k+1}$$.

However, hexagonal numbers are also increasing; and only make sense for integer coefficients. Therefore, there cannot be a hexagonal number $$h_n = t_{2k}$$ such that $$h_{k}<h_n<h_{k+1}$$, so $$t_{2k} \neq h_n$$.

Question 15c

i) Using Newton’s second law we have the equation

 \begin{align*} m \ddot x &= -mg – kv^2 \ (\textrm{where } m =1)\\ \ddot x &= -(g+kv^2) \end{align*}

Next, we use the relationship $$\ddot x = \frac{dv}{dt}$$ to find $$T$$ when $$v=0$$ at the maximum height of the object, with limits $$t=0,v=u$$ and $$t=T,v=0$$:

 \begin{align*} \frac{dv}{dt} &= -(g+kv^2) \\ \int_u^0 \frac{1}{k\left(\frac{g}{k}+v^2\right)} \ \mathrm{d}v &= – \int_0^{T} \ \mathrm{d}t \\ \frac{1}{k}\sqrt{\frac{k}{g}}\left[\tan^{-1}\left(u\sqrt{\frac{k}{g}}\right)\right]_u^0 &= – T \\ \frac{1}{\sqrt{gk}}\left(0-\tan^{-1}\left(u\sqrt{\frac{k}{g}}\right)\right) &= – T \\ \end{align*}

We obtain that the time taken to reach the maximum height is

 \begin{equation*} T = \frac{1}{\sqrt{gk}} \tan^{-1}\left(u\sqrt{\frac{k}{g}}\right). \end{equation*}

ii) This time we use the relationship $$\ddot x = v \frac{\mathrm{d}v}{\mathrm{d}x}$$, where we have that $$v=0$$ at the maximum height. Using these conditions, we integrate with limits $$x=0,v=u$$ at the bottom and $$x=H, v=0$$ at the maximum height:

 \begin{align*} v \frac{\mathrm{d}v}{\mathrm{d}x} &= – (g+ kv^2) \\ \int_u^0 \frac{v}{g+kv^2} \ \mathrm{d}v &= – \int_0^H \ \mathrm{d}x \\ \frac{1}{2k} \int_u^0 \frac{2kv}{g+kv^2} \ \mathrm{d}v &= – \int_0^H \ \mathrm{d}x \\ \frac{1}{2k} \left[\ln(g+kv^2)\right]_u^0 &= – H \\ \frac{1}{2k} \ln \left( \frac{g}{g+ku^2}\right) &= -H \\ H&= \frac{1}{2k} \ln \left( \frac{g+ku^2}{g}\right). \end{align*}

We obtain the maximum height as

 \begin{equation*} H = \frac{1}{2k} \ln \left( 1 + \frac{ku^2}{g} \right). \end{equation*}

Question 15d

For $$n=2$$, observe that $$2^2 + 3^2 = 13 < 25 = 5^2$$. We claim that $$2^n + 3^n < 5^n$$ for all $$n \ge 2$$. The base case $$n=2$$ has just been verified. Now, assuming the claim to be true for some $$n \ge 2$$, we observe that

 \begin{align*} 2^{n+1} + 3^{n+1} &= 2^n \times 2 + 3^n \times 3 \\ &< 3(2^n + 3^n) \\ &< 5^n \times 3 \\ &< 5^{n+1}. \end{align*}

Thus the statement is true for $$n+1$$. Hence our claim holds true by the principle of mathematical induction. Alternatively we could use a binomial expansion, and note that

 \begin{equation*} 5^n = (2+3)^n = 2^n +3^n +\sum_{k=1}^{n-1} \binom{n}{k} 2^{n-k} 3^k > 2^n + 3^n \ \text{for all } n \geq2. \end{equation*}

Question 16a

i) Since $$\mathbf{i}, \mathbf{j}, \mathbf{k}$$ are all unit vectors, the triangle inequality implies

 \begin{equation*} 1 = |\overrightarrow{OP}|=|x\mathbf{i} + y\mathbf{j} + z\mathbf{k}| \le |x\mathbf{i}| + |y\mathbf{j}| + |z\mathbf{k}|= |x|+|y|+|z| \end{equation*}

where $$|\overrightarrow{OP}|=1$$ since $$\overrightarrow{OP}$$ is a position vector of a point on the unit sphere. From the above, we obtain that

 \begin{equation*} |x|+|y|+|z| \ge 1 \end{equation*}

ii) Recalling the geometric dot product between vectors $$\mathbf{a}$$ and $$\mathbf{b}$$ we have that

 \begin{equation*} |\mathbf{a}\cdot \mathbf{b}| = ||\mathbf{a}| |\mathbf{b}| \cos \theta | \leq |\mathbf{a}| |\mathbf{b}| \end{equation*}

since $$0\leq |\cos \theta| \leq 1$$. Substituting for $$\mathbf{a}$$ and $$\mathbf{b}$$ gives

 \begin{equation*} |a_1 b_1 + a_2 b_2 + a_3 b_3 | \leq \sqrt{a_1^2+a_2^2+a_3^2} \sqrt{b_1^2+b_2^2+b_3^2} \end{equation*}

as required.

iii) Consider the inequality in (ii) with the vectors $$\mathbf{a} = (1,1,1)$$ and $$\mathbf{b} =(|x|,|y|,|z|)$$. We immediately obtain

 \begin{equation*} |x|+|y|+|y| \leq \sqrt{1+1+1} \sqrt{|x|^2+|y|^2+|z|^2} = \sqrt{3}\sqrt{x^2+y^2+z^2} = \sqrt{3}. \end{equation*}

Remark: We note that a more general version of this question appeared in the Matrix Term 2 Workbook in 2020 :).

