A-level Maths Pure Mathematics 2 9MA0_02 Unofficial markscheme Watch

Poll: What mark do you think you got?
97+ (19)
15.57%
94-96 (6)
4.92%
90-93 (13)
10.66%
86-89 (17)
13.93%
82-85 (14)
11.48%
78-81 (7)
5.74%
75-77 (6)
4.92%
71-74 (4)
3.28%
below 71 (36)
29.51%
RedGiant
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Vote what you think the A* mark will be here: https://www.strawpoll.me/15889159

Post your answers below, these are just my answers that I remember. If you disagree with any then let me know.

1) Functions
a) (2 marks) gg(5) = 40/7?

b) (1 mark) Range of g(x): 2 < g(x) <= 40/7

c) (3 marks) Inverse function for g(x), and state the domain: domain of f^-1(x) = range of f(x)

2) Vectors
a) (2 marks) Position vector: 6i-7j+10k

b) (3 marks) Work out the value of a: 2-2root2

3) Modulus & proof
a) Prove if m and n are irrational, then mn is not necessarily irrational (disprove with counter-example). Easiest method is m=root 2 and n=root 8

4) Sigma notation

a) (4 marks) Show that q: use the addition rule, seperate each term, you had to show the geometric and arithmetic sequences for each term.

b) (2 marks) Calculate sum of sequence: First determine the period, and the sum of the period (13/6). Then do (50*13/6). Answer was 325/3.

5) Quadratic equation for trajectory
a) (3 marks) Form the equation: 1 mark for using -x^2, 2 marks for complete equation: y= -0.03x^2 + 1.2x

b) (2 marks) Determine the longest distance when H=3: Equate your eq. from part a) to 3. Then calculate x.

c) (1 mark) Reason why it is unreliable: The model assumes that air resistance has a negligible/no effect on the trajectory, hence the actual height of the ball will be lower than what the model predicts. Or, you could state that the model allowed for negative values for H, which is impossible.

6) Polynomial function
a) i) (1 mark) Calculate f(2): f(2) = 0.

ii) (2 marks) Hence, express f(x) has a product of 2 algebraic factors: Due to factor theorem, as f(2)=0, (x-2) is a factor of f(x). So f(x)=(x-2)(-x^2-3x-5) (can't remember the second factor).

c) (2 marks) Explain why this equation has exactly 2 solutions: The right factor had no real solutions, as (b^2-4ac)<0. The left factor had 2 solutions, as it was y = +-sqrt(2).

d) (1 mark) State how many solutions this equation has: 3 solutions

7) Newton Raphson
a) (3 marks) Show that x=(...): x=(x-f(x)/f'(x)). Required you to find f '(x), and then combine the x and the fraction together.

b) (1 mark) Why won't this work when x=1: Because f '(0) = 0, so the denominator is 0 - hence, x is undefined.

8) Trigonometric equations
a) (2 marks) Show that (1-cos2x) = sin2xtanx: 1-cos2x = (1-(1-2sin^2x)) = 2sin^2(x) = tanxsin2x

b) (5 marks) Find all solutions to 5sinx - 5cosx = 2
2 marks for converting it into the form of Rsin(x+a) = 2
3 marks for finding all solutions (I got 2 solutions - from sin(x-45) = 2/5root2 )

c) (6 marks) Find all solutions to (sec^2x -5)(1-cos2x) = tan^2xsin2x
1 mark for converting the (1-cos2x) factor to sin2xtanx
1 mark for dividing through by sin2xtanx
1 mark for using tan^2x+1=sec^2(x), and forming an equation for tanx
3 marks for finding all solutions (i got 3 solutions, one being -pi/4).

9) Sucking mint question
a) (5 marks) Form an equation for the radius with respect to time: First define variables r and t. I got r^3=60/4(t) + 125 or something.

b) (3 marks) Calculate the time for the mint to dissolve: I got 5 minutes, 6 seconds.

c) (1 mark) Why is this model unrealistic: It allowed for infinitely small radii, which is unrealistic (it can never actually be of 0 radius).

10) Differentiation from first principles
a) (5 marks) Show that d/dx(cosx) = -sinx: Follow through standard method for differentiation from first principles

11) Area under graph
(10 marks) Calculate the area under the graph:

1 mark for getting gradient of normal
2 mark for getting equation of normal
1 mark for getting the x coords for the equations (x=1, x=e, x=2e)
3 marks for integrating y = xlnx by parts (from x=1 to x=e)
2 marks for integrating equation of normal (from x=e to x=2e)
1 mark for final answer: A = 5/4e^2 + 1/4

12) Differentiation
a) Partial fractions: A=3, B=4, C=-2
b) Show that function is always decreasing - For f'(x), the denominators are always positive (xyz)^2 and numerators are negative. Negative + Negative = negative, so f'(x) is always negative.

