2009 Mathematics 2U HSC Worked Solutions (1 Viewer)

[M@ster P]

just marked it i got 102/120. max is 105 and min 97 could make silly mistake that im not aware of. lol so hopefully enough for band 6

 

[addikaye03]

Can someone scan the paper? i just wanna see it/

 

[T.Goodhew]

Equation for 3. d) doesn't seem quite correct..

There are many variations of "the equation" to use for Simpson's Rule but I assure you the final answer is correct.

 

[bornjovied]

Very happy now. Thank you T. Goodhew. I got 117 . Should be good for 98??

 

[mR sinister]

Question 8 a)

it asks ,
For which value of x is the derivative, f ' (x), negative?

Well first derivative determines the gradient, i.e Slope.

Shouldnt the answer be x < 1
As after x = 1 the derivative begins to increase meaning +ive slope.

it is only negative for less than 1

 

[absorber]

very happy now. Thank you t. Goodhew. I got 117 . Should be good for 98??

99

 

[boxhunter91]

OMGGG i drew the graph right in Q8 wooot.
I should def be close to 90% raw.
What you rekn aligned?

 

[Renzel]

That was a shit read. I found a few questions to be ambiguous.

Eg. Q7 (b). Question states: Between 5am and 5pm on 3 March 2009, the height, h, of the tide in a harbour was given by:

h= 1+0.7sin π/6t for t between 0-12

Technically, you should be able to sub in any value for t, without differentiating, to get the height no? That's what the question states. Therefore somebody could go along and sub in t=0 all the way up to t=12 and find that the value for h will increase accordingly. Im not saying this is the correct way, I realise that you are to find the minimum value etc etc, but if you were to do it that way, you are technically correct as well. That's why i hate it.

 

[T.Goodhew]

Question 8 a)

it asks ,
For which value of x is the derivative, f ' (x), negative?

Well first derivative determines the gradient, i.e Slope.

Shouldnt the answer be x < 1
As after x = 1 the derivative begins to increase meaning +ive slope.

it is only negative for less than 1

At x=-1 the curve is stationary (f'(x)=0) then from x = -1 to x=1 the curve is decreasing (f'(x) =0) at an increasing rate, and from x=1 to x=3 the curve is still decreasing (f'(x)=0) however at a decreasing rate.

Remember the first derivative f'(x) will have to get from a large negative number (as the curve is decreasing quickly just after x=-1) down to zero before it can become positive (assuming the first derivative is continuous), considering the next stationary point after x=-1 is at x=3 there is no way that the first derivative could be positive before x=3.

There would be an exception if the point at x=1 is a horizontal point of inflexion (f'(x)=0 and f''(x)=0) but on face value it appears to just be a standard point of inflexion.

If you draw tangents to the curve at arbitrary points from x=-1 to x=3 you will see that all their gradients are negative (proof that f'(x) <0 and the curve is decreasing). Note if x=1 were a horizontal point of inflexion the tangent would be a straight line at this point, it is clearly not.

 

[bornjovied]

99

wow. even better. i hope you're right

 

[bornjovied]

That was a shit read. I found a few questions to be ambiguous.

Eg. Q7 (b). Question states: Between 5am and 5pm on 3 March 2009, the height, h, of the tide in a harbour was given by:

h= 1+0.7sin π/6t for t between 0-12

Technically, you should be able to sub in any value for t, without differentiating, to get the height no? That's what the question states. Therefore somebody could go along and sub in t=0 all the way up to t=12 and find that the value for h will increase accordingly. Im not saying this is the correct way, I realise that you are to find the minimum value etc etc, but if you were to do it that way, you are technically correct as well. That's why i hate it.

I hear alot of people saying the same thing, but i really enjoyed how alot of the problems were applied to the physical world. it was challenging, very satisfying.

It was definitely harder than previous years, but i loved it.

 

[Ruth Paske]

Geoff Wright sends his seal of approval to your solutions, Tim.
Such style and efficiency. Congratulations, you must have been taught by one of the best!
Now, any tips for the Melbourne Cup?

 

[verdades]

That was a shit read. I found a few questions to be ambiguous.

Eg. Q7 (b). Question states: Between 5am and 5pm on 3 March 2009, the height, h, of the tide in a harbour was given by:

h= 1+0.7sin π/6t for t between 0-12

Technically, you should be able to sub in any value for t, without differentiating, to get the height no? That's what the question states. Therefore somebody could go along and sub in t=0 all the way up to t=12 and find that the value for h will increase accordingly. Im not saying this is the correct way, I realise that you are to find the minimum value etc etc, but if you were to do it that way, you are technically correct as well. That's why i hate it.

That's what I did. I hadn't planned to when I first read it, the proper process for it simply went right out of my head as I tried to do the question normally.
Though I subbed 1 - 4, 6, 8, 9, 10 and 12, because I'd gotten the pattern by then >.>

From what little I can remember of my paper, I've lost a fair few x.x

 

[Thecorey0]

That's what I did. I hadn't planned to when I first read it, the proper process for it simply went right out of my head as I tried to do the question normally.
Though I subbed 1 - 4, 6, 8, 9, 10 and 12, because I'd gotten the pattern by then >.>

From what little I can remember of my paper, I've lost a fair few x.x

You did not have to differentiate it. By observation, you should know that sinx always lies between -1 and 1. Hence the minimum height occurs when sinx =-1. Its similar to shm in 3u.

 

[verdades]

You did not have to differentiate it. By observation, you should know that sinx always lies between -1 and 1. Hence the minimum height occurs when sinx =-1. Its similar to shm in 3u.

Yeah, I got that pretty easily, once I'd gathered that t=3 = 1. I did the other ones for appearance's sake.
I didn't do particularly well with 3u overall, never understood SHM much, so... >.<'

 

[random-1005]

im not going to read that. ill end up comitting suicide if i do LOL.

bye sydney uni, nice dreaming of you

i love this site, it has the perfect combination of asian try hards that cant stop thinking about scaling, and then you have the people playing with dolls that want to hang themselves

entertaining

 

[RuffWoof]

band 5 here i come