Hi,


I’m not taking A-level Maths this year, but interested to know the solution for part (b):

If x(t) and y(t) are variables satisfying the differential equations
dy/dt + 2 dx/dt = 2x +5 and dy/dt – dx/dt = 2y + t,
(a) Show that 3 d²y/dt² - 6 dy/dt + 4y = 2 – 2t.
(b) Find the solution x in terms of t for the second order of differential equation given that y(0) = y’(0) = pi.

Thanks.

Hi iFruit,


Thank you very much

Hi,


The question is a little off topic, please help to factorise x⁴+ 4

Thanks.

Hi Sir,


What is the least possible value of
(a^2 + 9)^(1/2) + [(b-a)^2 + 4 ]^(1/2) + [(8-b)^2 + 16]^(1/2)
for real numbers a and b?

Thanks.

Hi 高人 CoffeCat,


Thanks.

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.

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.

Hi CoffeeCat,


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

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: )

Hi Sir, thanks.


Hi Coffecat,
Got it, nice solution, thanks.
Just got to know what was the problem to my working,
=> f(1) = 2^(3n+1) + 2k - 1 = 0 (wrong, as the remainder can be 0 or multiple of 7).
=> f(1) = 2^(3n+1) + 2k - 1 = 2(7 +1)^n + 2k - 1.
Remainder of f(1) / 7 = 2 + 2k - 1 = 2k +1.
k = 3, 10, 17, 24, …, 94.
Total 14 number of positive integers k.