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2011-12-20T15:43:00.009+08:00

A level H1/H2 Maths/Physics Tuition

Making resolutions to better your grades and score As in year 2012 and beyond? Especially for those who are beginning JC studies in the year ahead, it is important to get a head start and pace yourself for the big major exams at the end of your marathon!

Or are you a student that is disillusioned, unsure of whether you could pass or score higher for your subject? Fret not! I will instill confidence and motivation in you by teaching you the right things!

Brief Profile of the Tutor
• Dynamic university grad with first class honours in Electrical Engineering from NUS
• 8 years of tutoring experience including students from RI and RJC
• Singapore’s Representative for 2001 Asian Physics Olympiad
===> Honourable Mention Award
• Excellence in A level Physics Award
===> Awarded only to top 10 scorers for A level Physics in Singapore
• Award winner of IME Poster Award
• Award winner for NUS Innovation Award
• 2 times NUS Dean’s List recipient
• 2001 A levels @ RJC with 4 As and 2 S papers (maths and physics) distinction
• 1999 O levels @ RI (E Maths -- A1, A Maths -- A1, Physics -- A1)
• Current Scholar working in an engineering company
• Trainer for Xinmin Sec Junior Physics Olympiad Team 2009, 2010 and 2011!
• Trainer for CHIJ Toa Payoh Junior Physics Olympiad Team 2011 ans 2012

My group tuition classes are currently conducted at ETON TUITION CENTRE.

Call 6385 3133 register now! Mention tutor Junwei for my classes!

Want to know more about me personally first? You can visit my homepage or read my helpful posts at Homework Forum.

I do not offer 1 to 1 tuition services at the moment as I'm working on creating educational apps and books at the moment. This is to benefit as many students as I can with my knowledge.

1 to 1 Tuition (at YCK road):

A levels:
2 hrs each session
Fee: $110 per hour (Sundays only)

A lvl H2 Maths: Permutations Combinations

2010-08-28T02:02:00.000+08:00

From http://www.sgforums.com/forums/2297/topics/409229

a) There is a committee of 5 boys and 3 girls. At a meeting, 2 boys are absent. Find the number of ways they can sit at a round table with 8 seats.

b) A deck of 10 cards is numbered from 0 to 9. Find the number of ways 5 cards can be chosen such that the sum of the chosen cards is larger than those not chosen.

*************************

Answer:

a) Factorial 8 different seats, but 2 are "repeated" as empty seats
Hence, the answer is

(8 - 1)! / 2! = 2520

b) There are ¹⁰C₅ ways of choosing 5 cards without restriction

The 5 cards chosen is either bigger than or lesser than the cards not chosen.

The probability that the chosen cards are greater than the "unchosen" cards is equal to the probability that the chosen cards are lesser than the "unchosen" cards

Hence we can just take ¹⁰C₅ divide by 2 = 126

A lvl H2 Maths: Vectors

2010-08-12T09:22:00.002+08:00

Question from http://www.sgforums.com/forums/2297/topics/407847

Relative to an origin O, the position vectors of points A and B are a and b respectively. Givent that angle AOB is 90deg, show that the position vector of the foot of the perpendicular from O to AB is

a + (|a|² / (|a|² + |b|²)) (b - a)

*************************

Answer:

Using the concept of vector projection, (i.e. AF = |AF| * unit vector AB),

OF = OA + AF = a + |a| cos θ * (b - a) / |b - a|, where θ = angle between OA and BA
= a + |a| [ {a.(a - b)} / {|a| |a - b| }] * (b - a) / |b - a|
= a + {a.a - a.b} / (|a|² + |b|²) * (b - a)
= a + |a|² / (|a|² + |b|²) * (b - a) since a.a = |a|² and a.b = 0

(shown)

A lvl H2 Maths: Applications of Differentiation (Rate of Change)

2010-08-10T23:14:00.002+08:00

Question from some tutorial:
A light shines from the top of a lamp post 15 m high. A ball is dropped from the same height from a point 9 m away from the light. It is known that the balls falls a distance s = 4.9t² m in t seconds. Find the speed of the shadow of the ball on the ground 0.5 second later.

*************************

Answer:

Let x be the distance of the shadow from the lamp post.

by similar triangles.
Differentiate implicitly with respect to t

Also,

When t = 0.5, s = 1.225, x = 110.20408, ds/dt = 4.9

Sub into equation

Hence, dx/dt = -440.8 ms^-1

Speed of shadow = 440.8 ms^-1

A lvl H2 Maths: Differential Equations

2010-07-07T09:07:00.000+08:00

Question from http://www.sgforums.com/forums/2297/topics/404564

Find the particular solution for
(4x+xy²)dx +(y+yx²)dy=0 , given that y(1)=2.

*A poly maths question, but nevertheless good for A levels as well!

*************************

Answer:
(4x+xy²) dx = -(y+yx²) dy

ln (4 + y²) = -ln (1 + x²) + C, where C = 2c
ln (4 + y²) + ln (1 + x²) = C
(4 + y²) (1 + x²) = A, where A = e^c

When x = 1, y = 2
Hence,
A = (4 + 4) * (1 + 1) = 16

Thus,
(4 + y²) (1 + x²) =16

O lvl A Maths: Trigonometric Identity

2010-06-24T16:28:00.001+08:00

From http://forum.channelnewsasia.com/viewtopic.php?p=3756239

Given cosec A + cot A = 3, evaluate cosec A - cot A and cos A.

*************************

Answer:

cosec A + cot A = 3
(cosec A + cot A) (cosec A - cot A)= 3 (cosec A - cot A)

cosec² A - cot² A = 3 (cosec A - cot A)
Since 1 + cot² A = cosec² A.....

3 (cosec A - cot A) = 1
cosec A - cot A = 1/3

cosec A + cot A = 3 ----- (1)
cosec A - cot A = 1/3 ----- (2)

Equation (1) + equation (2):
2 cosec A = 10/3
cosec A = 5/3
sin A = 3/5

Equation (1) - equation (2):
2 cot A = 8/3
cot A = 4/3

cos A = sin A * cot A
cos A = 4/5

O lvl A Maths: Quadratic Equations

2010-06-01T14:29:00.000+08:00

Question from http://www.sgforums.com/forums/2297/topics/401082

Given that y = (x² + 2x - c) / (x - 1) and x is real, find the range of values of c for which y can take all real values.

*************************

Answer:

y = (x² + 2x - c) / (x - 1)

y (x - 1 ) = x² + 2x - c

yx - y = x² + 2x - c

x² + 2x - yx - c + y = 0

x² + (2 - y) x + (y - c) = 0

Since x is real, discriminant ≥ 0

(2 - y)² - 4 (1) (y - c) ≥ 0

4 - 4y + y² - 4y + 4c ≥ 0

y² - 8y + (4c + 4) ≥ 0

Since the equation needs to be ≥ 0, it means it has either equal roots or no real roots

so (-8)² - 4 (1) (4c + 4) ≤ 0
64 - 16c - 16 ≤ 0
48 - 16c ≤ 0
48 ≤ 16c
c ≥ 3 (ans)

Graph example for clarity:

Plotted with
(i) c = 2
(ii) c = 4
(iii) c = 3

A lvl H2 Maths: Inequalities

2010-03-25T01:21:00.003+08:00

Solve |2 - 3x| > 2 + 3x - x²

*************************

Answer:

Using GC,

For |2 - 3x| > 2 + 3x - x²,

x < 0 and x > 2

A lvl H2 Maths: APGP

2010-03-20T23:31:00.000+08:00

AJC Prelims 2000

The n^th term of an infinite geometric series is equal to one fourth of the sum of all the terms after ( but not including ) the n^th term. Show that the sum to infinity of the geometric series is five times its first term.

*************************

Answer:

T_n = 1/4 (S_∞ - S_n)

4 - 4r = r
5r = 4
r = 4/5

Hence, S_∞ = a / (1 - r) = 5a (shown)

A lvl H2 Maths: APGP

2010-03-20T23:22:00.000+08:00

AJC Prelims 1999

An arithmetic sequence has first term a and common difference d. It is given that the sum of the first four terms is less than the sum of the next four terms by 8. Also, the first, third and sixth term of the sequence are three consecutive terms of a geometric progression. Find the exact values of a and d.

*************************

Answer:
Given first term a and common difference d

S₄ < S₈ - S₄
(4/2) (2a + 3d) < (8/2) (2a + 7d) - (4/2) (2a + 3d)
4 (2a + 3d) < 4 (2a + 7d)
4d > 0
d > 0 ------- (1)

T₁, T₃, T₆ are in G.P.
(T₃)² = T₁ * T₆
(a + 2d)² = (a)(a + 5d)
4d² - ad = 0
d (4d - a) = 0
d = 0 (reject since d > 0) or d = a/4

Choose a = 4 and d = 1

A lvl H2 Physics: Measurements

2010-03-01T10:32:00.000+08:00

From http://www.sgforums.com/forums/2297/topics/392121

To find the density of an aluminium sheet, the mass, length, and breadth were determined within 1%. Its mean thickness was found by folding it twice to give a stack of 4 sheets, the thickness of which was measured to be (11.2 +/- 0.4)mm and the zero reading of the screw gauge was recorded as (1.2+/-0.4)mm. Calculate the percentage uncertainty of its density.

*************************

Answer:

let t = thickness and zero error = z, length = l, breadth = b, mass = m, density = D

D = m / (0.25(t-z) * l * b)

ΔD / D = Δm / m + Δl / l + Δb / b + Δ(t-z) / (t-z)
ΔD / D = Δm / m + Δl / l + Δb / b + [Δt + Δz)]/ (t-z) ===> important part: Δ(t-z) = Δt + Δz

Sub in values

ΔD / D = 0.01 + 0.01 + 0.01 + 0.8/10
ΔD / D = 0.11

Thus, percentage uncertainty = 11%

A lvl H2 Maths: Integration

2010-02-23T01:25:00.002+08:00

Prove that cosec^-1 x = sin^-1 (1/x).
Hence find, for x > 0,

*************************

Answer:

Let y = cosec^-1 x
Thus,
cosec y = x
sin y = 1/x
y = sin^-1 (1/x)

Since y = cosec^-1 x, therefore
cosec^-1 x = sin^-1 (1/x)(proved)

O lvl A Maths: Trigonometric Identity

2010-02-18T00:19:00.002+08:00

a) Prove that (tan A + cot A)(sin A + cos A) ≡ 1/cos A + 1/sin A

b) Prove that sec A ≡ sin 2A / sin A - cos 2A / cos A

c) Prove that cot² A - cos² A ≡ cot² A cos² A

*************************

Answer:

a) LHS = (tan A + cot A)(sin A + cos A)
= tan A sin A + tan A cos A + cot A sin A + cot A cos A
= (sin A / cos A) sin A + (sin A / ~~cos A~~) ~~cos A~~ + (cos A / ~~sin A~~) ~~sin A~~ + (cos A / sin A) cos A
= (sin² A / cos A) + sin A + cos A + (cos² A / sin A)
= (sin² A / cos A) + (sin² A / sin A) + (cos² A / cos A) + (cos² A / sin A)
= (sin² A + cos² A) / cos A + (sin² A + cos² A) / sin A
= 1 / cos A + 1 / sin A
= RHS (proved)

b) RHS = sin 2A / sin A - cos 2A / cos A
= (2 sin A cos A) / sin A - (2 cos² A - 1) / cos A
= 2 cos A- 2 cos A + 1 / cos A
= sec A
= LHS (proved)

c) LHS = cot² A - cos² A
= cos² A / sin² A - (cos² A sin² A) / sin² A
= { cos² A ( 1 - sin² A) } / sin² A
= cos² A (cos² A / sin² A)
= cos² A cot² A
= RHS (proved)

O lvl A Maths: Indices and Logarithms

2010-02-14T02:18:00.002+08:00

Given that log_p(x²y) = 8 and log_p(y² /x) = 6, evaluate

(i) log_p(xy)
(ii) log_p(y / x)

if y/x = 9, find the value of p, of x and of y.

*************************

Answer:

log_p(x²y) = 8
===> log_px² + log_py = 8
===> 2log_px + log_py = 8 ------------------------- (1)

log_p(y² /x) = 6
===> log_py² - log_p x = 6
===> 2log_py - log_p x = 6 ------------------------- (2)

(1) + 2 * (2):
log_py + 4 log_py = 8 + 2(6)
5 log_py = 2o
log_py = 4 ------------------------------------------------------- (3)

Sub (3) into (1)
2 log_px + 4 = 8
log_px = 2 ------------------------------------------------------- (4)

(i) log_p(xy)
= log_px + log_py
= 4 + 2
= 6

(ii) log_p(y/x)
= log_py - log_px
= 4 - 2
= 2

If y/x = 9, from (ii)
log_p9 = 2
p² = 9
p = 3 or -3 (rej for log to be defined)
Thus, p = 3.

Thus, log_px = 2
x = p²
x = 9

log_py = 4
y = p⁴
y = 81

A lvl H2 Maths: Probability

2010-02-14T01:42:00.006+08:00

The car wash at a particular garage has a choice of 4 different programmes. A driver chooses a programme in the garage and then, proceeds to the car wash.
The probability of any driver choosing a particular programe is

Programme	Probability	Time taken (in mins)
A (Wash)	0.25	2
B (Wash and Dry)	0.5	3
C (Superwash)	1/6	3
D (Superwash and Dry)	1/12	4

At a particular instant, there are 3 cars queueing to use the car wash. Find

(a) the probability that the drivers of the first and second cars have chosen the same programme
(b) the probability that exactly two of the three drivers have chosen a programme which includes a dry
(c) the probability that at least one of three drivers have chosen programme C

A fourth car X joines the queue at the instant that the first car enters the car wash. Assuming that no time is lost between the time one car leaves the car wash and the programme for the next car beginning, find
(d) the probability that it will be at least 10 minutes before car X enters the car wash.
(e) the probability that it will be at least 10 minutes before car X enters the car wash, given that at least one of the previous three drivers chose programme C.

