HSC 2013 Maths Marathon (archive) (1 Viewer)

Sy123 · Aug 11, 2013

Re: HSC 2013 2U Marathon

anthony_wheels said:
View attachment 28510 I hope it's correct... may not be though lol

Yep well done, however for the last part I would have liked some better justification of how the term, $(n-1)x^{n+1}$ converges to zero

One way is that:

$\lim_{n \to \infty} (n-1)x^{n+1} = \lim_{n \to \infty} nx^{n+1} - \lim_{n \to \infty} x^{n+1}$

We know the second limit converges to zero, because $\lim_{n \to \infty}(1/2)^n = 0$ (sketch the graph of 2^(-n))

$=\lim_{n \to \infty} nx^{n+1} = x \times \lim_{n \to \infty} nx^n = x \times 0 = 0$

Sy123 · Aug 11, 2013

Re: HSC 2013 2U Marathon

$$i) Find real numbers A and B, such that$ \ \ \frac{A}{x} + \frac{B}{x+1} = \frac{1}{x(x+1)}$

$$ii) Hence find$ \ \int \frac{dx}{x(x+1)}$

$iii) Show by expanding the sum out$

$\sum_{k=1}^N \left(\frac{1}{k} - \frac{1}{k+1} \right ) = 1 - \frac{1}{N+1}$

$iv) Hence find the value of the sum$

$\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \frac{1}{4 \times 5} + \dots$

anthony_wheels · Aug 11, 2013

Re: HSC 2013 2U Marathon

Computer lagged majorly trying to attach that for some reason... haha

Sy123 · Aug 11, 2013

Re: HSC 2013 2U Marathon

anthony_wheels said:
View attachment 28514

Computer lagged majorly trying to attach that for some reason... haha

Yep that is correct well done.

$$Define a function$ \ y=F(x) \ \ $such that for some value$ \ x, \ \ F(x) \ $is the greatest integer less than or equal to $ \ x$

$$For example$ \ F(1.5) = 1 \ \ F(\pi) = 3 \ \ F(4.8) = 4 \ \ F(5) = 5$

$$i) Sketch the graph of$ \ y=F(x) \ \ $for$ \ x >0 \ \ $ and prove that$ \ \ \int_0^n F(x) \ dx = \frac{1}{2}n(n-1) \ \ $for some integer$ \ n$

$$ii) Prove that for integers$ \ k \ $and$ \ n \ \ \int_0^n F(kx) \ dx = \frac{1}{2} n(nk-1)$

HeroicPandas · Aug 25, 2013

Re: HSC 2013 2U Marathon

$\rightarrow \\ \\ $If $y = sin^x(x), \\ \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~$Find $\frac{dy}{dx}$

Menomaths · Aug 25, 2013

Re: HSC 2013 2U Marathon

HeroicPandas said:
$\rightarrow \\ \\ $If $y = sin^x(x), \\ \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~$Find $\frac{dy}{dx}$

ln y = x ln sinx
1/y = x (cosx/sinx) + ln sinx
1/y = x cotx + ln sinx
1 = y (x cotx + ln sinx)

Sy123 · Aug 25, 2013

Re: HSC 2013 2U Marathon

$$i) Let$ \ f(x) = (a+x) - 2\sqrt{ax}$

$$for positive constant$ \ a \ $and$ \ x>0$

$$Prove by differentiation$ \ \ (a+x) \geq 2\sqrt{ax}$

$ii) A farmer has$ \ P \ $metres of fencing to fence in a flat rectangular area of land$

$$Prove that if the area to be fenced in is a maximum, then the area is$ \ \ \frac{P^2}{16}$

Menomaths · Aug 25, 2013

Re: HSC 2013 2U Marathon

Sy123 said:
$$i) Define the function$ \ f(x) = (a+x) - 2\sqrt{ax}$

$$for positive constant$ \ a \ $and$ \ x>0$

$$Prove by differentiation$ \ \ (a+x) \geq 2\sqrt{ax}$

$ii) A farmer has$ \ P \ $metres of fencing to fence in a flat rectangular area of land$

$$Prove that if the area to be fenced in is a maximum, then the area is$ \ \ \frac{P^2}{16}$

Was my answer for the above question correct?
For ii)
P=2(x+y)
y=P/2 -x
A= x(P/2 -x)
A= Px/2 -x^2
A'= P/2 - 2x
Let A'= 0
2x = P/2
x=P/4
Sub back into A=xy, for area to be maximum it has to be square (bs skill over 9000) Therefore (P/4)^2 = P^2/16

Sy123 · Aug 25, 2013

Re: HSC 2013 2U Marathon

Menomaths said:
Was my answer for the above question correct?
For ii)
P=2(x+y)
y=P/2 -x
A= x(P/2 -x)
A= Px/2 -x^2
A'= P/2 - 2x
Let A'= 0
2x = P/2
x=P/4
Sub back into A=xy, for area to be maximum it has to be square (bs skill over 9000) Therefore (P/4)^2 = P^2/16

For the above answer, when differentiating ln(y) with respect to x you get: $\frac{y'}{y}$ instead of $\frac{1}{y}$

It is correct except for the 1 on the LHS, it should be y'

And for your solution, it is correct except for the part where you assume the area if a maximum if its a square, instead of subbing into A=xy, you can just sub it into A=x(P/2 - x), but that isn't how I intended the solution to come about, you are supposed to use the result from part (i) and not need to use differentiation =)

Menomaths · Aug 25, 2013

Re: HSC 2013 2U Marathon

Sy123 said:
For the above answer, when differentiating ln(y) with respect to x you get: $\frac{y'}{y}$ instead of $\frac{1}{y}$

It is correct except for the 1 on the LHS, it should be y'

