General tips

1. Never divide by something that can be zero - cause there is a scope of it being division
by 0
2. After finding a solution
a. After you find a solution, always look at adomain
b. Reinsert into equation and make sure there is no dividing by solution error
3. If (x2-7x+11)8 = 1, then x2-7x+11 = (+- 1)8
4. Always draw graph wherever possible
5. Never divide by x as x can be a solution
6. Look at highest power term for no of solutions
7. Degrees is rounded to 1 dp
8. Numerical answers to 3 sf
9. Write down each value so it’s the most exact
10. Radian’s is to 3 sf
11. Always mention formula
12. Disjointed - write ‘or’
13. Spend time thinking about question ⇒ don’t assume and jump in
14. No decimal points in denominator
15. No square roots in denominator
16. Don’t round values, just store and add and write to max decimal places
17. Show more steps!
18. To show that a point is outside the circle, show that the distance from the centre to that
point > radius
19. Diagonals of a parallelogram are not equal, but they bisect as the angle with which they
are made is different (they are the supplementary angles which make), opp angles equal
20. It is good practice to suggest that candidates read the question again after completing
their solution to ensure that they have fulfilled all the requests in the instructions, and
that their result seems sensible.

Things to check before submitting paper

1. Given answer in radians when domain is in radians
2. Horizontal transformations - stretch by 0.5 horizontally means 2x, so it becomes
compressed
3. Whether you answered what question asked for - underline

1. Whether you applied correct geometric distribution formula

2. Continuity correction
3. Greater than - you did 1 -
4. Conditional Probability

Graphs to know

Finding LCM:
Only primes

Pure Mathematics 1
When deciding whether to do a + or - sign for the root, think about what f-1(x) is giving. If it is
giving values before k and that would mean more negative values, -, otherwise, +

Graphs

Happy face = +ve,

Sad face = -ve
Where the equation is of this type only
Where a, b and c can be positive or negative

That means if it is (a/b/c - x), it is wrong as x has to be positive, so rearrange to take out the
negative sign

It has to be only 1 for the coefficient of x - so take any coefficient out if you need

In this equation:
(2x - 1)(2 - x)(x + 1)
2(x - ½) * -1(x - 2) * (x + 1)
-2(x - ½)(x - 2)(x + 1)
Here, k = -2, a = ½, b = 2, c = -1

If you have k(x - a)2(x - b), then where the curve turns at x-axis is this factor. For example -
here:
B = 1 as it turns at x = -1
Meet at x = 1
If a = odd, it is like this, if not the one above

Vcylinder = πr2h
Vcone = πr2h/3
SAcylinder = 2πr(r + h)
SAcone = πr(r + l), l = slant height
CSAcylinder = 2πrh
CSAcone = πrl
Focus on prism v pyramid - note the difference - prism has straight faces to another base, while
pyramid has angled faces to a tip
𝑚
𝑛
(xn)m = xnm ≠ 𝑥
xn * xm = xn + m
𝑛
𝑥 𝑛−𝑚
𝑚 = 𝑥
𝑥

When you take root in an inequality, for negative root, switch the sign
Always divide quadratic if possible
Always write down quadratic Critical values, and then draw it to see which part of the quadratic
is above the x axis

Question difficulty is proportional to the place in the paper

This means both are one function as it is a continuous intervals

Completing Square
(h, k) in completing the square formula is also:
2
−𝑏 −𝑏 −4𝑎𝑐 2
( 2𝑎
, 4𝑎
𝑜𝑟 𝑘 = − ℎ − 𝑐)
Sub value to find a
Asymptote
𝑎𝑥 + 𝑏
y= 𝑐𝑥 + 𝑑
Finding vertical asymptote:

cx + d = 0
−𝑑
x= 𝑐
- this is the equation of the vertical asymptote

Finding horizontal asymptote

𝑎
𝑐
= 𝑦 is equation of horizontal asymptote
 ●
○ In this question, plug in kx + 6 into equation 2 to solve
If there is no x term in a quadratic, turning point is at 0, c
^ = ax2 + c

𝑎𝑥 + 𝑏
To find inverse of functions of the form 𝑐𝑥 + 𝑑
, follow these steps
1. Collect y terms on one side and x terms on other
2. Isolate y by factorising

Look at domain of f(x) for range of f-1(x) when working with roots - to see if it’s ± or not - just do
inequality (>= as non complex) to 0 to find domain and
Self inverse → f(x) = f-1(x)

When a function is self inverse, it is symmetrical about y = x

(fg)-1(x) = g-1f-1(x)
Where f(x) intersects y = x, f-1(x) also crosses

(a, b) in f(x) = (b, a) in f-1(x)

Therefore when drawing use your fingers on f(x) and rotate it 90 degrees and draw the same for
f-1(x) and check if (a, b) = (b, a)

Transformations
Make this steps 1, 2, and 3 and then write answer with labels to those steps
f(-x) reflects in y axis
-f(x) reflects in x axis
f(x) + a is a translation to the graph y = f(x) by (0 𝑎)
f(x - a) is a translation by a, 0

f(x + a) is a translation by -a, 0

Horizontal translations change the x term - as there is now a new x that is there, and that x
requires a input of 2 to create a output same as for 0 for example and therefore, it will have to
be New x = (x - 2) to make it the output for 0

When they say describe, say the transformation in words

Say the graph is y = x2

If you make it (x - 2)2

Then, new x = x - 2, x = 2 for the same input of 0 to original function to have the same shape but
at a different spot.

For translations apply the translation to all x and then solve

Vertical translations to WHOLE f(x)

Translation of 2 to 1/x makes it 1/(x - 2)

Counter intuitive logic

To find horizontal translation from 2x to 2x - 2, do 2(x +b) = 2x + 2

2x - 2x - 2b = 2
-2b = 2
b = 1, but the translation is by -1 as it is counter intuitive

f(ax) Stretch Factor = 1/ a

Stretch & reflection at same level of precedence

● When two vertical transformations or two horizontal transformations are combined, the
order in which they are applied may affect the outcome.
● When one horizontal and one vertical transformation are combined, the order in which
they are applied does not affect the outcome.
● Vertical transformations follow the ‘normal’ order of operations, as used in arithmetic
● Horizontal transformations follow the opposite order to the ‘normal’ order of operations,
as used in arithmetic.

