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OCR

A2 Maths

Revision Notes

A2 Maths and Further
CORE 3
Algebra and Functions
Domain is the input, range is the possible set of
outputs and the function must be one-one for an
inverse to exist
...
e
...
rev
...
Some
iterative equations may fail to converge

The Simpsons rule is a more efficient numerical
calculation of the integral than the trapezium rule
previously used in c2 (equation in formula book)

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OCR

A2 Maths

Revision Notes

CORE 4
Algebra and Graphs
𝑓(𝑥)
𝑔(𝑥)ℎ(𝑥)

=

𝐴
𝑔(𝑥)

𝐵

+ ℎ(𝑥) hence 𝐴ℎ(𝑥) + 𝐵𝑔(𝑥) = 𝑓(𝑥)

solve for h(x)=0 and g(x)=0 for A and B
𝑓(𝑥)
𝑔(𝑥)ℎ(𝑥)2

And repeated root:

=

𝐴
𝑔(𝑥)

𝐵

𝐶

+ ℎ(𝑥) + ℎ(𝑥)2

following process above
Long division of polynomials, by finding the largest
value of x in descending orders of power

(1 + 𝑥) 𝑛 can be expanded using the formula in the
book where |𝑥| < 1, it must be in the form (1 +mx) in
order to work
Parametric equation are where 𝑦 = 𝑔(𝑡), 𝑥 = 𝑓(𝑡) to
convert to Cartesian rearrange in terms of t and
substitute into the other

Differentiation and Integration
𝑑(sin 𝑘 𝑓(𝑥))
𝑑𝑥

𝑑𝑣
𝑑𝑥

= 𝑘 × 𝑓 ′ (𝑥) × cos 𝑓(𝑥) × sin 𝑘−1 𝑓(𝑥)

Use partial fractions to integrate:∫ 𝑢
...
𝑑𝑥 𝑑𝑥 , ensure careful selection that results in an
integrable result i
...
not integrating the natural log

𝑑(tan 𝑘 𝑓(𝑥))
𝑑𝑥

= 𝑘 × 𝑓 ′ (𝑥) × sec 2 𝑓(𝑥) × tan 𝑘−1 𝑓(𝑥)

𝑑𝑢

sinx
cosx

-cosx
-sinx

With the arrows
differentiate and
against the
arrows integrate

∫ cos2 𝑥 𝑑𝑥 can be rewritten as a double angle formula
in terms of cos2x and then be solved

𝑑𝑥 = 𝑢𝑣 −

𝑑𝑥

∫ 𝑓(𝑥)𝑔(𝑥)𝑑𝑥 let 𝑢 = 𝑔(𝑥) then ∫ 𝑓(𝑥)
...

∫

𝑘𝑓′(𝑥)
𝑓(𝑥)

= 𝑘 ln 𝑓(𝑥)

Implicit differentiation allows differentiation of y in
terms of x:

𝑑𝑔(𝑥)𝑓(𝑦)
𝑑𝑥

= 𝑔′ (𝑥)𝑓(𝑦) + 𝑔(𝑥)𝑓 ′ (𝑦)
...

𝑑𝑦
𝑑𝑥

= 𝑓(𝑥)𝑔(𝑦) ⇒

1
𝑔(𝑦)

𝑑𝑦 = 𝑓(𝑥)𝑑𝑥

Once separated then integrate: ∫

1
𝑔(𝑦)

𝑑𝑦 = ∫ 𝑓(𝑥)𝑑𝑥

Then there will be a constant of integration on one side
only, which can be found by given coordinates and the
differential equation can be formed

Vectors

𝑥
( 𝑦) = 𝑥𝐢 + y𝐣 + z𝐤
𝑥
Vector equation of a line: 𝑟 = 𝒂 + 𝑡𝒑, where a is the
position vector (start point) and p the displacement
vector of the line (direction vector), you can find p by
subtracting two points on the line
...

To find if a line intersects set up simultaneous
equations for the i, j and k vectors to find t and the

other constant, if the third equation isn’t consistent
then lines are skew that is if they aren’t multiples of
each other as then they are parallel
...
𝒃 = |𝑎||𝑏|, |a| and |b| are the lengths of the
displacement vectors (p) = √𝑥 2 + 𝑦 2 + 𝑧 2
𝑎1
𝑏1
𝒂
...
( 𝑏2 ) = 𝑎1 𝑏1 +
𝑎3
𝑏3
𝑎2 𝑏2 + 𝑎3 𝑏3
a
...


𝐵𝑥+𝐶
𝑓𝑥 2 +𝑔

+

When the fraction is top heavy,
i
...


quadratic
linear

constant
linear

= linear +

When an equation is top heavy and divided asymptotes
occur at the whole part of the division, hence if linear
an oblique asymptote

Y=1 point is the same, y>1 graph is pushed closer to xaxis and y<0 pushed further from x-axis

Turning points are found by differentiation

Horizontal asymptote is the root, x intercept is the
same, stationary point at y is rooted and x is the same
...
e
...
𝑑𝜃
2

Hyperbolics Functions
sinh 𝑥 =

𝑒 𝑥 −𝑒 −𝑥
;
2

sech 𝑥 =

1
; csch
cosh 𝑥

cosh 𝑥 =

𝑒 𝑥 +𝑒 −𝑥
2

; tanh 𝑥 =

1
; coth
sinh 𝑥

𝑥=

𝑥=

𝑒
𝑒

𝑥 −𝑒 −𝑥
𝑥 +𝑒 −𝑥

1
tanh 𝑥

sinh−1 𝑥 = ln(𝑥 + √𝑥 2 + 1) ∀𝑥

cosh 𝑥 + sinh 𝑥 = 𝑒 𝑥
𝑑
sinh
𝑑𝑥

Derived by writing in terms of exponentials then solving
for y in terms of x
...
e
...

