2u Mathematics Marathon v1.0 (2 Viewers)

[Mountain.Dew]

okay, lemme finish this off...

aim: differentiate f(x) = 2x^2 + 3x + 3 by first principles.

f'(x) = lim h-> 0 [[ f(x+h) - f(x) ] / h ]...merely state formula
f'(x) = lim h-> 0 [[2(x+h)^2 + 3(x+h) + 3 - 2x^2 + 3x + 3] / h]...substitute values
f'(x) = lim h-> 0 [[2x^2 + 4xh + 2h^2 + 3x + 3h + 3 - 2x^2 + 3x + 3] / h]...expand
f'(x) = lim h-> 0 [[4xh + 2h^2 + 3h] / h]...cancel and collect like terms
f'(x) = lim h-> 0[4x + 2h + 3]...divide by h, then use the limit.

so therefore, f'(x) = 4x + 3

so thats my 2 cents.

 

[Mountain.Dew]

heres more of my 2 cents...

and to answer original question...

differentiate 2x^5 + 3x^3 + x + 3 BY FIRST PRINCIPLES

==============================================

f'(x) = lim h-> 0 [[ f(x+h) - f(x) ] / h ]...merely state formula
f'(x) = lim h-> 0 [[2(x+h)^5 + 3(x+h)^3 + x+ h + 3 - 2x^5 + 3x^3 + x + 3]/h]
f'(x) = lim h-> 0 [[2x^5 + 10x^4h + 20x^3h^2 + 20x^2h^3 + 10h^4 + 2h^5 +3x^3 + 9x^2 + 9xh^2 + 3h^3 + x + h + 3 - 2x^5 - 3x^3 - x - 3]/h]
f'(x) = lim h-> 0 [[10x^4h + 20x^3h^2 + 20x^2h^3 + 10xh^4 + 2h^5 + 9x^2h + 9xh^2 + 3h^3 + h]/h]
f'(x) = lim h-> 0 [10x^4 + 20x^3h + 20x^2h^2 + 10xh^3 + 2h^4 + 9x^2 + 9xh + 3h^2 + h]

therefore, f'(x) = 10x^4 + 9x^2 + 1

====================================

now, to continue the marathon, heres the next question...

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:
(a)
(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...)
(b) i dont need strict answers for this one, but for each of the following, from (i) - (ix), i am in a poker game with 4 players, i am seated to be dealt 2nd. what is the probability then, just for my hand? will the probability change IF the seating arrangment was different? REMEBER: NO NEED FOR ACCURATE, MATHEMATICAL ANSWERS...EXPLANATIONS THROUGH VERSE IS FINE.

 

[YBK]

Nice work mountain dew!

Can't do your question, we're not up to that topic yet

 

[Mountain.Dew]

oops sorry, i got a little bit carried away with combinatronics, these questions should be in the ext 1 and/or ext 2 section. but, the question is there for interest.

i will post up another question, when i can find one...

 

[Riviet]

Definitely too hard for 2 unit level.

See if any of you guys can do the last question posted in MX1 marathon.

 

[insert-username]

Definitely too hard for 2 unit level.

See if any of you guys can do the last question posted in MX1 marathon.

Dude, there's hard, very hard, and haven't done it yet...


I_F

 

[Mountain.Dew]

okay, i guess i have to post up the next question:

from 1995 HSC Mathematics 2/3U Common

Question 6(b)

we have a graph of y=4/x and y=5-x

(i) find the coordinates of the pts of intersection
(ii)find the area bounded by the curves.

===============================

 

[YBK]

Spoiler

(i) 5 - x = y (1)

5 - x = 4/x

5x - x^2 = 4
x^2 - 5x + 4 = 0
(x - 4) (x - 1)

.: intersects at x = 4 and x = 1
for y coordinate substitute into (1)
(4,1) and (1,4) are intersection points

(ii) Can't do this, haven't learnt integration yet.

 

[Riviet]

okay, i guess i have to post up the next question:

from 1995 HSC Mathematics 2/3U Common

Question 6(b)

we have a graph of y=4/x and y=5-x

(i) find the coordinates of the pts of intersection
(ii)find the area bounded by the curves.

===============================

Spoiler

ii) Area=

4
/
| (5 - x - 4/x) dx = [5x - x²/2 - lnx/4]₁->⁴
/
1

=(20 - 8 - ln4/4) - (5 - 1/2)

=(30-ln4)/4

Next question:

Solve 5x²-4x-2=0 by completing the square.

 

[YBK]

Spoiler

Solve 5x²-4x-2=0 by completing the square.[/B][/QUOTE]

(root(5)x + 4 / (2 root5))^2 - 2.8 = 0

root(5)x + 4 / (2 root5) = 7.84

root(5)x = 7.84 - 4 / (2 root5)
x = {7.84 - 4 / (2root(5))} / root 5

Looks confusing, but I *think* it's right.

 

[Riviet]

Not sure what you did YBK

Spoiler

Start by making a=1

 

[YBK]

Next question:

A truck is to transport goods over a distance of 500 km. The owner calculates that he has fixed costs of 45 dollars per hour and 22 cents per km, plus a variable cost of 0.8 v cents per km where v km/h is the average speed of the truck over the journey.

i) Show that the cost C of the journey is given by C = 22500/v + 4v + 110

ii) Find the speed v which minimises the cost C

iii) Calculate the minimum cost of the journey

 

[YBK]

Not sure what you did YBK

Spoiler

Start by making a=1

lol, darn... I posted up a new question.. I'll give your one another shot now

 

[Jono_2007]

Here's what i got;

Spoiler

5x^2-4x-2=0
5x^2-4x+4=2+4
(√5-2)^2=6
√5-2=±√6
x=2±√6/√5

 

[Jono_2007]

Is mine right or is his right?

 

[YBK]

Is mine right or is his right?

Mine's probably wrong

 

[Riviet]

Very close Jono...

Why did you guys forget to make a=1?

 

[SoulSearcher]

let me give this a shot

Next question:

Solve 5x2-4x-2=0 by completing the square.

Spoiler

5x² - 4x - 2 = 0
5x² - 4x = 2

now divide by 5

x² - 4x/5 = 2/5
x² -4x/5 + 16/100 = 14/25
(x - 4/10)² = 14/25

now take the square root of each side

x - 4/10 = +/- {(root)14/5}
therefore
x = 4/10 +/- {(root)14/5}
x = {2 +/- (root) 14} / 5

 

[Jono_2007]

This is what i was told;
1)remove any constant to the RHS of the equation.
2)Note the coefficient of x, and follow the rule
"Half IT",
"Square IT",
and "add it to BOTH sides".
3)Rewrite the LHS as a perfect square and siplify the RHS.
4)Take the square root of both sides.
5)solve for x

 

[insert-username]

Jono, your rules are almost there, but you've forgotten one thing: the coefficient of x² must be 1 to complete the square.

Next question:

Solve 5x²-4x-2=0 by completing the square.

Another complete the square? Anyways, SoulSearcher is correct

Spoiler

x² - 4x/5 - 2/5 = 0

x² - 4x/5 + 4/25 = 14/25

(x - 2/5)² = 14/25

x - 2/5 = ±(√14)/5

x = (±√14 + 2)/5

I_F