If it is (-x)^2 and you translate by (-4,0), you can either

a) make it x^2 and replace replace x by (x+4) to make it (x +4)^2

b) keep it as (-x)^ 2 and replace JUST X by (X+4) to make it ( - ( x+4)) ^ 2 - - > (X +4)^ 2. Also
upon
expanding it is the same. It islike (1)^2 = (- 1)^2

Complete square for f-1 of quadratics or use quadratic formula and then look at domain to see if
it’s + or -

Inverse of quadratics
1. f(x) = y
2. complete square
3. solve for y
a. sq. root of both sides and look @ domain

To find domain of (fg)-1(x), find range of (fg)(x), and then make it domain of inverse

To find (fg)(x) range, substitute lower bound of domain of g into g and substitute g(lowerbound)
into f(x)
Always find range of f(x) for f-1(x) domain - or draw graph

Always draw graph

In a question like this, you should do these steps. Remember since the scale factor is ½, the x
value for the same value becomes halved.

ALWAYS, ALWAYS
a) Either draw graph of f(x) with specified domain
b) Or find range of f(x)
To find domain of f-1(x) or even fg or gf

Use or for two diff values which can produce two different results and cannot exist together
If two translations, just write them in a vector form

2
∫ f(x) where f(x) is an odd function = 0
−2

In this question, sub in x into g(x) and make it <= 0 and then since x <= q, it has to be x <= -2
If square * a of a number = -ve, add, else subtract

Including the condition that a composite function gf can only be formed when the range of f is
within the domain of g

If root 1 = k, and root 2 = -2k, they both cannot be 0

For a curve and line meeting, equate and b2 - 4ac >= 0t

Two points are colinear if

a ) They Share a common point
D) The two lines have same gradient
IFA,B&C,then if colinear Gradient AB= Gradient BC

If you are given three vertices of a parallelogram and you are asked to find the fourth, remember
the rule, the midpoint of one diagonal is equal to the midpoint of the other diagonal

Diagonals are not equal in rhombus, but they meet at a common midpoint

Area of rhombus = d1d2/2

● Always start with grid in section first - so if you run out of time, you are fine as you can
guess multiple choice, but not in grid in
● Underline what they ask first and key terms!!
● Finish no calc in 10-12 mins, ideally 15 minutes - you can do it again and in second time
- look for traps and underline and do the nocalc section again
● Finish calc section in 40 minutes
● Equation of circle
○ (x - h)2 (y - k)2 = r2 - completing the square form
■ (h, k) = center
■ r = radius
○ X2 + 2gx + y2 + 2fy + c = 0 - expanded form
 ■ (-g, - f) = center
2 2
■ R- 𝑔 +𝑓 − 𝑐
○ If it is 2x2 + 2x + 2y2 - 4y = 6, divide everything by 2 and then you can use the
expanded form
● Cannot be roots of non square numbers
● The question numbers are juggled, so your number 2 may not be the same for the other
person
● For graph questions sub in values always - be smart enough to just take random values
for all options
● Use options and plug in for what are al the solutions for the given equation
● You can divide if you know that it is (kx + g) where g is a positive constant as you know it
will not produce a dividing by 0 error
● Power → 1 check where the power is 1 and that is the correct exponential function
● Always check the answer with the data given in the question
● Whatever they are asking underline always
● If x is increased by y % and then decreased by z%, then you can do this
○ [x][1 + (y/100)][1 - (z/100)]
● System of linear equations
○ No. of solutions
■ Infinitely many solutions
● Graph is on top of each other
● Infinite combinations of x and y will make lines coincide
● Same slope
● Same y int
■ No solution
● Parallel lines for example
● X+y=2
● X+y=5
● Same slope
● Different intercept
■ Unique solution
● Different slope
● Given that ⇒ Conditional probability
● Median
○ If odd number of observations, median = ((n + 1)/2)th observations
○ If even number of observations, median = ((n/2)th + (n + 2/2)th)/2 observations
● Whenever there are two right angled triangles
○ Solve using similarity of trigonometry
● Whenever there are two non right angled triangles
○ Solve using similarity
● If it is per square foot, divide by foot2
● Roots of form ax2 + bx + c
○ Sum = -b/a
○ Product = c/a
 ● Vertex - at midpoint of two roots
○ ax2 + bx + c - general form
■ Vertex = (-b/2a, f(-b/2a))
○ a(x-b)(x-c) - x-int
■ Vertex = ((b + c)/2, f((b + c)/2))
○ a(x - h)2 + b
■ Vertex = (h, b)
● A sample’s data can only be generalised to the population it has been conducted on
○ Ex: Some history professors are selected from 3 large universities in California
■ You can only generalise to the 3 large universities in california

Here, the height doesn’t necessarily need to be from the midpoint of BC

You can find the height by the following steps:

1. Find gradient of BC
2. Find gradient of the height using -1/BC
3. If you assume A is (x, y), use the formula y - y1 = m(x-x1) to find equation of the height
4. Now, you can find the point where the height intersects CB by finding the equation of CB
and equating it to the equation of the height
5. Next, you can find the height using the distance formula
Perpendicular bisector from midpoint and the gradient is -1/the equation

If two points are not endpoints of diameter, you cannot take midpoint as center and you cannot
take distance as the diameter. You can only assume only that this is a chord and thus the
perpendicular bisector of that contains the center of the circle.

In this question, you have already found perpendicular bisector of AB in part a, so just solve
simultaneously with that and 12x - 5y = 70
Here, since there are two possible circles, it intersects the circle at top and bottom. You can now
form two equations and solve:

We posit a center (𝑝,𝑞) and note we have squared radius (2√2=8.

