# P4 Edexcel 19th Jan 2004

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HI.has anyone done p4 today??how do you feel?could anyone drop down some answer for the paper.thanks

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#2

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HI.has anyone done p4 today??how do you feel?could anyone drop down some answer for the paper.thanks

**boss-dior**)HI.has anyone done p4 today??how do you feel?could anyone drop down some answer for the paper.thanks

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#4

hi.. I certainly don't expect you to post any answers, but could you possibly suggest which topics you would advise to revise for this exam as I'm doin revision at the moment for it on friday. If you could let me know I could focus on these areas. Thankyou very much anyone who can help.

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#5

Hmm I definitely didn't think it was easy, I thought the questions were harder than normal in places, and there was no 'show' question except for q1 i think.

My answers are:

2) (The NR one) third approximation was 0.8260

3) (Complex numbers one) zw^2 = 8root3i + 8. angle BOC = 90 degrees.

4) (1st order ODE) . y = x^-3e^-x (xe^x - e^x + C).

Solve under boundry counditions to get y = 1/8(1+1/e) when x = 2.

5) (inequalities). Solutions: x= 2 - root(2) and 4 - root(2).

6) (2nd ODE.) Solved to get y = e^-2x(Acosx + Bsinx) + sin 2x - 8cos 2x.

When x is large can approximate by 16 sin^2 (x) + sin 2x - 8.

7) (Polar coords). Integral = 3pi/16 - 9root3/32. (could be wrong, rushed this at end). Distance between lines = 2/3 . root(6).

What do other people think?

My answers are:

2) (The NR one) third approximation was 0.8260

3) (Complex numbers one) zw^2 = 8root3i + 8. angle BOC = 90 degrees.

4) (1st order ODE) . y = x^-3e^-x (xe^x - e^x + C).

Solve under boundry counditions to get y = 1/8(1+1/e) when x = 2.

5) (inequalities). Solutions: x= 2 - root(2) and 4 - root(2).

6) (2nd ODE.) Solved to get y = e^-2x(Acosx + Bsinx) + sin 2x - 8cos 2x.

When x is large can approximate by 16 sin^2 (x) + sin 2x - 8.

7) (Polar coords). Integral = 3pi/16 - 9root3/32. (could be wrong, rushed this at end). Distance between lines = 2/3 . root(6).

What do other people think?

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#6

(Original post by

Hmm I definitely didn't think it was easy, I thought the questions were harder than normal in places, and there was no 'show' question except for q1 i think.

My answers are:

2) (The NR one) third approximation was 0.8260

3) (Complex numbers one) zw^2 = 8root3i + 8. angle BOC = 90 degrees.

4) (1st order ODE) . y = x^3e^-x (xe^x - e^x + C).

Solve under boundry counditions to get y = 1/8(1+1/e) when x = 2.

5) (inequalities). Solutions: x= 2 - root(2) and 4 - root(2).

6) (2nd ODE.) Solved to get y = e^-2x(Acosx + Bsinx) + sin 2x - 8cos 2x.

When x is large can approximate by 16 sin^2 (x) + sin 2x - 8.

7) (Polar coords). Integral = 3pi/16 - 9root3/32. (could be wrong, rushed this at end). Distance between lines = 2/3 . root(6).

What do other people think?

**theone**)Hmm I definitely didn't think it was easy, I thought the questions were harder than normal in places, and there was no 'show' question except for q1 i think.

My answers are:

2) (The NR one) third approximation was 0.8260

3) (Complex numbers one) zw^2 = 8root3i + 8. angle BOC = 90 degrees.

4) (1st order ODE) . y = x^3e^-x (xe^x - e^x + C).

Solve under boundry counditions to get y = 1/8(1+1/e) when x = 2.

5) (inequalities). Solutions: x= 2 - root(2) and 4 - root(2).

6) (2nd ODE.) Solved to get y = e^-2x(Acosx + Bsinx) + sin 2x - 8cos 2x.

When x is large can approximate by 16 sin^2 (x) + sin 2x - 8.

7) (Polar coords). Integral = 3pi/16 - 9root3/32. (could be wrong, rushed this at end). Distance between lines = 2/3 . root(6).

