Mathematics Extension 1 Marathon HSC 09 (2 Viewers)

iMAN2 · Oct 14, 2009

Similar to the Mathematics Thread started by ForrestGump at here.

How To Play:
The first person asks a Mathmatics Ext 1 question (maybe one you've had trouble with). The next person answers it then posts another question for the next person then so on.

Tip:
If you feel an area is your weakness then try asking a question on this area to see others responses to it which may help you in your understanding.

1st Question: (2 Marks)
Let $f(x)=Ax^{3}-Ax+1$ where $A>0$
Show that f(x) has exactly one zero when $0<A<\frac{3\sqrt{3}}{2}$

jet · Oct 14, 2009

untouchablecuz · Oct 14, 2009

find the tangents to y=x³-2x²+1 which pass through the origin

hermand · Oct 14, 2009

peopleeeeeee. you need to ask questions back. i'm bored. give me something to do! haha.

untouchablecuz · Oct 14, 2009

i didddddddd

ninetypercent · Oct 14, 2009

Show that:
$2sin^{-1}\frac{3}{5} = sin^{-1}\frac{24}{25}$

AlexJB · Oct 14, 2009

@untouchablecuz: dont have pen+paper, doing phys review. tried doing your q, and failed first time on ms word. takes too long, ill try later with paper. answer me this though. i get it down to y-y1=m(x-x1), when x=0, y=0, hence y1 = mx1. Am I meant to solve for m?

---

^^^
I'm not that great with these kind of proofs :/

That an acceptable proof?

Aerath · Oct 15, 2009

AlexJB said:
^^^
I'm not that great with these kind of proofs :/

That an acceptable proof?

Correct.

Since you didn't post a question....

Sketch y = 4x^2 / (x^2 -9)

untouchablecuz · Oct 15, 2009

AlexJB said:
@untouchablecuz: dont have pen+paper, doing phys review. tried doing your q, and failed first time on ms word. takes too long, ill try later with paper. answer me this though. i get it down to y-y1=m(x-x1), when x=0, y=0, hence y1 = mx1. Am I meant to solve for m?

let the point x=a be where the tangent(s) intersect with the curve

at x=a, dy/dx=3a²-4a [1]

if the tangents pass through (0, 0), they are of the form y=mx

at x=a, dy/dx=m [2]

[1] and [2] are equal, since the gradient at x=a of the curve is equal to that of the line

.'. 3a²-4a=m [3]

on the curve,

when x=a, y=a³-2a²+1 [4]

on the line,

when x=a, y=ma [5]

[4] and [5] are equal

.'. ma=a³-2a²+1 [6]

sub [3] in [6]

a(3a²-4a)=a³-2a²+1

2a³-2a²-1=0

according to an online cubic solver, a~1.297156508177424

sub into the equation of the cubic and y~-0.1826150067

sub into [5]

m=-0.1407810126

.'. the equation of the tangent is y=-0.1407810126x

realistically, they'll give you an equation that factors nicely

jet · Oct 15, 2009

$y = \frac{4x^2}{x^2 - 9} \\ \text{x-intercepts: } 4x^2 = 0,x = 0 \\ \text{y -intercept } y = 0 \\ \frac{dy}{dx} = \frac{8x\left (x^2 - 9 \right ) - 2x \cdot 4x^2}{\left (x^2 - 9 \right )^2} = 0 \text{ for stationary points} \\ 8x^3 - 72x - 8x^3 = 0 \\ 72x = 0 \\ x = 0 \\ \text{At } x = 0^{-}, \frac{dy}{dx} > 0; x = 0^{+}, \frac{dy}{dx} < 0 \\ \text{Therefore } x = 0 \text{ is a Maximum} \\ \text{Vertical asymptotes: } y = \frac{4x^2}{\left (x - 3 \right )\left (x + 3 \right )} \\ \therefore \left (x - 3 \right )\left (x + 3 \right ) = 0 \Rightarrow x = 3, x = -3 \\ \text{Horizontal asymptotes } \lim_{x \to \pm\infty} \frac{4x^2}{x^2 - 9} \\ = \lim_{x \to \pm\infty} \frac{4}{1 - \frac{9}{x^2}} \\ = 4 \\ \text{To sum up, we have } x\text{-intercept and } y \text{-intercept 0} \\ \text{A maximum at } \left (0,0 \right ) \\ \text{Vertical asymptotes } x = \pm 3 \text{ and a horizontal asymptote } y = 4$

hermand · Oct 15, 2009

untouchablecuz said:
i didddddddd

i am deadset blind. sorry, my bad.