(𝑝−3)2+(𝑞−2)2=8, (𝑝−7)2+(𝑞−2)2=8

Subtracting,

(𝑝−7)2−(𝑝−3)2=0

That factors nicely,

(𝑝−7−(𝑝−3))(𝑝−7+(𝑝−3))=0

−4(2𝑝−10)=0

𝑝=5

So far no ambiguity. We’re left with

(5−3)2+(𝑞−2)2=8

(𝑞−2)2=8−4=4
𝑞−2=±2

𝑞=0 or 𝑞=4

Answer: (𝑥−5)2+𝑦2=8 and (𝑥−5)2+(𝑦−4)2=8

For questions with tangent to circle at a point, find equation of center to point, then do
perpendicular gradient and y-y1..
Show that two points is a diameter
● Find midpoint of the two points,
● Then find radius using the midpoint
● Then find diameter and equate it to the distanc ebetween the two points to prove
If PQR is in a circle, PR is a diameter
In this question, you can’t assume that 2x + y is perpendicular to the chord between the two
points. Therefore, you must form simultaneous equations like this:

In this question, the midpoint of AC is equal to D

Since you know ABC = 90 degrees, AC = diameter and therefore, D = midpoint of AC as it is a

semicircle
Area of sector = ½ r2theta where theta is in radians

If you have three points on a circle and you ant to find the equation of the circle, you have to
construct perpendicular bisectors from both of those lines and the intersection is the centre and
then you can find the radius using pythagoras theorem

To prove if something is a diameter, show that the angle opposite to it is 90 degrees

IN a question like this where you actually have the equation of hte circle, just make a dummy
equation with m as the gradient and sub into the equation of the circle and then use b2-4ac

Coordinate geometry
This is a very simple question

Since AB and AD intersect at A, 3x = (x+11)/4, so x =1, y = 3

Now, E is the midpoint of AC, so C - E = E - A, C = (12, 14)

Now, you know the gradient of BC = gradient of AD, so gradient = ¼, so

1 14−𝑏
4
= 12−𝑎
, 𝑠𝑜 12 − 𝑎 = 56 − 4𝑏, 𝑎𝑛𝑑 𝑏 = 3𝑎, 𝑠𝑜 12 − 𝑎 = 56 − 12𝑎, 𝑠𝑜 44 = 11𝑎, 𝑎 = 4, 𝑏 = 12, 𝑠𝑜 𝐵
Then, you know that gradient of CD = gradient of AB = 3, so
14−𝑑
3 = 12−𝑐
, 14 − 𝑑 = 36 − 3𝑐, 𝑠𝑜 4𝑑 = 𝑐 + 11, 𝑠𝑜 𝑐 = 4𝑑 − 11, 𝑠𝑜 14 − 𝑑 = 36 − 3(4𝑑 − 11) , 14 − 𝑑
c = 9, so D(9,5)

Parallelogram properties
1. Diagonals are not equal
 2. Diagonals bisect each other
3. Gradient of opposite sides are equal
4. Length of opposite sides equal
5. Opposite angles equal
6. Adjacent angles add up to 180
7. Midpoint of diagonal is the center

Trig

Here, c is like before going into cos-1 so it is like π/2π, so it is basically cosx, so do acosx - b =
0, cosx = b/a

The time for which the passenger is above a height of 90 m - it isn’t interval for which the
passenger is above a height of 90 m; it is the duration of time for which the passenger is above
a height of 90 m

There is a very simple way to solve questions like this:

Remember that the point where the circle meets the tangent is the midpoint - that is it, and then
you can find the centre of the other circle, and you already know radius
Write a formula for circle area/perimeter then solve to make sure you do not forget

To prove first part, make sure you state that the angles add up to 360 of all 6 sectors because
they make up a full circle
N = 3 as you already found theta = 30 degrees, and then since theta is from 0 to 360, 0 < theta
< 1080
For the largest solutions, use 150 degrees. So do 150 + 360 + 360 = 870 degrees, so theta =
870/3 = 290 degrees

In this question, range = 14, so d = ±7, as it can be down and then up or up and then down like
below, c = 3
Or you can solve by saying 10 = max point = c + d, -4 = c - d, so you get 2c = 6, c = 3, so d = 7,
but it can be ± as the order does not matter
Since CAB = 60 degrees (equilateral), so the angle alpha is 30 degrees - next to tgt
An arc equal in length to the radius of the circle subtends an angle of 1 radian at the centre of
the circle

The reference/basic angle is always positive

Without using angle if you want to find trigonometric ratios, you have to make a triangle

A function and its inverse meet at y = x

sin-1(-x) = -sin-1(x)
tan-1(-x) = - tan-1(x)
Transformations order applies in trig

sin2(x) + 3sinxcosx + 2cos2x = 0

This is already in quadratic form (ax2 + bx + c) → the b term is splittable as it has components of
the a and c.

sin(180 - x) = sin x
Sin (x +- 360) = sin x
cos(-x) = cos x
cos(x +- 360) = cos x
tan (x +- 180) = tan x

Can use above formulas to solve and directly do inverse of negative as well

Keep one more decimal place for all values before finding final answer

Range questions
→ Just sub in the equation from x to that equation and modify the range as you go

Proving → write LHS = ….

Until ….. = RHS

Sub into equation to check after solving

In proofs, if you have tan and sin or cos, almost always you can sub sin/cos for tan and you
should be fine

Sometimes, resetting proofs can be useful

If you know you need to make it into x and it is x and y, make y into x

Can cross check answer using the formula method

Drawing trig curves and their inverse - steps

1. Find domain and range of both functions

2. Draw a proper grid with a good scale which is equal
3. Draw y = f(x) with special attention to where it intersects the axises, and the start and
end points (both coords)
4. Draw y = x
 5. Draw f-1(x) with special attention to where it intersects the axises, and the start and end
points (both coords) - make sure it is symmetrical - check after you are done that the
coords match

2
cos θ = − 1 − 𝑘 as you rearrange for sin2x + cos2x = 1 and since it is quadrant 2 therefore,
it is negative

𝑘
tan θ = − 2
1−𝑘
sin(θ + π) = 𝑠𝑖𝑛 − θ as quad 4

When you have 0 <= x <= 360 and you have sin 2x = 1, and you want to sub u as 2x

So sin u = 1, and 0 <= 2x <= 720, so 0 <= u <= 720

When drawing graphs, make sure you consider the endpoints (for questions with hence solve)
and make sure you do the same domain for both

For one mark, if they ask you for the other point a line meets a trig graph at, then just state that
don’t try to solve it

If you have
2 3 3 3 −1 3 ° −1 3
𝑐𝑜𝑠 θ = 4
, 𝑐𝑜𝑠θ =± 2
, 𝑠𝑜 𝑖𝑓 𝑐𝑜𝑠θ =+ 2
, θ = 𝑐𝑜𝑠 ( 2
) = 30 , 𝑎𝑛𝑑 θ = 𝑐𝑜𝑠 (− 2
) = 150
You have to say the second statement if you are getting that θ for a negative cos value so you
have to say cos-1 for that -root3/2 for that as well when you say 150
Series
Say un rather than tn
Numbers a, b and c are in
a) Arithmetic progression if 2b = a + c
i) 5, 10, 15
1) 2(10) = 5 + 15, so in arithmetic progression
b) Geometric progression if b2 = ac
i) 5, 10, 20
1) 102 = 5 x 20 = 100

nCr = multiplication of r numbers down from n ( n included ) / r!

