AQA June 2016 Further Pure 2 Unofficial Mark Scheme

These are my answers to todays FP2 exam. Not all of them are necassarily correct, please feel free to correct me on any you think are wrong. Not sure about the question numbers.
FP2 June 2016
1.
a)

A=4

(2 marks)
b)

\frac{50}{609}

(4 Marks?)

2.
a) i)

\beta=1-2i

(1 Mark)
ii)

\alpha\beta = 5

(1/2 marks)
b)

\gamma = \frac{1}{3}

p=-5 q=-7

(3 Marks)

3. a) Show that s is the required integral
b)

\ln7- \frac{3}{4}

(5 marks?)

4. a)

\frac{\sqrt3}{2(1+3x)\sqrt(x)}

(2 Marks)
b)

\frac{\sqrt3\pi}{18}

(4 Marks)

5. a)

8

(2 Marks)
b)i) Sketch with a circle with centre

-4\sqrt3 + 4i

radius

4

(3 Marks)
ii)

2\pi/3

(2 Marks)
c) Roots are

2e^{\frac{ix\pi}{18}}

where

x=-7,5,17

(5 Marks)

6.a) Show that Q
b)

6\sqrt3 + 3arccosh2 + 5

7. Induction Q. (6 marks?)
Trivial in the case

n=1

since clearly

1+p \geq 1+p

Assume true for

n=k

(1+p)^{k+1} = (1+p)^{k}(1+p) = (1+kp)(1+p)[br]= 1 + kp + p + kp^2 \geq 1 + (k+1)p

kp^2 \geq 0

Thus true for

n=k+1

so true for all positive integers n by induction.

8. a) Show that Q (3/4 Marks?)
b) Roots were

i\tan(x\pi)

where

x= 1,3,5,7

(2 Marks)
c) i)

1

(4 Marks)
ii)

6

(4 marks)

(edited 7 years ago)

Scroll to see replies

Reply 1

Barbecuetime123

Grade boundary predictions?

Reply 2

girlruinedme

I thought the second last part of the last question you had to use the product of roots of the quartic equation so would equal -2?

Reply 3

girlruinedme

Original post by Barbecuetime123

Grade boundary predictions?

Probably same as last year. Maybe slightly lower.

so 59-61 for A* and about 53-56 for an A

Reply 4

Barbecuetime123

Original post by girlruinedme

I thought the second last part of the last question you had to use the product of roots of the quartic equation so would equal -2?

That's what I did... It seemed a bit wishy washy as I was doing it though

Reply 5

IrrationalRoot

Original post by jjsnyder

These are my answers to todays FP2 exam. Not all of them are necassarily correct, please feel free to correct me on any you think are wrong. Not sure about the question numbers.
FP2 June 2016
1.
a) A=4
b) 50/609
2.
a) i) B=1-2i
ii) ab = 5
b) Gamma is 1/3
c) p=-5 q=-7
3. a) Show that s is the required integral
b) ln7-3/4
4. a) (sqrt(3)/2)/(1+3x)sqrt(x)
b) (sqrt3)pi/18
5. Induction Q.
6.a) Show that Q
b) 6sqrt3 + 3arcosh2 + 5
7. a) 8
b)i) Sketch with a circle with centre -4root3 + 4i radius 4
ii) 2pi/3
c) Roots are 2e^i(xpi) where x=-7,5,17
8. a) Show that Q
b) Roots were itan(xpi) where x= 1,3,5,7
c) i) 1
ii) 6

Yep this is all correct. :smile:

Reply 6

jjsnyder

Original post by Barbecuetime123

That's what I did... It seemed a bit wishy washy as I was doing it though

Considering it was the product of two squares it has to be positive. You used the product of the roots which for a quartic is the positive of the last coefficient not the negative. And the last coefficient was 1 (you had to remember to divide by 2 as the term in z^4 had coefficient 2).

Posted from TSR Mobile

Reply 7

IrrationalRoot

Original post by girlruinedme

I thought the second last part of the last question you had to use the product of roots of the quartic equation so would equal -2?

Nope, leading coefficient of the quartic was in fact 2, not 1. Plus it's a quartic so the constant term is equal to the product of the roots (same sign). So you get 2/2=1.

Reply 8

wallpaperpaste

I got (sqrt3)pi/9 for Q4b

Original post by jjsnyder

Reply 9

IrrationalRoot

Original post by wallpaperpaste

I got (sqrt3)pi/9 for Q4b

Was 18 unfortunately, must have missed a factor of 1/2 somewhere.

Reply 10

jjsnyder

Original post by wallpaperpaste

I got (sqrt3)pi/9 for Q4b

Anyone else get this? Was fairly sure it was a denominator of 18

Posted from TSR Mobile

Reply 11

wallpaperpaste

Can anyone explain the proof by induction question?

Reply 12

jjsnyder

Original post by wallpaperpaste

Can anyone explain the proof by induction question?

I'll put an explanation in the MS now :smile:

EDIT: Done it
Posted from TSR Mobile

(edited 7 years ago)

Reply 13

Iridocyclytis

Original post by jjsnyder

Last coefficient was one and first coefficient was 2, so shouldnt the product of the roots be 1/2 ? Basically I'm wondering why you multiply by 2

Reply 14

girlruinedme

Original post by jjsnyder

Ohh i remember putting 2. So ill probably lose a mark for forgetting to divide by 2?

Reply 15

jjsnyder

Original post by Iridocyclytis

Last coefficient was one and first coefficient was 2, so shouldnt the product of the roots be 1/2 ? Basically I'm wondering why you multiply by 2

Last and first coefficients were 2 (this can be seen by symmetry of the expansion). Thus 2/2=1

Posted from TSR Mobile

Reply 16

tanyapotter

Original post by jjsnyder

These are my answers to todays FP2 exam. Not all of them are necassarily correct, please feel free to correct me on any you think are wrong. Not sure about the question numbers.
FP2 June 2016
1.
a) A=4
b) 50/609
2.
a) i) B=1-2i
ii) ab = 5
b) Gamma is 1/3
c) p=-5 q=-7
3. a) Show that s is the required integral
b) ln7-3/4
4. a) (sqrt(3)/2)/(1+3x)sqrt(x)
b) (sqrt3)pi/18
5. Induction Q.
Trivial in the case n=1 since clearly 1+p >= 1+p
Assume true for n=k
(1+p)^k+1 = (1+p)^k(1+p) = (1+kp)(1+p)
= 1 + kp + p + kp^2 >= 1 + (k+1)p as kp^2>=0
Thus true for n=k+1 so true for all positive integers n by induction.

6.a) Show that Q
b) 6sqrt3 + 3arcosh2 + 5
7. a) 8
b)i) Sketch with a circle with centre -4root3 + 4i radius 4
ii) 2pi/3
c) Roots are 2e^i(xpi) where x=-7,5,17
8. a) Show that Q
b) Roots were itan(xpi) where x= 1,3,5,7
c) i) 1
ii) 6

In 7c) aren't the roots 2e^(xpi) where x= -7/18, 5/18 and 17/18?