Q. A particle moves in a straight line and its position at time 't' is given by; $x=5+\sqrt{3}\sin 3t-\cos3t$
i. Express $\sqrt{3}\sin 3t-\cos3t$ in the form $R\sin(3t-\alpha )$ where $\alpha$ is in radians.
ii. The particle is undergoing simple harmonic motion. Find the amplitude and the centre of the motion.
iii. When does the particle first reach its maximum speed after time t=0?

jet · Oct 15, 2009

After hermand's question is done:

$\text{As a particle moves around a circle, its angular velocity is given by } \\\frac{d\theta}{dt} = \frac{1}{1 - t^2} \\ \text{You are given that the particle starts at } \theta = \frac{\pi}{4} \\ \text{Find t as a function of }\theta \text{ and prove that } \frac{\pi}{4} \leq \theta < \frac{3\pi}{4} \text{ and explain why the particle never moves through an angle of more than } \frac{\pi}{2}$

xFusion · Oct 15, 2009

$x-5=\sqrt{3}sin3t - cos3t \sqrt{3}sin3t - cos3t=Rsin(3t -\alpha )=Rsin3tcos\alpha -Rcos3tsin\alpha$
$Rcos\alpha =\sqrt{3}$
$Rsin\alpha =1$
$tan\alpha =1/\sqrt{3} \therefore \alpha =\pi /6$
R=2
$\therefore \sqrt{3}sin3t - cos3t = 2sin(3t-\frac{\pi }{6})$
(ii) amplitude=2, center of motion=5
iii) particle reaches maximum speed when x=5
$5-5=2sin(3t-\frac{\pi }{6})$
$\therefore t=\frac{\pi }{18}s$

addikaye03 · Oct 15, 2009

New question:

Find the values of r and n if in the expansion (1+x)^n in ascending powers of x, the (r+1)-th coefficient is twice the r-th and (r+10)-th coefficient is twice the (r+11)-th.

What is the greatest coefficient in this expansion?

-Onlooker- · Oct 16, 2009

addikaye03 said:
New question:

Find the values of r and n if in the expansion (1+x)^n in ascending powers of x, the (r+1)-th coefficient is twice the r-th and (r+10)-th coefficient is twice the (r+11)-th.

What is the greatest coefficient in this expansion?

Ok im supposed to answer but im not going to (dont know how to work the latex crap )

But just wanted to say : These questions are like Harder 3unti Q's , super awesome, I might add

jet · Oct 16, 2009

I had a question above. No one's answered it yet.

scardizzle · Oct 16, 2009

addikaye03 said:
New question:

Find the values of r and n if in the expansion (1+x)^n in ascending powers of x, the (r+1)-th coefficient is twice the r-th and (r+10)-th coefficient is twice the (r+11)-th.

What is the greatest coefficient in this expansion?

is n = 29 and r= 9?

untouchablecuz · Oct 16, 2009

i tried yesterday and today, i cant get that goddam upper bound on the angle

its probably something simple

eoklm11 · Oct 19, 2009

New question:

Find the values of r and n if in the expansion (1+x)^n in ascending powers of x, the (r+1)-th coefficient is twice the r-th and (r+10)-th coefficient is twice the (r+11)-th.

What is the greatest coefficient in this expansion?

given

Tr+1 = 2Tr

Tr+1/Tr = 2 ,

Simlilarly
Tr+11/ Tk+10 = 2

There fore
Tr+1/Tk = Tr+11/ Tk+10

now Tr+1/ Tr = (n-r)x/(r+1)

from Tr+1= x^(r+1)(n!)/(r+1)!(n-r-1)! = x.x^r(n!)/(r+1)r!(n-r-1)!

and Tr= x^r(n!)/r!(n-r)!= x^r(n!)/r!(n-r-1)(n-r)!

Similarly Tr+1/ Tr = (n-r-10)x/(r+11)

Thus
(n-r)x/(r+1) =(n-r-10)x/(r+11)
\......................................

idk i'm stuffed someone help me?:S

blaze.bluetane · Oct 19, 2009

scardizzle +1

Mathematics Extension 1 Marathon HSC 09 (2 Viewers)

Member

Banned

Active Member

je t'aime.

Active Member

ninety ninety ninety

Unmotivated

Retired

Active Member

Banned

je t'aime.

Banned

Member

The A-Team

Member

Banned

Salve!

Active Member

Member

Member

Users Who Are Viewing This Thread (Users: 0, Guests: 2)