*************************

Answer:

(a) Probability that drivers of first and second car chose the same programe
= P(A, A) + P(B, B) + P(C, C) + P(D, D)
= (1/4 * 1/4) + (1/2 * 1/2) + (1/6 * 1/6) + (1/12 * 1/12)
= 1/16 + 1/4 + 1/36 + 1/144
= 25 / 72

(b) P(includes dry) = P(B) + P(D) = 1/2 + 1/12 = 7/12
P (does not include dry) = P(A) + P(C) = 1/4 + 1/6 = 5/12

Probability that exactly two of the three drivers have chosen a programme which includes a dry
= 7/12 * 7/12 * 5/12 * ³C₂ ==> (choose 2 drivers out of 3 to have included dry)
= 245/576

(c) Probability that at least one of three drivers have chosen programme C
= 1 - Probability that none of three drivers have chosen programme C
= 1 - (5/6)³
= 91/216

(d) Probability that it will be at least 10 minutes before car X enters the car wash
= P (4 min, 4 min, 4 min) + P (4 min, 3 min, 3 min) + P (4 min, 4 min, 2 min) + P (4min, 4min, 3 min)
= (1/12)³ + ³C₂ * 1/12 * 2/3 * 2/3 + ³C₂ * 1/12 * 1/12 * 1/4 + ³C₂ * 1/12 * 1/12 * 2/3
= 1/1728 + 1/9 + 1/192 + 1/72
= 113/864

(e) P (at least 10 minutes before car X enters the car wash | at least one of the previous three drivers chose programme C)
= P (at least 10 minutes before car X enters the car wash and at least one of the previous three drivers chose programme C) / P(at least one of the previous three drivers chose programme C)
= { P (C, C, D) + P (C, D, D) + P (B, C, D) } / {91/216}
= { (³C₂ * 1/6 * 1/6 * 1/12) + (³C₂ * 1/6 * 1/12 * 1/12) + ( 3! * 1/2 * 1/6 * 1/12) } / {91/216}
= {1/144 + 1/288 + 1/24} / {91/216}
= 45 / 364

How to be an effective student

2010-02-12T23:27:00.002+08:00

The list below summarises the essential skills required to be an effective student. You can learn to be an effective and successful students following the list, but whether you will become one eventually will depend on whether you are prepared to make a sustained effort to do so.

1) Take responsibility. No ones owes you a living. Neither do you owe others. Losers make excuses for themselves; winners make their own successes. Don't blame others for your own mistakes.

2) Be organised. Plan and prioritise your time well. Work with weekly timetables and daily To-Do lists with a filofax. God is fair; everyone only has 24 hours a day, nothing more.

3) Be independent. Think and do things by yourself instead of asking others to do it for you.

4) Become an efficient learner. Effective study skills are important for life-long learning. Examples include mind-mapping, speed reading, etc.

5) Be proactive. Success is more likely to look for those who take the initiative to go and get it.

6) Be disciplined. It is very easy to lapse into compromises and postponed schedules. But remember, tomorrow never comes. Remind yourself constantly that you are committed to certain goals and will attain them, whatever effort it takes you. Focus!

7) Start well. Pitch in immediately. The first few weeks are critical. Having a good start will give you the momentum to carry you further.

8) Concentrate. Don't waste time on unfocussed efforts. You won't go anywhere far without concentrating; you will just fall off.

9) Stay healthy. Don't neglect recreation, family and friends. Plan for your social life so that they aren't disruptive to your studies.

Teachers' remarks that changed the history of physics

2010-02-08T01:17:00.001+08:00

Archimedes, you are late again. Don't tell me that you were locked again in the bathroom.

Copernicus, when will you understand that you are not the center of the world?

Galileo, if you will drop stones from the top of the tower one more time, you will be dismissed forever.

Kepler, till when will you stare at the sky?

Newton, will you please stop idling away under the apple tree?

Ohm, must you resist Ampere's opinions on current events?

Nikola Tesla, I see that everyone is attracted to your magnetic personality.

Einstein, a crocodile is greener or is it wider?

Schrödinger, stop abusing cats!

Heisenberg, when will you be sure of yourself?

O lvl A Maths: Indices and Logarithms

2010-02-08T01:05:00.002+08:00

Solve the equation log₃ 2 + log₃(x + 4) = 2 log₃ x.

*************************

Answer:

log₃ 2 + log₃(x + 4) = 2 log₃ x
log₃ 2 (x + 4) = log₃ x²
2 (x + 4) = x²
x² - 2x - 8 = 0
(x - 4)(x + 2) = 0
x = 4 or -2

For log₃ x to be defined, x = 4.

Note: x = -2 is rejected because log₃ x will be undefined.

O lvl A Maths: Indices and Logarithms

2010-02-08T00:47:00.005+08:00

(i) By using the substitution y = 2^x, find the value of x such that

4^x+1 = 2 - 7(2^x).

(ii) Express 3^x(2^2x) = 7(5^x) in the form a^x = b. Hence find x.

*************************

Answer:

(i) Since 4^x+1 = 4(4^x) = 4(2²)^x = 4(2^x)², the equation becomes

4(2^x)² = 2 - 7(2^x)

Substituting 2^x = y,
4y² = 2 - 7y
4y² +7y - 2 = -
(4y - 1)(y + 2) = 0
y = 0.25 or y = -2
2^x = 2^-2 or 2^x = -2 (N.A. as no solution)
Hence, x = -2

(ii) Since 3^x(2^2x) = 3^x(2²)^x = 12^x,
12^x = 7(5^x)
(12 / 5)^x = 7

Taking log on both sides:

x lg (12 / 5) = lg 7
x = lg 7 / lg (12/5)
x = 2.22

A lvl H2 Maths: Permutations Combinations

2010-01-13T00:41:00.002+08:00

(i) In how many ways can 5 different books be distributed among 10 people if each person can get any number of books?

(ii) How many 4 digit numbers greater than 2500 can be formed from digits 0,1,2,3 and 4 if no digit can be used more than once?

*************************

Answer:

(i) View it as each book has a choice of which person to go to.

So each book has 10 choices since each person can get any number of books
Hence, number of ways = 10⁵ = 100000

(ii) First digit must be 3 or 4, i.e. 2 choices
The next 3 digits can be any number other than the first digit.

So 2 * 4 * 3 * 2 = 48

A lvl H2 Maths: Permutations Combinations

2010-01-02T09:04:00.003+08:00

RJC 2001 Term III Common Test Q1

Four-digit numbers are to be formed from the digits {1, 2, 3, 4, 5, 6}. Find the number of ways this can be done if
i) no digit may appear more than once; [1]
ii) any digit may be repeated; [1]
iii) there is at least one repeated digit, but no digit may appear more than twice. [4]

*************************

Answer:

i) ⁶P₄ = 360

ii) 6*6*6*6 = 1296

iii) Case only 1 digit repeated: (⁶C₁ * ⁵C₂) * (4! / 2!) = 720
Case 2 repeated digits: ⁶C₂ * (4!/2!2!) = 90

Summing up:
720 + 90 = 810

O lvl A Maths: Roots of Equation

2009-11-21T18:57:00.002+08:00

Question from http://www.sgforums.com/forums/2297/topics/345695

Find the range of values of p for which the line y - x = 2 meets the curve y² + (x+p)² = 2

*************************

Answer:

y = 2 + x -----------------------(1)
y² + (x+p)² = 2 -----------------(2)

Sub (1) into (2)
2x² + (4 + 2p) x + (2 + p²) = 0

Discriminant ≥ 0
i.e. b² - 4ac ≥ 0
16 +16p + 4p² - 16 - 8p² ≥ 0
16p - 4p² ≥ 0
4p (4 - p) ≥ 0

Draw a n-shape curve with roots 0 and 4 and shade the above since coefficient of p² ≥ 0

Thus,
0 ≤ p ≤ 4

O lvl A Maths: Integration

2009-11-18T23:11:00.003+08:00

Question from http://sgforums.com/forums/2297/topics/345581

Find d/dx (sin³ 2x) and hence evaluate .

*************************

Answer:

d/dx (sin³ 2x)
= (3sin² 2x)(2cos 2x)
= 6(sin² 2x)(cos 2x)

Hence,

= (0.5 - 0) - (1/6 - 0)
=1/3

O lvl E Maths: Probability

2009-11-18T22:59:00.004+08:00

Question from http://sgforums.com/forums/2297/topics/345310

In a city, 20% of the citizens are carriers of a certain disease.If a citizen is a carrier, the probabbility that a blood test will indicate a positive result is 0.95. If citizen is a non-carrier, the probability that the blood test will indicate a postive result is 0.10. Two citizens are selected at random for the blood test. Find the probability that

(a) both citizens have positive results
(b) both citizens have negative results

*************************

Answer:

(a) P(positive results) = 0.2 * 0.95 + 0.8 * 0.1 = 0.27
P(both citizens have positive results)
= P(positive results) * P(positive results)
= 0.27 * 0.27
= 0.0729

(b) P(negative results) = 0.2 * 0.05 + 0.8 * 0.9 = 0.73
P(both citizens have negative results)
= P(negative results) * P(negative results)
= 0.73 * 0.73
= 0.5329

O lvl A Maths: Trigonometry

2009-11-17T22:16:00.003+08:00

Question from http://sgforums.com/forums/2297/topics/344664

It is given that x and y are acute angles such that sin(x-y) = 4/5 and sin x cos y = 9/10.
Show that

a)
(i) sin(x+y) = 1
(ii) tan x = 9 tan y

b) using the answers from part (a), find the value of tan x.

*************************

Answer:

sin(x-y) = 4/5
sin x cos y - cos x sin y = 4/5
9/10 - cos x sin y = 4/5
cos x sin y = 1/10

a)
(i) sin (x + y)
= sin x cos y + cos x sin y
= 9/10 + 1/10
= 1 (shown)

(ii) (sin x cos y) / (cos x sin y) = (9/10) / (1/10)
(sin x / cos x) / (sin y / cos y) = 9
tan x / tan y = 9
tan x = 9 tan y (shown)

b) From sin (x+y) = 1, x + y = 90 degrees since x and y are acute.
Hence, tan (x+y) = infinity

tan (x+y) = (tan x + tan y) / (1 - tan x tan y)

===> tan x tan y = 1
===> tan y = 1 / tan x

Using a)(ii), tan x = 9 tan y
Hence,

tan x = 9 / tan x
tan² x = 9
tan x = sqrt (9)
tan x = 3 (since x is acute, tan is positive)

O lvl E Maths: Probability

2009-11-17T22:13:00.001+08:00

Question from http://sgforums.com/forums/2297/topics/344458

The probabilities that Alif, Kannan and Caili will pass an art examination are 4/5, 5/7, and 2/3 respectively. Find the probability that

(a) at least 1 of them will pass the art examination

(b) Only Alif will pass the art examination

*************************

Answer:

a) P(all fail) = 1/5 * 2/7 * 1/3 = 2/105

P(at least one pass) = 1 - 2/105 = 103/105

b) P(only Alif will pass the art examination) = 4/5 * 2/7 * 1/3 = 8/105

O lvl A Maths: Plane Geometry

2009-07-29T01:30:00.004+08:00

Question from http://www.sgforums.com/forums/2297/topics/368889

In the diagram, XTPY is a tangent to 2 circles at P and ABTC is a tangent to the circle of centre O at C.
The line AP is a chord of one circle and produced to meet the other circle at Q.

(i) Explain why a circle passes through O,C, T and P

*************************

Answer:

Let angle COP = 2x

Hence angle CQP=x (angle at centre = 2 angle at circumference)

angle PCB = angle CQP =x (alt. segment theorem)

Also, angle CPT= angle CQP=x (alt. segment theorem)

Hence, angle CTP= 180 - 2x (sum of angles in triangle)

Since angle COP + angle CTP = 180, quadrilateral CTPO is a cyclic quad, at which a circle will pass through all of the points. (shown)

Thanks to Leekeewei at sgforums for providing the solution.

O lvl A Maths: Plane Geometry

2009-07-25T01:29:00.004+08:00

Question from http://www.sgforums.com/forums/2297/topics/368877

In the diagram, P is any point on the semicircle centre O, and PQ is perpendicular to AB. The inscribed circle centre C touches PQ, AB and the semicircle at D, E and F respectively. Prove that

(a) A,D and F lie on the same straight line

(b) AD * AF = AQ * AB

(c) AE² = AQ² + AQ * QB

(d) AE = AP

*************************

Answer:

(a) DC is perpendicular to PQ, hence DC is parallel to AO.
OCF is a straight line.

=> angle DCF = angle AOF
=> FDC and FAO are similar triangles.

Hence, A, D and F lies on the same straight line.

(b) Draw a straight line from F to B.

angle AFB = 90 degrees = angle AQD
angle DAQ = angle BAF

=> angle ADQ = angle ABF
=> ADQ and ABF are similar triangles.
=> AD / AB = AQ / AF

AD X AF = AQ X AB (proved)

(c)
AE² = AF * AD( tangent-secant theorem)

From part (b) AD X AF = AQ X AB
AE² = AQ * AB
AE² = AQ ( AQ + AB)
AE² = AQ² + AQ * AB (proven)

(d) Draw a full circle and extend a line to form a chord

AD = AQ * QD
PQ = QZ ( radius from centre bisect chord)
PQ * QZ = PQ²

Therefore, PQ² = AQ * QB ( Intersecting chords theorem)

From part (c) : AQ² = AE² - (AQ)(QB)
AP² = AQ² + PQ²

Therefore, AP² = [AE² - (AQ)(QB)] + [(AQ)(QB)] ( Replaced found eqn)
AP² = AE²
AP = AE (proved)

Thanks to forumers at sgforums for providing the question and answer :D

O lvl A Maths: Surds

2009-06-07T22:24:00.003+08:00

Question from http://www.sgforums.com/forums/2297/topics/344321

A rectangular block has a square base. The length of each side of the base is (√5 - √3) m and the volume of the block is ( 4√3 - 2√5). Find , without using a calculator, the height of the block in the form of a√3 + b√5.

