pure Flashcards
how to find coefficient eg of 4 when (3-x)^8
8C4 a(in this case 3)^4 b(in this case -x)^4
when is binomial theorem valid
when x lies between -1 and 1
what to do if you have a surd as a denominator in a fraction
rationalise the denominator by using a complete the square form eg times ur fraction by 1-root2/1-root2
the chain rule
define g(x) as thing u do first, f(x) as thing you do second and then f’(g(x))g’(x)
integrate
Add 1 to the power and divide by the power +1 and then substitute in values to find what you ahem to add(c) to get correct equation
product rule
use when u and v are both functions of x
u x dv/dx+ v x du/dx
quotient rule
use for a fraction with functions of x as top and bottom
u= numerator, v= denominator
dy/dx=(vu’-uv’)/v^2
remainder theorem/ find the remainder
to find the remainder when f(x) is divided by (x-b) do f(b)/ substitute b into the equation and see what the answer is
the factor theorem
the factors must be factors of the +c at the end of the equation
the discriminant
b^2-4ac- if =less than 0 there are no solutions, if =0 there is 1, if =more than 0 there are 2
stationary points
where dy/dx=0
classifying stationary points
- make a table and sub in the x values into dy/dx- if +, 0, - its a max point, if -, 0, + its a minimum point
- differentiate again- less than zero= max point, if =0 use the grid method, more than 0= minimum point
f(x) moved a upwards
f(x)+a
F(x)-a
moved a downwards
f(x) moved a leftwards
f(x+a)
F(x-a)
a rightwards
stretch by scale factor a parallel to the y axis
af(x)
f(ax)
f(x) stretched by scale factor 1/a parallel to the x axis
reflection on x axis
-f(x)
f(-x)
reflection on y axis
y=e^x reflected on y=x
y=loge^x
e^x differentiated
e^x
trapezium rule
1/2h(y1+last y coordinate)+2(all rest added together)
area under speed time graph
distance
gradient of distance time graph
speed
gradient of speed time graph
acceleration
area under acceleration time graph
change in velocity
prove k^3-k is divisible by 6
=k(k^2-1)
=k(k-1)(k+1)
=(k-1)k(k+1) 3 consecutive integers
at least one is even so will have a multiple of 2 and as there are 3 consecutive numbers one must be a multiple of 3, thus it must be divisible by both 3 and 2 and so 6
prove root 2 is irrational
assume it is rational and therefore can be represented as a/b where a and b are integers and a/b is in its simplest form
so 2=a^2/b^2
2b^2=2a^2- a must be even as its square is divisible by 2 so a=2k
hence b^2=k^2 so b must also be even and therefore a/b is not in its simplest form as both can be divided by 2