What do other people think?

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#7

(Original post by

Agree on all - however my mate thought (and i think hes right) that the sine approximation should have been better done by cosidering sin(2x-a) where a is arctan something - also 1st order ode should y = x^-3... presume thats what you meant as you got the same answer as me for the next part

**It'sPhil...**)Agree on all - however my mate thought (and i think hes right) that the sine approximation should have been better done by cosidering sin(2x-a) where a is arctan something - also 1st order ode should y = x^-3... presume thats what you meant as you got the same answer as me for the next part

I did it quite wierdly:

d(3sinxcos2x)/dx = 0.

Therefore 3cos2xcosx - 6sinxsin2x = 0.

Therefore 1/2 = tanx tan 2x. Therefore 1/2 = 2tan^2(x)/(1-tan^2(x)). so tan(x) = +/- 1/root5. Take positive: cos(x) = root(5/6). So cos2x = 2/3. So r = 2. So the distance between the point and the initial line = 2 sinx = 2/root(6). So total distance = 4/root(6) = 2/3 . root(6).

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#8

(Original post by

Incidentally, how did you do the polar coords one (the final question).

I did it quite wierdly:

d(3sinxcos2x)/dx = 0.

Therefore 3cos2xcosx - 6sinxsin2x = 0.

Therefore 1/2 = tanx tan 2x. Therefore 1/2 = 2tan^2(x)/(1-tan^2(x)). so tan(x) = +/- 1/root5. Take positive: cos(x) = root(5/6). So cos2x = 2/3. So r = 2. So the distance between the point and the initial line = 2 sinx = 2/root(6). So total distance = 4/root(6) = 2/3 . root(6).

**theone**)Incidentally, how did you do the polar coords one (the final question).

I did it quite wierdly:

d(3sinxcos2x)/dx = 0.

Therefore 3cos2xcosx - 6sinxsin2x = 0.

Therefore 1/2 = tanx tan 2x. Therefore 1/2 = 2tan^2(x)/(1-tan^2(x)). so tan(x) = +/- 1/root5. Take positive: cos(x) = root(5/6). So cos2x = 2/3. So r = 2. So the distance between the point and the initial line = 2 sinx = 2/root(6). So total distance = 4/root(6) = 2/3 . root(6).

i had 3cos2xcosx - 6sinxsin2x = 0.

=> cosx(co2x - 4sin^2x) = 0

cosx = 0 gives minima, so

3cos2x - 2 = 0 => cos2x = 2/3 => r = 2

also 1 - 6sin^2x = 0 so sinx = 1/root6 and the rest is the same as yours

I spent quite a long time on the first 3 or so q's then the miiddle to late ones were fine and i finished with loads of time to spare. I also spent a while deciding whether it was quicker to find zw^2 then find mod and arg or to simply use lzw^2l = lzllwllwl and arg(zw^2) = 2argw + argz. In the end i just found zw^2 which was the right choice cos i had to sketch it in the next part

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#9

Incidentally, my method has one flaw in it accounting for the fact that I've forgotten to include possibilities where cos(x) = 0 or cos(2x) = 0. The only possibility is that cos 2x = 0 and x = pi /4 +2npi. But if this is the case the RHS is positive, so it's not possible. (We can't have cos(x) = 0 since -pi/4 <= x <= pi/4).

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the inequality one,I got 2-root2<x<2+root2....

anyone got the same or just me?/

anyone got the same or just me?/

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#11

(Original post by

the inequality one,I got 2-root2<x<2+root2....

anyone got the same or just me?/

**boss-dior**)the inequality one,I got 2-root2<x<2+root2....

anyone got the same or just me?/

(The equation was |(x-2)(x-4)| = 6 - 2x.)

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#12

**boss-dior**)

the inequality one,I got 2-root2<x<2+root2....

anyone got the same or just me?/

you'd see that for the equation

+(x-2)(x-4) = 6 - 2x

the root had to be less than two,

and the root for

-(x-2)(x-4) = 6 - 2x

had to be less than 4.

I found today's paper rather difficult.

J.

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#13

(Original post by

Hmm I definitely didn't think it was easy, I thought the questions were harder than normal in places, and there was no 'show' question except for q1 i think.