Different objects from combination

nC0 = 1 = nCn

nCr = nC(n - r)

0! = 1 as

5! = 5 * 4!
4! = 4 * 3!
…. n!= (n -1)! * n

1! = 1 * 0! = 1

So 0! Has to be 1

Arithmetic series = Addition of terms

Arithmetic sequence = not addition

0! = 1

As
3 3! 3!
C3 = 3!(3−3)!
= 3!0!
=1
1/0! = 1
So 0! = 1/1 = 1

Solve equations with power using division

Here n = 41 as 200 is not counted twice - therefore, it needs to be 41

To find n just use a + (n-1)d = tn

Number of years:

Don’t try to use the formula - just look at how the cost changes and go to the year - just go
through manually - write it all down

If it is inclusive, you add both years

Differentiation
You can prove a function is increasing if it’s dy/dx minima is always > 0 (minima > 0)
To give a reason for this, say the value of gradient is approaching 2 as ꝺx approaches 0 -
gradient is the limit of the gradient of the chords as ꝺx approaches 0

This is because the gradient at E is the limit of the gradients of the chords as the x-value tends
to 3 or ꝺx tends to 0.

Increasing function is > 0, decreasing is < 0, neither is = 0

If you have dy/dx = 12/(4x - 8)2

Since the denominator is 2 you can never have 0 as the denominator can never be 0, divide by
zero error. So this is increasing function not neither

If you have f(x) = x2

And you have two points: x = x, and x = x + h

so f(x) = x2 and f(x+h) = (x+h)2
So gradient =
2 2 2 2 2
𝑓(𝑥+ℎ) − 𝑓(𝑥) (𝑥+ℎ) − 𝑥 𝑥 +2𝑥ℎ +ℎ −𝑥
𝑥+ℎ − 𝑥
= ℎ
= ℎ
= ℎ(2𝑥 + ℎ)/ℎ = 2𝑥 + ℎ, 𝑠𝑜 𝑎𝑠 ℎ −−> 0, 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 = 2𝑥
So first principles applies, and this gives the gradient at any point

To find equation of tangent to curve at a point, find the gradient at that point and use the points
at that point to find the equation using y - y1 = m(x-x1)
Use calculator to check

𝑑 𝑓'(𝑥) 𝑑 1 −1
𝑑𝑥
𝑓(𝑥) = = 2
2 𝑥 𝑑𝑥 𝑥 𝑥
2 𝑛 𝑛 𝑛 𝑛
(𝑎𝑥 + 𝑏𝑥) = (𝑥(𝑎𝑥 + 𝑏)) = 𝑥 (𝑎𝑥 + 𝑏)

If you have a straight line for which you need to find the angle with the x axis and you have the
gradient, you need to remember that Gradient =
∆𝑦 𝑂 −1
∆𝑥
= 𝐴
, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑡𝑎𝑛 θ = 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡, 𝑠𝑜 θ = 𝑡𝑎𝑛 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡

ALWAYS write f’(x) </>/= 0 when saying it is decreasing/increasing/at turning point

In an inequality like this:

3
4𝑥 −32
3 > 0, 𝑦𝑜𝑢 ℎ𝑎𝑣𝑒 𝑡𝑜 𝑔𝑜 𝑐𝑎𝑠𝑒 𝑏𝑦 𝑐𝑎𝑠𝑒
𝑥

Case 1:

+/+ > 0

Case 2:

-/- > 0

To solve, just equate the numerator and denominator to + or - and then look which is the
common region for the inequalities [based on which one doesn’t include the other and is
therefore correct

After solving:
X3 > 0, x3 > 8 ⇒ x > 2, x > 0 so x > 2 is the common region

X3 < 0 [Negative - so switch direction of inequality] x3 < 2 ⇒ x < 0 as that is the common region
on a number line

Keep all powers in index form when differentiating

𝑑𝑉 𝑑ℎ
If water leaks out, then both 𝑑𝑡
𝑎𝑛𝑑 𝑑𝑡
𝑎𝑟𝑒 𝑏𝑜𝑡ℎ 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒 𝑎𝑠 𝑤𝑎𝑡𝑒𝑟 𝑖𝑠 𝑙𝑒𝑎𝑘𝑖𝑛𝑔 𝑜𝑢𝑡
In part iv), you have this:

-2x + 8√x = k, so when you get the quadratic, the c coefficient becomes -k, not +k

In this question, try to find a way by either knowing that there is only one point where it meets -
so b2-4ac = 0 OR solving by equating both equations and proving - be creative

TRUST

If you have to give in terms of m, just give in terms of m, not in terms of numbers - so if you have
a solution with m, use it

Don’t find y of stationary point unless you have to

To find angle alpha, you must do tan-1m2 - tan-1m1

You cannot do tan-1m2-m1 as trigonometric functions don’t split like that. You have to find big
angle minus smaller angle Cannot do the difference as that would give the angle of a line with
gradient m2 - m1, not the angle between the two lines.
You are essentially finding b which is not the same as alpha -

Find tan-1 of one gradient, and subtract from the bigger one. This needs to be done as otherwise
you would just find gradient of a line with angle theta. This way, you are finding the angle
between two lines, which is angle of one line - the other

Integration

After subbing in for integration in limits, make sure after subtracting that you have made it - -
positive. Also, if you have k for integration take dummy value for k and do integration on
calculator
When finding area between curve for a complex equation like
2𝑥 + 3, 𝑗𝑢𝑠𝑡 𝑒𝑞𝑢𝑎𝑡𝑒 𝑡𝑜 0 𝑎𝑛𝑑 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 𝑧𝑒𝑟𝑜𝑠 𝑡𝑜 𝑠𝑘𝑒𝑡𝑐ℎ 𝑖𝑡 𝑎𝑛𝑑 𝑡ℎ𝑒𝑛 𝑠𝑜𝑙𝑣𝑒
The volume here is

