These are just predictions based on response, difficulty and last year’s G/Boundaries

                                                                    (ONLY FOR PAPER 2)

Grade

PREYING AQA ARE NICE AND GRADE BOUNDARY FOR AN A* IS NO HIGHER THAN 230.

A*

A

B

C

D

E

U

Raw

83± 2

64 ± 4

55 ± 5

40 ± 6

32 ± 7

18 ± 8

0

2023 (ref)

80u hi

62 that low

think it'll be 66 (Prob 63-65)

100

38

27

16

0

QuestionNumber

Question

marks

Answer

Comment

1.

What is the circle equation?

1

(x+1)^2+(y+3)^2 = 36

2.

What was the area under the curve?

it asked for the definite integral

(cant remember the values)

1

between 3 and -2, the very right if i remember is 265 or some sort

125

you’re meant to add the two areas together irrespective of whether the area on the left is negative.

although the two areas were positive, we were finding the integral - as one of of the areas was below the x axis it would be negative when integrating between the given limit no s so it’s 125

3.

What is the set notation?

(1-x)(x+4)<0

1

x: x<1 U x: x>4

10000% U, im given a scientific calc and i can solve inequality, and its union

4.

Differentiation question

3

(2sinx+3cosx)^2 + (6sinx-cosx)^2 = 30

wasn’t it differentiate x^3/sinx ? or x^2/sinx

5.

Logs question

3

5^x-2 = 7^1570

log base 5 5^(x-2) = log base 5(7^1570)

x-2= 1570log base 5 (7)

x= 1570log base 5(7) +2

x=1900.23

1900 (2 s.f.

Like i am so certain, for 1570, x came out as 1900.23

6.

fí

6

Sinx=root(6)/3

(2sinx+3cosx)^2+(6sinx-cosx)^2 = 30

Expand to get 40 sin^2x + 10cos^2x = 30

Use cos^2x= 1- sin^2x

40sin^2x + 10(1-sin^2x) = 30

30sin^2x=20

Sin^2x=⅔

Sinx=root(6)/3, can't be negative as that would be greater than 180 (needs  to be obtuse)

Ain workwer is sqrt6/3???same thing as root two over tjree

it is sqrt(⅔) :)

erm what the sigma

I did rsin rcos for this q and my dumbass forgot that you can expand brackets

7.

what is this question>>???????

Find the binomial expansion of (1+3x)-1 up to x2

2

(1 + 3x)^-1 ?  <first 3 terms of this

1 - 3x + 9x^2

was it (1 + 3x)^-1?

^^yhhhh it was nvm

aii)

Find the first 3 terms of the binomial expansion of (2-3x)^-1

2

½ + + 3/4x + 9/8x^2

7b.

binomial expansion

• 2

Guys wasn’t the second bi6x-9x^2nomial expansion fractional? I got fractions

Isn’t there a partial fractions after this

36x/ ( 1+3x)(2-3x)

Then the one after was expansion of 12x /(1+3x)(2-3x)

Can we add this q

There was a part to the question asking you to split it into partial fractions before finding the expansion of the annoying fraction – what were your numerators for both fractions? I bejeje and that’s missing here, it was around 3 marks for that in addition. I think i maybe got 6 and -4. But like I forgo

Heyyy for the partial fractions I got 8 and -4, then for the second part I divided those by 4 and then timed them by the expansions

yess 8 and -4 do sound more familiar. did your final answer have no constant term? just smth like 18x - ??x^2

Yes one of my terms cancelled out!! But I do remember one of the terms being a h fraction but idk if I did it wrong

yay i think this was a part c of question 7

First fraction was raised (1+3x)^-1 and second fraction was (2+3x)-½ Am I the only one that ended up with ½ at the start of the question where the 2 bion

8. Was

m

onkey question

1-3 maybe7 altogether I think

1st part -

Write in the form y = a +blogx

Show b = 5.6/log3(8)

