2014 Mathematical Methods Exam 2 Paper + Solutions (MCQ + ER)

[chuck981996]

Attached are my suggested solutions for the extended response section of today's exam. The (new) "2014 Exam 2 Q5 UPDATED.pdf" attached file should now be error-free. (Whereas the original, sadly, is not). Hopefully this means the solutions are now completely devoid of errors. (Although I can't 100% guarantee that, obviously: I'm not VCAA)

Multiple choice answers thanks to jumcakes:

• A
• D
• D
• E
• C
• B
• E
• A
• A
• E
• E
• C
• C
• E
• B
• C
• B
• B
• C
• D
• D
• E

Also attached is an entirely BLANK copy of the ENTIRE exam.

ERRATA:
5ci) d E [1,3)
5cii) d<1
5eii) u = 1/2(1 + sqrt(5)), v = 1/2(1 - sqrt(5))
The new file has been updated to include these errors.

UPDATED ERRATA:
2f,g) Units should be m/yr

N.B. The scan is randomly in colour on the last two pages because I'm at home and don't have access to my school's amazing super-scanner. (Clearly my scanner thinks "black & white" means greyscale) I also couldn't include it in the original file and reattach because of the filesize limit - again, don't have access to a scanner capable of compressing and pixelating the files.

-Charlie

[ivysan]

thanks alot, could you do it little bit faster

[RKTR]

5ci  my answer was 1<= d<3
cii mine was d<1
fi  u is the positive one while v is the negative one. since it says gradient at u is positive while gradient at v is negative

[Blondie21]

OMG I converted 5eii to decimals OMG WHY UGGHHHHG

[speedy]

5ci  my answer was 1<= d<3
cii mine was d<1
fi  u is the positive one while v is the negative one. since it says gradient at u is positive while gradient at v is negative

I got the same as you for cii and fi

Edit: Is 5b positive direction? Since we need to sub (x-1)

[RKTR]

I got the same as you for cii and fi

nice . how about ci? if d is 1, intercepts will be at (0,0) and (2,0) . so d can be 1?

[speedy]

nice . how about ci? if d is 1, intercepts will be at (0,0) and (2,0) . so d can be 1?

Mm, I didn't get the same as you but I think you're right. Since 0 isn't positive.

[Aman476]

In 5)c)I I if d =5 which is a real number not between 1 and 3, you'll shift it past.its rightmost intercept so doesn't that mean it has no solutions I said d was between.         (-inifnity, 1)

[Aman476]

In 5e I I thought u had to be positive since m is positive at u (positive gradient)?

[Mujteba]

For 5eii
You need to reject the negative u solution and the positive v solution because the gradient at u is positive and is negative at v.

For 5ci
The boundary d=1 is included as if it were 1 then the intercept to the left of the positive one would be zero.
For 5cii
Should be d<1. You can see the result if you set up two inequalities for the two intercepts and set them both to >0. You end up with d<1, and d<5 so you take the intersection between both which is d<1

Also for 4e
Should n=14 because it is asking for min value of n, not all values of n for which blah blah blah...

[chuck981996]

Okay I'm home, so I can reply now.

5ci  my answer was 1<= d<3
cii mine was d<1
fi  u is the positive one while v is the negative one. since it says gradient at u is positive while gradient at v is negative

You're right on all three counts. Sorry for the muck up!

Also for 4e
Should n=14 because it is asking for min value of n, not all values of n for which blah blah blah...

I think what I wrote is okay, because I'm stating n>=14 which implies n=14 is the minimum value, but I think VCAA would accept either

Is 5b positive direction? Since we need to sub (x-1)

We need to sub (x+1), so no.

[jumcakes]

I've posted some suggested solutions for multiple choice on the main thread which seem to be good (hopefully).

• A
• D
• D
• E
• C
• B
• E
• A
• A
• E
• E
• C
• C
• E
• B
• C
• B
• B
• C
• D
• D
• E

[GeniDoi]

e) ii) solution is partially incorrect, you needed to reject the negative solution for u (and hence the positive solution for v) as when u = (1-Sqrt(5))/2, g'(u) is negative and the question states that at g(u), the gradient is "m" where m is a positive number.

[chuck981996]

e) ii) solution is partially incorrect, you needed to reject the negative solution for u (and hence the positive solution for v) as when u = (1-Sqrt(5))/2, g'(u) is negative and the question states that at g(u), the gradient is "m" where m is a positive number.

Thanks! RKTR already pointed this out - it's in the OP

[Aman476]

Chuck do you reckon you'll be able to keep being the sick kent you are and post the spec solutions tomorrow and monday? (I'm praying yes - seeing the solutions completely gets rid of that uncertainty/anxiety post exam) haha

You're fkn amazing man, you did like both exams in 20 min - glad i'm not competing with you in methods this year haha (but you're in spec - so for that  )  hahaa