The black box - blue box so upper curve in this case is the line. If you rotate 90 degrees, you will
get it

Remember for volumes around y axis between two curves:

1. Equations have to be in terms of y, and it is x2

2. Coordinates are the y-coordinates
3. Upper curve is when you rotate axises what is the upper curve
4. Then after making in x2, make sure that you integrate after subtracting upper2-lower2
Past Paper Tracker
9709_s22_qp_12
71/75
Q10 a) domain of f-1 will be given using x not f-1(x)
Q10 b) mistake in last step (forgot to divide as a result of lack of space
Q10 c) ecf
Q11 a) missed two answers which were in the domain

9709_s22_qp_51
47/50
2 b) Forgot that Mrs. Lan was a woman, therefore, it could be 1 Mrs. Lan, 2 (other) women and
2 men also
2 c) Here, the values do not exceed the boundaries and the mean and frequency remains same
so sd remains the same

9709_w22_qp_12 70/75
Q5a) When you stretch horizontally by 0.5, it actually compresses, as it becomes f(2x) not f(x).
Think of it like this, when you are not plugging in 0.5x, you are plugging in 2x, not 0.5x .
Basically, stretches by 0.5 are compressions, while stretches by 2 are actual stretches.But their
effect on x is different, as 0.5 gives 2, while 2 gives 0.5 [2]
Q7b) If domain is in radians, give answers in radians [1]
Q9d) Read the question, it said g-1f(x), not just g-1(x) - underline in the question [2]

9709_m22_qp_52
Q2 b) The three weeks can be ordered in three different ways as it is binomial so 3C2 [2]
Q3 a) You can use a false origin starting from 0.5 rather than 0 [1]
Q6 b) They said after she has checked at least 6 chocolates which means when she has
checked 7 or more chocolates. You must have six chocolates - which are not lemon - so P(X >
6)[1]

● Continuity corrections
● Conditional probability
9709_w22_qp_13 74/75
Q5 c)

If it is (8, 6), then teh minimum point would not be at y = 0. Therefore, it has to be 8,2

Also, since it is multiplied by 3, if it is 8,6 , centre will come at 18, we want at 6

Simple solution for 11 b ii
Since AC is perpendicular to BC, the circle is something like this

Since angle BCA = 90, that means AB is diameter, so midpoint of AB = centre of circle, then you
can solve

9709_m23_qp_52
Q1 a) On the axis of the cf graph, you have to put cumulative frequency [1]
This question has binomial till trial 6. Then trial 7 is geometric

Remember that P(B) = ¼ * ⅓ * ½ * 8 different ways = ⅓

P(A ∩ 𝐵) = 1/4 * 1/3 * 1/2 * 1/1 * 4 = 1/6

6 b) Remember to finish the question - it says 4 randomly chosen cyclists so you have to do
answer4 - check number of marks before solving [1]

9709_s20_qp_52 - Full

9709_w22_qp_11

Q2 b) Finish the question - after you find c, represent with c [1]

Q8 a) Increasing function is > 0, decreasing is < 0, neither is = 0

In this question, since the denominator is 2 you can never have 0 as the denominator can never
be 0, divide by zero error. So this is increasing function.

Q9 d) Brain fade [1]

Q11 c) This can easily be solved using angle in circumference theorem

Angle in centre = 2 x angle in circumference

9709_w22_qp_53 42/50
Q1 Give variance to exact value
Q2 c) Write frequency in cf calculations separately so that you do not make mistake [2]
Q4 a) Show how you get 64 (43) [1]
Q4 d) P(Y > 4) = 1 - P(Y = 1, 2, 3, 4) ≠ 1 - P(Y = 1, 2, 3) ; P(Y ≥ 4) = 1 - P(Y = 1, 2, 3) [2]
Q6 c)
There is a key difference you must understand in this question. When you are just finding
arrangement where there are duplicate letters, you do not need to multiply (if 2 Ts and selecting
1 T) by 2 as it is the same arrangement. Whereas, when you are finding the probability, you
must multiply by 2 because you are finding the probability - so it multiplies as there are 2
different Ts. Also, the total arrangements is 9C5 because you can double count in this case as
selecting double of the same letters is fine as this is not arranging, you are selecting [3]

9709_s11_qp_63 45/50
Remember that the second part doesn’t have the same final score thing - it is separate [3]

There are no half marks - so the midpoints are( 0 + 9)/2 = 4.5, (10 + 19)/2 = 14.5, etc. - the
upper bound is ub - 1 as not equal to [2]

9709_w22_qp_51 46/50

Q2 b) There is a difference between > and ≥ in continuity corrections [2]

Q5 a) Just show outcome, not sum [1]
Q5 d) Dumb mistake [1]

Probability and Statistics

Sd is never negative
Mode is value with highest probability, mean is most likely

Write frequency in cf calculations separately so that you do not make mistake

When comparing two data sets, look at central tendency to compare magnitude, and then look
at spread, and compare how ‘spread out’ the data sets are

For questions with two dice, always draw a table like this then circle
If you do P(F) + P(CH) + P(F ∩ 𝐶𝐻), you are double counting, as the P(F) has P(CH), and P(F
∩ 𝐶𝐻) has P(F) and P(CH) in it as well. So to find P(F ∪ 𝐶𝐻) = P(F) + P(CH) - P(F∩ 𝐶𝐻) as F
and CH are non mutually exclusive events, ie P(F ∩ 𝐶𝐻) ≠ 0. Or, what you can also do is just
add P(F) + P(M ∩ 𝐶𝐻) as the cases are Female (so includes curry hot and not curry hot,
 second and third statement is ploy) or male and curry hot

The venn diagram is what I am talking about - as only F is the only red part so 1/20 (F and CH’),
and only CH is M and CH as left is F, so you want to find the sum of these three

Data representation Advantages Disadvantages

Stem and Leaf diagrams Contains all the original data values, so For larger data sets it can be
● One can easily find the range, median and quartiles accurately ● Very tedious to draw
● Mean and standard deviation are exact values ● Confusing to look at
Very useful for a small data set
One can see the shape of the distribution by rotating the diagram through 90 degrees anticlockwise
One can compare 2 data sets by drawing a back to back stem and leaf diagram

Histograms Shows the shape of the distribution Before drawing a histogram, the data needs to be assembled into a grouped frequency
One can group the data into classes of any width table. As a result:
Useful for larget data sets ● Some of the information in the original data set is lost
From the scale on the horizontal axis, one can more easily see the range as compared to a ● The values for the median, quartiles, mean and standard deviation (found from
stem-and-leaf diagram (where the key needs to be used to interpret the values) the frequency table/histogram) would be estimates rather than exact values.

