OCR FP2 2017 (non MEI)

I can't seem to edit the google docs thing but I got pi/4 for one of the first integral (complete the square one?). The polar graphs were sinx and cos2x i think....
For the last question i change she to the same sigma formula and n came out in the 2000s so I don't know if anyone's else got that?!
For q6, the last part I got aln(10+root3)(can't remember a) not sure if this answers the question though?
Hope some of this helps :smile:

Ok I'm here now.

You need to press the button at the top to request access.

Reply 21

beastmode1

12

Original post by rebirth61213

Wait, I thought positive quadrant meant the top right one?

That's what I meant lol

Reply 22

EleanorAngel

4

for 5)iii (the area between the two curves in the positive quadrant) I did the integral of sin theta using the polar integral formula with limits pi/6 and zero, then multiplied this by two to get pi/12 + root3/16
anyone else do/get something similar??

Reply 23

Fortox

12

So there's some debate at my college on the wording of the polar area question. It said find the area between the two curves. Does that mean find the area enclosed by both curves or find the area shared by both curves? It was a 50/50 split in my class.

Reply 24

Joeking1999

9

Original post by Fortox

So there's some debate at my college on the wording of the polar area question. It said find the area between the two curves. Does that mean find the area enclosed by both curves or find the area shared by both curves? It was a 50/50 split in my class.

Same here I didn’t know what it wanted out of the two so I just did the silicon o how first quadrant so pi/4

Reply 25

LemonsLoveLife

1

Original post by EleanorAngel

for 5)iii (the area between the two curves in the positive quadrant) I did the integral of sin theta using the polar integral formula with limits pi/6 and zero, then multiplied this by two to get pi/12 + root3/16
anyone else do/get something similar??

Yep, I worked out the Sin area and added half of the Cos2 area. I then took away the area you're talking about.

Reply 26

username1924691

2

that was so easy?? i reckon the grade boundaries will be brutal

Reply 27

beastmode1

12

Not quite sure what the answer for the area is but I reckon u will pick up 4 ish for working out various areas.
It wasn't bad apart from that question

Reply 28

Jojos bizaro

1

Did anyone get pi/8 - 3root3/32 for the area question?

Reply 29

10fishli

5

Original post by EleanorAngel

for 5)iii (the area between the two curves in the positive quadrant) I did the integral of sin theta using the polar integral formula with limits pi/6 and zero, then multiplied this by two to get pi/12 + root3/16
anyone else do/get something similar??

I think this is what I got yay

Reply 30

BaobaoKe

Original post by Jojos bizaro

Did anyone get pi/8 - 3root3/32 for the area question?

Thats what I got

Reply 31

beastmode1

12

Original post by EleanorAngel

for 5)iii (the area between the two curves in the positive quadrant) I did the integral of sin theta using the polar integral formula with limits pi/6 and zero, then multiplied this by two to get pi/12 + root3/16
anyone else do/get something similar??

Got exactly that ÷ 2

Reply 32

Jojos bizaro

1

Original post by BaobaoKe

Thats what I got

Thank God at least someone does! You and I are in this together now.

Reply 33

jadeemma

9

was the graph on the back page wrong? At x=1 y=0.37 but the graph was at a much higher value, like 0.7??

Reply 34

jadeemma

9

Also for 6ii, it asked for the root of the equation, but wasn't 1/3ln(3+10^0.5) the value of x not w?

Reply 35

jadeemma

9

I got the polar area to be pi/16 - 3root3/32
Did anyone else get this?

Reply 36

EleanorAngel

4

Original post by jadeemma

Also for 6ii, it asked for the root of the equation, but wasn't 1/3ln(3+10^0.5) the value of x not w?

pretty sure 1/3In(3+10^0.5) was the result of doing sinh^-1 x which would equal w

Reply 37

Fortox

12

Original post by EleanorAngel

pretty sure 1/3In(3+10^0.5) was the result of doing sinh^-1 x which would equal w

w=sinhx not sinh^-1 x. The question was definitely worded wrong. To get w you would have to do sinh(1/3ln(3+sqrt(10))) which would not give a nice answer.

Reply 38

fuzzball909

6

Original post by jadeemma

I got the polar area to be pi/16 - 3root3/32
Did anyone else get this?

yep same

Reply 39

Fortox

12

Original post by Jojos bizaro

Did anyone get pi/8 - 3root3/32 for the area question?

Yes but I think the actual answer is pi/16 - 3sqrt(3)/32. The mistake I made was integrating cos(2theta) from 0 to pi/2 to get the area under cos(2theta). I should have done it from 0 to pi/4.

OCR FP2 2017 (non MEI)

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you