*************************

Answer:

Volume of rectangular block = length * length * height
4√3 - 2√5 = (√5 - √3) * (√5 - √3) * height
4√3 - 2√5 = (5 - 2√15 + 3) * height
4√3 - 2√5 = (8 - 2√15) * height

Thus,
height = (4√3 - 2√5) / (8 - 2√15)
height = (2√3 - √5) / (4 - √15) =====> divide by 2 top and bottom. Easier to work with smaller numbers
height = {(2√3 - √5)(4 + √15)} / {(4 - √15)(4 + √15)} ===> Rationalisation
height = (8√3 - 4√5 + 6√5 - 5√3) / (16 - 15)
height = 3√3 + 2√5 (ans)

A lvl H2 Maths: Inequalities

2009-06-07T22:11:00.005+08:00

Question from http://www.sgforums.com/forums/2297/topics/343748

Find the set of values of x for which (e^x²) / (x-1) - 8x > 0.
Hence, solve (e^x²) / ( |x| - 1) - |8x| > 0.

*************************

Answer:

Plot

(i) y = (e^x²) / (x-1)
(ii) y = 8x

For (e^x²) / (x-1) > 8x,
-1.9587 <> 0.9999

(e^x²) / ( |x| - 1) - |8x| > 0
(e^|x|²) / ( |x| - 1) - 8|x| > 0
Sub x = |x|
-1.9587 < |x| < -0.1138 (reject) , |x| > 0.9999

O lvl A Maths: Applications of Differentiation (Tangents Normals)

2009-06-07T22:02:00.003+08:00

Question from http://www.sgforums.com/forums/2297/topics/343229

A curve has the equation y = 3x³ - 5x + 4

(a) Find the equation of the tangent to the curve at A(1,2)

(b) Find the coordinates of the other point on the curve at which the tangent is parallel to the tangent at A.

*************************

Answer:

(a) dy/dx = 9x² - 5

At x = 1, dy/dx = 9(1)² - 5 = 4
===> Gradient of tangent at x = 1 is 4

Hence, to find eqn of tangent:
y - 2 = 4 (x - 1)
y = 4x -2

(b) When dy/dx = 4, (parallel)
9x² - 5 = 4
9x² = 9
x² = 1
x = 1 or -1

The other point is when x = -1
when x = -1, y = 6

Hence, the other point is (-1, 6)

A lvl H2 Maths: Complex Numbers

2009-05-26T11:48:00.003+08:00

Question from http://www.sgforums.com/forums/2297/topics/361489

Show that the roots of the equation z⁵ - (z-i)⁵ = 0 are ½(cot (kπ/5) + i) where k = 1, 2, 3, 4

*************************

Answer:

z⁵ = (z-i)⁵

(z / (z-i))⁵ = 1

(z / (z-i)) = e^(2kiπ/5), k = 0,1,2,3,4

1 + i/(z-i) = e^(2kiπ/5)
i/(z-i) = e^(2kiπ/5) - 1
= e^(kiπ/5) (e^(kiπ/5) - e^(-kiπ/5))
= e^(kiπ/5) (2 i sin kπ/5)

Divide by i throughout
1/(z-i) = e^(kiπ/5) (2 sin kπ/5)
z-i = e^(-kiπ/5) [1/(2 sin kπ/5)]
z = ½ ( [ cos (kπ/5) - i sin (kπ/5) ] / sin (kπ/5) ) + i
z = ½ ( cot (kπ/5) - i + 2i)
z = ½ ( cot (kπ/5) + i ), k =1, 2, 3, 4 (k=0 is rejected because cot (0) is undefined) (shown)

A lvl H2 Maths: Vectors

2009-05-26T00:01:00.005+08:00

TJC 2007 P1 Q4

Relative to an origin O, position vectors of A, B and C are a, b and c respectively, where a, b and c are non-parallel vectors. M is the mid-point of AC and P is on the line AB produced such that AB:BP = 2:3. The line PM meets the line BC at a point S. Show that the position vector of S is ⅛ (5b + 3c) .

*************************

Answer:

By repeated use of Ratio theorem,

OM = ½(a + c) and OP = ½(5b - 3a)

OS = μOP + (1-μ) OM
= ½( μ(5b - 3a)+(1-μ)(a + c) )
= (½ - 2μ)a + (5μ/2)b + ((1-μ)/2) c

Also, by ratio theorem again (seeing that B, S and C are collinear),
OS = λc + (1-λ)b

Since a, b and c are non-parallel, we can compare coefficients of the vectors

½ - 2μ = 0 ====> μ = ¼
λ = ((1-μ)/2) = ¾/2 = ⅜

Hence, from OS = λc + (1-λ)b
OS = ⅜ c + (1-⅜) b
OS = ⅛ (5b + 3c) (shown)

O lvl A Maths: Plane Geometry

2009-05-24T12:13:00.003+08:00

Question from http://www.sgforums.com/forums/2297/topics/359741

A line PQ is drawn through the verex A of triangle ABC such that BP and CQ are perpendicular to PQ. If M is the midpoint of BC and MR is perpendicular to PQ, prove that MP = MQ.

*************************

Answer:

We can draw an auxiliary line (in red) to help us see easier.
Draw it such that CH is perpendicular to MR and CG perpendicular to BP
(Note: BP is parallel to MR and parallel to CQ since all are perpendicular to PQ)

We can see that triangle BCG and MCH are similar, with BM = MC
Hence, by intercept theorem, GH = HC

Since PGHR and RHCQ are rectangles, PR = GH = HC = RQ
PR = RQ

Since PR = RQ and MR is perpendicular to PQ, then triangle MPQ is isosceles.
Hence, MP = MQ (shown)

O lvl A Maths: Plane Geometry

2009-05-24T10:56:00.003+08:00

Question from http://www.sgforums.com/forums/2297/topics/361016

In the trapezium ABCD, AD is parallel to BC, BC = 3AD and the diagonals AC and BD meet at E. The line through A is drawn parallel to DB to meet the extended line of CB at F. Prove that

(a) FB = AD
(b) EC = 3AE

*************************

Answer:

(a) Since AF is parallel to BD, and AD is parallel to BC (because F is an extension of BC, hence making both FB and FC also parallel to AD ), ADBF must be a parallelogram, with FB = AD.

(Extra notes: The opposite sides of a parallelogram are always of equal lengths.)

(b) Since BC = 3AD, and AD is parallel to BC, triangle AED and triangle BEC must be similar triangles. As triangle AED and triangle BEC are similar triangles, all sides of triangle BEC must be 3 times the similar sides of triangle AED, hence EC = 3AE.

P.S. Credits to ForbiddenSinner for the solutions and explanations.

O lvl A Maths: Plane Geometry

2009-05-21T00:28:00.004+08:00

Question from http://www.sgforums.com/forums/2297/topics/360786

The diagonals of cyclic quadrilateral PQRS intersect at U. The circle's tangent at R meets PS produced at T. If QR = SR, prove that

PR * ST² = UR * RT²

*************************

Answer:

angle RSQ = angle SQR (isos)
angle SQR = angle SRT (alt. segment)
Hence angle RSQ = angle SQR
==> SQ is parallel to RT
==> PSU and PTR are similar triangles

Using intercept theorem,
UR / PR = ST / PT
UR / PR = ST² / (PT)(ST)

Using tangent secant theorem,
ST * PT = RT²

Hence,
UR / PR = ST² / RT²
PR * ST² = UR * RT²

O lvl A Maths: Plane Geometry

2009-05-13T23:46:00.003+08:00

Question from http://www.sgforums.com/forums/2297/topics/359768

In the triangle ABC, BF and CE are perpendicular to AC and AB respectively. D and G are the midpoints of EF and BC respectively. Prove that GD is perpendicular to EF.

*************************

Answer:

dot a perpendicular line from G to BE and label it as H
dot a perpendicular line from G to FC and label it as I

Note that by similar triangles (or by intercept theorem),

Considering triangles HBG and EBC
because BG = ½BC,
EH = ½BE
HG = ½EC

Considering triangles GIC and BFC
because BG = ½BC,
FI = ½FC
GI = ½BF

Thus,
EG² = EH² + HG²
= ¼BE² + ¼EC²
= ¼BC²

GF² = GI² + FI²
= ¼BF² + ¼CF²
= ¼BC²
= EG²

Thus, GF = EG

Since GF = EG, and ED = DF, then GEF is an isosceles triangle where GD is perpendicular to EF

A lvl H2 Phy: Quantum Physics

2009-05-09T01:04:00.002+08:00

The potential difference between the target and cathode of an X-ray tube is 50 kV and the current in the tube is 20 mA. Only 1% of the total energy is emitted as X-radiation.

(a) What is the maximum frequency of the emitted radiation?

(b) At what rate must heat be removed from the target in order to keep it at a steady temperature?

*************************

Answer:

(a) hf_max = eV
Thus,
f_max
= eV/h
= (1.6 * 10^-19) * (50 * 10³) / (6.63 * 10^-34)
= 1.2 * 10¹⁹ Hz

(b) Rate of heat removed
= electrical power supplied - rate of emitted X-ray energy
= 99% * electrical power supplied
= 0.99 VI
= 0.99 * 50000 * 20 * 10^-3
= 900 Js^-1

A lvl H2 Phy: Quantum Physics

2009-05-07T00:52:00.002+08:00

Suggest a reason why the work function of a metal is different from the energy required to remove the outer electron from an isolated atom of a metal.

*************************

Answer:

There are free electrons in a metal which are not attached to any atom. The work function of the metal is the minimum energy to remove on of these free electrons. This is different from the the energy required to remove the outer electron of an isolated atom.

A lvl H2 Maths: Sequences and Series

2009-05-03T12:13:00.006+08:00

TJC 2005 FM P1 Q1

The sequence of real numbers u₁, u₂, u₃, … is such that .

By using the method of differences, or otherwise, show that .

*************************

Answer:

Hence,

Also,

Hence,
(shown)

Compiled Topical Revision Checklist

2009-04-18T12:05:00.004+08:00

Hi all,

I have compiled some topical revision checklist you can use for your A levels. Will create some for O levels when there's time.

JC Maths Topical Checklist

JC Physics Topical Checklist

A lvl H2 Maths: Binomial Distribution

2009-04-09T00:59:00.002+08:00

Question from http://www.sgforums.com/forums/2297/topics/355089

A dancer performs a certain routine faultlessly 80% of the time. Using a suitable approximation, find the probability that in 160 such occasions, she will perform the routine faultlessly on at least 75% of the number of such occasions.

*************************

Answer:

Y ~ B(160, 0.8)
Let X be the Normal Approximation to Binomial, i.e. X ~ N (128, 25.6)

75% of 160 = 120

Thus,
P(Y > 120) = P(X ≥ 120.5) by Normal Approximation
= 0.931 using Graphic Calculator

A lvl H2 Maths: Binomial Distribution

2009-04-09T00:45:00.003+08:00

Question from http://www.sgforums.com/forums/2297/topics/355089

It is claimed that 90% of the children who were given flu jabs do not catch flu in the next 6 months. A paediatrician gave a flu jab to 150 children in October. Find the least integer r such that the probability of less than r children catching flu within the next 6 months is more than 3%.

*************************

Answer:

Let X be the number of children catching flu within the next 6 months.
Hence,
X ~ B(150, 0.1)

P(X < r) = P(X ≤ r - 1)
We can use the graphic calculator to find the answer.

L1: 1 to 10
L2: binomcdf(150, 0.1, L1)

We see that when L1 is 8, L2 is 0.03074 (first number of L1 for L2 to be greater than 3%)

Therefore r -1 = 8,
r= 9

Least integer r = 9

O lvl A Maths: Remainder Factor Theorem

2009-04-09T00:39:00.002+08:00

Question from http://www.sgforums.com/forums/2297/topics/355078

Given that 2x² + 3px - 2q and x² + q have a common factor x - a, where p, q and a are non-zero constants, show that 9p² + 16q = 0.

*************************

Answer:

Let f(x) = 2x² + 3px - 2q and g(x) = x² + q.

By Factor Theorem, f(a) = 0 and g(a) = 0

So
2a² + 3pa - 2q = 0 -- (1)
a² + q = 0 -- (2).

From (2), a² = -q
Substitute (2)into (1):
-2q + 3pa - 2q = 0
3pa = 4q
a = 4q/3p (remember that p is non-zero, and q is non-zero)

Substitute a into (2):
(4q/3p)² = -q
16q² = -9p²q
q(16q + 9p²) = 0

But q is non-zero, hence,
9p² + 16q = 0 (shown)

O lvl A Maths: Trigonometric Identity

2009-03-30T14:33:00.004+08:00

Prove the following trigonometric identity:

*************************

Answer:

O lvl E Maths: Data Analysis -- Standard Deviation

2009-03-30T00:34:00.004+08:00

A factory produces Krispy chocolate bars. Ten bars from the factory were randomly selected and weighed, and the mass of each bar, measured to the nearest grams, is given below.

25, 23, 24, 26, 24, 24, 27, 22, 25, 26

Find the mean and standard deviation for the mass of the chocolate bars selected from the factory.

*************************

Answer:

mean = ∑x / n
= (25 + 23 + 24 + 26 + 24 + 24 + 27 + 22 + 25 + 26) / 10
= 24.6

The mean chocolate bars produced is 24.6 g.

∑x² = 25² + 23² + 24² + 26² + 24² + 24² + 27² + 22² + 25² + 26²
= 6072

Standard deviation

= 1.43 g

It all happened so fast

2009-03-24T08:13:00.001+08:00

A turtle was walking down an alley in New York when he was mugged by a gang of snails. A police detective came to investigate and asked the turtle if he could explain what happened.

