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# C2 Maths AS aqa 2016 (unofficial mark scheme new)

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1. Thought this was one of the most challenging papers I've seen... on par with June 2012. First time I haven't been able to answer a question which was disappointing. Reckon about 67/75.
2. (Original post by jordanwu)
Hmm for 9a I got y=1/2m-6n... I'm pretty sure you had a negative indice on the bottom so it would become positive when you bring it to the top???
Yeah thats what I got, i had a m/2-6n but I left it as 3^(m/2-6n) since it didn't say write what y=" " it just said right this in the form 3^y where y is an expression in terms of m and n, but I don't think they would penalise anyone if they wrote y="" or 3^"".
3. [QUOTE=Eisobdxhsonw;65161255]
(Original post by Bosssman)

I'm still confused as to why it was 5+3cosx and not 3+5cosx also as to why the min value was -2, cheers tho
The question was show that (16 + 9sin^2x)/(5-3cosx) can be written in the form p + qcosx and state the minim value, and the exact value of x for this point.

Here's the method:

(16 + 9(1 - cos^2x))/(5 - 3cosx)

(25 - 9cos^2x)/(5 - 3cosx) then difference of two squares

(5 + 3cosx)(5 - 3cosx)/(5 - 3cosx)

The (5 - 3cosx) cancel to give 5 + 3 cosx
Minimum value of cosx is -1 so minimum value is 5 + 3(-1) = 5 - 3 = 2 when x = pi
4. (Original post by Myachii)
Gotta love AQA
Thanks for the clarification though, that was annoying me. Awful lot of work for 2 marks (never seen anything like it before either)
Yes, I do agree it should have been worth more marks!
5. [QUOTE=Bosssman;65162331]
(Original post by Eisobdxhsonw)

The question was show that (16 + 9sin^2x)/(5-3cosx) can be written in the form p + qcosx and state the minim value, and the exact value of x for this point.

Here's the method:

(16 + 9(1 - cos^2x))/(5 - 3cosx)

(25 - 9cos^2x)/(5 - 3cosx) then difference of two squares

(5 + 3cosx)(5 - 3cosx)/(5 - 3cosx)

The (5 - 3cosx) cancel to give 5 + 3 cosx
Minimum value of cosx is -1 so minimum value is 5 + 3(-1) = 5 - 3 = 2 when x = pi
I find it hilarious that the value for x was pi, it felt like they were trying to trick us and it ended with a simple answer. :P
6. (Original post by jordanwu)
Hmm for 9a I got y=1/2m-6n... I'm pretty sure you had a negative indice on the bottom so it would become positive when you bring it to the top???
Got the same.
Question was:

log3(c) = m
log27(d) = n

log3(sqrt(c)) = log3(3^1/2m)
log3(d^2) = log3(3^6n)

It wanted sqrt(c)/d^2 which is:
log3(sqrt(c)) - log3(d^2) = log3(3^1/2m) - log3(3^6n)

sqrt(c)/d^2 = (3^1/2m)/(3^6n)
= 3^1/2m-6n

...right? I've already lost enough marks on this paper

(Original post by Parhomus)
I find it hilarious that the value for x was pi, it felt like they were trying to trick us and it ended with a simple answer. :P
Better than -1648x^10 making you doubt yourself for the remainder of the paper :P
7. (Original post by collegeboy66)
And for 2c divided 5 by 0.2 and got a stretch of sf 25 in the y axis. Why isnt this correct?
This is incorrect as you you can't directly divide the base if it has an exponent, for example (3/2)^2 isn't the same as (3^2)/2
The solution to this question was to recognise that 0.2 = 1/5 = 5^-1 and so resulted in 5^x = 0.2^-x and hence the answer being a transformation of reflection in the y axis
8. (Original post by Myachii)
+1 OP pls fix

Another thing - 6c is confusing the hell out of me

y=\sqrt{x^2+9}
onto
y=3\sqrt{x^2+1}

Let f(x) = \sqrt{x^2+9}

f(x) --> 3f(x) = Stretch in y direction SF 3
However, your mark scheme says "Stretch in x direction SF 1/3"
For that to be the case, wouldn't the equation have to be the following:
f(x) --> f(3x) = Your answer
Which would lead to the "original equation" having to be:
y=\sqrt{3x^2+1}

My logic was that the 3 is multiplying the entirety of the sqrt function.
For example:
5^x --> 3 * 5^x = Stretch in y direction SF 3
5^x --> 5^3x = Stretch in x direction SF 1/3
This would be the case because the number 3 is only multiplying the x by 3, not the entire function (which in this case is 5^x)

You said you proved it was x direction SF 1/3 using differentiation, could you show this please? Using methods and formulae that I'd been taught, I got to a y stretch SF 3.