GUYS THE SECOND PART WAS TO WORK OUT THE VALUE OF A

I GOT LIKE A =3.44

Yes same^

2nd part-

x= ¼

y when x= negative so its not accurate

NOT ACCURATE FOR BABIES YOUNGER THAN A WEEK AS YOU GET NEGATIVE

Yo how much marks was part a ?4

3 marks

9.

determine whether there is a point of inflection where x = 0

4

NOT A POINT OF INFLECTION(\_/)

F(x)= x^2+2cosx

F`(x)=2x-2sinx

F”(x)= 2-2cos(x)

F”(0)=0

F’(-0.1) = Negative nu mber

F’(0) = 0

• F’(0.1) = Positive number

Then choose values of x around it and prove that there is n lb o change of sign

F””(-1) = 0.91

F””(1) = 0.911

no change of sign so not an inflection

i think ur val is too big, use -0.9 and 0.1 would be better tho imo

^^ i think -1 or +1 is ok, textbook/mark schemes use it, but i get what u mean

ok ok thx for clarifying^

Guys there was no change of sign 🙂

^^^yhh :)

im gonna delete

 the other one sos ok:)

^^keep the f””(-1) and f””(1) part

there was a change of sign from -ve to +ve so its a local minimum not a point of inflection

10.

Explain step 3

prove that there is no largest number in the range 3 < k < 4

1

•  4

x = smallest number

y= (3-x)/2

That means

x<y<3

3<y<x say that as x was the smallest value then you get y as the smallest value which contradicts the statement

Conclude with the correct statement

Same thing for large value

There exists a largest value

x = largest value < 4

y = (4+x)/2

hence x< y < 4

(do you not have to flip the positive to negative? You ave to make it (n-4)/2)

-GUYS I did n-4/2

(wdym?)6

Conclude with the statement

-No it works with + cause the difference between x and 4 is halved this way, so x<y<4

<<Yh seem correct, did something like this aswell, talked about it being a mid point

1

3

T3 = (T2 +50)* 1.002

I forgot the rest Rip

T3=(((50x1.002)+50)x1.002)+50)x1.002

      =50x1.002^3+50x1.002^2+50x1.002

Calculate the amount of money in her savings account after 10 years:

you had to factorise out 50 from the equation

T120 = 50(1.002 + 1.002^2 + 1.002^3 ect)

Was a geometric sequence but a was 50.1 :

=£6787.16

I used gemoetric sequence and got the same answer as you

n

<<agreed

part was just the sum to 120 using geometric equation you already get but a=50 cause thats what you start with no and r=1.002

I solved using a arithmetic sequence and got the correct answer is that fine?

U n+1 = (un +50) x 1.002

U1 = 50 x 1.002

U 120 = 6787.16

Possible alternative solution

Genuine question i sat and iterated the formula 120 times on my calculator. Am i cooked - pleasee I was thinking of doing this too

But was it really a geometric sequence??

Do you get any marks for just righting the answer 

12.

Circle answer: particle of mass 2kg has driving force of 10N and resistive force of 4N find a

1

3ms^-2

<<<yh

13

Tick box Q

Velocity time graph

1

Acceleration changes instaneously

14.

displacement question????

1

2

-8

0<t<3

14.

find the values of a and b

4

given two forces:

add then use F = MA where m was 2 or 3 kg

yeah probably, i got fractions over 11, probably incorrect

so did i

it looked wrong tho

same i got fractions

A=8, b=1???

15.

Apple Branch question

The two apples rest on two separate branches separated by a vertical distance d. The two apples fall at the same instant. The first apple takes 0.5 seconds to hit the ground. The second apple lands 0.1 seconds later. show that d = 54 cm

4

Use suvat for first value(1.225 or something)

use suv at for second value(1.77 something)

minus second value from first

u get 0.539m = 53.9 cm and round

16.

^^wasn't it

r = p + 2qe^-0.2t

^^YHHHHHH

when t=3

swr there was no 2 before the q and was just -q or was i tweaking???

 question

u had to show acceleration = 82

then find p?yeah

4 for part a

and 2 for b i think, marks wise

q=45/e^-0.2x3=81.982

HAH

4A it HA

change in displacement of Q is a scalar vector of P therefore they are parallel

both have same velocity when calculated to be (3i +4j)

B

17b.