The turtle looked at the detective with a confused look on his face and replied “I don't know, it all happened so fast.”

==> Speed is important <==

Tuition Announcements

2009-03-21T09:08:00.000+08:00

For the year of 2010,

REGISTRATION is now opened for Grp tuition for J1 students going into J2

for H1/H2 Maths!
and H1/H2 Physics!

Call 96756225 or email tan_junwei@hotmail.com to enquire! Spaces are limited.
Thanks!

Location and Rates:
Please visit http://www.freewebs.com/strategictuition/Rates.html

A lvl H2 Maths: Sequences

2009-03-15T20:13:00.010+08:00

The sequence of real numbers b₁, b₂, b₃, ..., is such that b₁ = 4 and b_n = b_n-1 + 2n for all n ∈ Z⁺, n≥2. Using the method of difference, show that b_n = n² + n + 2 for all n ∈ Z⁺.

*************************

Answer:

Using Method of difference,

b_n - b₁
= b_n - b_n-1
+ b_n-1 - b_n-2
+ b_n-2 - b_n-3
+ b_n-3 - b_n-4
+ ..... - .....
+ b₃ - b₂
+ b₂ - b₁

= 2(2) + 2(3) + ... + 2(n-2) + 2(n-1) + 2(n)
= an AP with first term 4, last term 2n, and (n-1) terms
= (n-1)/2 * (4 + 2n)
= (n-1)(n+2)
= n² + n - 2

So,
b_n - 4 = n² + n - 2
b_n = n² + n + 2 (shown)

O lvl A Maths: Applications of Differentiation (Rates of Change)

2009-03-15T09:24:00.004+08:00

A man 1.5m tall is walking at a speed of 2 ms^-1 away from a lamppost which has a lamp 5m above the ground, as shown in the figure.

Find
(a) the rate at which his shadow is increasing,
(b) the speed of the top of his shadow.

*************************

Answer:

Let the length of shadow = x
Let the distance from the lamppost to the man = y

Since the man is 1.5 m tall,
using similar triangles,

(x+y)/5 = x/1.5
y/5 = 7x/15
y = 7x/3

(a) Given: dy/dt = 2
To find: dx/dt

From y=7x/3, dy/dx = 7/3

dy/dt = dy/dx * dx/dt
2 = 7/3 * dx/dt
dx/dt = 6/7 ms^-1

Thus, the shadow is increasing at 6/7 ms^-1

(b) Let z = distance of top of shadow from lamppost
z = x + y

Thus,
speed of the top of his shadow
= dz/dt
= d(x + y) / dt
= dx/dt + dy/dt
= 20/7 ms^-1

A lvl H2 Maths: Sequences

2009-03-14T22:17:00.004+08:00

Question from http://www.sgforums.com/forums/2297/topics/348959

If S_n = 1 - 4 + 9 - 16 + ... + (n)² (-1)⁽ⁿ⁺¹⁾ and given that S_2r+1 - S_2r-1 = 4r + 1, show that S_2n+1 = (n+1)(2n+1).

*************************

Answer:

S₁ = 1
Using the method of differences,
S_2n+1 - S₁
= S_2n+1 - S_2n-1
+ S_2n-1 - S_2n-3
+ S_2n-3 - S_2n-5
+ ..... - .....
+ S₅ - S₃
+ S₃ - S₁

= A.P. of n terms with first term 5 and common difference 4 (T₁=5, T₂=9)
= (n/2) (2(5) + (n-1)(4))
= n (5 + 2n - 2)
= 2n² + 3n

Thus,
S_2n+1 - S₁ = 2n² + 3n
S_2n+1 - 1 = 2n² + 3n
S_2n+1 = 2n² + 3n + 1
S_2n+1 = (n + 1)(2n + 1) (shown)

Habits of Effective Students

2009-03-04T10:00:00.001+08:00

When I was in secondary school and JC, I scored in some subjects, didn't score in others. I often wonder why. Looking back later during my uni days, I realised that there are certain habits that effective students possess, and by adopting these habits for myself consciously (instead of unconsciously during my earlier years), I emerged with a stellar first class honours within the competitive NUS Electrical Engineering faculty. And I have done this journey with minimal tuition help.

So here I am, sharing a few tips and habits with all readers. Hopefully, you will be able to achieve better results than I did!

1. First and foremost, paying attention in class in important. During my time, I paid 100% attention to subjects I liked, and was usually half awake for subjects I didn't really like. Needless to say, I scored for those I paid attention to with minimal efforts. This is because teachers usually go through the different types of questions before your exams. They know what's going to come out, and what's not, and thus prepare you for the questions during class. Make sure you revise and internalise (not just memorise) the example questions beforehand.

2. Complete all your homework by yourself. Try to do as many as you can by yourself before seeking help. This helps to drive in concepts and methods taught in class, and assist you in mastering them well. I have seen people copying homework instead of doing it themselves. These people usually have to put in a much greater effort at the end of the exams to score better, stressing themselves out.

3. Practise even more questions from the same topic. The old adage "practice makes perfect" holds true. With more practice comes more confidence, leading to lesser mental blocks. Speed of doing questions will increase, and number of careless mistakes will decrease. These are extremely important criteria for you to achieve spectacular exam results. You can do this by using the Ten Year Series, past year prelim papers, assessment books, or even free websites like ExamWorld, which provide some free questions and solutions for students.

4. Learn from other students as well. You can do this by looking at good works that have been marked. Look at the different comments by the teacher, learn the different approaches to answering questions and essays, and understand the common pitfalls to avoid them in the future. Afterall, it's always easier to learn from others mistakes right? This was the main method I used to improve my A level economics essay from a fail to an eventual A. I photocopied A grade essays from my classmates to analyse, understand their points, their argument and analysis, etc.

5. Find and discover various ways to check your answers, especially for hard sciences like maths and physics. Looking through your answers and workings again is only one way of checking. Most questions have more than 1 approach (for maths) and by using a different approach and getting the same answer, you are more or less assured of being correct. Other methods can include working backwards to reach what the question wanted to you too. Or perhaps, for proving questions, substituting a value to find that both sides of the equation give you the same results. Or even for physics, by remembering some constants, you can more or less know whether your answer is reasonable a not. Example would be finding the density of oil to be 667 kg per metres cube. Obviously oil is less dense than water (1000 kg per metres cube), so the answer was reasonable. This habit is extremely important, and I usually teach my tuition students how to do so and where to look out for such "free marks".

6. Make short notes to read an hour before your exam. Realise that for subjects like physics, knowing and understanding the concepts well, memorizing the equations and definitions of the different terms, can suffice in scoring As for your exams. The knowing and understanding the concepts should have been covered in points 1 to 4 above. The equations and definitions on your short paper are merely to there to help refresh your memory so that you do not stumble in exams.

7. Teach other students freely when they ask you for help. If you can explain the question well, it truly means that you have internalised and understood perfectly the concepts being tested in the question. I realised during my JC time that questions other students ask are usually of higher standards than simple questions. By helping them, I trained my mind to be able to solve tougher questions as well. And by being able to explain to them, I became even more confident and sure of myself in those concepts. Take it as a surprise test/revision that occurs frequently.

Article also found at http://www.associatedcontent.com/article/1529295/habits_of_effective_students.html?cat=4

A lvl H2 Maths: APGP

2009-03-02T10:35:00.003+08:00

Question from http://www.sgforums.com/forums/2297/topics/348959

Each term of an AP is added to the corresponding term of a GP to form a 3rd sequence S, whose 1st 3 terms are -1, -2, 6. The common ratio of the GP is equal to the 1st term of the AP.

Prove that the 1st term of the AP is a root of the equation a³ - a² - a +10 = 0. Verify that a = -2 is a root of this equation. With this value of a, find the nth term of the sequence S and obtain an expression for the sum of the 1st n terms of S.

*************************

Answer:

Let first term of AP = a
Thus Common ratio of GP = a
Let first term of GP = b

First 3 terms of GP = b, ba, ba²
First 3 terms of AP = a, a+d, a+2d

Hence,
b + a = -1 ---(1)
ba + a + d = -2 ---(2)
ba² + a + 2d = 6 --- (3)

From (1), b = -1 - a ---(4)
From (2), a + d = -2 - ba
2a + 2d = -4 - 2ba ---(5)

Sub (5) into (3)
ba² - a + (-4 - 2ba) = 6
ba² - a - 2ba = 10 ---(6)

Sub (4) into (6)
(-1 - a)a² - a - 2(-1 - a)a = 10
-a² - a³ - a + 2a + 2a² - 10 = 0
-a³ + a² + a - 10 = 0
a³ - a² - a + 10 = 0

Thus, the 1st term of the AP is a root of the equation a³ - a² - a +10 = 0 (shown)

When a = -2,
(-2)³ - (-2)² - (-2) +10 = -8 - 4 + 2 + 10 = 0
Thus, a = -2 is a root of this equation (verified)

Sub a = -2 into (1)
b = -1 -a = 1
Sub b = 1, a = -2 into (2)
d = -2 - a - ba = -2 + 2 + 2 = 2

So,
nth term of S = nth term of AP + nth term of GP
= (-2) + (n - 1)(2) + (1)(-2)^(n-1)
= 2n - 4 + (-2)^(n-1)

Talking Clock

2009-02-27T09:55:00.001+08:00

A drunk was proudly showing off his new apartment
to a couple of his friends late one night, and led the way to his bedroom where there was a big brass gong and a mallet.

"What's that big brass gong?" one of the guests asked.

"It's not a gong. It's a talking clock," the drunk replied.

"A talking clock? Seriously?" asked his astonished friend.

"Yeah," replied the drunk.

"How's it work?" the other friend asked, squinting at it.
"Watch this," the drunk replied.

He picked up the mallet, gave the gong an
ear-shattering pound, and stepped back.

The three stood looking at one another for a moment.

Suddenly, someone on the other side of the wall
screamed, "You asshole... it's 3:15 in the morning!!!"

Moral of the story: Sound can travel through solids

O lvl E Maths: Probability

2009-02-22T14:19:00.002+08:00

From http://www.sgforums.com/forums/2297/topics/348803

Suppose that 45% and 40% of Singaporeans approve and disapprove the idea of building a casino respectively. If three Singaporeans are interviewed at random,find the probability that
a) all of them will either approve or disapprove of the idea

b) at least two of them will not approve of the idea

c) at most one of them will approve of the idea.

*************************

Answer:

Let A = Approve, D = disapprove

(a) P(all of them have opinions) = 0.85 * 0.85 * 0.85 = 4913/8000

(b) Let X = not disapprove
P(X) = 1 - 0.4 = 0.6

P(at least 2 disapprove) = P(DDX) + P(DXD) + P(XDD) + P(DDD)
= 3 (0.4 * 0.4 * 0.6) + 0.4 * 0.4 * 0.4
=44/125

(c) Let NA = not approve
P(NA) = 1 - 0.45 = 0.55

P(at most 1 approve) = P(all NA) + 3 P(A.NA.NA)
= 0.55 * 0.55 * 0.55 + 3 (0.45 * 0.55 * 0.55)
= 2299/4000

O lvl E Maths: Ratio Rate Proportion

2009-02-22T13:31:00.001+08:00

From http://www.sgforums.com/forums/2297/topics/348909

A rectangular pool can be filled by two pipes together in 6 hours. The smaller pipe alone takes 5 hours longer than the larger pipe to fill the pool. Find them time each pipe will take to fill the pool on its own.

*************************

Answer:

Let the time taken by small pipe alone = Ts
Let the time taken by bigger pipe alone = Tb
Let Volume of pool = V

Rate of flow of small pipe = V/Ts
Rate of flow of bigger pipe = V/Tb

Get the equations out

Ts - Tb = 5
Ts = Tb + 5 ------------- (1)

V / ( V/Ts + V/Tb) = 6
TsTb / (Ts + Tb) = 6
Ts Tb = 6Ts + 6 Tb
Sub (1) into equation
Tb² + 5Tb = 6Tb + 30 + 6 Tb
Tb² - 7Tb - 30 = 0

Solving, Tb = 10 hours or -3 (reject)
Hence, Ts = 15 hours

A lvl H2 Maths: APGP

2009-02-22T00:52:00.007+08:00

From http://www.sgforums.com/forums/2297/topics/348847

1) In an AP , 8th term is twice the 4th term and the 20th term is 40. Find the common difference and the sum of the terms from the 8th to the 20th term.

2) The r^th term of an AP is (1+4r). Find in terms of n, the sum of the first n terms.

3) An arithmetic series has first term 1000 and common difference -1.4. Calculate the 1st negative term and the sum of all the positive terms.

*************************

Answer:

1) Given: U₈ = 2U₄, U₂₀ = 40
Apply nth term formula:
a + 7d = 2(a +3d)
a - d = 0 ---(1)
Also, a + 19d = 40 ---(2)
solving simultaneously, d = 2
a = 2

Sum of terms = S₂₀-S₇
= (20/2)(2(2) + (19)(2)) - (7/2)(2(2) + (6)(2))
= (10)(42) - (7/2)(16)
= 364

2) U_r= 1 + 4_r
When r = 1, U₁ = 5
When r = 2, U₂ = 9
When r = 3, U₃ = 13
Thus, common difference = 4

Thus, S_n = (n/2)(2(5) + (n-1)(4))
S_n = (n)(2n + 3)

3) T_n = a + (n-1)d
a = 1000, d = -1.4
Thus, when T_n <> 1001.4
n > 715.286

Hence, the first negative n term is when n = 716

First negative n term = 1000 + (716 - 1)(-1.4) = -1

First negative term is when n = 716
so the last positive term is when n = 715
So, sum of all positive numbers
= (715/2) (2000 + 714 * -1.4)
= 357643

A lvl H2 Maths: APGP

2009-02-22T00:38:00.003+08:00

2002 A lvl Maths P1 Q10
Question from http://www.sgforums.com/forums/2297/topics/348847

An AP in which the common difference is twice the 1st term is called a special AP. Show that the sum of the first n terms of a special AP with 1st term a is n²a.

