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Add Maths June 2014

How did everybody find the exam today?
I've walked out knowing I left out 6 marks worth but aside from that hopefully still an A.
Also can anyone remember what they got for their objective function equation in the linear programming question?

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Reply 1
found it pretty tough ngl, but hopefully I've still managed an A-I can think of some pretty silly mistakes already though

The objective function I think the max volume was 35, as 10x would be 35? I think
Reply 2
i put 32 as the max volume because it was something like 3x+1.5y (or vice versa, cant remember)
Reply 3
what about the question about the remainder theorem, that was pretty tough
Reply 4
my two points for linear programming was (4,6)
Reply 5
overall, it was quite a difficult paper, what do you think the grade boundaries for an a will be
Reply 6
I walked out literally almost cried. The kid in my school who got 100% in gcse and always got like 95% in add maths mocks walked out and said he left so many blank and he hated it and it was horrible and he didn't understand it at all.
That was a horrific paper. The past papers seemed so much more easier than this
i found it really bad just like the rest of my class
but there was this one really clever girl who found it easy and just got stuck on one question
(M****, if you're reading this:colone:)
Original post by uuubbbub
what about the question about the remainder theorem, that was pretty tough


The function f(x)=x3−4x2+ax+b f(x)=x^3-4x^2+ax+b is such that:

•

x = 3 is a root of the equation f(x) = 0

•

when f(x) is divided by (x – 1) there is a remainder of 4.


(i) Find the value of a and the value of b.
(ii) Solve the equation f(x) = 0.

Part (i):
So, we know that if x = 3, f(x) = 0, so let's plug it in: this leads to
27 – 36 + 3a + b = 0
3a + b = 9
We also know (due to the factor theorem) that f(1) must produce 4.
1 – 4 + a + b = 4
a + b = 7
We now have simultaneous equations:
(equation 1): 3a + b = 9
(equation 2): a + b = 7
(1) – (2): 2a = 2
a = 1

1 + b = 7
b = 6

Part (ii):
Now, we can fill in a and b: f(x)=x3−4x2+x+6 f(x)=x^3-4x^2+x+6
We know that when f(x) = 0, when x = 3. This means that (x – 3) is a factor (factor theorem)
Now we can find the other factors and therefore solve for all roots.
x3−4x2+x+6=(x−3)(x2–x–2)x^3-4x^2+x+6=(x-3)(x^2–x–2)
(x−3)(x2–x–2)=(x−3)(x−2)(x+1)(x-3)(x^2–x–2)=(x-3)(x-2)(x+1)
So, x = –1, x = 2,x = 3
Reply 9
I found it pretty easy to be honest, the questions were perhaps a little more difficult than usual but I just put that down to the stress of the actual exam.

For the objective function, I put V(x, y) = 3.5x + 3y
With this, and a simultaneous equation of the two lines I had drawn, I got x and y as 4 and 6 and the maximum volume as 32
Original post by Maprunner
I found it pretty easy to be honest, the questions were perhaps a little more difficult than usual but I just put that down to the stress of the actual exam.

For the objective function, I put V(x, y) = 3.5x + 3y
With this, and a simultaneous equation of the two lines I had drawn, I got x and y as 4 and 6 and the maximum volume as 32


As did most of us, I believe :smile:

The graph was a bit funny – that little kink in it just pushed the objective function to that 'corner' instead of the bottom right. It was a nice spot.
Reply 11
Original post by danconway


The function f(x)=x3−4x2+ax+b f(x)=x^3-4x^2+ax+b is such that:

•

x = 3 is a root of the equation f(x) = 0

•

when f(x) is divided by (x – 1) there is a remainder of 4.


(i) Find the value of a and the value of b.
(ii) Solve the equation f(x) = 0.

Part (i):
So, we know that if x = 3, f(x) = 0, so let's plug it in: this leads to
27 – 36 + 3a + b = 0
3a + b = 9
We also know (due to the factor theorem) that f(1) must produce 4.
1 – 4 + a + b = 4
a + b = 7
We now have simultaneous equations:
(equation 1): 3a + b = 9
(equation 2): a + b = 7
(1) – (2): 2a = 2
a = 1

1 + b = 7
b = 6

Part (ii):
Now, we can fill in a and b: f(x)=x3−4x2+x+6 f(x)=x^3-4x^2+x+6
We know that when f(x) = 0, when x = 3. This means that (x – 3) is a factor (factor theorem)
Now we can find the other factors and therefore solve for all roots.
x3−4x2+x+6=(x−3)(x2–x–2)x^3-4x^2+x+6=(x-3)(x^2–x–2)
(x−3)(x2–x–2)=(x−3)(x−2)(x+1)(x-3)(x^2–x–2)=(x-3)(x-2)(x+1)
So, x = –1, x = 2,x = 3


Perfect i wrote exactly that
Original post by danconway
As did most of us, I believe :smile:

The graph was a bit funny – that little kink in it just pushed the objective function to that 'corner' instead of the bottom right. It was a nice spot.


I just tested every vertex - I had plenty of time
Reply 13
Original post by Maprunner
I found it pretty easy to be honest, the questions were perhaps a little more difficult than usual but I just put that down to the stress of the actual exam.

For the objective function, I put V(x, y) = 3.5x + 3y
With this, and a simultaneous equation of the two lines I had drawn, I got x and y as 4 and 6 and the maximum volume as 32


So did I :smile:


Posted from TSR Mobile


Haha, good to see some confirmation - I was a little worried I'd gone and said it was easy and then touted the wrong answer
Reply 15
What did you guys get for binomial answers (the last question)?

For ii) (likelihood the box is unsatisfactory)
I did 1- (10C0 x 0.95^10) + (10C1x 0.95^9 x 0.05^1)
and came out with something like 0.0861

For iii) I did (10C0 x 0.95^10) + (10C0x 0.95^10 x 10C1x 0.95^9 x 0.05^1)
and got 0.7874...

Similar answers?!

Oh also what did you get for the time difference question? I got 1hr 9 minutes but think I may have made a rounding error 😟


Posted from TSR Mobile
(edited 9 years ago)
@annabeltsr
TSR is playing up so I can't quote

Those Binomial answers look right for what I got

I think I put 70, because it was something like 1.16 hours which rounds to 70 mins
Original post by annabeltsr
What did you guys get for binomial answers (the last question)?

For ii) (likelihood the box is unsatisfactory)
I did 1- (10C0 x 0.95^10) + (10C1x 0.95^9 x 0.05^1)
and came out with something like 0.0861

For iii) I did (10C0 x 0.95^10) + (10C0x 0.95^10 x 10C1x 0.95^9 x 0.05^1)
and got 0.7874...

Similar answers?!

Oh also what did you get for the time difference question? I got 1hr 9 minutes but think I may have made a rounding error 😟


Posted from TSR Mobile


All above are correct, I posted some answers in previous thread
I thought it went pretty well. What were the conditions for the binomial distribution of the mugs - thought that was a bit of a strange question?

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