CORE 3 Revision Notes (Updated March 2013 – Exam Report Jan 13)

Formula to learn

Examinators comments

PREVIOUS PARTS OF A QUESTION a question of more than one part, one part is often a hint to help with a subsequent part.

SHOW / PROVE Any given result needs a full and complete proof to earn the marks which includes correct notation, algebra and accurate use of brackets throughout. Candidates must show sufficient and all relevant steps in the proof of the given answer.. It is expected that proofs be completed in the right direction.

HENCE means use previous answer (not all realised what ‘hence’ implied.)

ACCURACY / ROUNDING Where there is a specified degree of accuracy in the question this must be adhered to in the answer, as otherwise marks may be lost. (The main error was with candidates who wrote answers to three significant figures, 1.24 and 1.23, rather than three decimal places.) When rounded answers are requested, students should be encouraged to give non-rounded answers first in case they make a rounding error

Also, students should understand that if a numerical answer is to be to a particular degree of accuracy, the previous working must, where possible, be to a greater degree of accuracy. Students should not use rounded partial results in their subsequent calculations to obtain a final result.

FINAL ANSWERS Some questions specify a particular form of answer and full marks will only be obtained if the final answer is in this form. It is expected that numerical or algebraic fractions or expressions should be simplified for the final mark.

Students who do not like fractions and multiply through expressions are changing the question and need to explain what they are doing very carefully if they are to avoid losing marks.

NOTATION There were one or two questions where marks were often lost through poor notation.

PRESENTATION There was some poor presentation where students laid their work rather randomly in the space provided, omitted brackets in solutions and used wrong notations. Some students’ writing tended to be illegible.

Chapter 1 Functions

A function has each x value generates exactly one corresponding value of y

Domain – values that x can take x= or x> etc

Range –values that y [f(x)] can take f(x)= or f(x)< or y= y> etc

EgFunction Domain Rangef(x) = x2 All x f(x) ≥ 0f(x) = √x x ≥ 0 f(x) ≥ 0f(x) = _1__ x + 3

x ≠ -3 f(x) ≠ 0

f(x) = ex (or eg 2x) All x f(x) > 0f(x) = ln x (or log x) x > 0 All f(x)f(x) = sin x All x -1 ≤ f(x) ≤

1

A Many-one function is a function n where there are two or more different x values that generate the same y value. Eg y = x2

A Many-One function does not have an inverse as it is not one-one

One –one functions - Each y value can be generated by only one x-value. To have an inverse a function must be one-one

Composite functionsIf f(x) = x2 g(x) = 3x + 2

gf(x) = 3x2 +2 do f first then gfg(x) = (3x + 2)2 do g first then f

Inverse functions f -1 (x)

Eg If f(x) = 4x + 3

Put y = 4x + 3Swap x and y x = 4y + 3Rearrange x - 3 = y

4f -1(x) = x - 3

4Domain of f -1(x) = Range of f(x)Range of f -1(x) = Domain of f(x)

The graph of f -1(x) is the graph of f(x) reflected in line y = x

f -1(x)

f (x)y = x

Why f(x)=x2 not a function : Candidates should be advised to answer this question as “not 1 to 1”.

Domainmany lost the mark for incorrect notation by using f(x) rather than x

Chapter 2 Modulus Function

The modulus function is where we make the answer positiveEg

Graphs

The graph of

The graph of and are drawn

when x = 0 when y = 0 so x = 3

(You can draw a table of values to help you)

Eg 1a) Solve x= 2x – 6 or x = -(2x – 6)6 = x or x = -2x + 6 3x = 6 x = 2

b) when x < 2 or x > 6

eg 2Draw graph of

a) Solve

x2 – 17 = 8 or x2 – 17 = -8x2 = 25 x2 = 9x = ± 5 x = ± 3REMEMBER ± (so 4 solutions !)

b) When x ≤ -5 -3 ≤ x ≤ 3 x ≥ 5

3

6

-17

-√17 +√17

172 xy

-√17 +√17

172 xy

-√17 +√17

-5 +5+3-3

Marks were often lost due to the quality of the sketch : label point of contact with

axes draw correct shape, indicate the coordinates of

the minimum point.

Make sure sketches go into both the upper quadrants (cross the y-axis)

Quadratic -Draw correct shape eg not

or with straight lines

Know what modulus does –Changes sign to positive

Remember the negative (2nd) solution

Remember x and 2x are not parallel

Questions involving inequalities require careful attention as to whether the inequality is strict or not.

