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

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2 years ago
#101
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.
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2 years ago
#102
(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^"".
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2 years ago
#103
[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
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2 years ago
#104
(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!
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2 years ago
#105
[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
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2 years ago
#106
(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
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2 years ago
#107
(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
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2 years ago
#108
(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:
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!
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2 years ago
#109
(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
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2 years ago
#110
(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
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2 years ago
#111
(Original post by jake4198)
6c) Is that not a stretch along the y-axis SF 3?
jake
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2 years ago
#112
(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)
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2 years ago
#113
Could you have done sum of 3 terms?
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2 years ago
#114
(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
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2 years ago
#115
(Original post by Jjohnson919)
Could you have done sum of 3 terms?
Like using a sledgehammer to crack a nut :P
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2 years ago
#116
(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:
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
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2 years ago
#117
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).
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2 years ago
#118
(Original post by Myachii)
Like using a sledgehammer to crack a nut :P
Ummmm... Do you know what question I'm talking about?
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2 years ago
#119
(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
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2 years ago
#120
(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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