Cumulative frequency Usually drawn to estimate the median and quartiles of grouped data Since a cumulative frequency diagram also needs a grouped frequency table to be
diagrams Useful for estimating the number/percentage of data values that lie below/above a given value drawn, all disadvantages associated with grouped frequency table may also be thought of
as present with cumulative frequency diagram

Box and whisker plots Show the shape of the distribution in a compact way One can not find the mean and standard deviation from them
Gives the lowest and highest values, the median and the quartiles directly
Particularly useful in comparing several sets of related data
Stem and leaf diagrams, histograms, cumulative frequency diagrams and box and whisker plots
are the ways to represent data. Mean, median and mode are measures of variation. Range, IQR
and standard deviation are measures of spread
Box and whisker - LQ, Lowest, Median, UQ, Highest
If odd number of data set in stem and leaf diagrams, then remove the median in both halves
and take the middle value
If even number of data set in stem and leaf diagrams, then split into half and half (as even so
you can split)

Median is at (n+1)/2th value

On the axis of the cf graph, you have to put cumulative frequency

Class frequency = class width x frequency density

Frequency = area

The midpoint = x

Mean = ∑xf/∑f
Standard deviation = ∑x2f - mean2

This means in histograms, f can be found by multiplying class width x frequency density. X will
be found by using the midpoint.
This means values of mean and standard deviation are estimates as midpoints of each interval
are used, not actual raw data

You can use a false origin in histograms

Use 4 sf in between calculations and if follow on question

Always label graphs

In the b part, you have to use the smallest bag number for the smallest coin value and biggest
for biggest coin value

Always have key in stem-and-leaf diagram

Class/Row width = Number of data points possible in each class

- Really the number of data points possible in each class

Use table to crosscheck CF determined graph values

If you want to make a stem and leaf diagram of this above, use the key to make it effectively.
Instead of using decimals, just do this:
Leaf has to have only digits - no more

In comparisons between data sets, always say ‘than x’ and mention data structures that support
your point

For comparisons - generally compare - median and iqr or skew. WHen comparing iqr, say data
is more/less spread out

Last 20 values is total - 20 + 1 as, if it is 66, then 47 - 66 = 20

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66

If there are two cf graphs, where they intersect is the point below which the number of values is
the same

Frequency density = Frequency/Class width => Density of frequency across the class width
If your data starts from 0.5, you can use a false origin. If it is more, use a kink

In this question, after you adjust the class boundaries for the histogram, you don’t have to adjust
it when you have to find the highest possible IQR. This is because it is a number of chocolate
bars, therefore, it cannot be a decimal, so do not make it a decimal, keep it as a whole number

Give all answers, even mean, to 3 sf

𝑥 = 𝑚𝑒𝑎𝑛

Have to bridge gap for histograms by adding and subtracting as that effectively bridges as top
and bottom line up - this is present because the values are rounded

The modifying number is ½ * what the readings are corrected to or gap/2

No need to shade bars in histograms, but labelling is a must

Class widths need to go till the upper limit of the class no more - cannot go above that digit - can
be below though
Always label for box and whisker, stem and leaf, cumulative frequency

Measures =
● Iqr
● Range
● Mean
● Median
● Mode

Best methods of showing data - one of central tendency and one of variation

Use 3n/4 for q3

Use n/4 for q1
Use n/2 for q2

You can also find the q1 and q3 by first finding the median then splitting the data set across the
median and then finding the median of the two split sets and that is q1 and q3

If decimals, inspect and see data values around

Always include key for stem and leaf

If you have this:

And you want to find the highest possible iqr, just look at the range in which
50, and 150 lie and then take the lowest of the range
And highest of the other
And minus

Here you can take the highest of the 40-60 but not 60, so use 59.9. And for
the other use 10.

ALways remember to adjust the class boundaries if they do not line up as

continuous from one to the other
∑nx = n∑x where n is an integer

For n odd values, median is at ((n + 1)/2)th value

For n even values, it is at the mean of the middle two values

Mode < Median < Mean ⇒ data are positively skewed

Mean < Median < Mode ⇒ when the data are negatively skewed

Grouped data - groups (0-10; 11-20)

Ungrouped - no group just one value and it's frequency or no frequency

Cf graphs - draw line directly to graph line

CF → If you have boundaries like this, since there has to be a whole

number of pupils which is greater than 1, it starts from 1. So it is 1→ 100
and then since the range is inclusive of 100, the next one starts from 101
→ 150

In cf graphs if they give you a table like this,write the lower boundaries and
whether it starts from 1

Standard deviation
This is just coded data. So just find the coded mean and use that for coded sd

i) part is just that since sd = 0, so all rides must be at the mean.

