C3 Maths A2 AQA 2016 (unofficial mark scheme new)

That last italics bit isn't right - you need to sub in the value of x back into the original f(x) first, like matty9

That's what I had but I also had no clue.

If sin(x) is 2/3, you can draw a right angle triangle to show that the remaining side must be root 5 due to pythagorus, so tan x = 2/root 5 and cos x = root 5 / 3, so sec x is 3/root 5 I think?

For the stationary point, did you get sin x = 2/3? A few people I spoke to got this, and none of us had any clue how this became a root? Had to keep it as a decimal in the end

If sin(x) is 2/3, you can draw a right angle triangle to show that the remaining side must be root 5 due to pythagorus, so tan x = 2/root 5 and cos x = root 5 / 3, so sec x is 3/root 5 I think?

You need to flip the sign on 5a, you put greater than or equal to not less than or equal to

Just realised I crossed out correct working **** my life.

I redid question 5a and got the same answer as I did in the exam which was 8ln8-8

Can someone confirm my working?

y = 16x - e^(2x)
dy/dx = 16 - 2e^(2x) which = at a stationary point.
16 = 2e^(2x)
8 = e^(2x)
2x = ln(8)
x = ln(8)/2

d2y/dx2 = -4e^(2x)

when x = ln(8)/2, dy/dx is negetive, therefore it's a max point.

f(x) <= ln(8)/2

For the Area of OAB question:

Was it a case of just making x = 0 and y = 0 to find A and B? I'm pretty certain I got A and B correct, then had absolutely no idea what to do to find the area. I tried integrating but I didn't look to me getting anywhere so I moved on because I was running out of time already by that point.

It would help if you could link answers to questions as i cant remember them

Questions:

1) a)Asked to find dy/dx [2]
b) find dy/dx [2]
c) find dy/dx [2]

2) a) show alpha lied between values [2]
b) re-arrange to get into the form x=e^{(ln5 / x)} [3]
c) use iteration thingy where X₁ = 2, X₃ =2.054 [2]
d) simpsons rule on 5-x^x which was 4.49 to 2dp [4]
e) use d) to find approximate for x^x which was 1.51 (could someone explain how to do this please) [2]

3) a) solve x² > |5x-6| *NEED CONFIRMED ANSWERS* [5]

4) a) explain transformation of y=e^x onto y=e^2x+5 stretch by 1/2 in x-direction and translation by (2.5,0) [4]
b) cant remember

5) a) find range of f(x) = 16x - e^2x (differentiate, find stationary point) f(x)>8ln8 - e^8ln(8) (i got e8ln(8) but im not sure if it is right) [5]
b) find gg(x) g(x) = 1/x therefore gg(x) = x [2]
6) a) integrate ( ln3x ) / x² by using by parts twice [4]
b) volume of revolution, integrate ( ( ln3x ) /x² )² limits of 1 to 1/3 pi( (ln3)² + 2ln(3) -4 ) [7]

7) a) integrate using substitution (5-2x)(1-4x)^1/3 using u=1-4x limits of 1/2 and 1/4 you get 117/224 [7]

8) a) use chain rule to show that (cosx)^-1 is secxtanx [2]
b) cant remeber
c) i think it was a 6 marker for the m = 2 and n = 3

9)a) show secx-tanx = -5 goes to secx+tanx=-0.2 [2]
b) find the exact value of cosx cosx=-1/2.6 [3]
c) solve the equation but x is changed into 2x-70 with range of -90 to 90 [3]

i belive i jumbled up a few so pls correct and give CONFIRMED answers, ty.

1 quotient rule. I thin p = -2. (Cant remember if 2 or -2)

For 3a i got x <= 6, 1 <= x <= 2, 3 <= 3

5a) 8ln8 -8 i think

6a) It was only parts once I think, you used parts twice in part 2. In the second part you integrated [ln(3x)]^2/x^2 I think.
My final answer was -(ln3)^2 - 2ln3 + 4 for B.

7a) 177/224 not 117/224 (I think you made a typo there.)

1 quotient rule. I thin p = -2. (Cant remember if 2 or -2)

For 3a i got x <= 6, 1 <= x <= 2, 3 <= 3

5a) 8ln8 -8 i think

6a) It was only parts once I think, you used parts twice in part 2. In the second part you integrated [ln(3x)]^2/x^2 I think.
My final answer was -(ln3)^2 - 2ln3 + 4 for B.

7a) 177/224 not 117/224 (I think you made a typo there.)

If sin(x) is 2/3, you can draw a right angle triangle to show that the remaining side must be root 5 due to pythagorus, so tan x = 2/root 5 and cos x = root 5 / 3, so sec x is 3/root 5 I think?