The (M-2)th term of a special AP is 663 and the Mth term is 731. Find M and find also the sum of the first M terms.

The sum of the first N terms of another special AP is S. Find in the terms of S, the sum of the next N terms.

*************************

Answer:

Let first term = a, common difference = 2a

S_n = (n/2)*(2a + (n-1)(2a))
= (n/2)*(2a + 2an - 2a)
= (n/2)*(2an)
= n²a (shown)

Mth term - (M-2)th term = 2 * common difference
731 - 663 = 2 * 2a
a = 17, d = 34

731 = 17 + (M-1)(34)
M = 22

Sum of first M terms =22² * 17 = 8228

S_n = n²a
S_2n = 4n²a

Difference of next N terms = S_2n - S_n
= 3n²a
= 3S

O lvl A Maths: Trigonometric Identity

2009-02-16T00:31:00.001+08:00

Prove the following identity:

(tan x – cosec x)² – ( cot x – sec x)² = 2(cosec x – sec x)

*************************

Answer:

LHS
= tan² x - 2 tan x cosec x + cosec² x - cot² x + 2 cot x sec x - sec² x
= tan² x - 2 sec x + cosec² x - cot² x + 2 cosec x - sec² x
= ( tan² x - sec² x ) + (cosec² x - cot² x) + 2 cosec x - 2 sec x
= -1 + 1 + 2 cosec x - 2 sec x
= 2 (cosec x - sec x)
= RHS (proved)

A lvl H2 Phy: Electromagnetism

2009-02-08T23:57:00.004+08:00

A horizontal rod PQ of mass 10g and length 0.10m is placed on a smooth plane inclined at 60° to the horizontal. A uniform vertical magnetic field of value B is applied in the region PQ. Calculate B if the rod remains stationary on the plane when a current of 1.73A flows in the rod. What is the direction of the current in the rod?

*************************

Answer:

To balance the forces, the force on the rod due to the B-field, F_B, is in the direction as shown:

F_B = BIL

If the rod remains stationary, resolving forces,
F_B cos 60° = W sin 60°
B (1.73) (0.1) (0.5) = (0.01kg * 9.81) (√3 / 2)
B = 0.982 T

For F_B to be in the direction as shown, by Fleming's Left Hand rule, the direction of current in the rod should be out of the paper.

O lvl Chem: Acid Bases and Salts

2009-02-02T13:27:00.002+08:00

J89/P2B/Q1

(a) What is an acid?
(b) Describe what you would see when a little copper(II) oxide is warmed with dilute nitric acid.
(c) In what way is copper(II) oxide behaving when it reacts with dilute nitric acid?

*************************

Answer:

(a) An acid is a compound which, when dissolved in water,produces hydrogen ions
(b) The black oxide dissolves in the acid to form a bluish green solution.
(c) It is behaving as a base.

A lvl H2 Maths: Vectors

2009-02-01T15:45:00.009+08:00

TPJC 2000 P2 Q14

Referred to the origin O, the points A, B and C have position vectors given respectively by

Find
(i) the cartesian equations of the line l passing through A and B;
(ii) the length of projection of BC on l;
(iii) the image of C in the line l;
(iv) the points where l meets the plane z = 0;
(v) the 2 points P on l such that cos ∡POB =

*************************

Answer:

(i)

Hence, line l :

(ii)

Thus, length of projection

(iii) The graphic is as shown

So, point M is point B, moved along downwards along line l

Unit vector along line l =

Hence,

(iv) Let the point be

Meeting the plane z = 0 means 3 + 2λ = 0
λ = -1.5

Hence, substituting λ = -1.5, we get the point of intersection as

v) Let
for some k since P is a point on l.

Note: So we need to find out what is the value of k.

Since cos ∡POB = , and
cos ∡POB =

Square both sides.

169k² + 26k + 1 = 112k² -16k + 16
57k² + 42k - 15 = 0
19k² + 14k - 5 = 0
(19k - 5)(k + 1) = 0
k = -1 or 5/19

Hence, the two points P are
(-1, 4, -1) or (5/19, 4/19, 29/19)

What constitutes a good private tutor

2009-01-29T14:31:00.005+08:00

There are many factors parents and students consider when choosing a effective private tutor. Some considerations can include age, sex, experience, qualifications, reputation/track record, budget, etc.

Given that there are many tutors around, how then could you as a tutor stand out? I shall share a few factors which I think is important.

1) Passionate and knowledgable in the subject you are teaching

This is an important criteria that you must fulfill. Other than previous grades in school, a good tutor has to demonstrate a good track record, and be able to make references to the why and how of tackling the concept or question.

Most importantly, a good tutor must be able to put seemingly difficult concepts into simple terms, making sure that the student can absorb and assimilate the details.

Passion is also important because a good tutor must want to impart his skills, knowledge as well as passion to his/her students. He/she must want to target for full marks, and not just sufficiently high grades, for his/her students.

2) Active tutoring

There are tutors who are tutoring passively. That means, they sit around and watch the students do their homework/schoolwork. There are tutors who simply tell the students to try and try again, without offering much active help. In my opinion, this is clearly a waste of money by the student. Why pay so much to employ a guardian to watch over you do your own work?

Active tutors take the initiative to provide notes and questions to their students, and teach in such a way that it helps the students think, comprehend, analyse, and understand concepts and questions. They repeat this process again and again. They train the student to grasp concepts, to utilise effectively every tool in their hands (calculators, rulers, etc) and to make full use of exam strategies to reduce their careless mistakes or take advantage of the structure in which questions are set.

3) Motivating

This is to me the most important factor to be present in a good tutor. To do so, tutors have to be able to build good rapport with their students. They have to create the hunger, the want, the confidence to score in their students.

Many a times, I have come across students who are desperate, dejected and totally low in confidence. They seeked tuition from me as a last resort, as if I'm a miracle pill to at least help them pass a subject. And luckily for them, I managed to instill more confidence in everyone of them, give them hope, and finally, they achieve a grade that is way better than a pass.

There are also students who have totally no idea why they should study a particular subject. I have this student who plans to go into game design, and cannot think of any reason to study and do well in physics, other than to make the grade. That unmotivation remains, until I explained how mechanics and physics are also required in games, and went into some details on how the programmer needs basic knowledge in these aspects in physics to create a realistic game.

There are many more stories, but I would have to stop. In short, a good tutor will be able to help students break out of their rut and work hard for their goals.

4) Be responsible

A good tutor will take responsibility for all of his/her students. Being responsible as a tutor means
i) willing and want to take the student all the way to the end of the year
ii) do not teach the wrong things. When unsure, state so, and do extensive research before teaching the student the correct things. It is a hideous sin to make the student unlearn wrong things.
iii) ensure that the student improve in grades

To your success :D

Article also found here http://www.associatedcontent.com/article/1483720/what_constitutes_a_good_private_tutor.html?cat=4

A lvl H2 Maths: Vectors

2009-01-27T23:02:00.008+08:00

HCJC 1999 P1 Q19

Points A, B, C D have position vectors -3i + 4j, 3i + pj, 2i + 3j and i - 2j respectively.
(i) Given that AB·AD = 0, find the value of p.
(ii) Show that ∡BAC = ∡DAC.
(iii) Show that the points B, C, D are collinear and write down the vector equation of the line l through these three points.
(iv) Find the position vector of N, the foot of the perpendicular from the point E (5, 18, -4) to the line l. Find also the shortest distance from E to the line l.

*************************

Answer:

(i) AB = OB - OA = 6i + (p-4)j
AD = OD - OA = 4i - 6j

AB·AD =0
24 - (p-4)*6 = 0
24 - 6p + 24 = 0
p = 8

(ii) AC = OC - OA = 5i - j
AB = 6i + 4j
AD = 4i - 6j

Since cos^-1 ∡BAC = cos^-1 ∡DAC
=> ∡BAC = ∡DAC (shown)

(iii) BC = OC - OB = -i - 5j
BD = OD - OB = -2i - 10j = 2 BC

Thus, B, C and D are collinear (shown)

Line l for λ ε R

(iv) Since N is on l, let N =

Hence,

Since is perpendicular to the line l,

2 - λ + 50 - 25λ = 0
26λ = 52
λ = 2

Hence,

O lvl Chem: Periodic Table

2009-01-24T10:33:00.002+08:00

D90/P2/Q5

A new element Z ,thought to be a halogen,has been discovered.Its relative atomic mass is found to be 370.

Which of the following are likely properties of Z ?
A. dark green gas which is soluble in water
B. black liquid which boils giving a dark brown gas
C. white solid which reacts with an acid giving hydrogen
D. grey solid which explodes on contact with hydrogen

*************************

Answer:

Answer is D. Since Z is a halogen, it should be heavier than Iodine and black in colour with high metlting point

Recall that Colours of Group VII elements gets darker on going down the group, with increasing melting point

O lvl Chem: Periodic Table

2009-01-23T11:39:00.011+08:00

D92/P3/B3

The diagram shows some properties and reactions involving substances A, B, C, D and E.

(a) Identify A, B, C, D and E.

(b)
(i) Write an equation for the reaction between A and aqueous sodium hydroxide solution.
(ii) Write an equation for the reaction between D and carbon.

(c) Suggest a use for E.

*************************

Answer:

A - iron (II) Sulphate, FeSO₄
B – iron (II) Hydroxide,FE(OH)₂
C – barium sulphate,BaSO₄
D – iron (III) oxide,Fe₂O₃
E – iron ,Fe

(b)(i) FeSO₄(aq) + 2NaOH(aq) + Fe(OH)₂(s) +
Na₂SO₄(aq)

(ii) 2Fe₂O₃(s) + 3C(s) à 4Fe(s) + 3CO₂(g)

(c)Making Iron nails.

A lvl H2 Maths: Vectors

2009-01-17T00:56:00.007+08:00

HCJC Year 2000 Paper 1 Q11

Given that a = 3i + j + k, b = i - k, c = 4i - 3j + 2k are position vectors of the points A, B and C respectively, find the position vectors of the points and Q which divide BC internally and externally in the ratio 1:2, respectively.

Show that AP is perpendicular to BC.

*************************

Answer:

By Ratio Theorem,
OP = ⅓(OC + 2OB)
OP = ⅓(4i - 3j + 2k + 2i - 2k)
OP = 2i - j

By Ratio Theorem,
OB = ½(OQ + OC)
OQ = 2OB - OC
OQ = 2i - 2k - (4j - 3j + 2k)
OQ = -2i + 3j -4k

AP = OP - OA = -i - 2j - k
BC = OC - OB = 3i - 3j + 3k

AP · BC = -3 + 6 - 3 = 0
Hence, AP is perpendicular to BC (shown)

A lvl H2 Phy: Current of Electricity

2009-01-17T00:41:00.002+08:00

RJC 2001 Common Test 1 Q 17

The capacitance of a certain variable capacitor may be varied between limits of 120 pF and 600 pF by turning a knob attached to the movable plates. The capacitor is set to 600 pF, and is charged by connecting it to a battery of e.m.f. 250 V.

(a) What is the charge on the plate?

The battery is then disconnected and the capacitance changed to 120 pF.

(b) Assuming that no charge is lost from the plates, what is now the potential difference between them?

(c) Calculate the initial and final energy stored in the capacitor during this change.

(d) Account for the change in energy.

*************************

Answer:

(a) Q = C_iV_i
= 600 * 10^-12 * 250
= 1.50 * 10^-7 C (3 s.f.)

(b) V_f = Q / C_f
= 1.50 * 10^-7 C (3 s.f.) / 120 * 10^-12
= 1250 V (3 s.f.)

(c) initial energy = ½C_iV_i²
= ½ (600 * 10^-12) (250)²
= 1.88 * 10^-5 J

Final energy = ½C_fV_f²
= ½ (120 * 10^-12) (1250)²
= 9.38 * 10^-5 J

(d) Work is done by an external force in turning the knob to decrease capacitance. As a result, more energy is stored in the capacitor.

O lvl E Maths: Arithmetic

2009-01-15T23:52:00.003+08:00

From http://www.sgforums.com/forums/2297/topics/343862

1)
a) Express 560 as a product of its prime factors(in index notation)
b) State the smallest integer value of x for which 560x is a square number

2) Write down the largest factor of 1155 other 1155 itself

3) Find the least integer value of n if 936n is a perfect square

4) A three-digit number is the product of 4 prime number. The sum of its prime factors is 30. Given that the three digits of the number are all prime and different, find the three-digit number.

*************************

Answer:

1)
i) 560= 2⁴ x 5 x 7
ii) We need 5 x 7
Thus, x = 35

2) 1155 = 3 x 5 x 7 x 11
largest factor = 5 x 7 x 11 = 385

3) 936 = 2³ x 3² x 13
We need 2 x 13
n = 26

4) This is a brute force method...
Prime numbers are 2, 3 , 5, 7, 11, 13, 17, 19 and so on...

Since the 3 digits must all be prime, they can only be 2, 3, 5, 7. Since we only have 4 possible digits without repetition, there are only 24 combos (4P3= 24).