Question 16b

We have that the position vector is

 \begin{equation*} \underset{\text{~}}{r}(t) = \begin{pmatrix} ut\cos\theta \\ ut\sin\theta – \frac{1}{2}gt^2 \end{pmatrix} \end{equation*}

We differentiate each term with respect to time to get the velocity vector:

 \begin{equation*} \underset{\text{~}}{v}(t) = \begin{pmatrix} u\cos\theta \\ u\sin\theta -gt \end{pmatrix}. \end{equation*}

In two perpendicular vectors, as in two perpendicular lines, the gradients multiply to get -1:

 \begin{align*} \frac{u\sin\theta-gt}{u\cos\theta} \times \frac{ut\sin\theta-\frac{1}{2}gt^2}{ut\cos\theta} &= -1 \end{align*}

Multiplying both sides of the equation by $$2u^2\cos^2\theta$$ gives

 \begin{align*} (u\sin\theta-gt)(2u\sin\theta-gt)&=-2u^2\cos^2\theta \\ 2u^2\sin^2\theta-(3gu\sin\theta)t+g^2t^2 &= -2u^2(\cos^2\theta) \\ g^2t^2 – (3gu\sin\theta)t+ 2u^2\sin^2\theta + 2u^2 cos^2 \theta&= 0 \\ g^2t^2 -(3gu\sin\theta) t+2u^2 &= 0 \end{align*}

where we use the fact that $$\sin^2\theta+\cos^2\theta=1$$ in the last step. Observe that this is a quadratic in $$t$$ with discriminant

 \begin{equation*} \Delta = (-3gu\sin\theta)^2 – 4(g^2)(2u^2) = 9u^2g^2\sin^2\theta – 8u^2g^2. \end{equation*}

To get two solutions for $$t$$, we solve $$\Delta > 0$$ to obtain

 \begin{align*} 9u^2g^2\sin^2\theta – 8 u^2g^2 &> 0 \\ 9u^2g^2\sin^2\theta &> 8 u^2g^2 \\ \sin^2\theta &> \frac{8}{9} \\ \sin \theta &> \frac{2\sqrt{2}}{3}. \end{align*}

Finally, considering the particle cannot be projected at an angle more than 90 degrees (vertically upwards), we have

 \begin{equation*} \sin^{-1}\left(\frac{2\sqrt{2}}{3}\right) < \theta < \frac{\pi}{2}. \end{equation*}

Once the angle $$\theta$$ is determined, we must also check that these two solutions for $$t$$ physically make sense, i.e. that they are between $$0$$ and $$\frac{2u \sin \theta}{g}$$ (the time of flight). Solving for $$t$$ would give

 \begin{align*} t &= \frac{1}{2} \left( \frac{3u\sin\theta}{g} \pm \sqrt{ \left(\frac{u}{g}\right)^2 (9 \sin^2 \theta – 8)}\right) \\ &= \frac{u}{2g} \left(3\sin\theta \pm \sqrt{9\sin^2\theta-8}\right). \end{align*}

Both of the plus and minus variants are positive: the minus variant is positive since

</tableand so
The larger solution is less than the flight time since
Question 16cWe consider each quadrant separately.Quadrant 1: In this quadrant, $$x\geq0,y\geq0$$ and $$0\leq \text{Arg}(z) \leq \frac{\pi}{2}$$. We would then be solving the inequality:where we note that this region need only be sketched on $$0\leq x \leq \frac{\pi}{2}$$ since it is trivially always true for $$x > \frac{\pi}{2}$$.
Quadrant 2:  Since $$x<0$$ but $$\frac{\pi}{2} \leq \text{Arg}(z) \leq \pi$$ then it never be the case that $$\text{Re}(z)\geq \text{Arg}(z)$$
Quadrant 3: We have thatolving the inequality:where multiplying both sides by $$x<0$$ in the last step reverses the direction of the inequality. Noting that we only sketch this region for $$-\pi \leq x \leq -\frac{\pi}{2}$$ and then there will be no solution for $$x < -\pi$$ and all solutions for $$x>-\frac{\pi}{2}$$.
Quadrant 4: We have that $$x>0$$ but $$\text{Arg}(z)<0$$ and so all of Quadrant 4 will be part of the region.
Finally at $$z=0$$, $$\text{Arg}(z)$$ is undefined so we have an open circle at the origin.In total we have the following plot:

 $$\sqrt{9\sin^2\theta-8} < \sqrt{9\sin^2\theta} < 3 \sin\theta$$ $$t > \frac{u}{2g}(3\sin\theta-\sqrt{9\sin^2\theta}) = 0$$ \begin{align*} t &= \frac{u}{2g} \left(3\sin\theta + \sqrt{9\sin^2\theta-8}\right) \\ &< \frac{u}{2g} \left(3\sin\theta + \sqrt{9\sin^2\theta-{\color{red}{8\sin^2\theta}}}\right) \\ &= \frac{2u\sin\theta}{g}. \end{align*} \begin{align*} x &\geq \tan^{-1}\left(\frac{y}{x}\right) \\ \tan x &\geq \frac{y}{x} \\ y &\leq x \tan x \end{align*} $$\text{Arg}(z) = -\pi + \tan^{-1}\left| \frac{y}{x} \right| = -\pi + \tan^{-1}\left(\frac{y}{x}\right)$$. \begin{align*} x &\geq – \left( \pi – \tan^{-1}\left(\frac{y}{x}\right) \right) \\ \tan^{-1}\left(\frac{y}{x}\right) &\leq x + \pi \\ \frac{y}{x} &\leq \tan(x+\pi) \\ y &\geq x \tan(x+\pi), \end{align*}

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