gjooheng:

exactly one yellow tile and 2 blue tiles in the same row can happen in the 4th or 3rd row, or both rows.


So using the venn diagram property n(AUB) = n(A)+n(B)-n(AnB)
The circled part is n(AnB)

http://i59.tinypic.com/2uiacsk.jpg\">

OK Lor:

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\">

OK Lor:

Hi,


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

Thanks.

OK Lor:

iFruit:

[quote=\"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

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

OK Lor:

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.

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.

mramk:

iFruit:

[quote=\"mathqa\"]

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.

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

Hi MathQA,

Nice working. But I think there is a small mistake in it. You can't apply initial conditions to the homogeneous equation (equation 6) as they are initial conditions for non-homogeneous equation. That's why your solution doesn't tally back for initial conditions. Also particular solution C = -11/4

The solution after applying initial conditions to the general equation should be (pi = π)

x(t) = ((1-2π)/4) e^t cos(t/√3) - (√3(1+2π)/4) e^t sin(t/√3) - 11/4

Regards.

atutor2001:

Hi Chief & iFruit


Thank you for your clarifications. Must set resolution for next year - \"To study A-level math with my youngest one\". If don't know sure will post here. Piaseh, you all gonna be very busy. Many many thanks in advance first.

Cheers