1. 235 = 5 x 47 (NA) 2 primes
2. 237 = 3 x 79 (NA) 2 primes
3. 253 = 11 x 23 (NA) 2 primes
4. 257 (NA) 1 prime
5. 273 = 3 x 7 x 13 (NA) 3 primes
6. 275 = 5² x 11 (NA) 3 primes
7. 325 = 5² x 13 (NA) 3 primes
8. 327 = 3 x 109 (NA) 2 primes
9. 352 = 2⁵ x 11 (NA) 3 primes
10. 357 = 3 x 7 x 17 (NA) 3 primes
11. 372 = 2² x 3 x 31 (NA) sum not 30
12. 375 = 3 x 5³ (NA) sum not 30
13. 523 (NA) 1 prime
14. 527 = 17 x 31 (NA) 2 primes
15. 532 = 2² x 7 x 19 (NA) This is the one
16. 537 = 3 x 179 (NA) 2 primes
17. 572 = 2² x 11 x 13 (NA) sum not 30
18. 573 = 3 x 191 (NA) 2 primes
19. 723 = 3 x 241 (NA) 2 primes
20. 725 = 5² x 29 (NA) 3 primes
21. 732 = 2² x 3 x 61 (NA) sum not 30
22. 735 = 3 x 5 x 7² (NA) sum not 30
23. 752 = 2⁴ x 47 (NA) 5 primes
24. 753 = 3 x 251 (NA) 2 primes

A lvl H2 Phy: Thermal Physics/Ideal Gas

2009-01-15T23:24:00.003+08:00

A sealed container holds a mixture of nitrogen molecules and helium molecules at a temperature of 290K. The total pressure exerted by the gas on the container is 120 kPa.

(M) molar mass of helium = 4 x 10^-3 kg / mol
R= 8.31 J / K mol
Na= 6.02 x 10²³ /mol

i) calculate root mean square speed of the helium molecules.

ii) calculate average kinetic energy of a nitrogen molecule.

iii)If there are twice as many helium molecules as nitrogen molecules in the container, calculate pressure exerted on the container by helium molecules.

*************************

Answer:

i) Average kinetic energy per mole = (1/2) * Na*m v² = 3/2 RT
Since M = Na * m= 4 x 10^-3 kg / mol,
then, v² = ( (3 * R T) / M = 3 * 8.31 * 290 / (4 x 10^-3) =1.81 x 10⁶
v_rms = √(1.81 x 10⁶)
= 1346 m/s

ii) Average kinetic energy of a nitrogen molecule
= 3/2 KT, where K = boltzmann constant
= 3/2 * 1.38 * 10^-23 * 290
= 6.00 *10^-21 J

iii) pV = nRT
From here, we can see that R is constant, T and V is same for both helium and nitrogen. The only difference is n.
So, we can expect that helium molecules will exert twice the amount of pressure as the nitrogen molecules in this question

Hence, pressure exerted on the container by helium molecules is 80 kPa.

O lvl Phy: Light

2009-01-14T00:42:00.003+08:00

Dunman High 1999 Prelims P2 Q4

The diagram shows a ray of monochromatic light incident normally on a triangular glass prism of critical angle 39.5°.

(a) Find the refractive index of glass.
(b) Explain why the ray will not emerge into the air from the face XY.
(c) What is the angle of incidence at which the ray meets the face YZ.
(d) Hence find the angle of refraction when the ray emerges into the air again.

*************************

Answer:

(a) refractive index = 1 / sin 39.5° = 1.57

(b) The angle of incidence is 60°. This is clearly higher than the critical angle of 39.5°. Hence, the ray will undergo total internal reflection, and thus will not emerge into the air from face XY.

(c) Angle of incidence = (180° - 30° - 30°) - 90° = 30°

(d) refractive index = sin (angle of refraction) / sin 30°
sin (angle of refraction) = 1.57 * sin 30°
angle of refraction = 51.7°

A lvl H2 Maths: Maclaurin's Series

2009-01-11T00:54:00.007+08:00

This is on Maclaurin series

Find f(x), whose expansion is given by

*************************

Answer:

(done)

The Principles of Academic Success

2008-12-29T10:15:00.001+08:00

Written and contributed by Tan Jye Yee, NUS Law School Class of 2012

I can isolate three important ingredients to academic success: proactivity, consistency and knowledge of the assessment system.

a) Being proactive means to pay total attention in class, ask questions when you don't agree with the teacher or need clarification, probe deeper than what is "required for the syllabus" to enhance your understanding and not accept what you are taught unless you are genuinely convinced through self-learning and validation.

b) Consistency would strengthen your grasp of what you learn through continual reinforcement. I would recommend at least 3 hours of study every week day, even when examinations are not near, and you should increase the number of hours towards the examination period.

c) Knowledge of the assessment system allows you to tweak your learning and studying process to prepare yourself for the questioning style, content and difficulty of the papers. Find out what you are likely to be tested on through whatever means and try to understand an examiner's point of view: why is the question asked (i.e. what is being tested), what answer would suffice to fully solve the question, what is expected of reasonably competent students of your educational level and what are pitfalls to avoid that might reveal your shaky understanding of the subject at hand? The last part is a crucial balancing act between providing adequate explanation and not revealing what you don't truly understand.

Apart from these instrumentalist tips, it's important to get your mind in the correct paradigm. The "how" question is often asked of academic success, but the "why" question is rarely examined by many students. Do you think academic success is important for you to earn a lot of money in the future? Do you want to truly learn and develop an educated, highly cognitive, analytical yet creative mind? Do you enjoy the compliments and credibility that you get with excellent grades? What are you doing in the classroom, where is it taking you? Think really hard about the real reason, see the world a little more, then decide again on your motive. While your objectives might remain practically stagnant (to succeed), your reasons for succeeding are supposed to be dynamic.

I have always been outstanding at examinations (10 A1s at O Level's and 6 Distinctions at A Levels) but have never really understood what I wanted from studying so persistently. In primary school, I did well because I wanted to make my parents proud. I wanted to earn their respect and increase my worth in their eyes. Now, I study harder to enrich my mind, to broaden my world perspective, to attain crucial, indispensable knowledge and skills as well as to learn how the world works so that I can thrive in it.

It might seem like a waste of time to ponder on this, since it's not as urgent as actual preparation for the examinations. But what is not urgent may be more important than what seems to be so pressing at the moment. Learn to snap out of your hectic life once in a while and do a re-evaluation of your life objectives. Write a personal mission statement and change it from time to time.

Most people are more keen to learn the "how" of success because we all know doing well in examinations generally brings positive outcomes. Yet knowing "why" brings assurance, happiness and a sense of inner stability, and would shift our fundamental paradigm and transform our attitude towards learning. In a sense, knowing "why" ultimately underlies your "want" and determines how successful you will be at utilising the "how".

If you do conclude that there is no good reason for you to study, to learn, so be it. I, however, recommend that you speak to others first about how you feel, get some more exposure in life and be truly convinced that your thinking or decision is the right one for you, taking into account your friends' and family's possible perception and treatment of you, your would-be status in society, your range of probable opportunities in the future, etc. Don't delude yourself just because you don't like to study or are not good at it. Your dislike can be overcome by cultivating an interest, and your competence can improve over time through guidance and effort, but a lost opportunity for education is gone forever.

A lvl H2 Maths: Applications of Integration (Area and Volume)

2008-12-20T12:54:00.007+08:00

Question from http://www.sgforums.com/forums/2297/topics/341192

O is the origin and A is the point on the curve y = tan x where x = π/3

i) Calculate the area of the region R enclosed by the arc OA, x-axis and the line x = π/3

ii)The region S is enclosed by the arc OA, the y-axis and the line y=√3.

Find the volume of the solid of revolution formed when S is rotates through 2π rad about the x-axis, giving answers in exact form.

iii) Find

*************************

Answer:

i)
Point A = (π/3, √3)
Area of region R

= ln 2
= 0.693 units²

ii)
Rotate around x-axis (which is y=0 line)
So, volume = volume of cylinder - volume rotated under curve OA around x-axis

iii)

We can take the difference between the area of rectangle formed by x-axis, y-axis, x = π/3 and y = √3, and the area of region R.

Thus,

= ( π/3 )(√3) - 0.693
= 1.12 units²

O lvl A Maths: Trigonometry

2008-12-14T01:47:00.008+08:00

From http://www.sgforums.com/forums/2297/topics/339918

If ,

find

*************************

Answer:

Since ,
so that it is positive.

Since ,

A lvl H2 Maths: Integration

2008-12-11T01:11:00.011+08:00

Question from http://www.sgforums.com/forums/2297/topics/339990

(a) Evaluate

(b) Using the substitution , find

with x in the range 0 < x < 1.

(c) Find .

*************************

Answer:

(a)

Note: For , the numerator is a differential of the denominator.

(b)
From ,

Sub and

Since ,
Also,

Thus, sin 2u = 2 sin u cos u

Hence,

(c)

A lvl H2 Phy: Simple Harmonic Motion

2008-12-11T01:06:00.003+08:00

A massless spring with spring constant 19 N/m hangs vertically. A body of mass 0.20 kg is attached to its free end and then released. Assume that the spring was un-stretched before the body was released. Find

a) How far below the initial position the body descends, and the
b) Frequency of the resulting Simple Harmonic Motion.
c) Amplitude of the resulting Simple Harmonic Motion.

*************************

Answer:

a) At the point of release, GPE = 0, KE = 0, EPE of spring = 0
At max distance x below initial position,
loss in GPE = gain in EPE of spring (KE becomes 0 at that point)
mgx = ½kx²
mg = ½kx
(0.20)(9.81) = ½(19)(x)
x = 0.2065 m
x = 0.207 m (3 s.f.)

b)

= 1.55 Hz

c) Highest point of oscillation = initial position
Lowest point of oscillation = 0.2065 m below initial position.
Thus, amplitude of oscillation = 0.2065/2 = 0.103 m (3 s.f.)

A lvl H2 Phy: Dynamics

2008-12-07T01:32:00.003+08:00

An 18-g rifle bullet travelling 230 m/s buries itself in a 3.6-kg pendulum hanging on a 2.8-m-long string, which makes the pendulum swing upward in an arc. Determine the horizontal component of the pendulum's displacement.

*************************

Answer:

Using Principle of conservation of momentum,
m₁v₁ + m₂v₂ = (m₁ + m₂)V
(0.018)(230) + (3.6)(0) = (3.6 + 0.018)(V)
V = 1.14 ms^-1

Using conservation of energy,
mgh = ½mv²
h = 0.0667 m

Hence, from diagram above, horizontal displacement, d, is given by

= 0.608 m

A lvl H2 Phy: Measurements

2008-12-07T00:05:00.004+08:00

N75 Paper 1 Q1

A student measured the density of a metal cylinder by finding its dimensions with a ruler, which he could not read to better than ±1 mm because of parallax error, and its mass with a balance reading to ±0.1 g. He recorded his readings as follows:

diameter of cylinder = 20.0 mm
height of cylinder = 18.0 mm
mass of cylinder = 51.3 g,

From which he calculated:

density of cylinder = 9.072 * 10³ kg m^-3

(a) Explain what is meant by parallax error. How can it be reduced?
(b) With the given uncertainties in the readings, what is the percentage uncertainty in the value of the density?
(c) Comment on the student's choice of apparatus and the presentation of his results. (His arithmetic is correct.)
How would you improve on his determination?

*************************

Answer:

(a) Parallax error takes place when the line of sight of the experimenter is not perpendicular to the scale. Randomness in the errors may occur when the angle of the line of sight varies randomly when repeated measurements are taken.

To reduce parallax error, the experimenter should place an attached pointer or the object being measured as close to the scale as possible. Repeat the measurement and find the mean of repeated measurements.

(b) Density = Mass / Volume
, where p = density, M = mass, D = diameter, h = height.

(c) There's too much uncertainty when using a ruler to make measurements in the scale of millimetres; the percentage uncertainty will be big. The presentation of the results is also insufficient because the uncertainty was not stated as well. However, the usage of the balance to measure the mass is good because the percentage uncertainty is small.

The student should measure the lengths with a vernier calipers (uncertainty ±0.1 mm) or a micrometer (uncertainty ±0.01 mm) for greater accuracy.
In addition, he should present his final results for the density along with the ± errors.

Survival guide for the last-min Physics student

2008-12-04T09:53:00.003+08:00

*Written by a current secondary school physics teacher from a reputable school*

Your Physics Test is coming... not in a month, not in a week... but three days’ time! How do you make the most out of the precious hours? O.K. Here is some straightforward survival tips from an ex-last minute Physics student who has become a Physics teacher.

If you have the luxury of studying over the weekend for a Monday test, your chances of survival is very high. If you are caught in the midst of a school week and your test is a few school days away, you need more determination and discipline in order to make it! I will teach you the different ways to organise your preparation time effectively in both situations.

Survival kit:
Physics Text book
Physics Ten Years Series
Dig out as many physics notes and worksheets AS YOU CAN (Do not sweat if you have trouble doing this)
Calculator
Pencil case and Paper

What to do?
“I hear and I forget. I see and I understand. I do and I remember”
So, remain calm even if you have not been very attentive in class. Do not bother to stare at the complicated formulas and diagrams trying to print the images into your brain. The secret to success in learning is... DOING PHYSICS! In the remaining time you have, your goal is to DO as many questions as you can in the topics to be tested.

Begin with ten year series questions first. They are the most basic and direct questions. Here is the secret: Read through the questions in a topic first even if you don’t have the knowledge or understanding to do them. Just read through the questions first and pick out key words and terms used. You are to notice what stuff are commonly asked in the topic. Next, be determined to attempt the questions as if you are being made to sit for a test. You will probably feel anxious and will only be able to do very few questions. After you are done with this step, close your eyes and reflect what are the questions that gave you the greatest anxiety and which keywords appeared very frequently in the topic. Do this no longer than 10 minutes.
Now, you are allowed to flip open the Text and SCAN through the chapter for 15 min only. Your aim is to pick up simple information that you have forgotten so that you can complete more questions that you did not do earlier on. Do not read into chucks of details. After this, pick up the questions again and this time round, you should be able to do a few more questions.

Next, you are allowed to open up the Text and complete the most difficult questions along aside. You will find that you will be more motivated to read through sections of information in the Text in order to solve the more complicated problems. Also, you would have filtered out the sections which you have scanned through for the simpler questions earlier on.

OK. You have finished the Ten year series and feeling more confident and relieved. Great! Now, to achieve perfection, take out the notes and worksheets that your teacher has given you in class. This is the time to read through the notes slowly and carefully to consolidate and summarise the facts and concepts of the topic. When you are done, look at the worksheets and attempt the questions again.

