3u Mathematics Marathon V 1.1 (1 Viewer)

[Riviet]

The first one's getting old so i decided to start a fresh new one!

This thread is not intended as a place to ask for help. Please make a new thread for that.

Rules:

1) No spamming. Please try to stick to strictly answering and questioning. If you have an alternate way of doing a problem, you can post that, too.

2) If you answer a question, try to find a new question for the next person to try. Hard or uncommon textbook questions and and questions from past papers are good places to look.

3) Please surround your answer in spoiler tags: (spoiler) solution (/spoiler) replacing the round brackets with these square brackets: [

Have fun.

Q1 Find

/
| x³(x⁴+1)² - x²/(x³-1) dx
/

 

[pLuvia]

Is it all over (x³-1)??

 

[Slidey]

Integrate each on its own:

Spoiler

First substitution is u=x^4+1, du=4x^3

Second, u=x^3-1, du=3x^2 dx

So the answer is: (x^4+1)^3/12 - ln(x^3-1)/3 + C

Question 2)

Evaluate the integral of |x^2-9| from -1 to 5, w.r.t. x.

 

[pLuvia]

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

 

[pLuvia]

Next question

Using Newtons method twice, solve the equation
log_ex = cos x has a root near x = 1

 

[Riviet]

w.r.t.x = with respect to x

 

[pLuvia]

I know what it stands for but does it mean just to integrate with x's in the answer?

I should know this but hey

 

[icycloud]

Question 2)

Evaluate the integral of |x^2-9| from -1 to 5, w.r.t. x.

Since it is an absolute value, we must split the range.

Spoiler

i.e. we solve x^2-9 >= 0 and x^2-9 < 0
Thus the range -1 ---> 5 becomes -1 ---> 3 and 3 ---> 5 respectively to 9-x^2 and x^2-9

Thus, I = ∫(-1 ---> 5) |x^2-9| dx
= ∫(-1 ---> 3) (9-x^2) dx + ∫(3 ---> 5) (x^2 - 9) dx
= (-1 ---> 3) [9x - x^3/3] + (3----> 5) (x^3/3 - 9x)
= 27 -9 + 9 - 1/3 + 125/3 - 45 - 9 + 27
= 41 1/3 (or 124/3) #

 

[Riviet]

I know what it stands for but does it mean just to integrate with x's in the answer?

I should know this but hey

It's hard to explain it but it means to differentiate x, treating it as "the variable" and all other terms as constants etc.

 

[nick1048]

It's hard to explain it but it means to differentiate x, treating it as "the variable" and all other terms as constants etc.

lol I think people understand this concept better once they've studied inverse trig functions to identify the difference between having respect for a pronumeral rather than merely making it the subject. Don't fret if you don't understand this idea fully as yet.

 

[icycloud]

Using Newtons method twice, solve the equation
ln x = cos x has a root near x = 1

Spoiler

Let x_1 = 1
f(x) = ln(x) - cos(x) = 0
f'(x) = 1/x + sin(x)

First go:
x_2 = x_1 - f(x_1)/f'(x_1)
= 1 - (ln(1)-cos(1))/(1/1+sin(1))
= 1.9826.........

Second go:
x_3 = x_2 - f(x_2)/f'(x_2)
= 1.9826..... - (ln(1.9826...) - cos(1.9826...))/(1/1.9826... + sin(1.9826...))
= 2.2871......
= 2.287 (3dp) #

Question 4
Prove that [cos(x) - cos(5x) + cos(9x)] / [sin(x) - sin(5x) + sin(9x)] = cot(5x)

 

[Riviet]

Got a hint icycloud?

I had a go at it the other day, havent got anywhere with the proof yet.

 

[icycloud]

Got a hint icycloud?

I had a go at it the other day, havent got anywhere with the proof yet.

Sure.

Hint:

Spoiler

Group the LHS as:

[cos(x) + cos(9x) - cos(5x)] / [sin(x) + sin(9x) - sin(5x)]

Then use some clever transformation with expansions of cos(A+B), cos(A-B), sin(A+B) or sin(A-B).

If you need another hint, do reply again. =D

 

[Riviet]

Question 4
Prove that [cos(x) - cos(5x) + cos(9x)] / [sin(x) - sin(5x) + sin(9x)] = cot(5x)

After i learnt a new trig formula today, i suddenly knew what to do:

Spoiler

LHS=[cos9x + cosx - cos5x] / [sin9x + sinx - sin5x]

={2cos[(9x+x)/2]cos[(9x-x)/2]-cos5x} / {2sin[(9x+x)/2]cos[(9x-x)/2]-sin5x}

=(2cos5xcos4x-cos5x) / (2sin5xcos4x-sin5x)

=[cos5x(2cos4x-1)] / [sin5x(2cos4x-1)]

=cos5x/sin5x

=cot5x

=RHS!!! WOOT WOOT! ;D

QED

Next Question:

If tan a, tan b, tan y are the roots of the equation x³-(a+1)x²+(c-a)x-c=0 , show that a + b + y = n(pi) + pi/4 , where n is an integer.

P.S This one seems difficult. I personally haven't done it yet!

 

[YBK]

Next Question:

If tan a, tan b, tan y are the roots of the equation x³-(a+1)x²+(c-a)x-c=0 , show that a + b + y = n(pi) + pi/4 , where n is an integer.

P.S This one seems difficult. I personally haven't done it yet!

What's the relationship between tan and pi?

I replaced x with tan a which equals zero... and did the same for tan b and tan y.
This makes them all equal to each other...

Therefore (I think): tan a = tan b = tan y

I'm probably wrong... which topic is that question from? Haven't encountered anything similar... confusing :s

 

[Riviet]

It's polynomial rice with a touch of trigonometric sauce.

 

[Mountain.Dew]

Next Question:

If tan a, tan b, tan y are the roots of the equation x3-(a+1)x2+(c-a)x-c=0 , show that a + b + y = n(pi) + pi/4 , where n is an integer.

P.S This one seems difficult. I personally haven't done it yet!

here goes:

Spoiler

consider tan(A+B+Y)...

i will use this notation: TA = tanA, TB = tanB, TC = tanC

now, tan(A+B+Y) = tan(A+b) + TY / (1 - tan(A+B)*TY)

now, expand further: tan(A+B+Y) = [(TA + TB) / (1- TA*TB)] + TY / (1 - [TY[TA + TB]/(1-TATB)])

then, times numerator and denominator by 1-TATB

SO, tan(A+B+Y) = [TA + TB + TY(1-TATB)] / (1 - TATB - [(TA + TB)TY])...starting to look good now...
tan(A+B+Y) = [TA + TB + TY - TATBTY] / [1 - (TATB + TATY + TBTY)]...looks familiar?
tan(A+B+Y) = ([sum of single roots] - [product of roots]) / (1 - [sum of double roots])
tan(A+B+Y) = [a+1] - [c] / 1 - [c - a] = a+1- c / 1 - c + a = 1

therefore, tan(A+B+Y) = 1

THEREFORE, by the general solution, [tanA = M, A = n(pi) + tan^-1M]

(A+B+Y) = n(pi) + tan^-1(1) = n(pi) + pi/4, n an integer.

mmmmmm yummy rice and sauce Riviet!

 

[Mountain.Dew]

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...)

 

[Riviet]

Nice work with the trig/polynomial question, Mountain Dew; you made it look so much easier.

 

[Mountain.Dew]

thankee riviet, thankee *bows to riviet*

now, who will attempt this question of poker?