C3 Maths A2 AQA 2016 (unofficial mark scheme new)

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 state value of p, p =-2 [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^xlimits of 1.7 to 0.5, 6 strips, answer = 4.49 to 2dp [4]
e) use d) to find approximate for x^xwhich was 1.51 (could someone explain how to do this please) [2]

3) a) solve x²>|5x-6| x<6, x>3 2>x>1 [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) find area of triangle in the form (e²+1)^m / eⁿ m = 2, n = 3 [6]

5) a) find range of f(x) = 16x - e^2x (differentiate, find stationary point) f(x)<8ln8 - 8 [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 )²limits of 1 to 1/3pi( -(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 177/224 [7]

8) a) use chain rule to show that (cosx)^-1 is secxtanx [2]
b) im sure this was a 2 marker?
c) find stationary point to get x value then sub into something to get y value which was -root5 [4]

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.

(edited 7 years ago)

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Reply 1

lollypop101

Original post by beanigger

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 [3]
c) use iteration thingy [2]
d) simpsons rule on 5-x^x [4]
e) use d) to find approximate for x^x[2]

3) a) solve x²>|5x-6| [5]

4) a) explain transformation of y=e^x onto y=e^2x+5 [4]
b) cant remember

5) a) find range of f(x) = 16x - e^2x [5]
b) find gg(x) [2]
6) a) integrate ( ln3x ) / x² by using by parts twice [4]
b) volume of revolution, integrate ( ( ln3x ) /x² )²[7]

7) a) integrate using substitution (5-2x)(1-4x)^1/3 using u=1-4x limits of 1/2 and 1/4 [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.

Question 4 I think it was 2x-5 idno

Reply 2

JamesNuttall

I believe the answer to the inequality was x>4, 1<x<2, x<-6 Or something along those lines, where they're all strict inequalities?

Reply 3

irMike

I feel like the area of the triangle question, which you've put as 8c, was earlier in the paper. I think maybe 6 and 8 may have been the other way around although I'm not too sure

Reply 4

matty9

5a you can simplify anything in the form e^(a ln b) but I think I got f(x) <= 8 or something (not confirmed!)

Reply 5

irMike

Original post by JamesNuttall

I believe the answer to the inequality was x>4, 1<x<2, x<-6 Or something along those lines, where they're all strict inequalities?

Yeah this was pretty much the only >4 marker that I was confident about

Reply 6

irMike

Original post by matty9

5a you can simplify anything in the form e^(a ln b) but I think I got f(x) <= 8 or something (not confirmed!)

I was already stressed by the time this question came up, When I saw it I was ready to give up completely. Didn't even see the remaining 4 questions either.

Reply 7

username2592579

i swear it was 4>x>-6 and 1<x and x>2
i drew that graph like 5 times and enlarged it, found that the above worked ? maybe im wrong

Reply 8

3SwordStyle

The question you wrote for 8c is actually for 4b. It was differentiating the y equation then working out A and B, where they cross the y and x axes. After it was working out the area of triangle OAB. Something along those lines. The answer you have is right (m=2 and n=3)

I think 8b or 8c (don't know if there was a 3rd part) was something about stationary points. You differentiated the equation, then worked out sinx and consequently cosx. Then you subbed these values back in for sec and tan and you get the y. Answer was -root5 i think.

Reply 9

matty9

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

Can someone confirm my working?

(edited 7 years ago)

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Reply 10

irMike

Original post by 3SwordStyle

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

Reply 11

Swindelz

Original post by beanigger

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

Questions:

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

5) a) find range of f(x) = 16x - e^2x (differentiate, find stationary point)

Question 4 coming momentarily...

Reply 12

JamesNuttall

Original post by irMike

I was already stressed by the time this question came up, When I saw it I was ready to give up completely. Didn't even see the remaining 4 questions either.

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

Reply 13

matty9

Original post by irMike

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

Yes! I got sinx = 2/3 then used pythag to get cosx = root5/3

Then subbing these into the orginal equation gave me -1/5 root 5 but I don't know how correct that is

Reply 14

Swindelz

Original post by JamesNuttall

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

Reply 15

3SwordStyle

Original post by irMike

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

Yeah sinx=2/3. You use the fact that angle x is the same so cosx is of the same triangle. Hence, from sinx=2/3, opposite=2 and hypotenuse=3. Therefore, by pythagoras, adjacent^2=3^2 - 2^2 = 9-4 = 5 So adjacent=root5, which is where i got my root5.

Reply 16

pb123

Original post by matty9

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

Can someone confirm my working?

that is correct, i got exactly the same thing

Reply 17

Swindelz

Original post by matty9

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

Can someone confirm my working?

This is what I got - pretty sure it's correct

Reply 18

Parhomus

Original post by irMike

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

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

Reply 19

JamesNuttall

Original post by irMike

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?