3u Mathematics Marathon V 1.1 (1 Viewer)

[Smiley :D CvH]

What does the w.r.t.x actually mean? Make x the subject?

Spoiler

∫[-1 to 5] (x² - 9) dx
= [x³/3 - 9x] [-1 to 5]
= 12

i dunno gt da answer = -12 are u referin 2 da da number on the bottom of the integral symbol being -1 and the top being 5?
or ish it the other way around, bt if it is sorrii my mistake

 

[Pianpupodoel]

okay, now to continue the grand marathon:

this was posted in the 2U maths marathon thread, but it was deemed too hard for 2U, so instead, it is here:

Question: I am in a poker game, and I am dealt 5 cards from an ordinary 52-card deck. What is the PROBABILITY (hehehe i know people have P'lty) that I will get:
(i) one pair
(ii) a two pair
(iii) a triple
(iv) a straight
(v) a flush
(vi) a full house
(vii) a four-of-a-kind
(viii) a royal flush?
(ix) high card? (ie no poker combinations...i leave this till last cos i think this is the hardest one...)

Seeing as though no one else seems to want to try the question, the answers, as well as good mathematical explanations behind them, can be found here:
http://www.math.sfu.ca/~alspach/comp18/

Continuing the probability and counting theme, this question is taken from the Y12 3U Cambridge textbook.
There are ten basketballers in a team. Find how many ways they can be split into two teams of five.

 

[icycloud]

There are ten basketballers in a team. Find how many ways they can be split into two teams of five.

Spoiler

10C5 / 2

Explanation:

Let the two teams be X and Y.
First, choose five players for X -- there are 10C5 ways to do this (the rest goes to team Y).
However, we could have chosen Y first, and then X, so we double-counted and must divide 10C5 by 2.

Next Question:
Change of themes here . Find ∫dx/( sqrt(x) * (1+x)).

 

[icycloud]

Fairly easy integration question:

∫dx/(1 + e^-x)

Might as well do this as well .

Spoiler

I = ∫dx/(1 + e^-x)

u = e^x, du/u = dx

I = ∫du/(u(1+u^-1))
= ∫du/(u+1)
= ln(u+1) + C
= ln(e^x + 1) + C
#

Next question:
Find all real x such that:
|4x - 1| > 2 sqrt(x(1-x))

 

[Mountain.Dew]

Next question:
Find all real x such that:
|4x - 1| > 2 sqrt(x(1-x))

more of my 2 cents...

Spoiler

first of all, we have a domain restriction -->x(1-x) >= 0, since a sqrted number cant be negative IN the real number system.

so, domain is restricted (so far) to 0 <= x <= 1

|4x-1| > 2sqrt(x(1-x)) ----> (4x-1)^2 > 4(x)(1-x)
16x^2 -8x + 1 > 4x(1-x)
16x^2 - 8x + 1 > 4x-4x^2
20x^2-12x+1 > 0 --> after factorising...
(10x-1)(2x-1) > 0, clearly observed from graph of y=(10x-1)(2x-1), that the solutions are --> x < 10, x > 1/2

but, we have restriction in domain... 0 <= x <= 1

FINAL solution is --> 0 < x < 1/10, 1/2 < x < 1.

 

[Mountain.Dew]

Next Question:
Change of themes here . Find ∫dx/( sqrt(x) * (1+x)).

more 2 cents thrown around here.

Spoiler

I = ∫dx/( sqrt(x) * (1+x))
let u = sqrt x.

du/dx = 1/2sqrt x, dx = 2sqrt x du = 2u du

so, I = ∫ 2u du / u(1+u²) = 2 ∫ du/(1+u²)
I = 2 tan^-1u + c
I = 2 tan^-1sqrt(x) + c

 

[Templar]

The answer to the inequality question could be found here.

 

[Mountain.Dew]

im sorry Templar, didnt know that u already had answered it. my bad.

now, heres the next question:

From 2001 Ext 1 HSC paper

6(b)

Consider the variable pt P(2at, at^2) on the parabola x^2=4ay

(i) prove that the equation of the normal at P is x + ty = at^3 + 2at
(ii) find coordinates of pt Q such that the normal at Q is perpendicular to the normal at P.
(iii) show that the two normals intersect at R, whos coordinates are:

x= a(t - 1/t) and y=a(t^2 + 1 + 1/t^2)
(iv) find the equation in cartesian form of the locus of the point R given in part (iii)

 

[Templar]

im sorry Templar, didnt know that u already had answered it. my bad.

I didn't answer it, and that wasn't my point. I just noticed that your solution is different to Keypad's, so could you please check your working, in particular with the inequality for x when solving the quadratic.

