Microsoft Word - C1.doc Notes 2018.docx · Web viewMicrosoft Word - C1.doc

Additional Maths

Revision Notes

May 2018

Revision website: http://nhsmaths.co.uk

Revision sessions: NC - Monday 18th 10.00am

JB - Thursday 14th 10.00am

Contents

1 Algebra...........................................................................................................3Indices....................................................................................................................................3

Rules of indices................................................................................................................................3Surds.......................................................................................................................................4

Simplifying surds..............................................................................................................................4Rationalising the denominator..............................................................................................4Factorising quadratics...........................................................................................................5Completing the square............................................................................................................6Solving quadratic equations...................................................................................................6

by factorising....................................................................................................................................6by using the formula........................................................................................................................6

Simplifying algebraic expressions..........................................................................................7Solving algebraic equations...................................................................................................7Remainder Theorem...............................................................................................................9Factor Theorem......................................................................................................................9Algebraic proof - Identities..................................................................................................10One linear and one quadratic equation............................................................................10

2 Coordinate geometry..................................................................................12Parallel and perpendicular lines..........................................................................................13Intersection of a straight line and a curve...........................................................................14

3 Differentiation.............................................................................................15General result.......................................................................................................................15Tangents...............................................................................................................................16Differentiation from 1st principles........................................................................................17Stationary points...................................................................................................................18

4 Integration...................................................................................................19Indefinite integrals...............................................................................................................19Definite integrals..................................................................................................................20

Area under a curve..........................................................................................................................20

5 Trigonometry..............................................................................................213D Pythagoras......................................................................................................................22Trigonometric graphs...........................................................................................................23

6 MENSURATION........................................................................................26Length of an arc...................................................................................................................26

Page | 2 Newtown High School Additional Maths Revision Notes May 2018

1 Algebra

Indices

Rules of indices

Examples:1.

¿20 x

−54 +13

4

x32

=20 x

84

x32

=20 x2− 3

2=20 x12

2.When there is a + or - as part of the fraction split it into parts and then use indices rules

¿18x

15

6x15

+6 x

25

6 x15

=3+x15


Surds

Simplifying surds

Example: To simplify

we notice that 50 = 25 2 = 52 2

= 5 .Example: To simplify

we notice that 40 = 8 5 = 23 5

3√8x 3√5=2 3√5

Rationalising the denominatorRationalising means getting rid of surds (at least from the demoninator).

We remember that multiplying (a + b) by (a – b) gives a2 – b2

Example:


50

403

50 25 2

25 2 2

403 3 8 5

Factorising quadratics

If you are asked to factorise an expression, then use one of techniques below to factorise but don’t use the quadratic formula and work backwards. The answer is the factorised expression, not a solution for x (or whatever variable is used).If you are asked to factorise and hence solve, then you need to use a factorising technique and then solve.If you are just asked to solve, and in particular if you are asked to find a solution to a certain number of decimal points, then that is the time to use the quadratic formula.

Examples of factorise questions: -

This is an example of (a + b)(a-b) = a2 – b2 . Also note you can divide through by 3 before factorising: - 3(x+4)(x-4)

Factorise

This is standard quadratic so just use the method you are happy withCross method:-

Factorise This is standard quadratic so just use the method you are happy withFrom ax2 + bx + c find 2 numbers that multiply to give ac and add to give b and then split bx into 2 bits and factorise:-12d2 + 5d – 2 What 2 numbers multiply to make -24 and add to make 5 :- 8 and – 3

12d2 + 5d – 2 = 12d2 +8d – 3d -2 = 4d(3d + 2) - (3d + 2) = (4d – 1) (3d + 2)


Completing the square. You might be asked to complete the square or the question might ask you to find the minimum value of a quadratic expression. Use completing the square to find the minimum value.

ax2 + bx + c. For this exam a will always just be 1 so the quadratic will be of the form x2 + bx + c.

x2 + bx + c = (x + b2¿¿2 - ( b

2)

2

+ c

Example: Complete the square in x2 – 6x + 7.Solution: The coefficient of x is –6, halve it to give –3. Squaring (x−3¿ gives x2−6 x+9so to just create x2−6 x you need to subtract 9 from (x−3)2

x2 – 6x + 7 = (x−3)2−9+7= (x – 3)2 – 2.Notice that the minimum value of the expression is –2 when x = 3, since the minimum value of (x – 3)2 is 0. Be careful to give the answer the question is looking for – either the minimum value, the x value required to give the minimum or both.