13) Mice population
a) (1 mark) State the initial population: 90

b) (4 marks) Show that dP/dT = T(300-T)/1200:
c) (3 marks) Find the time at which dP/dt is at a maximum: Find the derivative of dP/dT, then equate it to 0.
time was -4ln(3/7)

d) (1 mark) State the largest possible population: 300
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number0
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I think 9 C was a little different, think it wanted you to find when dP/dT was a maximum not the population
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number0
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Could be wrong tho
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RedGiant
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#4
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(Original post by number0)
I think 9 C was a little different, think it wanted you to find when dP/dT was a maximum not the population
Can't really remember that question, what did u do for it?
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squadt_
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Yea got those answers, any idea what the grade boundaries will be for an a*? 240?
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number0
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(Original post by RedGiant)
Can't really remember that question, what did u do for it?
So once you showed what they wanted you to show for dN/dt i found the maximum value of dN/dT by differentiating it again, then found the maximum by setting to zero and solved to get N = 150 i think, so then subbed that back into the orginal equation for the value of T that corrosponded to it. I
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RedGiant
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#7
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(Original post by number0)
So once you showed what they wanted you to show for dN/dt i found the maximum value of dN/dT by differentiating it again, then found the maximum by setting to zero and solved to get N = 150 i think, so then subbed that back into the orginal equation for the value of T that corrosponded to it. I
Ah gotcha, if it's a maximum surely the second derivative should just be less than 0?
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Notnek
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RedGiant your poll is public by the way. You'd probably get more votes if it was private.
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RedGiant
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(Original post by Notnek)
RedGiant your poll is public by the way. You'd probably get more votes if it was private.
Oops, how do I change it?
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number0
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(Original post by RedGiant)
Ah gotcha, if it's a maximum surely the second derivative should just be less than 0?
Nah i mean it wanted like the actual value of dN/dt that was its maximum value, i think.So dN/dT = xyz.. , so you had to find the largest value of dN/dT i think. Nothing to do with the population being maximum. it can't have asked for to find the maximum population in that question because the population gets closer and closer to 300 so there isnt like a proper maximum
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Notnek
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(Original post by RedGiant)
Oops, how do I change it?
I've changed it for you. I'm not sure if you have the "edit poll" option.
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RedGiant
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(Original post by number0)
Nah i mean it wanted like the actual value of dN/dt that was its maximum value, i think.So dN/dT = xyz.. , so you had to find the largest value of dN/dT i think. Nothing to do with the population being maximum. it can't have asked for to find the maximum population in that question because the population gets closer and closer to 300 so there isnt like a proper maximum
Cool cool, so N was 150? what did you get for the time question after it?
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number0
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(Original post by RedGiant)
Cool cool, so N was 150? what did you get for the time question after it?
Think part b was just the show that then, part c was doing the derivative of dN/dt finding its maximum then finding the corrosponding time. It was a four marker and I think the time came out to be -4ln(3/7) or like 3.39... It could easily be wrong I was sorta guessing what I was doing so dont take it like its right
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squadt_
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How many solutions were there to the equation one (after the polynomial)?
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have
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RedGiant
Q1 a) It was gg(5) not gg(2)
b) The lower bound you wrote is right, the upper bound is not 5, it's the answer to 1a)

4c) A better comment imo would be to say that the model implies the ball has negative height after 40m.

5c) It didn't say explain why this equation has exactly two factors,
it had the same quadratic, but in terms of y^2 and not x, and it said show that there are exactly two solutions.
You're right to use the discriminant to show the right hand factor has no roots, but you have to say y^2 = 2, so y = +-sqrt(2) which is two roots.
For 5d) it had -f(tan theta) = 0, and you do a similar thing, showing when tan theta = 2 for theta between 7pi and 10pi, so there's 3 solutions,

For 6c) I remember, there being two solutions -pi/4 and like 1.3 something something. Can't help much there.

7c) Similar to 4c, The model also implies that the mint will have negative radius after 5 minutes 6 seconds. It can have infinitely small radius (it can completely dissolve and be 0)
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thekidwhogames
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Challenging exam, I think I got 96. Bloody he'll compared to old spec
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have
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Just to also add you forgot the Newton-Raphson question
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RedGiant
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(Original post by have)
RedGiant
Q1 a) It was gg(5) not gg(2)
b) The lower bound you wrote is right, the upper bound is not 5, it's the answer to 1a)

4c) A better comment imo would be to say that the model implies the ball has negative height after 40m.

5c) It didn't say explain why this equation has exactly two factors,
it had the same quadratic, but in terms of y^2 and not x, and it said show that there are exactly two solutions.
You're right to use the discriminant to show the right hand factor has no roots, but you have to say y^2 = 2, so y = +-sqrt(2) which is two roots.
For 5d) it had -f(tan theta) = 0, and you do a similar thing, showing when tan theta = 2 for theta between 7pi and 10pi, so there's 3 solutions,

For 6c) I remember, there being two solutions -pi/4 and like 1.3 something something. Can't help much there.

7c) Similar to 4c, The model also implies that the mint will have negative radius after 5 minutes 6 seconds. It can have infinitely small radius (it can completely dissolve and be 0)
Thanks for the input, I'll add these in.

Although I got 3 solutions for the tanx equation. I tested them as well.
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thekidwhogames
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(Original post by have)
Just to also add you forgot the Newton-Raphson question
How did you find it; 95+? It was quite challenging and nice
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thekidwhogames
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If i remmeber there were 3 solutions to some trig one?

It took me a while since I over complicated but I may have made a silly mistake there
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