Remember if x2 = 25, x=± 5

Chapter 3 Transforming graphs

Single transformationsReflection in x-axis y = 2x → y = -2x y = f(x) → y = -f(x)

Reflection in y-axis y = 2x → y = 2-x y = f(x) → y = f(-x)

Stretch parallel to y-axis SF ½ y = ½ x2 y = f(x) → ½ f(x)

Stretch parallel to y-axis SF 2 y = 2x2 y = f(x) → 2f(x)

Stretch parallel to x-axis SF ½ y = (2x)2 y = f(x) → y = f(2x)

Stretch parallel to x-axis SF 2 y = (½ x)2 y = f(x) → y = f(½ x)

Translation y = x2 → y = (x – 2)2 y = f(x) → y = f(x -2)

Translation y = x2 → y = (x+2)2 y = f(x) → y = f(x +2)

Translation y = x2 → y = x2 + 2 y = f(x) → y = f(x) + 2

Translation y = x2 → y = x2 – 2 y = f(x) → y = f(x) - 2

Translation y = x2 → y = (x - 2)2 + 3 y = f(x) → y = f(x -2) + 3

Translation y = x2 → y = (x – 2)2 – 3 y = f(x) → y = f(x - 2) - 3

Translation y = x2 → y = (x + 2)2 + 3 y = f(x) → y = f(x + 2) + 3

Translation y = x2 → y = (x+2)2 – 3 y = f(x) → y = f(x + 2) - 3

Combining transformations

y = sin x to y = 3 sin (x – 40) translation followed by stretch in y direction SF 3

y = x5 to y = (-x)5 + 7 reflection in y-axis followed by translation

y = cos x to y = - cos 3x stretch in x direction SF 1/3 followed by reflection in x-axis

special case (both in brackets)(other way around to whats expected!)

y = x3 to y =(2x + 3)3 translation followed by stretch SF ½ in x-direction

Use correct transformation words translations / stretch / reflectionFor translation need vector

Chapter 4 Trigonometry

Sin -1, cos -1 , tan -1

If y = sin x x = sin-1 yRemember the graphs and the start / end points of sin-1 x (-1, -π/2) (1, π/2)cos-1 x (-1, π) (1, 0)

sec x = 1___ Cos xcosec x = 1___ sin xcot x = 1__ = cos x tan x sin x

y = cos -1 x

y = sin -1 x

y = tan -1 x

cosecy cosecy

secy

secy

coty

coty

These graphs need to be learnt End points Asymptotes Where graphs cross the axes Coordinates In radians

In sec x and cosec x graphs, there must not be variable heights

Write coordinates correct way around and in radians

Remember

sin2 x + cos2x =1

sin x = tan xcos x

Eg It is given that 3(cosec2x + 2) = 17 – 10cot x

a) Show that this equation can be written in the form

3cot2 x + 10 cot x – 8 = 0

b) Show that tan x = 1.5 or tan x = - ¼

c) Hence solve 3(cosec2x + 2) = 17 – 10cot x for 0 < x < 2π

Answers

a) 3(cosec2x + 2) = 17 – 10cot x3cosec2x + 6 = 17 – 10cot x3(1 + cot2x) + 6 = 17 – 10cot x using cosec2x = 1 + cot2x3 + 3cot2x + 6 = 17 – 10cot x3cot2x +10 cotx – 8 = 0

b)Factorising(3cot x – 2)(cot x + 4) = 0Cot x = 2/3 cot x = - 4

So tanx = 3/2 (=1.5) ot tan x = -1/4 as tan x = 1/cot x

Tanx = 3/2 x = 56.3 0 or 236.3 0 Tan x = -1/4 x = -14.00 or 166.0 0 or 346.0 0

Make sure your calculator is in the correct mode

Make sure there are no extra values and the values are in the correct range

Sin-12x is not

Sin2x is not sin x2

You must use the correct formula – no marks if you get these wrong

Eg sin2x do not separate the sin from 2x do not multiply 2x by anything

(eg 4x sin 2x is not sin8x2)

It may not factorise – use the quadratic formula if this happens

Chapter 5 Natural logarithmse and ln

if y = ex then ln y = x

(on formula sheet)

ln x follows the rules of logs

ln 1= 0ln 3 + ln 2 = ln 6 ln a + ln b = ln (ab)ln 8 – ln 2 = ln 4 ln a - ln b = ln (a/b)ln 16 = ln 24 = 4 ln 2 ln ab = b ln a