My answers are:

2) (The NR one) third approximation was 0.8260

3) (Complex numbers one) zw^2 = 8root3i + 8. angle BOC = 90 degrees.

4) (1st order ODE) . y = x^-3e^-x (xe^x - e^x + C).

Solve under boundry counditions to get y = 1/8(1+1/e) when x = 2.

5) (inequalities). Solutions: x= 2 - root(2) and 4 - root(2).

6) (2nd ODE.) Solved to get y = e^-2x(Acosx + Bsinx) + sin 2x - 8cos 2x.

When x is large can approximate by 16 sin^2 (x) + sin 2x - 8.

7) (Polar coords). Integral = 3pi/16 - 9root3/32. (could be wrong, rushed this at end). Distance between lines = 2/3 . root(6).

What do other people think?

**theone**)Hmm I definitely didn't think it was easy, I thought the questions were harder than normal in places, and there was no 'show' question except for q1 i think.

My answers are:

2) (The NR one) third approximation was 0.8260

3) (Complex numbers one) zw^2 = 8root3i + 8. angle BOC = 90 degrees.

4) (1st order ODE) . y = x^-3e^-x (xe^x - e^x + C).

Solve under boundry counditions to get y = 1/8(1+1/e) when x = 2.

5) (inequalities). Solutions: x= 2 - root(2) and 4 - root(2).

6) (2nd ODE.) Solved to get y = e^-2x(Acosx + Bsinx) + sin 2x - 8cos 2x.

When x is large can approximate by 16 sin^2 (x) + sin 2x - 8.

7) (Polar coords). Integral = 3pi/16 - 9root3/32. (could be wrong, rushed this at end). Distance between lines = 2/3 . root(6).

What do other people think?

Also, I agree that the paper was harder than usual. When doing past papers I found that I could finish a paper quite quickly. I found this one quite intensive.

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(Original post by

I agree on those apart from Newton raphson, I think I got 0.8177. I am glad you got 3pi/16 - 9root3/32 for the integral, I was worried that I messed that up for some reason. I was rushing towards the end.

Also, I agree that the paper was harder than usual. When doing past papers I found that I could finish a paper quite quickly. I found this one quite intensive.

**mikesgt2**)I agree on those apart from Newton raphson, I think I got 0.8177. I am glad you got 3pi/16 - 9root3/32 for the integral, I was worried that I messed that up for some reason. I was rushing towards the end.

Also, I agree that the paper was harder than usual. When doing past papers I found that I could finish a paper quite quickly. I found this one quite intensive.

I got 0.8177 as well.....quite happy.....

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(Original post by

If you drew a graph,

you'd see that for the equation

+(x-2)(x-4) = 6 - 2x

the root had to be less than two,

and the root for

-(x-2)(x-4) = 6 - 2x

had to be less than 4.

I found today's paper rather difficult.

J.

**hornblower**)If you drew a graph,

you'd see that for the equation

+(x-2)(x-4) = 6 - 2x

the root had to be less than two,

and the root for

-(x-2)(x-4) = 6 - 2x

had to be less than 4.

I found today's paper rather difficult.

J.

if I got the roots wrong,will I get follow through marks for it in part c.???

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#16

(Original post by

if I got the roots wrong,will I get follow through marks for it in part c.???

**boss-dior**)if I got the roots wrong,will I get follow through marks for it in part c.???

This is because it is unfair to penalise you more than once for a single error.

J.

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#17

I agree with your third approximation for the Newton Raphson one, i did it wrong in the exam.

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#18

(Original post by

I agree with your third approximation for the Newton Raphson one, i did it wrong in the exam.

**theone**)I agree with your third approximation for the Newton Raphson one, i did it wrong in the exam.

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#19

(Original post by

I almost did but went back and checked my number crunching. I hate questions like that - like linear interpolation, iteration, and most of S1 where its just typing numbers in a calculator - and i usually do it wrong somewhere

**It'sPhil...**)I almost did but went back and checked my number crunching. I hate questions like that - like linear interpolation, iteration, and most of S1 where its just typing numbers in a calculator - and i usually do it wrong somewhere

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