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OCR C2 Exam May 18th 2012

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From what I can remember the area they gave for the sector was 45pi. I got the angle as 2pi over 5. If you use the area formula 0.5*r²*angle you get 225. If you square root that you get 15.

I did that :smile:

Reply 341

issyconnor

Original post by ingeek2010

The radius is definitely 15cm.

If you substitute into A= 1/2R^2THETA.

you get A= 0.5 * 15^2 * (2/5)Pi
A = 45Pi

It was 15 and the area of segment was 34.4

:smile:

So relieved. Someone just wrote that it was something like 15 over square root pi. I was panicking for a moment there.

Reply 342

J DOT A

Original post by ingeek2010

The radius is definitely 15cm.

If you substitute into A= 1/2R^2THETA.

you get A= 0.5 * 15^2 * (2/5)Pi
A = 45Pi

It was 15 and the area of segment was 34.4

:smile:

A=45 not 45pie.

If you substitute 0.5*(15/pi)^2*2/7pi or whatever the radian thing was it was 45

Reply 343

kenann

Original post by cameron95

What did people get for question 4 -
4cos^2(x) - sin(x) - 7 = 0. (or something like that)

i got sinx=-3/4 and -1, but I think they were meant to be positve. I don't know what i did wrong as i went through the question twice as i thought they were wrong.

yea, i got 3/4 and 1.

Original post by J DOT A

A=45 not 45pie.

If you substitute 0.5*(15/pi)^2*2/7pi or whatever the radian thing was it was 45

nope, incorrect, it's clearly stated the area of the segment = 45pi.

(edited 11 years ago)

Reply 344

Bohla

Finished comfortably, went back through to check answers, originally got 38.4 or whatever for the area between arc and chord question. Checked in Degree by accident, meaning i've changed to the wrong answer of 130 something..

also for some reason wrote Sum to Infinity as 1/(a-r).

I am the king of ridiculous mistakes.

Hopefully made the A though, all the questions seemed fairly uniform bar 7 and 9

Reply 345

J.C

Original post by J DOT A

A=45 not 45pie.

If you substitute 0.5*(15/pi)^2*2/7pi or whatever the radian thing was it was 45

No I'm pretty sure the area was 45pi.

Reply 346

dirrane

Here is the questions paper;

q1)i)find binomial expansion of (3+2x)^5

ii)hence find the binomial expansion of (3+2x)^5 + (3-2x)^5

Q2) i) Find ∫(x^2-2x+5)

ii)hence find the equation of the curve for which dy/dx=x^2-2x+5, through point (3,11)

Q3)i)express 70 degree exactly in radians

ii)if area of sector is 45pi cm^3, then find value of r

iii)find are of the segment-to 3 s.f

Q4) solve 4cos^2+7sinx-7=0, between 0 and 360

Q5)a)i)write down u2 and u3, when u1 is 4 and the equation is , un+1=2/un

ii)describe the behavior of the sequence

b)i)in an arithmetic progression the ninth term is 18, and the sum of the
first nine term is 72, find the first term and the common difference

Q6)i)using trapezium rule,with, 2 strips each of width 4, show that an approximate value of ∫4rootx dx is 32+16root5

ii) use a sketch graph to explain why the actual value is greater than 32+16root5

iii)use integration to find the exact vaue of ∫4rootx dx

7a)i)given that a is acute angle such that tan a=2/5, find the exact value of cos a

ii)given that b is the obtuse angle such that sin b=3/7,find the exact value of cos b.

b)traingle ABC with AC=6cm,BC=8cm and angle BAC=60, and angle ABC=y, find the value of sin y, simplifying your answer.

Q8)two polynomial are difined by

f(x)=x^3+(a-3)x+2b, and g(x)=3x^3+x^2+5ax+4b
where a and b are constant

i) given that f(x) and g(x0 have a common factor of (x-2), show that a=-4 and find the value of b

ii)using the value of a and b. factorize f(x) fully, and show that f(x) and g(x) have two common factors.

Q9)a)i) An arithmetic progression has first term log27 base 2, and common difference log x base 2.

i)show that the fourth term can be written as log 27x^3 base 2.

ii)given that the fourth term is 6, find the exact value of x.

b) a geometric has the first term log27 base 2, and common ratio of log y base 2.

i) find the set of value of y for which the geometric progression has the sum of infinity

ii) find the exact value of y for which the sum to infinity of the geometric progression is 3.

Hope you all did well :smile:

(edited 11 years ago)

Reply 347

issyconnor

For the question describe the behaviour of the sequence, did anyone else write that it was periodic ?

Reply 348

samjj8

Probably lost at least 6 marks in total. :\

Reply 349

kenann

Original post by issyconnor

For the question describe the behaviour of the sequence, did anyone else write that it was periodic ?

yes, and that is the correct answer

Reply 350

erniiee

Original post by issyconnor

You might find it helpful, when you do core three or four, to bring 2 calculators in the exam. One in degree mode, one in radian mode. To distinguish between the two just put a sticky label at the time with 'degree' on the calculator in degree mode. It seems like a stupid idea, but my teacher recommended it as one of my worries about core 2 was getting the trig questions wrong because I'm quite forgetful and would not remember to change from degree to radian.

:smile:

Thanks for the advice!

Original post by cleverslacker

I actually found it to be harder than most past papers, I had done like 2 before, and I defo found it harder than jan 2012

Oh snap, that means I probably screwed up a lot of that question. Don't really rememeber tbh though

Ah you'll be fine :smile:

Reply 351

J DOT A

OH **** SAKE it was 45pi.
So I think I would of lost 3+3-6 marks.

so 66/72...any guesses on how many UMS that is?

Reply 352

om-nom ._.

Original post by issyconnor

For the question describe the behaviour of the sequence, did anyone else write that it was periodic ?

yep i did

hopefully it's right!

Reply 353

J.C

Original post by kenann

yes, and that is the correct answer

Do you think they would allow "repeating"?

Reply 354

erniiee

Original post by J DOT A

A=45 not 45pie.

If you substitute 0.5*(15/pi)^2*2/7pi or whatever the radian thing was it was 45

Reply 355

J DOT A

Original post by erniiee

lololol fml. Lost stupid marks.

Reply 356

Dude Ranch

Original post by J DOT A

OH **** SAKE it was 45pi.
So I think I would of lost 3+3-6 marks.

so 66/72...any guesses on how many UMS that is?

bare

Reply 357

needtosucceed=)

Original post by om-nom ._.

yep i did

hopefully it's right!

is that the same as recurrence relations?

Reply 358

issyconnor

Original post by J.C

Do you think they would allow "repeating"?

Maybe. Someone in my class asked my teacher (he went to to an OCR conference thing in April) and told us that OCR are quite exact might be looking for the actual term 'periodic'. It depends, really.

Reply 359

KS8

I'm a 2nd year student and this is my last chance to get over 90UMS in C2 which I need and omg I couldn't concentrate at all during the exam. I went back to the acute angle and obtuse ones at the end and finished with 5 secods left lol but unfortunatley had not time to go back and check through my answers.

For the segment do people realise you can just use the formula:

Area = (1/2)r^2(feta - sin(feta))

I think I potentially messed up on the anle ones and the moderator besides me was watching as I was writing at like 100mph lol.

Also on the last log I got y = 2/3 (this worked when subbing it back in)

but on the one prior to that I wrote down -1 < r < 1 and forgot to implement it into the log.

Oh well, hopefully I have done as well as I wanted.