Edexcel C4 June 2016 Unofficial Mark Scheme

This thread is more accurate - please go here
http://www.thestudentroom.co.uk/showthread.php?t=4184191

Question 1 - Binomial Expansion

Spoiler

Question 2 - Trapezium Rule, Integration

Spoiler

Question 3 - Differentiation

Spoiler

Question 4 - Differentiation - time for decay

Spoiler

Question 5 - Differentiation, Parametric Equations

Spoiler

Question 6 - Integration

Spoiler

Question 7 - Integration

Spoiler

Question 8 - Vectors

Spoiler

Unsure on 3 and 6; I would say standard boundaries. I might do working for questions if someone can post them but I still have 3 exams left so I can't guarantee it

(edited 7 years ago)

Scroll to see replies

Reply 1

veldt127

Original post by k.russell

I think the binomial is wrong no? (last term)
3 x 4 x -5 x (5/2) ^ 3 x 1/6 = -625/4

Yes, and the whole binomial was multiplied by 1/8, giving -625/32

Reply 2

ninjass

4 you needed x in terms of t and 6a was 3ln(3y+2) - 2 ln y +c . But the rest i think is corect

Reply 3

ebolaboy

what was the question for 8a?

(edited 7 years ago)

Reply 4

k.russell

Original post by veldt127

Yes, and the whole binomial was multiplied by 1/8, giving -625/32

oh yeah lol, think I did do that anyway actually

Reply 5

veldt127

Original post by ninjass

4 you needed x in terms of t and 6a was 3ln(3y+2) - 2 ln y +c . But the rest i think is corect

6 was a typo, thanks for spotting that.

It was t in terms of x I believe - hence why you found t when x=20 at the end

Not sure on that one

Reply 6

Smithy1597

How did you do the last part to q8

Reply 7

BellaKa

What about Q6c? I got 4pi/3 - sqrt(3)/2

Reply 8

family'sfailure

how many marks was question 4?

Reply 9

JLegion

For question 5, I ended up with two values for t, being pi/3 and pi/6, depending on whether you used y or x to find t.

I found dy/dx using both values of t, so found two different values of dy/dx.

Anyone else have this problem?

Reply 10

stephaniechan

Do you lose a mark for not saying +c

Reply 11

username2378153

Original post by BellaKa

What about Q6c? I got 4pi/3 - sqrt(3)/2

I think the question goes something like this:

\displaystyle 8\int_0^\frac{\pi}{3} {sin^{2}\theta} \ d\theta

= \displaystyle 4\int_0^\frac{\pi}{3} {1-cos2\theta} \ d\theta

=4\left[ x - \dfrac{1}{2}sin2\theta \right]_0^\frac{\pi}{3}

=4(\dfrac{\pi}{3}-\dfrac{1}{2} (\dfrac{\sqrt{3}}{2}))

=\dfrac{4\pi}{3} - \sqrt{3}

Reply 12

saira.p

Does anyone remember question 5?

Reply 13

username2378153

Original post by saira.p

Does anyone remember question 5?

I think it was this:

x=4\tan t , y=5\sqrt{3} \ sin2t

Work out

\dfrac{dy}{dx}

P(4\sqrt{3},\dfrac{15}{2})

Then it asked to find point Q where

\dfrac{dy}{dx}=0

Edit: Oh and

0\leq t < \dfrac{\pi}{2}

(edited 7 years ago)

Reply 14

m9698

Original post by MagneticFlux

I think the question goes something like this:

\displaystyle 8\int_0^\frac{\pi}{3} {sin^{2}\theta} \ d\theta

= \displaystyle 4\int_0^\frac{\pi}{3} {1-cos2\theta} \ d\theta

=4\left[ x - \dfrac{1}{2}sin2\theta \right]_0^\frac{\pi}{3}

=4(\dfrac{\pi}{3}-\dfrac{1}{2} (\dfrac{\sqrt{3}}{2}))

=\dfrac{4\pi}{3} - \sqrt{3}

sin squared x is equal 1/2 - 1/2 cos 2x

Reply 15

ombtom

Original post by m9698

sin squared x is equal 1/2 - 1/2 cos 2x

They took (1/2) out as a factor. 8/2 = 4.

Reply 16

Dot.dot.

Yeah, i got two different values for question 5 too! I subbed both into the equation and one gave a minus number I think, so I used the other, idk?

(edited 7 years ago)

Reply 17

m9698

Original post by ombtom

They took (1/2) out as a factor. 8/2 = 4.

Apologies

Reply 18

m9698

Original post by m9698

Apologies

I made a silly mistake in the question before where i put Dx by Dtheta as 6sinthetacostheta , would i get 3/5 marks considering everything else was correct ?