Solving equationseg 1solve e2x – 8ex + 15 = 0(ex – 5)(ex – 3) = 0ex = 5 x = ln 5ex = 3 x = ln 3

eg 2solve ex + 12e-x = 7 multiply by ex

e2x + 12 = 7ex

e2x – 7ex + 12 = 0(ex – 4)(ex – 3) =0ex = 4 x = ln 4ex = 3 x = ln 3

eg 3solve (ln x)2 + 6ln x + 8 = 0(ln x + 4)(ln x + 2) = 0ln x = 4 x = e4

ln x = 2 x = e2

Finding interceptsEg y = ex - 2 - 10

when x = 0 y = e-2 – 10 (0, e -2 – 10) when y = 0 ex - 2 – 10 = 0 ex – 2 = 10 x – 2 = ln 10 x = 2 + ln 10 (2 + ln10, 0)

y = ln (x + 1) - 2when x = 0 y = ln 1 - 2 = -2 (0, -2)when y = 0 ln (x + 1) - 2= 0 ln(x+1)= 2 x + 1 = e2 x = e2 - 1(x = e 2 - 1, 0)

e-2x = 3-2x = ln 3 (not x = ln -1.5)x = -1/2ln 3x = ln 3-0.5

x = ln ( 1/√3)

ln(5x + 3) x (5x + 3)3 is notln(5x + 3)4

Errors

ln(x – e) expanded as ln x – ln e

Chapter 6 Differentiation

Basic differentiationLEARN

ON FORMULA SHEET

Special case d/dx (ln 2x) = 2 = 1 2x xProduct ruleLEARN

Eg f gy = x2 cos x

f ’ g f g’

dy/dx = 2x cos x - x2 sin x

Quotient rule

ON FORMULA SHEET

Egy = __x 2 __ (f) Sin x (g)

f’ g f g’

dy/dx = 2x sin x – x 2 cos x sin2x g2

dxdy

(not 2ex, e2x, or 2xe2x)

Only use product rule for differentiation

Only use quotient rule for differentiation

Use the correct formula (correct way around)

Use brackets

x = 4y + y2

dx = 4 + 2ydy

so dy = 1 dx 4 + 2y

dx/dy is not gradient of normal ( -dx/dy is)

dx = 4 + 2y

d/dx = - ¼ not

Chain rule

LEARN

Eg y = (4x + 3)10

dy/dx = 4x2 10(4x + 3)9 = 40x2(4x + 3)9

Eg y = sin (6x2 – 3)dy/dx = 12x cos(6x2 – 3)

Eg 3 y = ln (3x2 + 5x) dy/dx = 6x + 5 X ___1___ 3x2 + 5x = 6x + 5 3x2 + 5x

GENERAL DIFFERENTATION FACTS

Stationary points when dy/dx = 0

Gradient of line = dy/dx (gradient of tangent)

Nature of stationary points

If > 0 minimum point

If < 0 maximum point

Equation of line y – y1 = m(x – x1) m= gradient Point on line (x1, y1)

Perpendicular of normal = -1/Perp of tangent

Use bracketseg d/dx(4x2 + 9x)3 = 3(8x + 9)(4x2 + 9x)2

d/dx ln(5x -2) =

(5 was frequently missed out)

d/dx(ln 6x) = 6/6x = 1/x (not 1/6x or 6/x)

Candidates need the derivative to find a gradient and also finding stationary points (=0)

If asked for the gradient at a point (or gradient of normal) need to substitute x in dy/dx

A number of candidates added a ‘+ c’ when they were differentiating.

When finding the equation of the tangent (a line) it is essential to find the gradient (a constant) at the requisite point first.

Chapter 7 Integration

Basic integrals

LEARN

ON FORMULA SHEET

Integration by substitution

Remember to change limits as well!

EASY use substitution v = x2 + 3

dv/dx = 2x so dv = 2x dx dv/2x = dx

=

Remember +c Integral sign dx in integral

Don’t use product or quotient rule when integrating

Above x-axes – area positive, below x-axes – area negative

A major error was dv/dx = sin x leading to v=cos x (instead of v = -cosx)

Make sure that you finddv and substitute dx in original integraldx

Original limits are x =

Error – integrating each value in expression

instead of simplifying

change all x to v before integrating (make sure there are no x)

Put back in terms of x

(with x replaced by θ) Many candidates fell at the first

hurdle by not replacing dx by an expression in θ and dθ.

Unfortunately, many confused

with and of course made no headway after that.

Candidates who produced the correct numerical answer from their calculators without correct working did not gain marks.

