HSC 2013 Maths Marathon (archive) (4 Viewers)

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) (4 Viewers)

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