Pure

Question 1

How to find a perpendicular vector from two vectors

Answer 1

(vector 1) . (x y z) = 0
(vector 2) . (x y z) = 0
Dot product

Question 2

Roots (alpha, beta, gamma) of polynomials - x^2 term

Answer 2

alpha + beta = - b/a

alpha x beta = + c/a

Question 3

How to find the modulus and argument of complex numbers (+ different quadrants)

Answer 3

Modulus = |z|
Argument = arctan(y/x)

Quad. 1: arg(z) = arctan(y/x)
Quad 2: arg(z) = pi - arctan(y/x)
Quad 3: arg(z) = arctan(y/x) - pi
Quad 4: arg(z) = - arctan(y/x)

Question 4

Solving simultaneous equations using vectors

E.g. intersection of 2 lines

Answer 4

ax + by = c
dx + ey = f
(a b) ( x ) = ( c )
(d e) ( y ) ( f )

Question 5

Meaning of matrix determinant:
2x2 and 3x3
Negative

Answer 5

2x2: determinant represents the area scale factor of the transformation
3x3: the determinant represents the volume scale factor of the transformation
Negative: A negative determinant means the orientation of the image is reversed

Question 6

Proof by induction steps

Answer 6

1. Prove for n=1 (or smallest number)
2. Assume true for n=k. Write it out. This is target expression
3. Prove for n=k+1. Will need to use the n=k formula depending on the question
4. Conclusion

Question 7

Proof by induction conclusion

Answer 7

If true for n=k, then it is true for n=k+1.

Since it is true for n=1, it is true for all integers where n>= 1

Question 8

Use of differential equations

Answer 8

To model situations involving rates of change

Question 9

Method of separation of variables (differential equations)

Answer 9

dy/dx = f(x) g(y)
(1/g(y)) (dy/dx) = f(x)
integral [1/g(y) dy/dx dx] = integral [f(x) dx]
integral [1/g(y) dy] = integral [f(x) dx]

Question 10

Polar coordinates: polar => cartesian conversion

Answer 10

x = rcos(theta)
y = rsin(theta)

Question 11

Polar coordinates: cartesian => polar conversion

Answer 11

r = sq root (x^2 + y^2)
tan theta = y/x
Draw diagram as theta may be negative

Question 12

How to find roots of complex polynomials
e.g. f(x) = z^4 + az^3 + bz^2 + cz + d
when you are given 2 roots

Answer 12

If x+yi is a root, then x-yi is also a root

Given: x+yi and a+bi

f(x) = (z - (x+yi))(z - (x-yi))(z - (a+bi))(z - (a-bi))
Expand out

Question 13

Partial fractions: rhs of equation?

1/(x+4)(x+1)

1/(2x^2 +3)(x+1)

1/((x+2)^2) (x+3)

1/((x^2 + 2)^2) (x+3)

Answer 13

A/(x+4) + B/(x+1)

(Ax+B)/(2x^2 +3) + C/(x+1)

A/((X+2)^2) + B/(x+2) + C/(x+3)

(Ax+B)/((x^2 + 2)^2) + (Cx+D)/(x^2 + 2) + E/(x+3)