Solving quadratic equations.

by factorising.Example: x2 – 5x + 6 = 0 (x – 3)(x – 2) = 0

x – 3 = 0 or x – 2 = 0 x = 3 or x = 2.

Note that a quadratic could have just 2 terms, but it is solved in the same way.

Example: x2 + 8x = 0 x(x + 8) = 0 x = 0 or x + 8 = 0 x = 0 or x = –8

N.B. Do not divide through by x first: you will lose the root of x = 0.

Example: x2 - 4 = 0 (x+2)(x - 2) = 0 x=2∨x=−2

by using the formulaalways try to factorise first.

ax2 + bx + c = 0 x =2a

Example: 3x2 – x – 5 = 0 will not factorise, so we use the formula with

a = 3, b = –1, c = –5

x =2 3

= –1.135 or +1.468


4acb 2

b

1) ( 1) 2 4 3 ( 5)

Simplifying algebraic expressions

There can’t be a solution if there isn’t an equation (i.e. an = sign)

Example:

Get a common denominator

x+2 yx+2 y

−3 x− yx+2 y

¿x+2 y−(3 x− y )

x+2 y

¿ −2x+3 yx+2 y

Note: This question hadn’t been asked for a few years till last year when they asked it like this, which caught people out:-

Note they said they wanted an expression – there is nothing to solve and what

was required was to simplify 362 x+5

+ 553 x−1 (i.e. 1 hat and 1 umbrella)

Solving algebraic equationsExample:

This is an equation requiring solving but same principle applies – get a common denominator which will be 6 x (x−3)


3 x 6 x(x−3)6 x(x−3)

+ x−63 x x

2(x−3)2(x−3) =

3 x+12(x−3) x 3x

3x

Now there is a common denominator we can multiply both sides by that common denominator which just leaves the top line of the fractions:-

18 x ( x−3 )+2 ( x−6 ) ( x−3 )=3 x (3 x+1 )

18 x2−54 x+2 ( x2−3 x−6 x+18 )=9 x2+3 x

20 x2−72 x+36=9 x2+3 x

11 x2−75 x+36=0(Clue: answer to 2 dec places = Formula!!)

x=−(−75 ) ±√752−4 (11)(36)22

=75±√5625−158422

¿ 75± 63.5722

=6.30∨0.52


Remainder TheoremIf a polynomial f (x) is divided by x−a ,the remainder is f (a) Example:

f (−3 )=6 (−3 )3−13 (−3 )2−3+2¿−162−117−3+2

¿−280Note: I strongly advise working out each term as shown above rather than putting the whole thing into a calculator and possibly making a mistake with a minus sign.Although not seen in a previous exam paper the coefficient of xmay not be 1 . The specification gives

this example

The method is the same – calculate with x=12

2¿

Factor Theorem

If the remainder when f (x) is divided by x−a is zero then x−a is a factor of f (x)

Questions often start by asking to show that x−a is a factor of f (x) but then usually continue with “ hence factorise f ( x )”

Example:

23+6(2)2−2−30=8+24−2−30=0 hence x−2 is a factor


Algebraic proof - Identities

Example:-

You need to make changes to the left hand side till you get it to look like the right hand side.Write it as below:-

2x7

− x−32

+ 3 x+221 LCM of 7, 2, and 21 is 42

¿ 12x42

−21 ( x−3 )42

+ 2(3 x+2)42

¿ 12x−21 x+63+6 x+442

¿ 67−3 x42

One linear and one quadratic equation.

Find x (or y) from the Linear equation and substitute in the Quadratic equation.

