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OCR C2 4712 14 Jan 2011 Discussion

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Reply 60
Avatar for AFK
AFK
Original post by rogersnm
Another answer: Q7ii) (second of the trigonometric questions) (again I might of remembered this wrong)

3cos⁡2x+2sin⁡x−3=03\cos^2x + 2\sin x - 3 = 0

sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1

cos⁡2x=1−sin⁡2x\cos^2 x = 1 - \sin^2 x

3(1−sin⁡2x)+2sin⁡x−3=03(1 - \sin^2 x) + 2\sin x - 3 = 0

3−3sin⁡2x+2sin⁡x−3=03 - 3\sin^2 x + 2\sin x - 3 = 0

3sin⁡2x−2sin⁡x=03\sin^2 x - 2\sin x = 0

sin⁡x(3sin⁡x−2)\sin x (3\sin x - 2)

Unparseable latex formula:

\sin x = 0$ or $\frac{2}{3}



x=0∘,41.8∘,138.2∘,180∘x = 0^{\circ}, 41.8^{\circ}, 138.2^{\circ}, 180^{\circ}


7i) 3tan2x=1
tan2x =1/3
2x=18.43
x=9.22 (3SF)
Reply 61
Avatar for rogersnm
rogersnm
4
x also equals 99.2 degrees because it's tan 2x so it doubles the range from 0 - 180 to 0 - 360 thus allowing tan 2x = 198.4 degrees
Reply 62
Avatar for nomzie
nomzie
lol wat is the working to the one that says deduct x^2 from the binomial expansionn??
Reply 63
Avatar for rogersnm
rogersnm
4
Along the lines of calculate the coefficient of x2x^2 in the equation (x+1)(1+2x)7(x+1)(1+2x)^7 (I have no idea what the first bracket's contents actually was), it came to 98x298x^2
(edited 13 years ago)
Reply 64
Avatar for ben_smith
ben_smith
9
Original post by rogersnm
Along the lines of calculate the coefficient of x2x^2 in the equation (x+1)(1+2x)7(x+1)(1+2x)^7 (I have no idea what the first bracket's contents actually was), it came to 98x298x^2


This will sound really picky but remember you only needed to put the 98 down as the coefficient.
Reply 65
Avatar for rogersnm
rogersnm
4
I know, I did :tongue: (do they actually penalize you for putting 98x298x^2?)

Edit: Checked the mark scheme for the jan 2010 paper which had a similar question, in that case putting the equivalent of 98x298x^2 instead of 9898 was deemed to be acceptable.
(edited 13 years ago)
Answers are here now:

http://www.thestudentroom.co.uk/showthread.php?p=29387148#post29387148
Reply 67
Avatar for CJAW
CJAW
9
Hi Mr M.
For Q8 part iii), what were the individual parts for the perimeter? Hopefully that will enable me to remember whether or not I got that right.
And also, for Q9 part iii), I factorised getting solutions of -1 and 1/4. Is that correct/ doing whar the question asked.

Thanks for the answers, much appreciated.
Reply 68
Avatar for ben_smith
ben_smith
9
Why is everybody coming out with different answers for this last question. I though it was actually quite easy.

40.8 (3s.f)]
Original post by shan735
I got 32 for the area of the integral

Find the area from -1 to a 1/4

then find the area from 1/4 to 3 :smile:

For the obtuse angle i also got 2.44 or something similar


Also for that 3Tan2x=1 question really scared me :L I panicked and used the 2tanA/1-tan^2A

In the end, i remembered, and I found out...IT BOTH WORKS :biggrin:



yay its reassuring knowing atleast one person in the country got 32, maybe we are correct, I used the same method as you
Reply 70
Avatar for rogersnm
rogersnm
4
An area of 32 is incorrect, that's what you get if you integrate y=(x+1)(x-3)(-4x+1) from -1 to 3, however that will give you the area of 0.25 to 3 minus the area of -1 to 0.25. You actually want to add the two areas, therefore you have to integrate it twice with different limits and then add the areas together.

You should still get a reasonable amount of the marks but you won't get all of them.

Correct Solution: http://www.wolframalpha.com/input/?i=%28integrate+y%3D%28x%2B1%29%28x-3%29%28-4x%2B1%29+from+0.25+to+3%29+-+%28integrate+y%3D%28x%2B1%29%28x-3%29%28-4x%2B1%29+from+-1+to+0.25%29
Incorrect Solution: http://www.wolframalpha.com/input/?i=integrate+y%3D%28x%2B1%29%28x-3%29%28-4x%2B1%29+from+-1+to+3
(edited 13 years ago)

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