Shapes can he split into components of simpler shapes,
the centre of mass of these shapes ca be found
independently

Then by taking moments in x and y directions the
coordinates of the centre of mass can be found
respectively from a vertex
If the shape has a line of symmetry in the x or y
direction then the centre of mass will be along that line
and taking moments locates whereabouts on that line
the centre of mass lies
...
O
...
E
...


Work done by the force – gain in energy

On a slope must take into account the resolved weight

1
2

Work done against force – loss in energy

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OCR

A2 Maths

Revision Notes

MECHANICS 3
Equilibrium of Rigid Bodies in Contact
Connected rigid objects can be split into components
...
P
...
E
...
O
...
5 and 31
...
5
Use a sketch to find which value to use

used to calculate the required probability of value

The Poisson Distribution
𝑋~𝑃𝑜(𝜆) where 𝜆 = 𝐸(𝑋) = 𝑉𝑎𝑟(𝑋)
𝑃(𝑋 = 𝑥) =

𝑒 −𝜆 𝜆 𝑥
𝑥!

The tables contain a list of the cumulative Poisson
probabilities for various x and 𝜆 values in the form
𝑃(𝑋 ≤ 𝑥)
A Poisson distribution models an event that is:
- Randomly occurring
- Doesn’t occur simultaneously
- Independent
- At a constant rate of occurrence

Passion can be approximated to the normal when 𝜆 >
15
Again since Poisson is too a discrete variable
continuity correction must also be used
Continuity correction
Binomial
𝑋~𝐵(𝑛, 𝑝)
n>50

np >5
nq>5

np =𝜇
npq=𝜎 2
Normal
𝑋~𝑁(𝜇, 𝜎 2 )

np<5
np≈npq
𝜆 > 15

Poisson
𝑋~𝑃𝑜(𝜆)

Poisson can be an approximation to binomial when 𝑛 >
50 and 𝑛𝑝 < 5

Sampling and Hypothesis Tests
A sample taken from a population which has a mean
and variance like the parent

𝐸(𝑥̅ ) = 𝜇 – mean remains the same
𝑉𝑎𝑟(𝑥̅ ) =

𝜎2
𝑛

–n is the number in the sample

A sample must be equally representative of each
feature, which is shown to be an unbiased selection
...


Central limit theorem states approximately if n>30,
sample can be approximated to normal no matter the
parent population distribution

sample can be written; ̅ ~𝑁(𝜇,
𝑋

𝜎2
)
𝑛

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OCR

A2 Maths

The variance of sample, either raw or summarised, can
be used to calculate an unbiased estimate of the parent
variance
population variance =

𝑛
×
𝑛−1

sample variance

A hypothesis test consists of a null hypothesis H0 and
an alternative hypothesis H1
Written in the form H0: 𝑝 = something,
something or 𝜆 = something

𝜇=

Two tailed test – H1≠ 𝜇
One tailed test - H1> 𝜇 H1< 𝜇
The significance level gives the probability of a test
statistic falling into the rejection region
The rejection region gives values for which test statistic
for which H0 is rejected, the boundary values are called
critical values
The acceptance region gives values for which test
statistic for which H0 is accepted
Test statistic is calculated from a sample and is used to
decide whether H0 should be rejected
To carry out a hypothesis test:
-

Define null and alternative hypotheses
Decide significance level (if two tailed it is halved
for each side of the distribution)
Determine critical values
Calculate test statistic

Revision Notes
-

Decide outcome on whether value is in acceptance
or rejection region
State conclusion, with relation to context of test

Discrete data has a nominal significance level, where
the critical value is the nearest whole number with a
probability of the significance level
...

Remember to still use a continuity correction
Hypothesis test on a sample is the same ensuring use
of the sample variance, using a large sample using
central limit theorem and an unbiased estimate of the
variance or a single discrete value using cumulative
probabilities in the tables
Type I error – is where the true null hypothesis is
rejected
Type II error- is made when a false null hypothesis is
accepted
𝑃(𝑡𝑦𝑝𝑒 𝐼 𝑒𝑟𝑟𝑜𝑟) = significance level of the test (for
discrete it’s the true significance level not the
approximate nominal level)
𝑃(𝑡𝑦𝑝𝑒 𝐼𝐼 𝑒𝑟𝑟𝑜𝑟) is found by using the new mean
value and the previous critical values:
e
...
Previous rejection region 𝑍 < 𝑎 and 𝑍 > 𝑏
𝑃(𝑡𝑦𝑝𝑒 𝐼𝐼 𝑒𝑟𝑟𝑜𝑟) = 𝑃(𝑎 < 𝑍 < 𝑏)
The same for discrete data again using actual critical
values

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