A little reminder to students doing Calculus now

When $\frac{dy}{dx} = 0$, it implies we have a stationary point.

To determine the nature of the stationary point, we can do either the first derivative test or the second derivative.

The first derivative test:

Students should write the actual values of $\alpha^-, \alpha, \alpha^+$ and $\frac{dy}{dx}$ in the table.

We use this under these two situations:
1. $\frac{d^2y}{dx^2}$ is difficult to solve for, that is, $\frac{dy}{dx}$ is tough to be differentiated
2. $\frac{d^2y}{dx^2} = 0$

The second derivative test:

Other things students should take note is concavity and drawing of the derivative graph.

Vectors Question #2

If $c = |a| b + |b| a$, where $a$ , $b$ and $c$ are all non-zero vectors, show that $c$ bisects the angle between $a$ and $b$.

2016 A Level H2 Physics (9646) Paper 1 Suggested Solutions

All solutions here are SUGGESTED. Casey will hold no liability for any errors. Comments are entirely personal opinions.

1. B
2. C
3. B
4. D
5. D
6. C
7. D
8. D
9. B
10. C
11. A
12. D
13. D
14. C
15. C
16. A
17. D
18. A
19. B
20. D
21. Question 21 is a flawed question. When unpolarised light goes through a polarizer, the I is halved while the A is reduced by a factor of root 2. But based on the information Cambridge provides, the answer is C.
22. D
23. D
24. C
25. C
26. B
27. C
28. A
29. B
30. B
31. A
32. C
33. D
34. B
35. B
36. B
37. D
38. C
39. C
40. C

Note to all: Casey will not respond to most of the comments as he is busy. You may contact him by SMS at  +65 9474 5005 if you have a burning question.

Feel free to explain the answers, if you are confident. Many thanks.

Random Questions from 2016 Prelims #12

HCI/1/6

A group of boys want to set up a camping tent. They lay down a rectangular tarp OABC on the horizontal ground with OA = 3 m and AB = 1.5 m and secure the points D and E vertically above O and B respectively, such that .

Assume that the tent takes the shape as shown above with 6 triangular surfaces and a rectangular base. The point O is taken as the origin and the unit vectors i, j and k are taken to be in the direction of , and respectively.

(i) Show that the line DE can be expressed as $r = 2k+\lambda(2i+j), \lambda \in \mathbb{R}$.

(ii) Find the Cartesian equation of the plane ADE.

(iii) Determine the acute angle between the planes ADE and OABC. Hence, or otherwise, find the acute angle between the planes ADE and CDE.

Note: Question can be made harder and trickier should Origin, O be placed in the center of the base OACB.

2016 A-level H1 General Paper (8807) Paper 1 Suggested Solutions

All solutions here are SUGGESTED. Christine will hold no liability for any errors. Comments are entirely personal opinions.

Please leave a comment/ reply here if you need to discuss with anything with me. Try not to spam the number. It will all be replied by midnight. Thanks!

Question 1: ‘Any adaptation of a novel for film, television, or the theatre is never as effective as the original/.’ Discuss.

Question 2: Assess the view that tradition buildings have no future in your society.

Question 3: ‘Longer life expectancy creates more problems than benefits.’ Discuss

Question 4: Considering the money involved, should developing countries be allowed to host major sporting events?

Question 5: ‘Human need, rather than profit, should always be the main concern of scientific research.’ Discuss

Question 6: ‘Countries experiencing conflict should be left to sort their own problems.’ How far do you agree?

Question 7: How far has modern technology made it unnecessary for individuals to possess mathematical skills?

Question 8: ‘People who do the most worthwhile jobs rarely receive the best financial rewards.’ To what extent is this true of your society?

Question 9: Evaluate the claim that equality of opportunity for females is a desirable, but unrealistic goal.

Question 10: Assess the view that most natural disasters are the result of human activity.

Question 11: Is competition always desirable?

Question 12:’Everyone has an opinion, but not everyone’s opinion is of equal value.’ What is your view?

Problem Solving Tips #1

So these past months, I have been focusing on harnessing students’ abilities to interpret questions properly. Every line in a question is there for a reason and hold little pieces of information. As a student, you need to piece these information together.

Take the following question as an example:

We have a, b and c=a+2b.
Given further that M is on OC, and point A, B, and M are collinear. Find the ratio of OM:OC.

Now this question looks rather short. Many students will first start by drawing to help them see. To be honest, I was tell students that drawing out vectors is not necessary since it doesn’t yield any marks and we spend 10 minutes trying to figure out how it is supposed to look. We are better learning how to read questions.

Let’s start dissecting

M is on OC tells us that $\vec{OM} = \lambda \vec{OC}$.

A, B, and M are collinear tells us that the points are parallel with a common point. NOTE: Collinear is different from parallel. The former is a proper subset of the latter actually.
This tells us that $\vec{AB} = \mu \vec{AM}$

Students should have no trouble continuing to solve for $\lambda$ and $\mu$.

