4756 OCR MEI FP2 22nd June 2015

the answer to the definite integral was 1/2root3 * ln(3)

Reply 125

Genty Boy

3

Original post by lizard54142

What did everyone get for the maclaurin series of ln((1+t)/(1-t))?

2t + (2t^3)/3

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Reply 126

OP

Original post by Genty Boy

What about the question where it said in terms of powers of cos? I can't remember what the coefficients were but the powers were 1 3 and 5... It looked familiar too, may have been in a past paper...

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16cos^5x - 20cos^3x + 5cosx I think?

Reply 127

Original post by Genty Boy

What about the question where it said in terms of powers of cos? I can't remember what the coefficients were but the powers were 1 3 and 5... It looked familiar too, may have been in a past paper...

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You had to use double angle formulae to obtain cos5theta = 16cos^5 theta - 20 cos^3 theta + 5costheta

Reply 128

Genty Boy

3

Original post by chizz1889

the answer to the definite integral was 1/2root3 * ln(3)

(Sqrt3)/6 looks prettier m8

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Reply 129

Genty Boy

3

Original post by lizard54142

16cos^5x - 20cos^3x + 5cosx I think?

Yeah looks familiar

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Reply 130

Original post by Genty Boy

(Sqrt3)/6 looks prettier m8

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True but oh well

Reply 131

123BOBBY

For the last integal I got 1/2root3

Reply 132

OP

Original post by 123BOBBY

For the last integal I got 1/2root3

You mean the second last one?

You're missing a ln3

Reply 133

123BOBBY

No I mean the for the last last part of q4

Reply 134

123BOBBY

I got 1/2root3ln(1+x/1-x)+1/2ln(1-x^2)

And +c

Done a level further maths in one year everything went well apart from d1

Reply 137

OP

Original post by 123BOBBY

I got 1/2root3ln(1+x/1-x)+1/2ln(1-x^2)

Yeah there's no root3 for that one

Reply 138

1. A) I) centre was (a,b)
Radius was Sqrt(a^2 + b^2)
Ii) show that...
B) f(t) = ln(1+t) f'(t) = 1/(1+t) f''(t) = -1/(1+t)^2 f'''(t)= 2/(1+t)^3
ln((1+x)/(1-x)) = 2x + 2/3 * x^3 ...

2. A) I) |w| = sqrt2
Arg (w) = 0, 2pi/5, 4pi/5, -2pi/5, -4pi/5
ii) |v|= 2. Arg(v) is +pi/10 to each before
v^5 = 32j
B) I) show that
Ii) cos5theta = 16cos^5theta - 20cos^3theta + 5costheta

3.i) eigenvalues are -1,0,7
Ii) show that *2 and then next vector was (4,1,3)
Iii) P is 3x3 matrix consisting of each eigenvector, D is matrix of each corresponding eigenvalue in its respective position with regards to its eigenvector
M^4 = PD^4P^-1 therefore when u express in terms of PDP^-1 your new D is the original one with both numerical components eigenvalues raised to the power 4.
Iv) M^4 = 43 M^2 + 42M

4. I) 1 - tanh^2x = sech^2 x
Ii) make tanhy = x differentiate rearrange to get 1/1-X^2
Iii) split into partial fractions and integrate each to obtain 1/2 * ln((1+x)/(1-x))
iv) sqrt3/6 * ln3
V) x/2 * ln ((1+x)/(1-x)) + 1/2 ln(1-x^2) + c

Feel free to make any corrections 😀

(edited 8 years ago)

Reply 139

OP