And for your solution, it is correct except for the part where you assume the area if a maximum if its a square, instead of subbing into A=xy, you can just sub it into A=x(P/2 - x), but that isn't how I intended the solution to come about, you are supposed to use the result from part (i) and not need to use differentiation =)

I knew the 1 was wrong lol
The only reason why I used differentiation for that question was because I don't know how to solve part i -.-''

Sy123 · Aug 25, 2013

Re: HSC 2013 2U Marathon

Menomaths said:
I knew the 1 was wrong lol
The only reason why I used differentiation for that question was because I don't know how to solve part i -.-''

HINT:

If you look at the function given and the inequality that you need to prove, it seems that through differentiation we need to prove that $f(x) \geq 0$

RealiseNothing · Aug 25, 2013

Re: HSC 2013 2U Marathon

Sy123 said:
HINT:

If you look at the function given and the inequality that you need to prove, it seems that through differentiation we need to prove that $f(x) \geq 0$

Is there a reason you want them to prove it using differentiation exactly?

Sy123 · Aug 25, 2013

Re: HSC 2013 2U Marathon

RealiseNothing said:
Is there a reason you want them to prove it using differentiation exactly?

Not really, I just thought since maxima/minima is a big part of the 2U course, they might as well utilise it.
And I'm divided in 2U whether to just ask it straight away or to give a hint (Using (a-b)^2 > 0 etc)

Menomaths · Aug 25, 2013

Re: HSC 2013 2U Marathon

Sy123 said:
HINT:

If you look at the function given and the inequality that you need to prove, it seems that through differentiation we need to prove that $f(x) \geq 0$

y = (a+x) - 2(ax)^1/2
y' = 1- a(ax)^-1/2
y' = 1 - a/(ax)^1/2
y' = ((ax)^1/2 - a)/(ax)^1/2
Let y' = 0
(ax)^1/2 -a = 0
a^2 = ax
x = a
a+x = 2(ax)^1/2
2a = 2(a^2)^1/2
2a = 2a
Therefore a+x=2(ax)^1/2
Is this even...an answer?

Sy123 · Aug 25, 2013

Re: HSC 2013 2U Marathon

Menomaths said:
y = (a+x) - 2(ax)^1/2
y' = 1- a(ax)^-1/2
y' = 1 - a/(ax)^1/2
y' = ((ax)^1/2 - a)/(ax)^1/2
Let y' = 0
(ax)^1/2 -a = 0
a^2 = ax
x = a
a+x = 2(ax)^1/2
2a = 2(a^2)^1/2
2a = 2a
Therefore a+x=2(ax)^1/2
Is this even...an answer?

Yes x=a is the minimum of the function, this means that f(x) > f(a)
However we aren't proving equality, we are proving the inequality here. Try and use the fact that f(x) > f(a)

Menomaths · Aug 25, 2013

Re: HSC 2013 2U Marathon

Sy123 said:
Yes x=a is the minimum of the function, this means that f(x) > f(a)
However we aren't proving equality, we are proving the inequality here. Try and use the fact that f(x) > f(a)

How is that possible when we're essentially subbing in the same thing?

Drongoski · Aug 25, 2013

Re: HSC 2013 2U Marathon

Menomaths said:
ln y = x ln sinx
1/y(dy/dx) = x (cosx/sinx) + ln sinx
1/y = x cotx + ln sinx
1 = y (x cotx + ln sinx)

Where is your dy/dx?

Sy123 · Aug 25, 2013

Re: HSC 2013 2U Marathon

Menomaths said:
How is that possible when we're essentially subbing in the same thing?

Since x=a is a minimum of the function f(x) for any value of x > 0
Then if we pick say x=1 (if a=/=1) then f(1) > f(a) right?

If you sketch the graph of f(x), then when x=a, it is a minimum value and the rest of the function is greater than or equal to it.
Its like saying

(x^2+3) > 3

but 3 = f(0) anyway, its a constant, so if f(x) = x^2 + 3, since when x=0 is a minimum, then f(x) > f(0).

I hope I've explained it properly

Menomaths · Aug 25, 2013

Re: HSC 2013 2U Marathon

Drongoski said:
Where is your dy/dx?

I know, should've put y' instead of 1/y

Sy123 said:
Since x=a is a minimum of the function f(x) for any value of x > 0
Then if we pick say x=1 (if a=/=1) then f(1) > f(a) right?

If you sketch the graph of f(x), then when x=a, it is a minimum value and the rest of the function is greater than or equal to it.
Its like saying

(x^2+3) > 3

but 3 = f(0) anyway, its a constant, so if f(x) = x^2 + 3, since when x=0 is a minimum, then f(x) > f(0).

I hope I've explained it properly

Oh, I see now. Thanks ^^

Sy123 · Aug 26, 2013

Re: HSC 2013 2U Marathon

$A standard fair die is rolled$ \ n \ $times, the product of the scores is calculated$

$Show that the probability that this product is even is$

$\frac{2^n-1}{2^n}$

HSC 2013 Maths Marathon (archive) (1 Viewer)

This too shall pass

This too shall pass

Member

This too shall pass

Heroic!

Exaı̸̸̸̸̸̸̸̸lted Member

This too shall pass

Exaı̸̸̸̸̸̸̸̸lted Member

This too shall pass

Exaı̸̸̸̸̸̸̸̸lted Member

This too shall pass

what is that?It is Cowpea

This too shall pass

Exaı̸̸̸̸̸̸̸̸lted Member

This too shall pass

Exaı̸̸̸̸̸̸̸̸lted Member

Well-Known Member

This too shall pass

Exaı̸̸̸̸̸̸̸̸lted Member

This too shall pass

Users Who Are Viewing This Thread (Users: 0, Guests: 1)