AQA Core 4 Exam Discussion 14/06/2012 !Poll, paper and unofficial MS (first post)!

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Original post by Dangerous Theory

Again, explain to me how x can have a range of between -4 to 4, while if I sub -5 into (4-x)^-0.5 I get a solution??

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http://en.wikipedia.org/wiki/Power_series

To put your theory to rest, check that page.

The expansion of 1/(1-x) is valid for |x|<1, which coincides what I wrote earlier.

/end.

Reply 901

Dangerous Theory

Original post by f1mad

I don't follow...x=-4, (4+4)^-0.5 = 0.354 x= -5, (4+5)^-0.5 = 0.3 ...I could continue...?

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Reply 902

Zoe_mj

Would -4 < x < 4 be right for 3bii?

Reply 903

f1mad

Original post by Dangerous Theory

I don't follow...x=-4, (4+4)^-0.5 = 0.354 x= -5, (4+5)^-0.5 = 0.3 ...I could continue...?

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x cannot equal -4.

Quite clearly x is either LESS than 4 or GREATER than -4.

There is no mention of EQUAL to.

Reply 904

LGrosvenor101

Original post by SomeoneIveNeverMet

Oh I meant I do AS Further Math this year too :tongue:

Wait so what papers do you do for it?! We had to do FP1 (in Jan), S1 and FP3 (gah).

Well, at least we're the lucky few who would gladly go to our insurances if we don't make it to our firms. I like to look at it that way. :smile:

Oh ok :L and I did FP1, M1 and D1 xD and exactly! I need Us in all my subjects to get Bs overall xD so im happy :L

Reply 905

Dangerous Theory

Original post by f1mad

x cannot equal -4.

Quite clearly x is either LESS than 4 or GREATER than -4.

There is no mention of EQUAL to.

I'm trying to prove MY point. You're saying x can equal anything, between - 4and 4, except I've just proved it can go lower than -4...

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Reply 906

f1mad

Original post by Dangerous Theory

I'm trying to prove MY point. You're saying x can equal anything, between - 4and 4, except I've just proved it can go lower than -4...

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What are you talking about?

Can you read?

|x|<4 is the same as -4<x<4.

You can carry on debating this with yourself (waste your time): it is correct.

You haven't proved anything, you're only embarrassing yourself. Do you know what "<" means? It doesn't mean it's equal to that number you're suggestion that supposedly proves your point.

http://en.wikipedia.org/wiki/Absolute_value

Read the above, instead of wasting both mine and your own time.

(edited 11 years ago)

Reply 907

1232584769

also the vector question, i left my answer for AP^2, as square root 20 (20^0.5), and i had used this in the next 6 mark question to find the 2 coordinates of B, do i get ecf and howmuch marks would i loose in the 6 mark question. Also the differential equation when i factorised 15 out of the equation i had forgotten to multiply the entire differential equation by 15 again, how much do i loose for that and do i get ecf for using this differential equation for the question after it.

Reply 908

Wolwen

Original post by Dangerous Theory

I'm trying to prove MY point. You're saying x can equal anything, between - 4and 4, except I've just proved it can go lower than -4...

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Yes but if it goes lower wouldnt the terms become bigger and bigger, making the expansion invalid as it's no longer a correct approximation?

Reply 909

Dangerous Theory

Original post by f1mad

What are you talking about?

Can you read?

|x|<4 is the same as -4<x<4.

You can carry on debating this with yourself (waste your time): it is correct.

IM TRYING TO PROVE THAT X<4 NOT |x|<4... I've already stated I KNOW |x| < 4 is the same as -4 < x < 4, this is EXACTLY what I'm arguing against. How can it possibly get an answer for x less than -4, while your range implies x can NEVER equal -4 or less?

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Reply 910

Raj K

how do people get A* for maths man..geniuses !!!

Reply 911

f1mad

Original post by Dangerous Theory

For the series to converge, it must hold for that inequality. Do you get it now or not?

Reply 912

Oromis263

Home.

Question 7.

a)

Unparseable latex formula:

l_1 = l_2 [br]\[br][br] \begin{bmatrix} 2\lambda \\ -2 \\ q - \lambda \end{bmatrix} = \begin{bmatrix} 8 + 2\mu \\ 3 + 5\mu \\ 5 + 4\mu \end{bmatrix}[br][br]\[br] [br] [br][br]so\ 2\lambda = 8+2\mu[br]\[br]-2 = 3 + 5\mu[br]\[br]q-\lambda = 5 + 4\mu[br]\[br]\Rightarrow \mu = -1 \Rightarrow \lambda = 3[br]\[br]q = 3 + 5 -4 = 4[br]\[br]P (6, -2, 1)

Unparseable latex formula:

\begin{bmatrix} 2 \\ 0 \\ -1\end{bmatrix} \bullet \begin{bmatrix} 2 \\ 5 \\ 4 \end{bmatrix}[br]\[br]\Rightarrow 2(2) + 0(5) -1(4) = 0 \Rightarrow cos\theta = 0, \theta = 90\ \therefore\ perpendicular

I think I'll do the rest of this one by hand and upload a picture >.<

Thus coordinates of B were

(4\dfrac{2}{3}, -5\dfrac{1}{3}, -1\dfrac{2}{3})\ and\ (7\dfrac{1}{3}, 1\dfrac{1}{3}, 3\dfrac{2}{3})

(edited 11 years ago)

Reply 913

Rotravis

Original post by Oromis263

Question 6.

Unparseable latex formula:

9x^2 - 6xy + 4y^2 = 3[br]\[br]implicit\ differentiation[br]\[br]\18x - 6y -6x\dfrac{dy}{dx} + 8y\dfrac{dy}{dx} = 0[br]\[br]\Rightarrow \dfrac{dy}{dx} = \dfrac{6(3x-y)}{8y-6x}[br]\[br]\Stationary\ at \dfrac{dy}{dx} =0[br]\[br]3x-y = 0 \Rightarrow y = 3x[br]\[br]\Sub\ this\ into\ original\ equation[br]\[br]27x^2 = 3 \Rightarrow x = \dfrac{-1}{3}\ or \dfrac{1}{3} \Rightarrow y = 1\ or -1

For dy/dx I got (-18x+6y)/(-6x+8y). Why the positive 18x on top?

Reply 914

Dangerous Theory

Original post by f1mad

For the series to converge, it must hold for that inequality. Do you get it now or not?

Yes. I thought it was asking for values for which the function would work. Although then it would be <= 4 anyway

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Reply 915

SomeoneIveNeverMet

Original post by LGrosvenor101

Oh ok :L and I did FP1, M1 and D1 xD and exactly! I need Us in all my subjects to get Bs overall xD so im happy :L

Wow seriously?! You must have done amazing last year! :O

Reply 916

f1mad

Original post by Dangerous Theory

Yes. I thought it was asking for values for which the function would work. Although then it would be <= 4 anyway

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It wouldn't. It'd be <4 only, try it. You will not get zero (and it wouldn't converge to zero, it'd stop where it is since the successive terms have limited to zero).

Reply 917

Dangerous Theory

Original post by f1mad

It wouldn't. It'd be <4 only, try it. You will not get zero (and it wouldn't converge to zero, it'd stop where it is since the successive terms have limited to zero).