HSC 2013 Maths Marathon (archive) (3 Viewers)

Sy123 · Jun 9, 2013

Re: HSC 2013 2U Marathon

$$Solve for$ \ \ x$

$\ln(1+x^2) + 1 = \ln(4+x^2)$

braintic · Jun 9, 2013

Re: HSC 2013 2U Marathon

braintic said:
For a given function f(x), we know that f(1)=3, f(3)=-2, f(6)=5, f ' (1)=f ' (3)=f ' (6)=0, and f '' (x) > 0 for 2 < x < 4, and -ve elsewhere.

Find the area of the region bounded by y=f ' (x) and the x-axis.

[Do not attempt to fit an equation to f(x) or any of its derivatives]

No-one?

(I should have said the function and its derivative are continuous throughout.)

aerowhale · Jun 18, 2013

Re: HSC 2013 2U Marathon

Sy123 said:
$$Solve for$ \ \ x$

$\ln(1+x^2) + 1 = \ln(4+x^2)$

$\log(1+x^2) + 1 = \log(4+x^2)\\$
$\log(1+x^2) + \log(e) = \log(4+x^2)\\$
$\log((1+x^2)e) = \log(4+x^2)\\$
$(1+x^2)e = 4+x^2\\$
$e + ex^2 = 4+x^2\\$
$ex^2-x^2 = 4-e\\$
$(e-1)x^2 = 4-e\\$
$x^2 = \frac{4-e}{e-1}\\$
$x = \pm \sqrt{\frac{4-e}{e-1}}$

omgiloverice · Jun 22, 2013

Re: HSC 2013 2U Marathon

Since I am doing probability in class the rest of you must also suffer

$The probability that a cityrail train arrives on 1:08 pm Saturdays is 0.19.$

$If$ \ n \ $is the number of weeks that pass$

$$How many weeks must pass until it is 0.98 certain that the train will arrive on time$

asianese · Aug 6, 2013

Re: HSC 2013 2U Marathon

Show that the tangent to the curve $y=\log_e x$ at $x=a$ is given by

$x-ay-a(1-\log_ea)=0$ .

If this tangent passes through the origin, find the value of $a$ .

Hence find the area of the region enclosed by the curve $y=\log_e x$ , the tangent to the curve which passes through the origin, and the $x$ -axis.

anthony_wheels · Aug 7, 2013

Re: HSC 2013 2U Marathon

This is my first time on one of these marathon things and I don't know how to use 'LaTex', but does a=e, and is the area (e/2)-1? I'm probs wrong but worth a try

omgiloverice · Aug 9, 2013

Re: HSC 2013 2U Marathon

$The displacement $x$ metres at time $t$ seconds of a particle moving in a straight line is given by:\\ \\ $x = 1+2\cos^{2}t \\\\ $If $v$ and $a$ are the respectively the velocity and acceleration of the particle at time $t$, $find:$\\ i) $v$ and $a$ in terms of $t$.\\ ii) Hence, show that $a^{2}=16-4v^{2}$

JT145 · Aug 9, 2013

Re: HSC 2013 2U Marathon

yay threads like this exist

anthony_wheels · Aug 9, 2013

Re: HSC 2013 2U Marathon

omgiloverice said:
$The displacement $x$ metres at time $t$ seconds of a particle moving in a straight line is given by:\\ \\ $x = 1+2\cos^{2}t \\\\ $If $v$ and $a$ are the respectively the velocity and acceleration of the particle at time $t$, $find:$\\ i) $v$ and $a$ in terms of $t$.\\ ii) Hence, show that $a^{2}=16-4v^{2}$

Should part.ii be: "Hence, show that a^2 = 16 -2v^2" ?

AnimeX · Aug 9, 2013

Re: HSC 2013 2U Marathon

anthony_wheels said:
Should part.ii be: "Hence, show that a^2 = 16 -2v^2" ?

ehh nope, I just had a go at it and part ii) is correct.

you should recheck your working, perhaps you forgot to square the 2 in (-v/2 )^2.

omgiloverice · Aug 10, 2013

Re: HSC 2013 2U Marathon

$$Consider the circle$ \ x^{2}+y^{2} -2x-14y+25=0$

$$ Show that if the line $y=mx$ intersects the circle in two distinct points, then: (1+7m)^{2}-25(1+m^2)>0$

asianese · Aug 10, 2013

Re: HSC 2013 2U Marathon

anthony_wheels said:
This is my first time on one of these marathon things and I don't know how to use 'LaTex', but does a=e, and is the area (e/2)-1? I'm probs wrong but worth a try

that's correct.