Example: Solve x – 2y = 3 x2 – 2y2 – 3y = 5

Solution: From first equation x = 2y + 3

Substitute in second (2y + 3)2 – 2y2 – 3y = 5

4y2 + 12y + 9 – 2y2 – 3y = 5

2y2 + 9y + 4 = 0 (2y + 1)(y + 4) = 0

y = –½ or y = –4

x = 2 or x = –5 from the linear equation.

Check in quadratic for x = 2, y = –½

L.H.S. = 22 – 2(–½)2 – 3(–½) = 5 = R.H.S.

and for x = –5, y = –4

L.H.S. = (–5)2 – 2(–4)2 – 3(–4) = 25 – 32 + 12 = 5 = R.H.S.


These questions are often start with you having to create the two equations to start with.Example a)

Often helps to draw a diagram: -

It is clearly easier to substitute for y rather than x in the second equation so put y=27−x into 2nd equation and solve for x.

2 x2+x (27−x )+18 x+9 (27−x )=703

2 x2+27 x−x2+18 x+243−9 x=703

x2+36 x−460=0( x+46 ) ( x−10 )=0

X can’t be -46 so must be 10 so dimensions are 11 x 17 and 19 x 37

Example b)

AB and BC are 45 and 60


2 Coordinate geometry


Parallel and perpendicular linesTwo lines are parallel if they have the same gradientand they are perpendicular if the product of their gradients is –1.Example 1:

Gradient of FG = 6−14

4−(−2)=−8

6=−4

3 Gradient of perpendicular = 34

Midpoint of FG ¿(−2+42

, 14+62 )=(1,10 )

Equation of perpendicular is y=mx+c m=34 and perpendicular goes through (1,10)

So 10 = 34 .1 + c c=10−3

4=37

4

Therefore equation of perpendicular is y=34

x+ 374

3 x−4 y+37=0

Example 2: Find the equation of the line through (4½, 1) and perpendicular to the line joining the points A (3, 7) and B (6, -5).

Solution: Gradient of AB is −5−76−3

=−123

=−4

gradient of line perpendicular to AB is ¼ , (product of perpendicular gradients is –1) so we want the line through (4½, 1) with gradient ¼ .

Using y = mx + c y = ¼ x + c 14

. 92+cc=1−9

8=−1

8

y = 14

x−18 or – 2x + 8y + 1 = 0.


Intersection of a straight line and a curve

If a line and a curve intersect then the coordinates at the point of intersection are the same.I.E. If line y=mx+d intersects with y=ax2+bx+c then the values of y (¿ x ) will be equal so mx+d=¿ ax2+bx+c which results in a quadratic in x that can be

solved.

Example:

Rearrange x+ y=10 to give y=10−x

Now put the 2 equations for y equal to each other

x2−6 x+14=10−x

x2−5 x+4=0

( x−4 ) ( x−1 )=0

x coordinates are 4∧1

When x=4 y=6∧when x=1 y=9

Coordinates of point of intersection are (4,6) and (1,9)


3 Differentiation

General resultDifferentiating is finding the gradient of the curve.


Example: Finding the equation of the tangent to a curveTangentsExample:

Find the equation of the tangent to y = 3x2 – 7x + 5 at the point where x = 2.

Solution:

We first find the gradient when x = 2.

so the gradient when x = 2 is 5. The equation is of the form y = mx + c where m is the gradient so

we have

y = 5x + c.

To find c we must find a point on the line, namely the point on the curve when x = 2.

When x = 2 the y value on the curve is 3 22 7 2 5 3 ,

i.e. when x = 2, y = 3.

Substituting these values in y = 5x + c we get 3 = 5 2 + c c = –7

the equation of the tangent is y = 5x – 7.


Differentiation from 1st principles

Example

¿ limh

→

0

(x+2)2+3 ( x+h )−(x2+3 x)h

¿

limh

→

0

x2+2 xh+h2+3 x+3h−x2−3x ¿

h ¿

¿ limh

→

0

2 xh+h2+3 hh

¿ limh

→

0

2 x+h+3

¿2 x+3

dydx

=2 x+3


Stationary points

A stationary point is when the gradient to a curve is zero. The gradient of a curve at any point is found by differentiating. Hence to find a stationary point differentiate (to get gradient) and equate it to zero.

