Cambridge Extension 2 solutions available (1 Viewer)

tywebb · Mar 31, 2020

Foundation solutions for Chapter 6 are available now, i.e.,

6A Q1,2,3,4,5,6
6B Q1, 2, 3, 4, 5, 6, 7, 8
6C Q1, 2,3,4,5,6
6D Q1, 2,3
6E Q1, 2, 3
6F Q1, 2, 3, 4, 5, 6
6G Q1, 2, 3, 4, 5, 6

tywebb · Jun 11, 2020

Chapter 1 Development solutions and Chapter Review Exercise 1H solutions are now available:

1A Q10, 11, 12, 13, 14, 15, 16
1B Q5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16
1C Q6, 7, 8, 9, 10, 11, 12, 13, 14
1D Q11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22
1E Q7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21
1F Q5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18
1G Q6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18
1H Q1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 18, 19, 20, 21, 22

(1H Q17 missing)

Also added:

3C Q5d

5G Q1, 2, 3, 4, 5, 6, 7, 8, 9

(meaning that if you previously downloaded Chapters 3 and 5 Foundation, then you need to redownload them)

tywebb · Feb 24, 2023

Added:

1H Q17
2F Q17, 18, 19, 20, 21
3C Q11d
3D Q10
4A 5e
6F Q7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17
6G Q7, 8, 9, 10, 11, 12, 13, 14, 15, 16
6H Q1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29

Still missing:

3E Q3e
6G Q17

So that's it. Just 2 questions to do.

3E Q3 (whole question so you can see 3e in context)

6G Q17

This refers to results of 6G Q14

Answers to 6G Q14

So we only have to put the solutions to 3E Q3e and 6G Q17 here to get complete solutions for Cambridge Extension 2.

tywebb · Feb 24, 2023

Here is the solution to 3E Q3e which I did because it is really easy. Not sure why Cambridge skipped the solution to this one.

First we see the answers to 3E Q3d:

$e^{i\pi/2},e^{-i\pi/6},e^{-5i\pi/6}$

To put them in Cartesian form

$e^{i\pi/2}=\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}=0+i=i$

$e^{-i\pi/6}=\cos\frac{-\pi}{6}+i\sin\frac{-\pi}{6}=\frac{\sqrt3}{2}-\frac{i}{2}$

$e^{-5i\pi/6}=\cos\frac{-5\pi}{6}+i\sin\frac{-5\pi}{6}=-\frac{\sqrt3}{2}-\frac{i}{2}$

So now that only leaves 6G Q17 (somewhat harder than 3E Q3e so maybe I'll put the solution up later if nobody else does).

tywebb · Feb 24, 2023

I reckon we can also use the results of 6G Q8 to assist with 6G Q17:

Here is 6G Q17 again:

$\text{w.l.o.g. let }A=(0,0),P=(a,b), S\text{ be the focus}.$

$\text{Now we have that }\alpha=\frac{\pi}{4}+\frac{1}{2}\tan^{-1}\frac{b}{a}\text{ from 6G Q8b}$

$\text{So }\sin2\alpha=\sin\left(\frac{\pi}{2}+\tan^{-1}\frac{b}{a}\right)=\frac{a}{\sqrt{a^2+b^2}}$

$\text{ and }\cos2\alpha=\cos\left(\frac{\pi}{2}+\tan^{-1}\frac{b}{a}\right)=\frac{-b}{\sqrt{a^2+b^2}}$

$\text{whereupon }S=\left(\frac{V^2\sin2\alpha}{2g},-\frac{V^2\cos2\alpha}{2g}\right)=\left(\frac{aV^2}{2g\sqrt{a^2+b^2}},\frac{bV^2}{2g\sqrt{a^2+b^2}}\right)\text{ from 6G Q14biii }$

$\text{Now to solve 6G Q17a - geometric argument}$

$\overrightarrow{AS}=\frac{aV^2}{2g\sqrt{a^2+b^2}}\vec{i}+\frac{bV^2}{2g\sqrt{a^2+b^2}}\vec{j}\text{ and }\overrightarrow{AP}=a\vec{i}+b\vec{j}$

$\text{Now }\overrightarrow{AS}\times\overrightarrow{AP}=\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\\frac{aV^2}{2g\sqrt{a^2+b^2}}&\frac{bV^2}{2g\sqrt{a^2+b^2}}&0\\a&b&0\end{vmatrix}=\vec{0}$

$\text{and so }\overrightarrow{AS}\text{ and }\overrightarrow{AP}\text{ are collinear }$

$\text{Hence }S\text{ lies on }AP$

$\text{6G 17b - algebraic argument}$

$AP: y=\frac{b}{a}x$

$\text{For S, }\frac{b}{a}x=\frac{b}{a}\frac{aV^2}{2g\sqrt{a^2+b^2}}=\frac{bV^2}{2g\sqrt{a^2+b^2}}=y$

$\text{So }S\text{ lies on }AP$

tywebb · Feb 24, 2023

Some may say I went outside the syllabus with my proof of collinearity when I used the cross product. Clearly I don't care

.

However it can also be done like this (within the syllabus):

$\overrightarrow{AS}=k\overrightarrow{AP}\text{ where }k=\frac{V^2}{2g\sqrt{a^2+b^2}}$

$\text{Hence }\overrightarrow{AS}\text{ and }\overrightarrow{AP}\text{ are collinear and so }S\text{ lies on }AP$

Also there may be other methods without making use of Question 8 but I think that will be much harder.

Feel free to post alternative solutions to Q17 here.

However now my attention shifts to Extension 1 solutions. I think some of the Cambridge solutions for that are missing too (and some even incorrect). That is a far more tedious task than for Extension 2 because the solutions for Extension 1 are currently 1780 (yr11) + 2592 (yr12) = 4372 pages - whereas the Extension 2 is a mere (!) 1345 pages ( + 1 now that I've solved 3E Q3e and 6G Q17).

tywebb · Feb 24, 2023

But before I do that, have you ever wondered why collinear is usually spelt with two l's?

Co-linear or colinear may be more sensible.

Colinear is an accepted form in the US.

However there is a reason for the two l's in the more common spelling.

Collinear has its roots in Latin, with the prefix col- meaning "with" and līneāris, "belonging to a line."

tywebb · Feb 26, 2023

In summary so far I have put together the solutions to 3E Q3e and 6G Q17 into a single pdf here

https://www.angelfire.com/ab7/fourunit/Cambridge-e2-3EQ3e-6GQ17.pdf

EDIT: better there so guests can download too rather than the attachment I had before - also better in case I have to edit it later.

arnavtyagi0507 · Mar 24, 2023

tywebb said:
Chapter 2 Development questions and Review solutions are available now, i.e.,

2A Q7, 8, 9, 10, 11, 12, 13, 14, 15
2B Q7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19
2C Q6, 7, 8, 9, 10, 11, 12, 13, 14, 15
2D Q7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18
2E Q3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21
2F Q6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16
2G Q1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25

hey how can i get access to these?

Cambridge Extension 2 solutions available (1 Viewer)

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