For those who have the luxury of studying over weekend:
Set up your survival kit on Friday night and place the physics textbook under your pillow and sleep early. Wake up at 6am on Saturday morning, eat a LIGHT breakfast and finish your ten year series before noon. Take a break at around 9 am and eat some snacks to replenish glucose for your brain. After lunch, if you have not finished your ten year series, continue to work on it. You must finish before 3pm in the afternoon. Spend the rest of the time before dinner reading through notes and running through worksheet questions. After dinner, pick up the phone and call up the class geek to ask for help on questions that you really cannot do. Ask him to give you a quick tutorial on the concepts as well to clear things up even more. Thank him and buy him dinner when the test is over. Then, continue practising the questions well into midnight. Go to bed and wake up around late morning on Sunday, go to church if you are a Christian. If you are not, meditate and pray according to your religious inclinations. Spend the rest of the day to consolidate definitions and formulas, drawing mindmaps and practising questions for speed.

For those who do not have the luxury of a weekend:
You need to block out all the after school commitments and so-called “study groups” with peers. Go home as early as you can, explain to people who want your time that you are rushing home to study for a test. Take your meal the very moment you reach home and start the process as prescribed above. But you probably can only finish ten year series on the first day. Then go to bed as usual, do not burn midnight oil. On second day, work on notes and worksheets. The next day, catch the class geek and schedule some time after school to consult him. Then go home and consolidate your stuff and practise for speed and accuracy. You would be ready for test the next day!

This guide applies not only to those studying three days before a test. It also works for those who are not studying last minute. It is just a tested and proven method to prepare for a test in the shortest time possible. I wish you all the best in Physics Tests!

ExamWorld's addition: This methodology can also be applied to other hard science subjects like mathematics, chemistry, and biology as well.

O lvl A Maths: Applications of Differentiation (Maxima and Minima)

2008-12-03T00:13:00.009+08:00

RI 1999 Sec 4 Common Test Q11

ABCD is a trapezium where AB is parallel to DC, AB = 10 cm, AD = BC = 8 cm and ∡ADC ∡BCD = θ radians. Prove that the area of the trapezium, S cm² is given by S = 16 sin θ (4 cos θ + 5).

Hence, calculate the value of θ for which S has a stationary value. Determine whether this value of S is a maximum or minimum. (12 marks)

*************************

Answer:

Area of trapezium
= area of the 2 side triangles + area of centre rectangle
= 2 * (½ * 8 cos θ * 8 sin θ) + 10 * 8 sin θ
= 64 sin θ cos θ + 80 sin θ
= 16 sin θ (4 cos θ + 5)
(shown)

Differentiate S w.r.t. θ

Set dS/dθ = 0 to find stationary value

cos θ = 0.4605823 or -1.0855823 (reject)
θ = 62.6°

When θ = 62.6°,

Hence, the value of S is a maximum.

O lvl E Maths: Number Patterns

2008-12-02T12:54:00.004+08:00

(a) Complete the number pattern by writing out the values of a, b, c and d.

Pattern	Sum, S	(S + 2)
2	2	4
2, 3, 2	7	9
2, 3, 4, 3, 2	a	c
2, 3, 4, 5, 4, 3, 2	b	d

(b) What sort of numbers do you have in the (S + 2) column?

(c) Find the sum of 2, 3, 4, ......, 99, 100, 99, ......, 4, 3, 2.

(d) If the middle number of the pattern of positive numbers is M, write a formula relating S and M.

(e) Find the sum of -2, -3, -4, ..., -66, -67, -66, ..., -4, -3, -2.

(f) If 2 + 3 + 4 + 5 + ... + (k - 1) + k + (k - 1) + ... + 5 + 4 + 3 + 2 = 223, find the value of k.

*************************

Answer:

(a)

Pattern	Sum, S	(S + 2)
2	2	4
2, 3, 2	7	9
2, 3, 4, 3, 2	14	16
2, 3, 4, 5, 4, 3, 2	23	25

(b) Perfect squares

(c) Sum = 100² - 2 = 9998

(d) S = M² - 2

(e) sum of -2, -3, -4, ..., -66, -67, -66, ..., -4, -3, -2
= negative of sum of 2, 3, 4, ..., 66, 67, 66, ..., 4, 3, 2
= - (67² - 2)
= -4487

(f) k² - 2 = 223
k² = 225
k = 15

A lvl H2 Chem: Atoms Molecules and Stoichiometry

2008-12-01T13:11:00.002+08:00

From RJC Chem Tutorial (2000/2001)

Explain the following observations as fully as you can:

(a) When a beam of alpha particles is fired at a thin sheet of gold foil, the majority pass through the foil and are either not deflected at all or deflected through only a small angle; however a few are deflected through large angles, some even greater than 90 degrees.

(b) The first ionization energy of Nitrogen is higher than that of either the elements immediately preceding or following it in the Periodic Table.

(c) Once the 3s and 3p orbitals are filled, subsequent electrons enter the 4s orbital first, rather than 3d orbitals. However, when transition metals form ions, electrons are lost from 4s orbital first then 3d orbitals

*************************

Answer:

(a) There are large spaces in the gold atoms for the alpha particles to pass through undeflected of only slightly repelled by the nucleus. ( alpha particles are positively charged ) However, when some of the alpha particles come into contact with the nucleus, they will be deflected at various angles depending on the angle of collisions.

(b) The effective nuclear charge on the valence electrons is greater in N than in C, therefore more energy required to ionize N.

*In O , the removal of 1 electron in the paired electrons would lower electron-electron-repulsion, enabling a more stable configuration. Thus, less energy is required to achieve this preferred configuration. Therefore, more energy required to ionize N in comparison.

(c) 4s is at a lower energy level than 3d, therefore they are filled up first. However, 3d is closer to the nucleus than 4s. Once both 3d and 4s are occupied by electrons, 3d electrons repel the 4s electrons even further away from the nucleus and up to a higher energy level, becoming more unstable. Therefore 4s electrons will be removed first during ionization.

O lvl A Maths: Applications of Differentiation (Rates of Change)

2008-11-30T01:04:00.005+08:00

The figure shows the curve y² = 2x. The points O (0,0) and P(x, y) lie on the curve. Given that the coordinates of Q is (x, 0),

(a) express the area, A cm², of the triangle OPQ in terms of x.
(b) If x is increasing at a rate of 4 units per seconds, calculate the rate of increase of A when x = 2.

*************************

Answer:

(a) Area of triangle OPQ, A = ½(OQ)(PQ) = ½xy
y² = 2x

Since area of triangle is positive, we take the positive value of y

Hence,

(b) Using product rule,

If x is increasing at a rate of 4 units per seconds, then dx/dt = 4.
(chain rule)

When x = 2, dA/dt = 6(2) / √(2)(2)
= 6

O lvl A Maths: Applications of Differentiation (Rates of Change)

2008-11-29T01:22:00.006+08:00

The lengths of the base and height of a right-angled triangle are (x - 1) cm and (x + 2) cm respectively. The length x cm at time t seconds is given by the expression x = 4t + 1, where t ≥ 0.

(a) Show that the area of the triangle in terms of t is given by A = 2t (4t + 3)
(b) Hence, find the rate of change of the area at the instant when t = 2.

*************************

Answer:

(a) Area of triangle, A = ½(x - 1)(x + 2)

Substituting x = 4t + 1 into the equation
A = ½(4t + 1 - 1)(4t + 1 + 2)
= ½ (4t)(4t + 3)
= 8t² + 6t
= 2t (4t + 3)
(shown)

(b) A =8t² + 6t
dA/dt = 16t + 6

When t = 2, dA/dt = 16(2) + 6 = 38

Thus, the area of the triangle is increasing at a rate of 38 cm² per second at the instant when t = 2.

O lvl A Maths: Trigonometry

2008-11-28T00:49:00.002+08:00

Solve each of the following equations for the domain 0° ≤ x ≤ 360°.

(a) 3 cos 2x = cos x + 2
(b) tan 2y = 4 tan y

*************************

Answer:

(a) 3 cos 2x = cos x + 2
Sub cos 2x = 2 cos² x - 1

3 (2 cos² x - 1) = cos x + 2
6 cos² x - 3 - cos x - 2 = 0
(6 cos x + 5) (cos x - 1) = 0
cos x = -5/6 or cos x = 1

For cos x = -5/6, x = 146.4°, 213.6°
For cos x = 1, x = 0°, 360°

Thus, x = 0°, 146.4°, 213.6°, 360°

(b) tan 2y = 4 tan y
Sub tan 2y = (2 tan y)/(1 - tan² y)

(2 tan y)/(1 - tan² y) = 4 tan y
2 tan y = 4 tan y (1 - tan² y)
2 tan y = 4 tan y - 4 tan³ y
4 tan³ y - 4 tan y + 2 tan y = 0
4 tan³ y - 2 tan y = 0
2 tan³ y - tan y = 0
tan y (2 tan² y - 1) = 0
tan y = 0 or tan² y = 0.5

For tan y = 0, y = 0°, 180°, 360°

For tan² y = 0.5,

or

y = 35.3°, 215.3° or y = 144.7°, 324.7°

Combining,
y = 0°, 35.3°, 144.7°, 180°, 215.3°, 324.7°, 360°

O lvl A Maths: Applications of Differentiation (Maxima and Minima)

2008-11-27T00:20:00.003+08:00

A closed cylindrical can with base radius r cm and height h cm is constructed from a thin sheet of metal to hold 50π cm³ of liquid.

(a) Show that the total area, A cm², of material needed to make this can is given by A = 2πr² + 100π/r.
(b) Find the values of r and h if the area of material used is to be the least.

*************************

Answer:

(a) Volume of the cylindrical can = πr²h
πr²h = 50π
h = 50/r²

Total surface area of cylindrical can,
A = 2πr² + 2πrh
= 2πr² + πr(50/r²)
= 2πr² + 100π/r
(shown)

(b) A = 2πr² + 100π/r
dA/dr = 4πr - 100π/r²

For maximum or minimum values,
dA/dr = 0
4πr - 100π/r² = 0
4πr = 100π/r²
r³ = 25

When ,

= 12π
>0

==> A is a minimum when

Substituting into h = 50/r²,

O lvl E Maths: Trigonometry Bearings

2008-11-26T00:47:00.002+08:00

In the diagram, the bearing of A from B is 320° and the bearing of B from C is 240°. Given that AB = BC = 10 km, calculate
(i) ∡BCA,
(ii) the bearing of C from A.

*************************

Answer:

(i) ∡QCB = 240° - 180° = 60°
∡NBC = 60° (alternate angles)
∡NBA = 360° - 320° = 40° (angles at a point)
∡ABC = 40° + 160° = 100°
∡BCA = (180° - 100°) / 2 = 40° (angles in an isosceles triangle)

(ii) ∡ACQ = 40° + 60° = 100°
Bearing of C from A = 100° (alternate angles)

O lvl E Maths: Ratio Rate Proportion

2008-11-26T00:28:00.002+08:00

RI 1998 Sec 3 EOY Q2

3 men, working 8 hours a day, can finish a job in 6 days. How many days would it take 2 men working 9 hours a day to complete the same job, assuming that they work at the same rate?

*************************

Answer:

Total number of man-hours when 3 men are working 8 hours a day for 6 days
= (3 * 8 * 6) man-hours

Days taken to complete when 2 men are working 9 hours a day
= (3 * 8 * 6) / (2 * 9)
= 8 days

O lvl A Maths: Equation of Circle

2008-11-25T00:51:00.002+08:00

(a) Show that the line 4y = x - 3 touches the circle

x² + y² - 4x - 8y + 3 = 0

(b) The straight line x = 3 intersects the circle
x² + y² - 8x - 10y - 9 = 0 at the points P and Q.
Find the equations of tangents at P and Q.

*************************

Answer:

(a) 4y = x - 3
x = 4y + 3

Sub x = 4y + 3 into x² + y² - 4x - 8y + 3 = 0
(4y + 3)² + y² - 4(4y + 3) - 8y + 3 = 0
16y² + 24y + 9 + y² - 16y - 12 - 8y + 3 = 0
17y² = 0
y = 0
Hence, x = 3

The line 4y = x - 3 touches the circle x² + y² - 4x - 8y + 3 = 0 at (3, 0)
(shown)

(b) When x = 3,
3² + y² - 8(3) - 10y - 9 = 0
y² - 10y - 24 = 0
y = -2 or y = 12

From equation, centre of circle = (4, 5)

Point P (3, -2)
Gradient between P and centre of circle = (5 - [-2]) / (4 - 3) = 7
Thus, gradient of tangent = -1/7
y + 2 = (-1/7)(x - 3)
7y + 14 = -x + 3
7y = -x - 11

Point Q (3, 12)
(I'm going to do a short cut here)
Basically, if we reflect any straight-line graph across a horizontal line, the gradient of it reverses.

So gradient of tangent at P is -1/7
Gradient of tangent at Q is 1/7.

Thus,
y - 12 = (1/7)(x - 3)
7y - 84 = x - 3
7y = x + 81

Alternative
Use implicit differentiation on the curve at x = 3 to find the gradient.

O lvl E Maths: Trigonometry

2008-11-25T00:31:00.004+08:00

In the diagram, ADC is a straight line. AB = 11 cm, BD = 10 cm and CD = 3 cm. Using as much of the information given below as is necessary, calculate
(i) sin ∡BAD,
(ii) BC²,
(iii) the area of triangle BCD

given that sin ∡ADB = 0.88, cos ∡ADB = -0.47, tan ∡ADB = -1.88

*************************

Answer:

(i) Using sine rule,
sin ∡BAD / 10 = sin ∡ADB / 11
sin ∡BAD = 0.88 / 11 * 10
sin ∡BAD = 0.8

(ii) cos ∡BDC = - cos ∡ADB = 0.47

Using cosine rule,
BC² = BD² + CD² - 2 (BD)(CD)(cos ∡BDC)
BC² = 10² + 3² - 2(10)(3)(0.47)
BC² = 80.8

(iii) sin ∡BDC = sin ∡ADB = 0.88

Area of triangle BCD
= ½(BD)(CD) sin ∡BDC
= ½(10)(3) (0.88)
= 13.2 cm²

O lvl E Maths: 2D Vectors

2008-11-23T00:34:00.006+08:00

In ∆ABC, D, E and F are the midpoints of BC, AB and AC respectively. The lines AD and CE intersect at point K.

