O-Level Additional Math

[OK Lor]

Hi CoffeeCat,


Thanks. The method was suggested by my dad . From the Remainder Theorem, when f(x) is divided by x-1 (or 7), the remainder will be f(1).
Modular arithmetic -> lx+4l >5? ok lor.
Congruences -> plane geometry? (needs to use a lot of right brain :lol: )

[CoffeeCat]

OK Lor:

Hi CoffeeCat,


Thanks. The method was suggested by my dad . From the Remainder Theorem, when f(x) is divided by x-1 (or 7), the remainder will be f(1).
Modular arithmetic -> lx+4l >5? ok lor.
Congruences -> plane geometry? (needs to use a lot of right brain :lol: )

no la modular arithmetic is merely a branch of mathematics with notation to deal with remainder, just like how logarithms is invented just to deal with exponents.
The cool thing about modular arithmetic and congruences is it make our life easier when expressing our ideas.

[OK Lor]

Hi CoffeeCat,


Cool! It's useful for SMO, will put in some efforts to understand the concept. Thanks.
a = b (mod m).

[OK Lor]

Hi Sir,


Please help (SMO 2010):

1. Find the value of (14^3 + 15^3 + 16^3 + … + 24^3 + 25^3)^(1/2).
2. Find the least possible value of f(x) = 9/(1+cos2x) + 25/(1-cos2x), where x ranges over all real numbers for which f(x) is defined.

Thanks.

[CoffeeCat]

OK Lor:

Hi Sir,


Please help (SMO 2010):

1. Find the value of (14^3 + 15^3 + 16^3 + .... + 24^3 + 25^3)^(1/2).
2. Find the least possible value of f(x) = 9/(1+cos2x) + 25/(1-cos2x), where x ranges over all real numbers for which f(x) is defined.

Thanks.

for smo u need a lot of tricks up your sleeve, for this you need to know the formula for sum of consecutive cubes.
The sum of consecutive cubes from 1 to n^3 is given by [ n(n+1)/2]^2.
so plug it in, [25*26]^2 - [13*14]^2 = 13^2 (50-14)(50+14)= ...
factorization rules always comes in handy.

[YLH88]

[quote]
mrswongtuition wrote :
My sis' Maths was terrible too. We got a HOD Maths from a Secondary school to coach her 1-to-1 for E & A Maths. She got A2 for E Maths and B3 for A Maths. Very good for my sister who used to fail A Maths and borderline pass E Maths.

If you are willing to spend on a good tutor, it's not too late now. Still have 9+ months to prepare [/quote]Hi mrswongtuiton,

I would like to have the contact for the Maths tutor. Have just pm you.

Thank you.

[OK Lor]

Hi CoffeeCat,


Thanks for Q1. Please help to comment on Q2:

f(x) = 9/(1+cos2x) + 25/(1-cos2x) = 9/[2(cos x)^2] + 25/[2(sinx)^2]
= [9(sec x)^2 + 25(cosec x)^2] / 2
f'(x) = 9[(sec x)^2](tan x) - 25[(cosec x)^2](cot x) = 0
9[(sec x)^2](tan x) = 25[(cosec x)^2](cot x)
(tan x)^4 = 25 / 9
tan x = (5/3)^(1/2); cos x = (3/8 )^(1/2), sin x = (5/8 )^(1/2).
Least possible value of f(x) = [9(8/3) + 25(8/5)] / 2 = 32 ?

Thanks.

[Muffins]

Just a question to ask. For my exam, I had gotten a question like this:


a) Express 756 in its index form

b) Hence, find the value k where 756k is a perfect square.

I had gotten part a) correct, but I had put 2⅓ for part b), but they marked me wrong, saying that the answer was 21. Could anyone explain why? Thanks.

[Vanilla Cake]

Muffins:

Just a question to ask. For my exam, I had gotten a question like this:


a) Express 756 in its index form

b) Hence, find the value k where 756k is a perfect square.

I had gotten part a) correct, but I had put 2⅓ for part b), but they marked me wrong, saying that the answer was 21. Could anyone explain why? Thanks.

756=2²x3³x7
(a) 756 in its index form = 2²x3³x7

Perfect square =2²x3⁴x7²
k = 3x7 = 21

(b)Value of k where 756k is a perfect square = 21

Pls do not :offtopic:

[atutor2001]

Muffins:

Just a question to ask. For my exam, I had gotten a question like this:


a) Express 756 in its index form

b) Hence, find the value k where 756k is a perfect square.

I had gotten part a) correct, but I had put 2⅓ for part b), but they marked me wrong, saying that the answer was 21. Could anyone explain why? Thanks.

There is probably a requirement in the question that \"k is a whole number, find the smallest value of k\", which is why your answer is not acceptable