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Edexcel FP3 June 2015 - Official Thread

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Reply 700

trooken

Original post by imusti96

Same i got 5/3

1/4 and 5/3 I got

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Reply 701

Ilovemaths96

Original post by elliemoriarty

I got 5/3 for q

dont say that!

Reply 702

piddym7

can someone post how to do the proof on the second part of the reduction formula question???????

Reply 703

cogito.

Original post by piddym7

can someone post how to do the proof on the second part of the reduction formula question???????

The -sin^n-1 x cos x under limit [pi/2,0]will become 0 eventually regardless of n since cos(pi/2)=0=sin(0), so Ln will left with (n-1)/n Ln-2 eventually, and then easy stuff

Not a hard question but strange, it will definitely hit some less able students comparing to 2009 one, (see my UMS comparison above)

Reply 704

blankk

Does anybody know how to do the revolution area part? I don't know think it is like the normal ones...Do you just differentiate the whole thing with the square root?

Reply 705

Ilovemaths96

8/105 for reduction part c???

Reply 706

Ilovemaths96

Original post by trooken

1/4 and 5/3 I got

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how?

Reply 707

username1432489

Original post by Ilovemaths96

8/105 for reduction part c???

Yes!

Reply 708

cogito.

Original post by blankk

Does anybody know how to do the revolution area part? I don't know think it is like the normal ones...Do you just differentiate the whole thing with the square root?

it's C4, not FP3 (i hope this message will not hit you so much....)

Reply 709

cogito.

Original post by Chazatthekeys

Yes!

Pls help me with a MS if you can remember some answers

Reply 710

ThatPerson

Original post by piddym7

can someone post how to do the proof on the second part of the reduction formula question???????

Let

J_n = \int^{\pi/2}_{0} \sin^{n} x \ dx = -\frac{1}{n}\sin^{n-1} x \cos x |^{\frac{\pi}{2}}_{0} + \frac{n-1}{n}J_{n-2}

Therefore

J_n = \frac{n-1}{n}J_{n-2}

. Also

Unparseable latex formula:

J_1 = \int^{\frac{\pi}{2}}_{0} \sin x \dx = 1

J_n = \frac{(n-1)(n-3)(n-5)...4.2}{n(n-2)(n-4)...5.3}J_1 \implies J_n = \frac{(n-1)(n-3)(n-5)...4.2}{n(n-2)(n-4)...5.3}

.

Think that was it. My memory after an exam isn't very good.

(edited 8 years ago)

Reply 711

trooken

Original post by Ilovemaths96

how?

Don't fully remember the question, but it was the integral of y², multiplied by π
You had to use one of the integrals from the formula sheet that don't involve hyperbolic or trigonometric functions, just partial fractions, although you didn't have to do the partial fractions yourself

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Reply 712

blankk

Original post by cogito.

it's C4, not FP3 (i hope this message will not hit you so much....)

But you need to do the square root of 1+(dy/dx)^2 thing don't you...

Reply 713

siuyuk233

anyone feel sure about their answer of the triangle area in hyperbola?

(edited 8 years ago)

Reply 714

username1553813

Original post by Chazatthekeys

I got OQR=1

Had like Q(e^t, e^t) and R(e^-t, -e^-t) and find OQ and OR by Pythagoras and then area = 1 therefore independent of variable 't'

Thank GOD!!!

Reply 715

cogito.

Original post by blankk

But you need to do the square root of 1+(dy/dx)^2 thing don't you...

It asked volume of solid not the surface area, sorry

Reply 716

Goos

Expected boundaries anyone?
I found the paper was fairly hard, made a few stupid mistakes but hope it's enough for an A.

Reply 717

username1432489

Original post by cogito.

Pls help me with a MS if you can remember some answers

I remember the answers, but not so much the questions! Lol

Reply 718

ThatPerson

Original post by cogito.

It asked volume of solid not the surface area, sorry

If I put 2pi instead of pi for the volume of revolution formula, do you think that's only one mark lost? (The integration and everything else was correct).