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OCR MEI C3 Maths June 2015

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Original post by ETRC
you can also say area is the same but under x-axis instead of above it
or you can say modulus area is the same


I finished the paper early so during the last few minutes after I triple checked every answer I decided to write the magnitude of the integral of g(x) between 0 and 3 is 1528ln2\frac{15}{2}-8\ln2, the same as the area found in ii)
Original post by iAmanze
Let me get this straight.

Integrals measure the area under a graph right?

This being said, under the graph (if I recall the question accurately) where were two parts, like 2 triangles.

So, I did 3 x 1 for the rectangle. Took away the two triangles from it to get my answer.

I wouldn't even know how to do this differently.

But yes, I was accurate with my limits so should be able to get it like you said.

Going over exams after they are finished is a bad idea for me lol. From coming out thinking I have an A*, to thinking I have a high A now, somehow if I was wrong on that, I may have a low A...

With M1 being a low A/ high B it all rests on the shoulders of C4 to see if I get overall an A* an A or even a damn B! O_O! Hate you C4!


You only needed to do the rectangle minus one integral if I remember correctly.
Original post by Connorbwfc
I think I have got between 59 and 64 on this paper. Is it still possible to get an A*?

Definitely as long as you smash C4
Original post by RemainSilent
what ques was that ? I just found the are of integral with limits 4 and 1 not sure if its the same ques as you tho


That is the easiest/quickest method, and is also what I did. I believe it's question 8.
Original post by SH0405
That is the easiest/quickest method, and is also what I did. I believe it's question 8.


so you did no take away from rectngle or sumthing? I got the same answer as other ut just minus
Reply 505
Original post by Primus2x
I finished the paper early so during the last few minutes after I triple checked every answer I decided to write the magnitude of the integral of g(x) between 0 and 3 is 1528ln2\frac{15}{2}-8\ln2, the same as the area found in ii)


that should be fine
how many mark would i get if i didnt subtract three from the rectangle?
Original post by RemainSilent
so you did no take away from rectngle or sumthing? I got the same answer as other ut just minus


I did the 3*1 rectangle minus the value of the integration between 1 and 4. Thought that was the easiest way to find the shaded area?
Reply 508
Original post by Primus2x
I did it like this
Unparseable latex formula:

[br]\arcsin{x}=\arccos{x}[br]\\[br]x=\sin\left ( \arccos{x} \right )[br]\\[br]x=\sqrt{1-\left ( \cos\left ( \arccos{x} \right ) \right )^2} \because \sin^2 \theta+\cos^2 \theta =1 \Rightarrow \sin \theta=\sqrt{1-\cos^2 \theta}[br]\\x=\sqrt{1-x^2}[br]x^2=1-x^2[br]\\[br]2x^2=1[br]\\[br]x^2=\frac{1}{2}[br]\\[br]x=\sqrt{\frac{1}{2}}[br]\\[br]\therefore x=\frac{\sqrt{2}}{2} \because \text{rationalise denominator}[br]



you should get marks for this. hopefully a mathematician marks the papers because there are lots of ways of doing some questions that are correct
Assuming marks are only lost completely on Q4, a little in terms of accuracy for the last parts of the section B questions and -3 marks for general section A questions could I still get an A in C3?

Paper was fine, I made lots of silly mistakes as I prepped for NM till 3AM.
NM went gracefully tho:tongue:

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Original post by RemainSilent
so you did no take away from rectngle or sumthing? I got the same answer as other ut just minus


I did RECTANGLE TAKE AWAY INTEGRAL.
Question 8. (iv) said 'write down the value of...' for the integral - not the area defined by it. Hence, the answer should be negative.
Original post by SH0405
Question 8. (iv) said 'write down the value of...' for the integral - not the area defined by it. Hence, the answer should be negative.


True, luckily I wrote most of the answers twice to make sure, In question one I wrote the coordinates as expressions involving arctangents, exponentials and square roots as well as writing down the approximate decimal coodinates. And in question 8 I wrote both the integral and the magnitude of the integral
I think I might have gotten full marks but I am still worried about question 5 because with all the other questions I could check my answer by using different methods or working backwards (e.g. I checked if my integrals were right by seeing if I got back to the original expressions by differentiating) but question 5 can only be solved by differentiating implicitly ( could not even plot it on my calculator since it only plots Cartesian, polar and parametric).

Can I still get at least 98% if I do question 5 wrong? (I definitely verified that (1,1) is on the curve so I will get marks for it) I might have written the correct answer of 1/2 but I really don't remember.
(edited 8 years ago)
Original post by Primus2x
I did it like this
Unparseable latex formula:

[br]\arcsin{x}=\arccos{x}[br]\\[br]x=\sin\left ( \arccos{x} \right )[br]\\[br]x=\sqrt{1-\left ( \cos\left ( \arccos{x} \right ) \right )^2} \because \sin^2 \theta+\cos^2 \theta =1 \Rightarrow \sin \theta=\sqrt{1-\cos^2 \theta}[br]\\x=\sqrt{1-x^2}[br]x^2=1-x^2[br]\\[br]2x^2=1[br]\\[br]x^2=\frac{1}{2}[br]\\[br]x=\sqrt{\frac{1}{2}}[br]\\[br]\therefore x=\frac{\sqrt{2}}{2} \because \text{rationalise denominator}[br]



Wow!... I just knew that sin45=cos45= 1/root2.. took me about 10 seconds to answer.
With the rectangle area question i used substitution and ended up with a weird answer.. How many marks would i lose.
Original post by Duskstar
//wip

3.12x3lnxdx[br]integratebyparts:[br]dvdx=x3,u=lnx[br]v=x44,dudx=1x[br]followthrough,and[br]=16ln215163. \,\, \displaystyle \int^2 _1 x^{3}\ln x \, dx[br]\mathrm{- integrate by parts:}[br]\dfrac {dv}{dx} = x^{3}, u = lnx[br]v = \dfrac {x^4}{4}, \dfrac{du}{dx} = \frac{1}{x}[br]\mathrm{- follow through, and}[br]= 16 \ln 2 - \frac{15}{16}



That answer isn't right, it is 4ln(2) not 16: http://www.emathhelp.net/calculators/calculus-2/definite-integral-calculator/?f=x%5E3+*+ln%28x%29&var=x&a=1&b=2&steps=on
How many marks do you think could be lost for writing that 1/3 + 1 = 2/3 instead of 4/3 by accident in Q2?

Think I win the award for most stupid marks lost.


Yeah you're right, thanks - I couldn't be bothered to work it out again lol.
Original post by Sophielouuu


Think I win the award for most stupid marks lost.


I'm sorry but I think I'll be taking that :rolleyes:

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