C2 Maths AS aqa 2016 (unofficial mark scheme new)

can I just say I did that for 9b at first and got plus/minus 5/2
but then I thoughts
you can't just "say" that you can't put logs at negative so +5/2 would be the answer so I crossed it out and used b^2-4ac= 0 on the quadratic that it formed - will I still get the marks???? FMLLLLLLL

Would do you guys predict an A will be? More than or less than 60?

It shouldn't have been possible to get +/- as the factorised quadratic was (2x-5)^2 = 0 which just gives x = 5/2 repeated root

Mtangent was=1/2 therefore Mnormal=-2. Therefore you just write y-5=-2(x-4) or whatever. I think he just expanded and rearranged it which is fine too.

Wasn't it 13.8 for perimeter? Some are saying its 13.6 but pretty sure it was 13.8 if you didn't round before the final answer

Thank you! I get how you would get that equation from there but where was the gradient of the tangent from

****.

i got 2x+5 and 2x-5 somehow......

I er factorised it and there were negatives in the quadratic so one has to be positive and one has to be negative right?
oh well I crossed it out though and used b2-4ac so that proves it right?

I dk if crossed out x=5/2 though so kengko;weo[fkdpknsgkm im panicking n stressing.

Using the first derivative, inputting the x value of P

Ah that's what I did! So I think the X coordinate in the mark scheme is wrong because X would have to be 4 instead of 5 in order for it to work!

Yes, the coordinate of P was (4,5)

The answer to the log question was definitely 5/2

The question was
Show that log4(2x+3) + log4(2x+15) = 1 + log4(14x + 5)

Also the trig question, it resulted in 5 + 3cosx, and so when x = pi, 3cosx = -3 so minimum value is 2

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

+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

The differentiation was for the normal through P translated to pass through maximum point M (different question)

To get stretch parallel to x-axis s.f. 1/3 I did this

Squared both sides of original equation to give y^2 = x^2 + 9
Then I squared the new equation to give y^2 = 9(x^2 + 1)
Expanded the bracket to give y^2 = 9x^2 + 9
Now the only difference between the two equations is there's a 9x^2 instead of x^2
So then you can have y^2 = (3x)^2 + 9 hence giving the transformation


Your idea of multiplying the entire function is a misread of the equation as y = 3root(x^2 + 9) when it is y = 3root(x^2 +1)

QUESTION 8B:

(Copied from first post):

8) find the value of tan(x) tan(x) = -5/4 [2]
b) re-arrange 16+9sin²x / 3 - 5cosx into the form p + qcosx and find the min value = 2, this is when x is equal to pi [4]

For part B, wouldnt x= 0, 2pi?

As it was 5 - 3cos(theta)

If theta = pi then its:

5 - (3*-1) = 8

hence theta should be 0 or 2pi as this gives:

5 - (3*1) =2 = Minimum value?

Also i think the question had it as plural??

Or am i missing something here?

Unoffical Mark scheme for C2 AQA 2016

It would help if you could correct the amount of marks in each question if it is wrong please

Questions:

1)a) integrate something to get -36x^-1 + ax² /2 [3]
b) find the value of a when the answer to 1)a) is equal to 16 with limits of 3 to 1, a = - 2 [2]

2)a) Draw the graph of (0.2)^x and state where it crosses axis (0,1) [2]
b) use logarithms to solve (0.2)^x = 4 , x = -0.861 [2]
c) describe transformation of (0.2)^x onto (5)^x reflection in y axis [1]

3)a) differentiate something to get 3x^-1/2 -1 [2]
b) find y co-ordinate of the maximum point M M(9,6) [3]
c) equation of normal to curve at P i think P was (5,6) y=-2x+13 [2]
d) the normal to curve at P is translated by (0,k) find value of k, k = 5.5 [3]

4)a) show that a+10d=8 [2]
b)sum to 2nd and 3rd term is 50, work out the 12th term of the series D=-2 and A=28, u12 = 6 [3]
c) work out the sum of the 21st term minus sum of 3rd term = 90 [3]

5)a) show that thetha (in radians) was 0.568 [2]
b) work out area of triangle 19.9 [2]
c) work out perimeter of shaded area, first piece of info given -> area of sector = area of shaded area

you had to realise that the area of sector = area of triangle - area of sector
re-arrange to get 2(area of sector) = 19.9 and solve for r to get 5.36?
perimeter of shaded = (9-r) + (8-r) + 5 + (length of arc = r*theta)
perimeter of shaded = 13.8 [6]

6)a)trapezium rule to get 65.6 [4]
b) describe transformation of graph y=(x² +9)^1/2 onto y = 5 + (x² +9)^1/2
Translation by (0, 5) [2]
c) describe transformation of graph y=(x² +9)^1/2 onto y=3(x² +1)^1/2 Stretch by scale factor 1/3 in x direction [2]

7)a) expand (2+x)⁵ work out p q r, i think p was 10 q was -80 and r was 80 (correct me if im wrong) [4]
b)find the coefficient of x¹⁰ in the expansion of (2+x)⁵ * (something else)⁷
it was something like -1648 [5]

8)a)find the value of tan(x) tan(x) = -5/4 [2]
b) values of tanx in the range of 0 to 360 degrees 45, 129, 225, 309 [3]
b) re-arrange 16+9sin²x / 3 - 5cosx into the form p + qcosx and find the min value = 2, this is when x is equal to pi [4]

9)re-arrange something into the form (c)^1/2 / d^2 to get 3^y y = 1/2 ( m + 12n) [4]
b) find the one and only solution of something, x = 5/2 ( you should get two equal roots i think) [4]

And for 2c divided 5 by 0.2 and got a stretch of sf 25 in the y axis. Why isnt this correct?

yup, I most definitely failed. with dignity x)