Hi Sir,


:welcome: :celebrate:

Hi,

Pls help on (ii):
A curve has parametric equation x = 2t – 1, y = 1/(t² + 1).
(i) Prove that the equation of the tangent at the point with parameter t is (t² + 1)² y + tx = 3t² - t + 1.
(ii) The tangent at point where t = 3 meets the curve again at the point where t = q. Find the value of q.
Ans: -4/3

Thanks.

mathqa:

OK Lor:

Hi,


Please evaluate
http://www.postimage.org/

Thanks.

Let u=tan(x). Rewrite it in term of u, it would become an integral of an elementary function in form of

http://lh6.ggpht.com/_nr85VD4DdiA/TPjKaT7U5HI/AAAAAAAAAE0/jSSE2KNWb9E/s800/elementary-function-integral.png\">

Hi mathqa, thanks .

Hi,


Please evaluate
http://www.postimage.org/

Thanks.

iFruit:

OK Lor:

Hi,


If y = eˣ / cos x, show that d²y/dx² = 2 tanx dy/dx + 2y.

Thanks.

y = eˣ / cos x = eˣ sec x

dy/dx = eˣ sec x + eˣ sec x tan x = eˣ sec x (1+ tan x) = y (1+ tan x)

d²y/dx² = dy/dx (1 + tan x) + y ( sec² x) = dy/dx (1 + tan x) + y ( 1 + tan² x) = y(1+ tan x)(1+tan x) + y (1 + tan² x) = 2y (1 + tan x + tan² x) ----(1)

2 tanx dy/dx + 2y = 2 tanx . y (1+ tan x) + 2y = 2y ( 1 + tan x + tan² x)---- (2)

From (1) and (2), d²y/dx² = 2 tanx dy/dx + 2y.

HTH

Hi iFruit,

Thanks for your help. Was stuck and coundn't continue from (1) to get the answer without (2), which it was just one more step to go, the reverse of (2) :lol:

Hi,


If y = eˣ / cos x, show that d²y/dx² = 2 tanx dy/dx + 2y.

Thanks.

mathqa:

OK Lor:

Hi,


Function f is defined by f(x) = { ax² + bx – 2, x ≤1; 2 – 1/x², x > 1
where a and b are constants. If f(x) is differentiable at x = 1, find the values of a and b.
Ans: a = -1, b = 4

Thanks.

1. f(x) is differentiable at x=1 -> f'(1-) = f'(1+) -> 2a+b=2
( 1+ denotes right hand side of 1,
1- denotes left hand side of 1,)

2. since f(x) is differentiable at x=1, it is continous at x=1 (please note that the converse is not true)
f(1-)=f(1+) -> a+b-2=1

3. In summary, 2a+b=2, a+b=3 -> Ans.

Thank you mathqa

Hi,


Function f is defined by f(x) = { ax² + bx – 2, x ≤ 1; 2 – 1/x², x > 1
where a and b are constants. If f(x) is differentiable at x = 1, find the values of a and b.
Ans: a = -1, b = 4

Thanks.

mathqa:

iFruit:

[quote=\"OK Lor\"]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 OK Lor,

Is this A-level question? AFAI can see, The equation is a non-homogeneous linear differential equation, so it needs some background to understand the solution..

iFruit is right to point out that this question is non-homogeneous linear differential equation. But it is nice to have for A level since it would test students's competence in many areas at once.

The solution is is posted at my blog. Cannot post it here due to oversize images.

http://mathqa.blogspot.com/2010/11/nonhomgeneous-differential-equations.html

MathQA[/quote]Hi,

Thanks to all the Maths Guru here

Hi iFruit,

This question is from Malaysia A level paper which I came across while browsing the web :!: .
Thanks.