Sy123 · Aug 10, 2013

Re: HSC 2013 2U Marathon

$$i) Explain why$ \ \ 1+ x + x^2 + x^3 + \dots + x^{n-1} = \frac{x^{n} - 1}{x-1}$

$ii) By differentiating, prove that$

$x + 2x^2+3x^3+ \dots + (n-1)x^{n-1} = \frac{(n-1)x^{n+1} - nx^{n} + x}{(x-1)^2}$

$$iii) You may assume$ \ \lim_{n \to \infty} nx^n = 0 \ $if$ \ -1<x<1$

$$Find the sum$ \ \ \ \frac{1}{2} + 2\left(\frac{1}{2} \right )^2 + 3 \left( \frac{1}{2} \right)^3 + 4 \left( \frac{1}{2} \right )^4 + \dots$

anthony_wheels · Aug 10, 2013

Re: HSC 2013 2U Marathon

asianese said:
that's correct.

Eshayssz haha

RealiseNothing · Aug 10, 2013

Re: HSC 2013 2U Marathon

Sy123 said:
$$i) Explain why$ \ \ 1+ x + x^2 + x^3 + \dots + x^{n-1} = \frac{x^{n} - 1}{x-1}$

$ii) By differentiating, prove that$

$x + 2x^2+3x^3+ \dots + (n-1)x^{n-2} = \frac{(n-1)x^{n+1} - nx^{n} + x}{(x-1)^2}$

$$iii) You may assume$ \ \lim_{n \to \infty} nx^n = 0 \ $if$ \ -1<x<1$

$$Find the sum$ \ \ \ \frac{1}{2} + 2\left(\frac{1}{2} \right )^2 + 3 \left( \frac{1}{2} \right)^3 + 4 \left( \frac{1}{2} \right )^4 + \dots$

You've typo'd part (ii) I think, should it be $(n-1)x^_{n-1}$ ?

Sy123 · Aug 10, 2013

Re: HSC 2013 2U Marathon

RealiseNothing said:
You've typo'd part (ii) I think, should it be $(n-1)x^_{n-1}$ ?

Yep thanks, changing it now

anthony_wheels · Aug 10, 2013

Re: HSC 2013 2U Marathon

Sy123 said:
$$i) Explain why$ \ \ 1+ x + x^2 + x^3 + \dots + x^{n-1} = \frac{x^{n} - 1}{x-1}$

$ii) By differentiating, prove that$

$x + 2x^2+3x^3+ \dots + (n-1)x^{n-1} = \frac{(n-1)x^{n+1} - nx^{n} + x}{(x-1)^2}$

$$iii) You may assume$ \ \lim_{n \to \infty} nx^n = 0 \ $if$ \ -1<x<1$

$$Find the sum$ \ \ \ \frac{1}{2} + 2\left(\frac{1}{2} \right )^2 + 3 \left( \frac{1}{2} \right)^3 + 4 \left( \frac{1}{2} \right )^4 + \dots$

Can I have a bit of a hint for Part ii) ?? Cheers

Carrotsticks · Aug 10, 2013

Re: HSC 2013 2U Marathon

Differentiate both sides and multiply both sides by something.

Sy123 · Aug 11, 2013

Re: HSC 2013 2U Marathon

$$i) Explain why$ \ 1 + 2 + 3 + \dots + n = \frac{n}{2} (n+1)$

$$ii) Show that$ \ (k+1)^3 - k^3 = 3k^2 + 3k + 1$

$$iii) By taking the sum of both sides from$ \ k=1 \ $to$ \ n$

$of the result in part (ii), and expanding the sum out, show that$

$1^2+ 2^2+ 3^2 + \dots + n^2 = \frac{n}{6} (n+1)(2n+1)$

anthony_wheels · Aug 11, 2013

Re: HSC 2013 2U Marathon

I hope it's correct... may not be though

lol

HSC 2013 Maths Marathon (archive) (3 Viewers)

This too shall pass

Well-Known Member

New Member

Member

Σ

Member

Member

ON is my homeboy

Member

Member

Member

Σ

This too shall pass

Member

what is that?It is Cowpea

This too shall pass

Member

Retired

This too shall pass

Member

Users Who Are Viewing This Thread (Users: 0, Guests: 3)