Example

Find expression for gradient by differentiating dydx

=3 x2−6 x

Stationary point when gradient is zero so 3 x2−6 x=0

3 x ( x−2 )=0x=0∨2

Find ycoordinate from original equation: When x=0 y=11when x=2 y=8−12+11=7

Coordinates are (0,11) and (2,7)

“Find the nature” means are the stationary points maximum or minimum. To find which differentiate a second time.

d2 yd x2 =6 x−6

When x=0 d2 yd x2 =−6 which is < 0 so (0,11) is a maximum

When x=2 d2 yd x2 =12−6 which is > 0 so (2,7) is a minimum


4 Integration

Indefinite integrals

N.B. NEVER FORGET THE ARBITRARY CONSTANT + C.

Example:

Start by getting all x ' s to the top = ∫21 x6−3 x2−x−2+6 dx

¿ 21x7

7−3x3

3− x−1

−1+6 x+c

¿3 x7−x3+ x−1+6 x+c


Definite integrals

Questions might be asked in the form of Example 1

Integrate as per indefinite integral but because we are integrating between 2 points ( 5 and 2) there is no need for a constant.

¿¿

¿ (250+50 )−(16+8 )=300−24=276

Area under a curveRather than be given a definite integral as per above you may be asked to find the area under a curve (the area between the curve and the x axis)

When x=2 y=− (2 )2+6 (2 )−8=−4+12−8=0 hence (2,0) is on the curve

When x=4 y=− (4 )2+6 ( 4 )−8=−16+24−8=0hence (4,0) is on the curve

The 2 points on the x axis that the curve cuts are at 2 and 4 so to find the area under the curve:-


∫2

4

−x2+6 x−8 dx=¿¿

= ¿(−643

+48−32)−(−83

+12−16)=−163

−(−203 )=4

3


5 Trigonometry

Knowing the values of sine, cosine, tangent for 0, 30, 45 and their multiples

Example

I recommend drawing the diagrams above but at the very least you need to write separately

sin 30=12

tan 60=√3

sin 30tan 60

=

12√3

= 12√3

=√36


3D Pythagoras

Most problems either involve finding the vertical height of a pyramid shape or the diagonal length of a 3D object. There are usually 2 steps to solving these problems, both usually involving pythagoras. It is best explained by following these examples.

Example 1

The base is a square area 64cm2 so each side is 8cm. We also know the vertical height is 12cm. We have been asked to find angle ABC. We know the height DC so if we can find length DB we can work out the angle.Length DB is half of AB and from Pythagoras we knowAB2 = 82+82 =128

AB = √128=8 √2Therefore DB = 4 √2

tanϴ= 124√2

ϴ¿ tan−1 3√2

=64.77 °


Trigonometric graphs


Example

They want the graph of minus 3cos x so the normal cos graph is flipped the other way up. There is no coefficient in front of the x and so one period is 360°. It is 3cosx though and so the amplitude is 3. The whole graph is shifted by +5 though so rather than going from +3 to -3 it will be 8 to 2.

Max is 8 and min is 2


Example 2

sin 5 x=1

5 x=sin−1(1)=90°

x=905

=18°

Next period begins at 360

5=72 so 2nd solution between 0 and 100 is at 72+18 = 90°

Example 3

3 sin 2 x=1sin 2 x=13

2 x=sin−1 13=19.47 ° x=9.7 °

Period is 360

2 so next x=1802

−9.7=80.3

Solutions are 9.7 and 80.3


6 MENSURATION

Volume of a 3D object that comes to a point at the top (e.g. cone or pyramid)

13

x area of base x vertical height

Length of an arc

Length = ϴ

360 x circumference of circle

Length of AB

¿ 45360 x24π

= 3π

These questions are sometimes given turning a 2D shape into a 3D shape

Example


The card will be folded such that point B joins with point C to form a circular base with a circumference of length BC

Length BC = 140360 x 36π ¿14 π

Circumference of the base of the cone is 14π = 2πr where r is the radius of the base

14π = 2πr so r = 7cm


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