2u Mathematics Marathon v1.0 (2 Viewers)

[insert-username]

Express the number 0.98765432109876543210 (the digits 9 through to 0 repeating infinitely) as a fraction.

Spoiler

Let 0.98765432109876543210 = @.

10000000000@ - @ = 9999999999@

Therefore 9876543210.9876543210 - 0.9876543210 = 9876543210

Therefore @ = 9876543210/9999999999

= 1097393690/1111111111

Next question: Find any stationary points on the curve f(x) = 3x⁴ - 7x³ + 2x² + 5x - 12 and distinguish between them.


I_F

 

[pLuvia]

Next question: Find any stationary points on the curve f(x) = 3x⁴ - 7x³ + 2x² + 5x - 12 and distinguish between them.

Spoiler

f'(x) = 12x³ - 21x² + 4x + 5
Stat points when f'(x) = 0
12x³ - 21x² + 4x + 5 = 0
Using long division etc etc

(x-1)(3x-5)(4x+3) = 0
x = 1, 5/3, -3/4
y = 5, -7¹⁰/₂₇, -10<sup>185/₂₅₆

f"(x) = 36x2 - 42x + 4

At x = 1
f"(x) > 0 Max TP
At x = 5/3
f"(x) > 0 Min TP
At x = -3/4
f"(x) > 0 Min TP</sup>

 

[Riviet]

OMG pLuvia! What a coincidence! I was doing the question! >_<
Oh well here's my solution...

Next question: Find any stationary points on the curve f(x) = 3x⁴ - 7x³ + 2x² + 5x - 12 and distinguish between them.

Spoiler

f'(x)=12x³-21x²+4x+5=0
f'(1)=0
by the factor theorem (3u), x-1 is a factor of f'(x)
by long division (3u), f'(x)=(12x²-9x-5)(x-1)
.: a stat point at x=1, and solving 12x²-9x-5=0
x=(-9+/-sqrt321)/24
.:y=-11,-10.2,-0.47
.: stat points are (1,-11) [(-9+sqrt321)/24,-10.2] and [(-9+/-sqrt321)/24,-0.47)
now f"(x)=36x²-42x+4
f"(1)=-2
<0
.:concave down at (1,-11)
.: local max at (1,-11)
f''[(-9+sqrt321)/24]=-6.6
<0
.: concave down at [(-9+sqrt321)/24,-10.2]
.: local max at [(-9+sqrt321)/24,-10.2]
f"[(-9+/-sqrt321)/24]=96.4
>0
.: concave up at [(-9+/-sqrt321)/24,-0.47)
.: local min at [(-9+/-sqrt321)/24,-0.47)

Damn, stuffed up factorising bit. Damn damn damn!!

Can i post next question?

Question: If p and q are the roots of the equation 3x²-x+9=0, without finding the roots, evaluate:
i) p+q
ii) pq
iii) p²+q²
iv) 1/p + 1/q
v) (1+p)(1+q)
vi) 6p²-3pq+6q²
vii) p³-q³ [harder]

 

[pLuvia]

lol, I'm faster than you

Spoiler

3x² - x + 9 = 0
i) 1/3
ii) 9/3 = 3
iii) (p+q)² - 2pq
= (1/9) - 6
= -5⁸/₉
iv) 1/p + 1/q = p+q/pq = 1/9
v) 1 + p + q + pq
= 1 + 1/3 + 3
= 4¹/₃
vi) 6(p²+q² - 3pq
= 18 - 1
= 17
vii) p³ - q³ = (p-q)(p² + pq + q²
= (p-q)(3¹/₃

I'm stuck hehe

 

[Riviet]

Hmm... i think i meant p³-q³, then you would get p+q in the factorisation, so that should be easy from there.
But how would you do p-q?
Oh yeah, don't forget to post in a new question pLuvia.

 

[pLuvia]

a) Find the area of the region bounded by the curve y = e^x, the coordinate axes and the line x=a, a>0
b) Find the limit of this area as a --> infinity
c) Find the volume of the soild generated by rotating the region in (a) about the X-axis and find the limit of this volume as a --> infinity

Have a go of this

 

[Riviet]

a) Find the area of the region bounded by the curve y = e^x, the coordinate axes and the line x=a, a>0
b) Find the limit of this area as a --> infinity
c) Find the volume of the soild generated by rotating the region in (a) about the X-axis and find the limit of this volume as a --> infinity

Have a go of this

a) Integral from 0 to a of e^xdx= e^a-1
b) as a->infinity, e^a-1->e^a
c) V= pi x integral from 0 to a of e^2xdx=pi(e^2a-1)/2
as a->infinity, v->(pi)e^2a/2?

 

[Riviet]

Next Question: Differentiate [2(9x²+x-2)^3/2]/3

 

[pLuvia]

Next Question: Differentiate [2(9x²+x-2)^3/2]/3

Spoiler

[2(9x²+x-2)^3/2]/3
= 2/3 x 3/2 (9x² + x - 2)^1/2 x 18x + 1
= (18x+1)(9x² + x -2)

Next Question: Integrate From [pi/2 to 0] (sinx) / (2 - cosx) dx

 

[.ben]

Next Question: Integrate From [pi/2 to 0] (sinx) / (2 - cosx) dx[/QUOTE]

is it =(ln2)-(ln1)
=0.69314718

?

 

[Riviet]

Integral from 0 to pi/2 of sinx / (2-cosx) dx = [ln(2-cosx)] from 0 to pi/2
=ln2-ln1
=ln2 or 0.69314718...
.: ben is correct, lol.

Question (from my 2u assessment): P(x,y) is a random point in the first quadrant of the cartesian plane. A is another point with co-ordinates (5,0). O is the origin.

i) Draw a diagram with the above information.
ii) From your diagram, write the gradients of OP and AP in terms of x and y.
iii) Show that the locus of all points P, such that OP is perpendicular to AP is x²-5x+y²=0
iv) Deduce that the locus given in (iii) is a circle and wite down its centre and radius.

 

[Holsworthy]

b) gradient of OP = y/x, gradient AP = y/(x-5)
c) gradient of OP x gradient of AP = -1 => (y/x) * (y/(x-5)) = -1 => x^2 -5x + y^2 = 0 => circle
d)
(x- 5/2)^2 + y^2 = (5/2)^2 => center = (5/2, 0), radius = 5/2

 

[Riviet]

Correct holsworthy! Now you may post in a question for the next person to answer (please keep to 2u level).

 

[Riviet]

Bump!

Let's get this thread up and running again...

Question: Use the substitution u=x⁴+1 to find

/
| x³(x⁴+1)² dx
/

 

[bananasmoothy]

What do the forward-slashes mean?

 

[klaw]

it's supposed to be an integral sign

 

[klaw]

Bump!

Let's get this thread up and running again...

Question: Use the substitution u=x⁴+1 to find

/
| x³(x⁴+1)² dx
/

u=x^4+1
du/dx=4x^3

Int [x^3(x^4+1)^2] dx =


1/4 Int (u^2) du
=1/4*(u^3)/3 +c
=[(x^4+1)^3]/12+c

 

[pLuvia]

Integration

 

[pLuvia]

Evaluate ∫[4 to 0] root(1+x⁴ dx using Simpsons rule with 5 functions values

 

[Riviet]

I reckon

/
| f(x) dx
/

is way cooler than ∫

especially when you can do

2
/
| f(x) dx
/
0