Use n = 17 + f here

Watch out for frequency, remember to root in sd

Look for coded means

 ∑𝑥 + ∑𝑦
Combined mean = 𝑛𝑥+𝑛𝑦

2 2
∑𝑥 +∑𝑦 ∑𝑥 + ∑𝑦
Combined variance = 𝑛𝑥+𝑛𝑦
− 𝑛𝑥+𝑛𝑦

2 2
∑𝑥 +∑𝑦 ∑𝑥 + ∑𝑦
σ= 𝑛𝑥+𝑛𝑦
− 𝑛𝑥+𝑛𝑦

If 19 knew capital of cuba, 20 knew capital of hungary and 7 knew both,

the venn diagram is:

P(just C or just H) = P(C) + P(H) - 2P(𝐶 ∩ 𝐻)

Exactly two types of repair:
= 𝑃(𝐸 ∩ 𝑀) + 𝑃(𝐸 ∩ 𝐵) + 𝑃(𝑀 ∩ 𝐵) − 3𝑃(𝑀 ∩ 𝐵 ∩ 𝐸) 𝑎𝑠 𝑡ℎ𝑎𝑡 𝑖𝑠 𝑡ℎ𝑟𝑒𝑒 𝑡𝑦𝑝𝑒𝑠 𝑜𝑓

As this is exactly two 𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝑐𝑜𝑢𝑛𝑡𝑒𝑑 𝑡ℎ𝑟ice

1 head when two coins are flipped:

In one year, only once can one fault happen. Since these are independent
events - it is only and, not or. Therefore, the choice is only between mech
and no mech, not mech or electrical. As it is independent and only happens
once in a year
So tree diagram looks like this

At least is also ‘not none’

At least is two and three over here

If only two outcomes, then outcome and outcome’

In this question

The tree diagram is like this:

a) Both teachers - ⅖ * ¼ = 0.1

b) In a business partner with each other:
i) There are three possible outcomes:
1) 3 * ⅕ * ¼ = 3/20
c) In a business partnership without each other
i) There are three possible men you can choose, and since it has
to be different - it has to be two women
1) ⅗ * 2/4 = 3/10

Simultaneous pick/draws = W/o replacement

Multiple along branches and add vertically

Remember to count twice in cases where picking can be in two orders

In this question:

Two choose B means two choose and the third doesn’t

P(B only) * P(A only) ≠ P(A ∩ B) for independent events

P(B) * P(A) = P(A ∩ B) for independent events

Mutually exclusive events - no common outcomes:

𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵), 𝑃(𝐴 ∩ 𝐵) = 0

Non mutually exclusive events - at least one common outcome:

𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∩ 𝐵) [𝑡ℎ𝑒 𝑐𝑜𝑚𝑚𝑜𝑛 𝑜𝑢𝑡𝑐𝑜𝑚𝑒(𝑠)]

𝑃(𝐴∩𝐵)
P(A|B) = 𝑃(𝐵)

Two independent events - one cannot be affected by the other occurring:

𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴) × 𝑃(𝐵) 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝑃(𝐴|𝐵) = 𝑃(𝐴|𝐵')

C x/+ C / C method - if you are selecting objects from different groups of
objects use this

Sometimes P(A ∩ B) = P(A)

Permutations and combinations

There are three different outcomes

1*****3
2*****2
3*****1
Since 3,1 and 1,3 no need to do 2 x

You only need to decide arrangements for the other five dice - so there are 6 different outcomes
for one dice, and for that 6 other for second dice, and 6 other … 65 • 3. You do not need to think
about outcomes of the dice at the end as those are fixed - and already done - no need to swap
with other dice, that is done using 65 which accounts for different possibilities
Remember that the total arrangements with R at beginning and R at end is 6!/3! = 120 - not the
total arrangements as it is looking for arrangements with two conditions

Arrangements with 3 Os together = 4! = 24, so 120-24

Remember that for probability, you multiply by the number of different ways you can select one
O
The scenarios are
OR _ _ = 3C1 x 2C1 x 3C2 = 18
ORR _ = 3C1 x 2C2 x 3C1 = 9
OOR_ = 3C2 x 2C1 x 3C1 = 18
OORR = 3C2 x 2C2 = 3
OOOR = 3C3 x 2C1 = 1 x 2 = 2
Total = 50, total possible = 8C4; probability = 50/8C4

Do table for selection of letters questions and writing of letters with gaps
The violinists, cellists and double bass players are already selected, so you do not need to do
arrangements for that

n
Pr and nCr only for n distinc objects

Two As together and Two Ls together does not mean Two As and two Ls next to each other,
these two blocks can move

n
Pr = n! x nCr

n 𝑛! 𝑛!
Pr = (𝑛−𝑟)!
, nCr = 𝑟!(𝑛−𝑟)!

Multiplying vs adding

Multiplying = for each scenario, there are multiple scenarios - and is used

Adding = Or included

If identical objects, swapping changes nothing (distinct)

Question: Find the number of permutations in ALLIGATOR if the two As have to be separated
by six letters and the Ls must not be separated

A______A_
Or
_A_______A

Since none of the Ls can be at the edge, and they are not distinct objects, you do not need to
multiply by 2.

So the only moving that can happen is between GTORI - so 5!

The two Ls can be in five positions

A LL _ _ _ _ A _
A_ LL _ _ _ A _
A_ _LL _ _ A _
A_ _ _LL _ A _
A_ _ _ _LL A _

So Number of permutations = 5! x 5 x 2 = 1200, cannot by 6! * 2 because the LL cannot be

move outside

For c part and d part,

You can do this:

R + B = 11
R + Y = 10
B + Y = 09

For c)

Ncombinations = 11C10 + 10C10

d)

Ncombinations = 11C9 + 10C9 + 9C9

A mathematics teacher has 10 different posters to pin up in their classroom but there is enough
space for only five of them. They have three posters on algebra, two on calculus and five on
trigonometry. In how many ways can they choose the five posters to pin up if:
a) There are no restrictions
b) They decide not to pin up either of the calculus posters
c) They decide to pin up at least one poster on each of the three topics algebra, calculus
and trigonometry
i) Cannot do 3C1 x 2C1 x 5C1 x 7C2 as the 7 could include the ones selected
ii) Have to list all possible combos and add them up
iii)
Always list combos for combinations

Check if identical copies

 For questions like this

You can either list or you can do this:

No. of chocolate biscuits picked Combinations

4
0 C0 x 16C6
4
1 C1 x 16C5
4
2 C2 x 16C4
You can only do above when you remove the total chocolate pieces to form 16, not otherwise

You cannot do it when you have to select 4 from 4 milk chocolate and 8 white and 4 dark and
you want to select at least one white and dark, you cannot do 8 x 4 x 14C2 as the 14 can
include the ones you've selected. Can only do if you did 4Cr

Spend more time on listing questions

For questions which one cannot be next to each other, use insertion method like this:

For iv) part, since there is repeating Es, just list each of the different outcomes with different
number of Es, but don’t go for the same Es and start doing with that and selecting 2 Es from 4
and all, cause that will just make it complicated - it is no longer unique when the two Es are
switched - it makes no difference if they are the other two Es or the same two Es or the same
two Es switched all the same thing

Just list all the ways with different number of Es and add them up
 Here you cannot do

EL_ _ _ = 5 x 4 x 3
As:

Select Select Select Select Select Select

C A A C T T

As you can see here, using this logic these two cases will be the same even tho the same
letters are selected. Therefore, you must do 5C3 for this case since the Es and Ls can be
switched, doesn’t matter it is the same and they are anyway selected in 1E, 1L and no
arrangements are done.