 

[KeypadSDM]

From 2001 Ext 1 HSC paper

6(b)

Consider the variable pt P(2at, at^2) on the parabola x^2=4ay

(i) prove that the equation of the normal at P is x + ty = at^3 + 2at
(ii) find coordinates of pt Q such that the normal at Q is perpendicular to the normal at P.
(iii) show that the two normals intersect at R, whos coordinates are:

x= a(t - 1/t) and y=a(t^2 + 1 + 1/t^2)
(iv) find the equation in cartesian form of the locus of the point R given in part (iii)

This wasn't such a bad question, it stretches your manipulative algebra a bit near the end.

Spoiler

P(2at, at²); x² = 4ay

i) Differentiating wrt x:
4ay' = 2x
y' = x/(2a)

Thus the gradient of the tangent @ P is:

m_T = (2at)/(2a) = t

And hence the gradient of the normal @ P is:

m_N = -1/m_T = -1/t

Substituting into the equation: m = (y - y₁)/(x - x₁) yields:

(y - at²)/(x - 2at) = -1/t
at³ - yt = x - 2at

x + ty = at³ + 2at ... (1)

As required.

ii) Set: Q(2at_Q,at_Q²)

Which means the gradient of the normal at Q is:

m_NQ = -1/t_Q

Now, as these normals are perpendicular, we require:
m_NQm_N = -1
I.e. t_Q = -1/t ... (2)

Thus we have found the point Q:
Q(-2a/t,a/t²)

iii) Substituting (2) into (1), we find that the normal at Q has equation:

x + t_Qy = at_Q³ + 2at_Q
(*t³)
t³x + t²(-1)y = (-1)³a + 2at²(-1)

Thus:

t²(y - tx) = a + 2at²
(y - tx) = (a + 2at²)/t² ... (3)

To find the points of intersection:

(1)t + (2):
y + t²y = at⁴ + 2at² + a/t² + 2a

y(1 + t²) = a([t⁴ + t²] + [t² + 1] + [1 + 1/t²])

Note that each of those square brackets is directly dividable by 1 + t²

y = a((t⁴ + t²)/(t² + 1) + (t² + 1)/(t² + 1) + (1 + 1/t²)/(1 + t²))
=a(t² + 1 + 1/t²)

(1) - (2)t:
x + t²x = at³ + 2at -(a/t)(1 + 2t²)
x(1 + t²) = (a/t)(t⁴ + 2t² - 1 - 2t²)
=(a/t)(t⁴ - 1)
=(a/t)(t² - 1)(t² + 1)

I.e.
x = (a/t)(t² - 1)
x = a(t - 1/t)

As required.

iv) Note: (t - 1/t)² = t² - 1 + 1/t²

Thus:

(x/a)² = y/a - 3
x² = ay - 3a²

I love those little tricks.

 

[Mountain.Dew]

KeypadSDM, please post up the next question.

 

[Mountain.Dew]

I didn't answer it, and that wasn't my point. I just noticed that your solution is different to Keypad's, so could you please check your working, in particular with the inequality for x when solving the quadratic.

thank you templar for pointing out the flaw in my working out.

the original post has been edited.

 

[Templar]

KeypadSDM, please post up the next question.

Firstly he doesn't bother with posting questions much, and secondly he's away until January 21st, so feel free to post your question.

 

[Riviet]

Mountain Dew, feel free to post the next question.

 

[Mountain.Dew]

Mountain Dew, feel free to post the next question.

damn u peoples!

okay, try this:

evaluate ∫t^2 - t - 2 / (t+1)(2t+5) dt

 

[pLuvia]

Spoiler

∫t² - t - 2 / (t+1)(2t+5) dt

= ∫ (t-2)/(2t+5) dt

Let u = 2t+5 [t = (u-5) /2]
du = 2dx [dx = du/2]

= ∫ [(u-9)/2] / u du/2
= 1/4 ∫ u-9/u du
= 1/4 ∫ 1 - 9/u du
= 1/4 [u - 9ln u] + C
= 1/4 [(2t+5) - 9ln(2t+5)] + C

Next Question:

Evaluate

∫ sec³x tanx dx

 

[Mountain.Dew]

Next Question:

Evaluate

∫ sec³x tanx dx

*another 2 cents thrown around*

I =∫ sec³x tanx dx

let u = secx
du/dx = secxtanx --> dx = du/secxtanx

I =∫ u³ tanx * du/ u*tanx
I = ∫ u² du
I = u³/3 + c = sec³x/3 + c

of course, this is the proper way of doing it, but can get same result from being merely observant, considering the implications of the chain rule.

 

[pLuvia]

Ok maybe that was a tad bit too easy

 

[Riviet]

Don't forget to post your question after you answer one.

 

[pLuvia]

You can post one Riviet