# Pure Core 3

A key point Summary of what's involved in each part of core three

- Created by: Ellen Shercliff
- Created on: 25-05-11 17:55

## Functions 1/2

- A function is a one-one or many-one mapping
- The set of numbers for which a function is defined is called the domain
- A function of f consists of:defining a rule, its domain
- The set of values a function takes is called the range
- When the domain of f is a continous interval, the range can be found by considering the graph of y=f(x). The range consists of the values y can take and is written as an inequality using f(x)

## Functions 2/2

- the composite function fg means f(g(x))
- a function only has an inverse where it is one to one, and is obtained by reflecting f(x) in y=x to get f^-1(x)
- A reverse flow diagram can be used to obtain the inverse of a function when x occurs only once in f(x).
- The inverse of f can be found my writing y=f(x), rearranging the equations so you get x= something and substituting x and y

## Tranforming graphs 1/2

- a translation of [a]

[b] translates the graph of y= f(x) to y=f(x-a) +b

- The graph of y=f(x) is transformed into the graph of y=-f(x) by a reflection in the x axis
- The graph of y=f(x) is transformed into the graph of y=f(-x) by reflection in the y axis

## Transforming graphs 2/2

- The graph of y=f(X) is transformed onto the graph of y=df(x) by a stretch scale factor d in the y-direction
- the graph of y=f(x) is transformed onto y=f(x/c) by a stretch of scale factor c in the x-direction
- The modulus function |x| is defined by:

>|x|=x when x>/= 0

>|x|=-x when x<0

## Inverse Trigonometric equations 1/2

- Sin^-1x has a domain -1 to 1 and a range -pi/2 to pi/2
- Cos^-1x has a domain -1 to 1 and a range 0 to pi
- Tan^-1x has a domain all real number and a range -pi/2 to pi/2

## Inverse Trigonometric equations 2/2

- secx=1/cosx
- cosecx=1/sinx
- cotx= 1/tanx or cosx/sinx

- cos^2X + sin^2X=1
- 1+tan^2X=sec^2X
- cot^2X+1=cosec^2X

## e 1/2

- The number e is irrational and its value is approximately 2.718
- The curve with the equation y=e^x has derivative dy/dx= e^x
- d/dx e^kx=ke^kx
- ∫ekx= 1/ke^kx (+c)

## e 2/2

- The natural logarithm ln x is definied for x>0 A useful equivelent statement is:

x=e^x <==> y=lnx - The derivative of ln x is 1/x
- ∫1/xdx= lnx +c

## Further differentiation and the chain rule 1/2

- y=sinx dy/dx= cos x
- y=cosx dy/dx=-sinx
- the chain rule can be used to differentiate composite functions:

dy/dx= dy/du X du/dx

## Further differentiation and the chain rule 2/2

- The chain rule can be extended to:

dy/dx= dy/dv X dv/du X du/dx

- dy/dx= 1/dx/dy and dx/dy=1/dy/dx

## Differentiation using the product and quotient rul

- where y=uv,

dy/dx=udv/dx+vdu/dx - where y=u/v,

dy/dx=(vdu/dx-udv/dx)/v^2

## Differentiation using the product and quotient rul

- y=tanx dy/dx= sec^2X
- y=cotx dy/dx= -cosecx
- y=secx dy/dx= secxtanx
- y=cosecx dy/dx= -cosecxcotx

## Numerical solution of equations and iterative meth

- If the graph of y=f(x) is continuous over the interval a to b and f(a) and f(b) have different signs, then at least one root of the equations f(x)=0 must lie in the interval a to b
- Cobweb and/or staircase diagrams can be drawn to illustrate whether convergence takes place or not for iterations of form x(n+1)=g(xn)

## Numerical solution of equations and iterative meth

An iteration of the form x(n+1)=g(xn) converges when the gradient of y=g(x) at the point of the intersection with the line y=x satisfies the condition |g'(x)|<1, provided a suitable value for x1 is chosen.

## integration by inspection and substitution 1/2

- for constants a,b,n where n =/=-1

∫(ax+b)^(n)dx=1/a(n+1) (ax+b)^(n+1) (+c)

- ∫e^(ax+b)dx= 1/ae^(ax+b) (+c)
- for x in radians,

∫cosxdx= sinx (+c)

∫sinxdx= -cosx (+c)

∫sec^2xdx= tanx (+c)

- for constants a,b and x

∫cos(ax+b)dx= 1/a sinx(ax+b) (+c)

∫sin(ax+b)dx= -1/a cos(ax+b) (+c)

∫sec^2(ax+b)=1/a tan(ax+b) (+c)

## integration by inspection and substitution 2/2

- for constants a and x

∫cosecax cotax dx =-1/a cosecax (+c)

∫secax tanax dx=1/a secax (+c)

∫cosec^2ax dx=-1/a cotax (+c)

∫tanax dx= 1/a ln |secax| (+c)

∫cotax dx= 1/a ln|sinax| (+c)

∫secax dx= 1/a ln|secax+tanax| (+c)

∫cosecax dx=-1/a ln|cosecax + cotax| (+c)

- ∫f'(x)/f(x)dx= ln|f(x)| (+c)

## Integration by parts and standard integrals 1/2

- ∫u dv/dx dx= uv- ∫v du/dx dx is known as the integration by parts formula and is given in the formula book.
- It is used to solve:

∫x^n sinmx dx

∫x^n cosmx dx

∫x^n e^mx dx

∫x^n ln mx dx

## Integration by parts and standard integrals 2/2

∫1/(a^2+x^2)dx= 1/atan^-1(x/a) (+c)

∫1/√(a^2+x^2)dx=

sin^-1 (x/a) (+c)

## Volume of revolution and numerical integration 1/2

- When a region bound by the curve y=f(X), the x axis, the lines x=a and x=b is rotated through 2pi radians about the x-axis the volume, v of the solid generated is called the volume of revolution and is given by:

v=∫(between a and b) piy^2dx= ∫(between a and b)pi[f(x)]^2dx

- When a region bound by the curve x=g(y), the y axis, the lines y=c and y=d is rotated through 2pi radians about the y-axis the volume, v of the solid generated is called the volume of revolution and is given by:

v=∫(between c and d) pix^2dx= ∫(between c and d)pi[g(y)]^2dx

## Volume of revolution and numerical integration 2/2

- In all the numerical methods of integration an improvement on the estimate can be obtained by increasing the number if strips
- The mid-ordinate rule for n strips is:

∫(between a and b) ydx =h(y1/2+y3/2+...+y(n-3/2)+y(n-1/2))

where h=b-a/n

- Simpsons rule for n strips and (n+1) ordinates where n is even is:

∫(between a and b) ydx =1/3h[(y0+yn)+4(y1+y3+...+y(n-1))+2(y2+y4+...+y(n-2)]

where h=b-a/n

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