Forgive me if I'm talking out my ass too
Ok I'll try help. I managed to get stretch 1/3 in x... but it's quite hard to explain. And even I'm not totally sure how I got there.
So we started with y=sqrt(x^2+9)
and went to y=3*sqrt(x^2+1)
Ignore the y for now.
Square both equations to get x^2+9 -> 9*(x^2+1)
Then we get x^2+9 -> 9x^2+9
ignore the 9 and look what happened to x
We went from x^2->9x^2
We want to see what happened to x, not x^2
so next sqrt both, and we end up with:
x-> 3x, therefore a stretch by a third in x direction!
9. (Original post by Myachii)
Got the same.
Question was:

log3(c) = m
log27(d) = n

log3(sqrt(c)) = log3(3^1/2m)
log3(d^2) = log3(3^6n)

It wanted sqrt(c)/d^2 which is:
log3(sqrt(c)) - log3(d^2) = log3(3^1/2m) - log3(3^6n)

sqrt(c)/d^2 = (3^1/2m)/(3^6n)
= 3^1/2m-6n

...right? I've already lost enough marks on this paper

Better than -1648x^10 making you doubt yourself for the remainder of the paper :P
I got the coefficient for the power of ten wrong; just because I messed up the addition of the coefficients at the end; even though I expanded both of them correctly. :C
10. (Original post by Myachii)
Got the same.
Question was:

log3(c) = m
log27(d) = n

log3(sqrt(c)) = log3(3^1/2m)
log3(d^2) = log3(3^6n)

It wanted sqrt(c)/d^2 which is:
log3(sqrt(c)) - log3(d^2) = log3(3^1/2m) - log3(3^6n)

sqrt(c)/d^2 = (3^1/2m)/(3^6n)
= 3^1/2m-6n

...right? I've already lost enough marks on this paper

Better than -1648x^10 making you doubt yourself for the remainder of the paper :P
Yes this is correct
11. (Original post by jake4198)
6c) Is that not a stretch along the y-axis SF 3?
jake
12. (Original post by Bosssman)
Yes this is correct
Considering you've been correct in every post of yours I've read this is a huge relief, tysm

OP pls fix:

9a) 3^1/2m-6n (or if you like to factorise, 3^1/2(m-12n)
13. Could you have done sum of 3 terms?
14. (Original post by Parhomus)
I got the coefficient for the power of ten wrong; just because I messed up the addition of the coefficients at the end; even though I expanded both of them correctly. :C
At least you'll be rolling in method marks
15. (Original post by Jjohnson919)
Could you have done sum of 3 terms?
Like using a sledgehammer to crack a nut :P
16. (Original post by Myachii)
+1 OP pls fix

Another thing - 6c is confusing the hell out of me

y=\sqrt{x^2+9}
onto
y=3\sqrt{x^2+1}

Let f(x) = \sqrt{x^2+9}

f(x) --> 3f(x) = Stretch in y direction SF 3
However, your mark scheme says "Stretch in x direction SF 1/3"
For that to be the case, wouldn't the equation have to be the following:
f(x) --> f(3x) = Your answer
Which would lead to the "original equation" having to be:
y=\sqrt{3x^2+1}

My logic was that the 3 is multiplying the entirety of the sqrt function.
For example:
5^x --> 3 * 5^x = Stretch in y direction SF 3
5^x --> 5^3x = Stretch in x direction SF 1/3
This would be the case because the number 3 is only multiplying the x by 3, not the entire function (which in this case is 5^x)

You said you proved it was x direction SF 1/3 using differentiation, could you show this please? Using methods and formulae that I'd been taught, I got to a y stretch SF 3.

Forgive me if I'm talking out my ass too
There is another way to prove the stretch by rearranging the first equation given. See attached working, I went through this with a friend after the exam
Attached Images

17. I don't think the test was that bad, pretty decent; but it was very hard, I know I lost a few marks. For the trapezium rule question I even checked it with my calculator and it said 65.3 or something so I was like phew I got the answer correct. (Answer was 65.6 but calculator can do integration to more precision).
18. (Original post by Myachii)
Like using a sledgehammer to crack a nut :P
Ummmm... Do you know what question I'm talking about?
19. (Original post by Parhomus)
I don't think the test was that bad, pretty decent; but it was very hard, I know I lost a few marks. For the trapezium rule question I even checked it with my calculator and it said 65.3 or something so I was like phew I got the answer correct. (Answer was 65.6 but calculator can do integration to more precision).
Best thing about C2 is checking Integration. Did the exact same as you
20. (Original post by Jjohnson919)
Ummmm... Do you know what question I'm talking about?
Allow me to explain myself:
You could do the sum of the first three numbers using the formula (which is what I presume you used), or you could just figure them out (26, 24, 22) because it was an arithmetic series so subtracting two wasn't that hard :P

I said it was like using a sledgehammer to crack a nut because if you used the formula to calculate the sum of the first three, you made yourself a hell of a lot more work and more room for error.

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