If you have to select questions with one letter and other letters
If you want one S and other letters

Then 6C3 = 20, 6C2 = 15 - with different letters

For the ii) part,

You cannot do _ E _ E _ A _ I _ as the vowels are not on the edges for these cases, you need to
do:
_M_M_N_C_T_
This allows you to count edge and in the middle case

For this question,

P(PP) = 2/11 * 1/10 = 2/110

P(SS) = 4/11 * 3/10 = 12/110
P(II) = 12/110
Total probability = 26/110 = 13/55

Here you have to go case by case where 4 attackers, 2 defenders, 1 midfielder / 3 attackers, 3
defenders, 1 midfielder / 3 attackers, 2 defenders, 2 midfielders

You cannot just multiply by 9C1 as you would be double counting (the 9 could include anybody
that was already selected
For the word Excellent,

If you want the Es at least one letter apart:

_X_C_L_L_N_T_

There are 7 spots, and it can go in any 3. Since the Es are identical, you do 7C3 then arrange
the rest of the letters

Now if you want to make a selection of four letters with only 1 L,

You need to go case by case - with one E, two Es, Three Es and no Es.

Now you cannot just do 7 x 6 x 5 / 3! As this implies that this is arrangements, and you cannot
do 7C3/3! As this rule does not apply for combinations

P n C questions to practice:
For this question, there are three different cases to consider
1*****3
3*****1
2*****2
So 3 different cases x 6 for first * x 6 for second * x 6 for third * x 6 for fourth * x 6 for fifth *, so 3
x 65 = 23328
Probability distributions

Remember that the second part doesn’t have the same final score thing - it is separate

There are no half marks - so the midpoints are( 0 + 9)/2 = 4.5, (10 + 19)/2 = 14.5, etc. - the
upper bound is ub - 1 as not equal to

Whenever without replacement is mentioned, underline it

When you have to do a probability distribution table, do easy ones first and subtract from 1 for
the hard one

Draw big tree diagrams - and their sum so you can see each one clearly. Spend time planning
this - draw sections
 In tree diagrams, keep the ordering the same

So keep it like this:

Rather than TCCT

Distribution - Table with data values and their corresponding probabilities

Simultaneous picks → C x C/C method

If you have P(R=r) =
𝑘(𝑟+1)
𝑘(𝑟+2)
𝑎𝑛𝑑 𝑦𝑜𝑢 ℎ𝑎𝑣𝑒 𝑅 ∈ {1, 3, 5, 7} 𝑡ℎ𝑒𝑛 𝑦𝑜𝑢 𝑐𝑎𝑛 𝑑𝑜 𝑃(𝑅 = 1) + 𝑃(𝑅 = 3) + 𝑃(𝑅 = 5) + 𝑃(𝑅 = 7) = 1

If 𝑃(𝑅 = 𝑟)∝𝑟, then P(R=r) = kr

R=r 10 20 70 100

P(R=r) 10k 20k 70k 100k

So add them up and equate to 1

E(X) changes based on number of trials

With replacement - Probability remains the same

If events G, R and Y are adding to 3 total, E(G) + E(Y) + E(R) = 3

Ratio of expected value = ratio of number of objects

Binomial and Geometric Distributions

P(Y > 4) = 1 - P(Y = 1, 2, 3, 4) ≠ 1 - P(Y = 1, 2, 3) ; P(Y ≥ 4) = 1 - P(Y = 1, 2, 3)
X ~ B(n,p)

Here, X is the number of successes

Least possible value of n usually, it will be > at the end

Greatest possible value of n, usually it will be < at the end

If you use logs with bases b in 0 < b < 1, it is a decreasing function like this

The red one is the log0.96x

So, in this if 𝑥<𝑦⟹logb𝑥>logb𝑦

So if you have an inequality with logs, then you flip sign because of this fact
So it is easier to take ln

Expectation is multiplied by number of trials

Geometric distributions

Here, keep in mind for the last part that the probability is the sum of all the different outcomes,
so use sum to infinity, as 0.52 < 1
In this question, in the (ii) part, you have just found the probability of box having one cracked
egg. Now, they are telling you about n boxes that a shop is selling, not n eggs. So, in 2 boxes
the probability of having one cracked egg is 0.766 x 0.766. So for n boxes, the probability of at
least one cracked egg is 0.766n. They want to find how many boxes it takes for this to be less
than 0.01

Binomial
● Fixed number of trials
● Only two outcomes
● Trials are independent
● Same constant probability

Remember 1 - qr for P(X <= r) while qr used for P(X > r)

It is not >= here as it is minus of equal to one, so if it were also = to it would not make sense
Normal Distributions
Remember to convert to z before doing phi-1 or phi

Check if a answer seems logical. Also, make sure for greater than that you do 1 -

Positively skewed:

Longer tail to the right

Low values are more likely than high values

Negatively skewed:

Longer tail to the left

High values are more likely than low values

Symmetrical

High values are as likely as low values

If you are combining two distributions and one has the peak at a higher value than the other,
and that is to the right of the other, they combine to form a negatively skewed distribution

This means when you draw it in a range using the kink, you draw one wider than the other, but
this isn’t really the case. Just make one shorter and wider and one thinner and taller

Make sure area is roughly the same

 4 dp for answer from normal table
3 dp for answer to take into table

Remember to do continuity whenever taking approximation → write it down

Discrete Continuous Reason

X>6 X > 6.5 You do not include any thing

which rounds to 6

X≥6 X > 5.5 You include anything which

rounds to 6

X<6 X < 5.5 You do not include any thing

which rounds to 6

X≤6 X < 6.5 You include everything which

rounds to 6

If you are doing phi-1 of a number and it ends up being between two intervals then use the
values in the table below

Give answer for number as an integer