(a) Given that AB = 6p and AC = 10q, express in terms of p and q:
(i) BC
(ii) AD
(iii) CE

(b) Given that AK = h AD and by considering ∆ABK, express BK in terms of h, p and q.

(c) Given that CK = k CE and by considering ∆BCK, express BK in terms of k, p and q.

(d) Using these 2 expressions for BK, find the values of h and k.

(e) Prove that BK, when produced, will pass through F.

*************************

Answer:

(a)
(i) BC = BA + AC = -6p + 10q

(ii) AD = AB + BD
= AB + ½BC
= 6p + ½(-6p + 10q)
= 3p + 5q

(iii) CE = CA + AE
= CA + ½AB
= -10q + ½(6p)
= 3p - 10q

(b) BK = BA + AK
= BA + hAD
= -6p + h(3p + 5q)
= (3h - 6)p + 5hq

(c) BK = BC + CK
= BC + kCE
= -6p + 10q + k(3p - 10q)
= (3k - 6)p + (10 - 10k)q

(d) Comparing coefficients of p,
3h - 6 = 3 k - 6
h = k

Comparing coefficients of q,
5h = 10 - 10k
Sub h = k
5k = 10 - 10k
15k = 10
k = h = 2/3

(e) BF = BA + AF
= BA + ½AC
= -6p + ½(10q)
= -6p + 5q

BK = [3(⅔) - 6] p + 5(⅔)q
= -4p + (10/3) q
= ⅔(-6p + 5q)
= ⅔BF

Since BK = ⅔BF, hence, BK produced will meed AC at F.

O lvl E Maths: Probability

2008-11-16T00:27:00.003+08:00

Given that
X = { x : x is an integer and 5 ≤ x ≤ 10 }
Y = { y : y is a multiple of 3 and 1 ≤ y ≤ 10 }

An element is chosen at random from each set.
Write down the total number of possible outcomes.

Using a probability diagram or otherwise, find the probability that:
(i) the product xy is a perfect square.
(ii) the product xy is an odd number.
(iii) the product xy > 20.

*************************

Answer:

X = { x : 5, 6, 7, 8, 9, 10}
Y = { y : 3, 6, 9}
Total number of outcomes = 6 numbers * 3 numbers = 18

(i) Perfect squares = 6² and 9²
so, probability that product xy is a perfect square = 2/18 = 1/9

(ii) xy is an odd number means both x and y must be odd numbers
Number of odd numbers in x = 3
Number of odd numbers in y = 3
so, probability that product xy is an odd number = 3 * 3 / 18 = 1/2

(iii) Number of terms of xy < 20 : 5 * 3, 6 * 3
===> only 2 terms.

So, 16 terms are greater than 20
Probability of product xy > 20 = 16/18 = 8/9

O lvl E Maths: Probability

2008-11-16T00:08:00.002+08:00

Two bags each contains 91 balls. In each bag, there are 3 red balls, 1 white and the rest blue.

(i) One ball is drawn from the first bag. Find the probability that it is not blue.

(ii) One ball is drawn from the second bag. Find the probability that it is blue.

(iii) One ball is drawn from each bag. Find the probability that the two balls are
(a) both blue,
(b) each of a different colour.

*************************

Answer:

In each bag, there are 3 red balls, 1 white and 87 blue.

(i) P(ball is not blue) = 4/91

(ii) P(ball is blue) = 87/91

(iii)
(a) P(both balls are blue) = 87/91 * 87/91
= 7569/8281

(b) P(both balls same colour)
= P(both red) + P(both white) + P(both blue)
= 3/91 * 3/91 + 1/91 * 1/91 + 7569/91
= 7579/8281

P(both balls are different colour) = 1 - P(both balls same colour)
= 1 - 7579/8281
= 702/8281

O lvl E Maths: Probability

2008-11-15T00:47:00.003+08:00

A student attends either a Mathematics class or an Additional mathematics class everyday. The probability of him bringing the wrong textbook is 3/5 if he attends the Mathematics class and 3/10 if he attends the Additional Mathematics class.

(a) If he attends Mathematics class on three consecutive days, find the probability that he will bring the wrong textbook
(i) on all the three days
(ii) on just one of the three days.

(b) If he is equally likely to attend the Mathematics class or the Additional Mathematics class, find the probability that he will bring the correct textbook to class on any given day.

*************************

Answer:

(a)
(i) P(wrong textbook on all three days)
= P(1st day wrong textbook) * P(2nd day wrong textbook) * P(3rd day wrong textbook)
= 3/5 * 3/5 * 3/5
= 27/125

(ii) P(just one of three days wrong textbook)
= P(1st day wrong, 2nd, 3rd day correct) + P(2nd day wrong, 1st, 3rd day correct) + P(3rd day wrong, 1st, 2nd day correct)
= 3/5 * 2/5 * 2/5 + 2/5 * 3/5 * 2/5 + 2/5 * 2/5 * 3/5
= 36/125

(b) P(correct textbook for Mathematics or for Additional Mathematics)
= P(attend Mathematics and bring correct txtbook for Mathematics) + P(attend Additional Mathematics and bring correct txtbook for Additional Mathematics)
= 1/2 * 2/5 + 1/2 * 7/10
= 11/20

O lvl Phy: Light

2008-11-15T00:09:00.002+08:00

Contributed by Javid
St Gabriel’s Sec/SA1/2008

The graph shows how the image distance change as an object is moved from a distance gradually towards a converging lens.

(a) What does the graph tells you about the nature of the image in region C? [1]

(b) There is a position along the principal axis where the image size is equal to the object size. Make use of this fact to find the focal length of the lens. [2]

(c) If the object is placed at P, describe the change in the size of the image observed if the object is moved closer to the lens. [1]

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Answer:

(a) The image is virtual, enlarged, upright and on the same side as the object [1].

(b) When object distance u = image distance v, the object and image should be of the same size. This occurs at 20 cm, which is actually twice the focal length [1].
Hence, focal length = 10 cm [1].

(c) The image increases in size as it gets closer to the lens [1].

O lvl A Maths: Applications of Differentiation (Maxima and Minima)

2008-11-14T20:02:00.005+08:00

Contributed by Javid
Adapted from an outdated textbook

A man rows 30 metres out to sea from point P on a straight coast. He reaches a point M such that MP is perpendicular to the coast. He then wishes to get as quickly as possible to a point Q on the coast 400 metres from P. If he can row at 40 m/min and cycle at 50 m/min, how far from P should he land?

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Answer:

Let the man land at the point R, x metres from P.

Using Pythagorean Theorem,
the distance he needs to row =

The distance he needs to cycle = (400 – x) m

Total time taken:

For a stationary value of T,

25x² = 16 (900 + x²)
9x² = 14400
x² = 1600
x = 40 (since x > 0)

When x = 39,
is negative.

When x = 41,
is positive.

=> the value of T = 40, is indeed a minimum point.

Thus, the man should land 40 m from P in order to reach Q as quickly as possible.

A lvl H2 Maths: APGP

2008-11-12T20:06:00.003+08:00

H2 Maths 2008 P1 Q10

(i) A student saves $10 on 1 January 2009. On the first day of each subsequent month she saves $3 more than in the previous month, so that she saves $13 on 1 February 2009, $16 on 1 March 2009, and so on. On what date will she first have saved over $2000 in total? [5]

(ii) A second student puts $10 on 1 January 2009 into a bank account which pays compound interest at a rate of 2% per month on the last day of each month. She puts a further $10 into the account on the first day of each subsequent month.

(a) How much compound interest has her original $10 earned at the end of 2 years? [2]
(b) How much in total is in the account at the end of 2 years? [3]
(c) After how many complete months will the total in the account first exceed $2000? [4]

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Answer:

(i)
first term = 10
2nd term = 13
3rd term = 16
so this is an AP, first term 10, common difference 3

When sum of AP ≥ 2000
n/2 (2*10 + (n-1)*3) = 2000
n (20 + 3n - 3) = 4000
3n² + 17n - 4000 = 0
Use your graphic calculator: n=33.79 or -39.46 (reject)
So on the 34th month, she will save over $2000
Date is 1st Oct 2011

(ii)
(a) end of 2 years = 24 months
Amount of compound interest = 10(1.02)24 - 10 = $6.08

(b) end of 2nd month = 10(1.02)2 + 10(1.02)
end of 3rd month = 10(1.02)3 + 10(1.02)2 + 10(1.02)
end of 4th month = 10(1.02)4 + 10(1.02)3 + 10(1.02)2 + 10(1.02)

So end of 24th month = 10(1.02)24 + 10(1.02)23 + ... + 10(1.02)2 + 10(1.02)
= sum of GP with first term 10(1.02) and common ratio 1.02 for 24 terms
= $10(1.02) (1.0224 - 1) / (1.02 - 1)
= $310.30

(c) Supposed at the end of n months, the account will first exceed $2000

sum of GP with first term 10(1.02) and common ratio 1.02 for n terms ≥ 2000
10(1.02) (1.02n - 1) / (1.02 - 1) ≥ 2000
(1.02n - 1) ≥ 3.92157
1.02n ≥ 4.92157
n ≥ lg (4.92157) / lg (1.02)
n ≥ 80.475

End of 81st month, amount = 10(1.02) (1.0281 - 1) / (1.02 - 1) = $2026.20
Start of 81st month = $2026.2 / 1.02 = $1986.47 (not needed, show for fun only)

The account first exceed $2000 after 81 complete months.

A lvl H2 Maths: Vectors

2008-11-12T19:58:00.006+08:00

H2 Maths 2008 P1 Q11

The equations of three planes, p1, p2, p3 are

2x – 5y + 3z = 3,
3x + 2y – 5z = -5,
5x + λy + 17z = μ,

respectively, where λ and μ are constants. When λ = -20.9 and μ =16.6, find the coordinates of the point at which these planes meet. [2]

The planes p1 and p2 intersect in a line l.

(i) Find a vector equation of l. [4]
(ii) Given that all three planes meet in the line l, find λ and μ. [3]
(iii) Given instead that the three planes have no point in common, what can be said about the values of λ and μ? [2]
(iv) Find the Cartesian equation of the plane which contains l and the point (1, -1, 3). [4]

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Answer:

Use Graphic calculator to solve for 3 unknowns
x = -4/11, y=-4/11, z = 6/11

(i) 2x – 5y + 3z = 3 ---- (1)
3x + 2y – 5z = -5 ---- (2)

line parallel to both planes = (2, -5, 3) x (3, 2, -5)
= (19,19,19) = 19 (1,1,1)

Let z = 0
2x - 5y = 3
3x + 2y = -5
Solving, x = -1, y = -1
(-1,-1,0) is a point on both planes

So, equation of vector l = (-1,-1,0) + t (1,1,1)
Note: Alternatively, use the GC to solve.

(ii) (5,λ,17) dot (1,1,1) = 0
5 + λ + 17 = 0
λ = -22

5x + -22y + 17z = μ
Sub (0,0,1)
μ = 17

(iii) Three planes have no points in common, means p3 does not intersect vector l
This means p3 must be parallel to vector l, but have no points on vector l
Thus, λ will still be -22, and μ can be anything except 17

(iv) Plane must also contain points (0,0,1) and (1,-1,3)
and be perpendicular to (1,1,1)

Vector linking points (0,0,1) and (1,-1,3) is (1,-1,2).

So plane must be perpendicular to both (1,1,1) and (1,-1,2)
Using cross product,
(1,1,1) x (1,-1,2) = (3,-1,-2)

so, cartesian equation of plane is 3x -y -2z = D
sub in (0,0,1)
D = -2

So final answer: 3x -y -2z = -2

A lvl H2 Phy: Thermal Physics/Ideal Gas

2008-11-12T00:44:00.004+08:00

State, in words, the relation between the increase in the internal energy of a gas, the work done on the gas, and the heat supplied to the gas.

(a) A quantity of 0.200 mol of air enters a diesel engine at a pressure of 1.04 * 10⁵ Pa and at a temperature of 297 K. Assuming that air behaves as an ideal gas, find the volume of this quantity of air.

(b) The air is then compressed to one twentieth of this volume, the pressure having risen to 6.89 * 10⁵ Pa. Find the new temperature.

(c) Heating of the air then takes place by burning a small quantity of fuel in it to supply 6150 J. This is done at a constant pressure of 6.89 * 10⁵ Pa as the volume of air increases and the temperature rises to 2040 K. Find
(i) the volume of the gas after burning the fuel,
(ii) the work done by the air during this expansion,
(iii) the change in the internal energy of the air during this expansion.

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Answer:

The increase in internal energy of a gas is equal to the sum of the work done on the gas and the heat supplied to the gas.

(a) PV = nRT
V = nRT/P
= (0.200 * 8.31 * 297) / (1.04 * 10⁵)
= 4.75 * 10^-3 m³

(b) Given V' = V/20, and P' = 6.89 * 10⁵ Pa,

= 98.4 K

(c)
(i) T₁/T₂ = V₁ / V₂ (when pressure is constant)
V₂ = 2040 / 98.4 * V'
V₂ = 2040 / 98.4 * 2.37 * 10^-4
= 4.92 * 10^-3 m³

(ii) Work done by air
= p ΔV
= 6.89 * 10⁵ * (4.92 - 0.237) * 10^-3
= 3230 J

(iii) ΔU = Q supplied + WD on air
= Q supplied - WD by air
= 6150 - 3230
= 2920 J