HARDER

b) use substitution v = 4x – 6

dv/dx = 4 so dv = 4dx dv/4 = dx

x = ¼ (v + 6)

Integrating f’(x) f(x)

Eg1

dx= ln(x2 + 3x)

Eg 2

dx= 2ln(x3 + 3x) as the bottom differentiated = 3x2 + 3x = half top

Make sure that you can spot these

Error

= (should be )

Candidates had difficulty in simplifying

Integration by parts

ON FORMULA SHEET

u = ln x or x

eg 1

As u= x dv/dx = cos x du/dx = 1 v = sin x

Eg 2

Derivative of a function

f(x) + c

Eg

Beware negative sign in formula

Choose correct functions for u and dv/dx

When using integration by parts candidates must realise that at the outset they MUST integrate one term AND differentiate the other.

ERRORx sin (2x-1) Putting u = x sin v = 2x - 1

Put u = lnx v = 1

Do integration by parts twice

Error

Chapter 8 Solids of revolution

LEARN

About the x-axis – remember limits are x =..– find y2 = ………….

About the y-axis – remember limits are y =..– rearrange equation so x2 =………

very common error was in

(100-y²)dy , where the result was often (100x – y³/3), again with candidates very unsure as to whether they were integrating with respect to x or y.

Learn these formulaInclude π

= x + 3

the initial mark here requires a fully correct integral, simplified in terms of x, including dx, with the limits and also π.

The function needed to be squared and this proved to be the downfall of many candidates as they needed to use correct notation.

ERROR

Is not correct

Revise indices

Chapter 9 Numerical methods

Locating roots see separate sheet

Eg 1f(x) = x4 - 3x

Show that f(x) has a root α between 1 and 2f(1) = -2f(2) = 10

change of sign so root between x=1 and x=2 (1 < α < 2)

Eg 2x4 = 6x2 – x

Show that f(x) has a solution between 2 and 3When x =2 LHS = 16 RHS = 22 LHS < RHSWhen x =3 LHS = 81 RHS = 51 LHS > RHS

So LHS = RHS solution root between x =2 and x = 3 2 < α < 3

Recurrance relationship

Eg xn+1 = _3__ 1 + xn

If x0 = 2x1 = 1x2

= 1.5x3

= 1.2

Staircase and cobweb diagrams

Staircase diagram Cobweb diagram

The midordinate rule

x0 x1x2 x3x0 x1 x2

in questions on numerical methods, working should be to a greater degree of accuracy than the final answer

Candidates should ensure that their calculators are in the correct mode in questions on numerical methods (radians)

There are still many candidates who still then write ‘change of sign’ therefore a root without clarification of where the root lies. Many candidates just stated “change of sign therefore root” without stipulating that –0.33 < x< –0.32.

Many lost the second mark by failing to state their conclusion and that α was in the required interval.

The main error was the incorrect labelling of the axes.

Simpsons rule

The midordinate rule

Eg

estimate by midordinate rule (6 strips)

Width of each strip = 5 – 2 = 0.5 62 2.5 3 3.5 4 4.5 5

x 2.25 2.75 3.25 3.75 4.25 4.75y 0.811 1.012 1.179 1.322 1.447 1.558

Area = width of strip x sum of midordinates

=0.5 x (0.811 + 1.012 + 1.179 + 1.322 + 1.447 + 1.558)= 3.66 to 3sf

Simpsons rule

EgUse Simpsons rule with 8 strips to calculate an approximation

to

Width of each strip = 7 – 5 = 0.25 8

x 5 5.25 5.5 5.75 6 6.25 6.5 6.75 7y 1.42 2.69 3.90 4.95 5.76 6.25 6.35 6.03 5.28

Approx area = 1/3 x width x (sum of ends + 4 x odds + 2 x evens)= 1/3 x 0.25 x [1.42 + 2.69 + 4 x(2.69 + 4.95 + 6.25 + 6.03) + 2 x (3.90 + 5.76 + 6.35)]

= 9.9

Use radians If the student had listed the functions not evaluated, they wouldn’t have been

penalised. Don’t round in middle of question-make sure keep all digits in calculator if asked for answer accurate to three significant figures. Candidates must

then work to a greater degree of accuracy. Errors occurred in working with three significant figures (when answer was to 3sf) and in writing the final answer to an inappropriate degree of accuracy. Numerical method questions need students to carefully study the required degree of accuracy and then work to a greater degree of accuracy.

When the start is 0 - a few candidates missed x = 0

The main error within the brackets for Simpson’s rule was to have the 2 and the 4 reversed in the formula

The most common error in the application of Simpson’s rule was to confuse “odd” with “even”.

A few made a slip in writing down digits from their calculator, and a few, after writing the correct expression failed to multiply by their h

Students should have used large brackets round the expression to help avoid the common error of only multiplying the first term by . 1/3