Claim: “Q also travels at a constant speed of 3i+4j”

Explain why it is not the case

1

He may be incorrect cause ur taking average, in 3 seconds u travelled with an average speed, the particle could be accelerating/ deaccelerating that in 3s travels the same tim

eA

18.

Moments - show that L-x=k and find the value of k

 4

K = 0.2( i think?)

Length = 2L

A = x

B = x+0.1

CWM = ACWM about A

(L-x)mg = (2L-x-0.1)mg

L-x = 2L-X-0.1

L-2x=0.1??

im not sure

Resultant force at B = 2A right where A is resultant force at A? yes think so

 mg(L-x)=Fb(2L-2x-0.1)

mg(L-x-0.1)=Fa(2L-2x-0.1)

2Fa=Fb

mg(L-x)=2mg(L-x-0.1)

L-x=2L-2x-0.2

L-x=0.2

geometry

How do you do this one??? 😭😭😭

Agreed, k = ⅕

Take moments around the centre of mass

Use the fact the Resistance force at B was double RF at A- the RF should then cancel

I ALSO GOT 0.2 OTHER PEOPLE SAID THEY HAD 2 though….

I got -0.2 OHH we both got the same thing DAMN I FORGOT IT COULDNT BE -0.2

IT WAS +

Oof I got -0.2 :(

IT WAS POSITIVE 0.2

2L and x was a distance wasn’t it. Frick. Must have messed up with the signs somewhere. Prob only 1 or 2 marks lost YES IT WASAND THEN 2L WAS DISTANNCE OF WHOLE THING PROB MAX 1 MARK LOST

I put I don’t like the topic

17c

Question 19? what was it?

i think the toy one was Q19 and the vectors on was Q20

What else on mechanics not on this google doc? Anyone know?Yes

5

,  the speed calculate the distances travelled, get a 13-12-5 sided triangle which has a right angle

Displacement from X forusing R=12i, P= 4i+3j?

Distance of P from X is 5

ye RHS, use pythagorus :))

12^2 + 5^2 =

 13^2

these man think were isaac newton

^lmao

^lmao x2

this question was to flag the individuals who had the mark scheme before hand im jk

20.a

4

7cos(11) upwards

s= k

u= 7cos(11)

v= 0

a=-9.8

t= dont give a fuck

v^2 = u^2 +2as

0 = (7cos11)^2 - 19.6k

19.6s = (7cos11)^2

k

Would I get the mark if I put 2.40 or nah tbf its max I loose like one mark

2.41* = k

K < y < 2.5

Or you can use 7sin(79)

(My dumbass must have used 7sin(49) on my calculator exam stress bruh)

I used 7sin(79)

im so stupid i used 7sin(11) and got 0.09

this question was 4 marks - how many marks do you think I would lose if I used 11 degrees as the angle?

^did the same mistake, i guess ill be 2/4 ish - thanks!

just a bit confused guys - I though max height given by formula U squared sin squared theta / 2G

might be to 2 sig figs implied by use g=9.8 in question?

yeah i put 2.4 cuz of g*

 21.

Show that a was >= to that

a>= (5(11+sqrt(3)/26)?

6

Had to make the friction coefficient (u) = 1, as this is when acceleration is smallest.

Then combine the two equations of motion for particles m and n.

n: 80a = 80g - t

m: 50a = t - 50gsin(60) - friction

130a = 80g -25sqrt(3) -25g

then rearrange to the final

Friction  = uR = 50gcos(60)

ye vertical, i think thats the last

if i did everything except the u=1 how many mRKS DO I GET? I did the sam

e 😓

^^Will probs only lose one mark

can you talk about using Mew as 1 due to Fmax as that would be the maximum value friction would take

21 b

What assumptions have you made about the model?

1

The rope is light and inelastic

I said the pulley was smooth? it already said that in the q