So this little exercise is to simply illustrate the importance of learning how to read questions and of course, writing it out.

2015 A-level H2 Mathematics (9740) Paper 2 Question 12 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
Let X and Y denote the mass of a apple and pear in grams respectively.
$\mathbb{E}(X_1 + \ldots X_5) = 5(300) = 1500$

$\mathrm{Var}(X_1 + \ldots X_5) = 5(20^2) = 2000$

$X_1 + \ldots X_5 \sim \mathrm{N}(1500, 2000)$

$\mathrm{P}(X_1 + \ldots X_5 > 1600) \approx 0.0127$ (3 SF)

(ii)
$\mathbb{E}[X_1 + \ldots X_5 - (Y_1 + \ldots Y_8)] = 5(300) - 8(200) = -100$

$\mathrm{Var}[X_1 + \ldots X_5 - (Y_1 + \ldots Y_8)] = 5(20^2) + 8(15^2) = 3800$

$X_1 + \ldots X_5 - (Y_1 + \ldots Y_8) \sim \mathrm{N}(-100, 3800)$

$\mathrm{P}(X_1 + \ldots X_5 - (Y_1 + \ldots Y_8) > 0) \approx 0.0524$ (3 SF)

(iii)
Let A and B denote the mass of a peeled apple and peeled pear in grams respectively.
$\mathbb{E}(A)= \mathbb{E}(0.85X) = 0.85(300)=255$

$\mathbb{E}(B)= \mathbb{E}(0.90Y) = 0.90(200)=180$

$\mathrm{Var}(A) = \mathrm{Var}(0.85X) = 0.85^2(20^2) = 289$

$\mathrm{Var}(B) = \mathrm{Var}(0.90Y) = 0.90^2(15^2) = 182.25$

$\mathbb{E}[A_1 + \ldots A_5 + B_1 + \ldots B_8)] = 5(255) + 8(180) = 2715$

$\mathrm{Var}[A_1 + \ldots A_5 + B_1 + \ldots B_8)] = 5(289) + 8(182.25) = 2903$

$A_1 + \ldots A_5 + B_1 + \ldots B_8 \sim \mathrm{N}(2715, 2903)$

$\mathrm{P}(A_1 + \ldots A_5 + B_1 + \ldots B_8 ~ \textless ~ 2750) \approx 0.742$ (3 SF)

2015 A-level H2 Mathematics (9740) Paper 2 Question 11 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
Number of ways $= \frac{8!}{2!2!} = 10080$ ways

(ii)
Number of ways $= 10080 -1 = 10079$ ways

(iii)
Number of ways $= 6! = 720$ ways

(iv)
Case 1: 2 A’s together and B’s separated
$5! \times ^6 C_2 = 1800$
Case 2: 2 B’s together and A’s separated
$5! \times ^6 C_2 = 1800$
Case 3: 2 A’s together and 2 B’s together
720

Number of ways =10080 – 1800 – 1800 – 720 = 5760 ways

2015 A-level H2 Mathematics (9740) Paper 2 Question 10 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
Graph to be inserted
(ii)
(a) $r \approx -0.9807$ (4 DP)
(b) $r \approx -0.9748$ (4 DP)
(c) $r \approx -0.9986$ (4 DP)

(iii)
Using GC, $P = -0.14686 \sqrt{h} + 34.78895$
$\therefore, P = -0.147 \sqrt{h} + 34.8$ (3 SF)

(iv)
$\therefore, P = -0.147 \sqrt{3.28h_{metres}} + 34.8$

$\therefore, P = -0.266 \sqrt{h_{metres}} + 34.8$ (3 SF)

2015 A-level H2 Mathematics (9740) Paper 2 Question 9 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
$\mathrm{P}(B|A) = \mathrm{P}(B)$ since A and B are independent.

$\therefore, \mathrm{P}(B|A) = \mathrm{P}(B) = 0.4$

(ii)
$\mathrm{P}(A' \cap B' \cap C')$

$= 1 - \mathrm{P}(C \cup B \cup A)$

$= 1 - (0.45 + 0.02 + 0.2 + 0.145)$

$= 0.185$

(iii)

If $\mathrm{P}(A' \cap B' \cap C) = 0$,

$\mathrm{P}(A' \cap B' \cap C')$

$=1 - [0.935 - (0.3 - 0.035 )]$

$= 0.33$ (maximum)

If $\mathrm{P}(A' \cap B \cap C) = 0$,

$\mathrm{P}(A' \cap B' \cap C')$

$= 1 - \mathrm{P}(A \cup B \cup C) - 0.22$

$= 0.165$ (minimum)

$0.165 \le \mathrm{P}(A' \cap B' \cap C') \le 0.33$