Part-worked answers and hints for FP1 June 2014 paper

Tips-for-students-exam

QUESTION 1. This just calls for remembering how to divide complex numbers, and how to find the modulus |z|, and keeping your head when it looks complicated.

z1/z2 = (p + 2i) / (1 – 2i)

and that

= [(p+2i)(1+2i)] / (1−2i)(1+2i)

which

= [(p−4)+2(1+p)i] / 5

then you just have to simplify that a tiny bit.

Remember |z1/z2] = |z1| / |z2|

so |z1/z2| = 13 means |z1| / |z2| = 13

Since |z2| = √ (5)

(because in general |a+bi| = √(a2+b2); or, alternatively, |a+bi| is the “r” you get when you calculate Pol(a,b) on your calculator)

|z1| = 13. √ (5)

so √ (p2+4) = 13. √ (5)

Square each side, and you have a quadratic in p.


QUESTION 2. You can get the Newton-Raphson formula from the formula sheet.

f(x) = x3 − 5/(2x3/2) + 2x − 3

The hard bit about differentiating here is sorting out the 5/(2x3/2) term

But that is just (5/2) x−3/2

When you differentiate that, you get (5/2)(−3/2) x−5/2


QUESTION 3. The third root must be the conjugate, 1 + 5i

Then −p is the sum of the roots (and you know all three roots)

−q is the product of the roots (and you know all three roots)


QUESTION 4. The first part of question 4 is just standard matrix multiplication. Look it up and practise it a bit if you don’t remember it.

https://mathsmartinthomas.wordpress.com/2015/12/13/multiplying-matrices-sing-along-video-clip/

AB and BA must be different because they are different sizes.

There are three rows in A, each of length two.

There are three columns in B, each of length two.

To form the elements of the matrix AB, each of the three rows in A multiplies each of the three columns in B.

For example, the 1st−row, 1st−column element of AB is got by multiplying the first row of A

1 2

by the 1st column of B

2

1

to get 1×2 + 2×1, or 4.

So AB has three rows and three columns.

But when you work out the elements of BA, for each you multiply one of the two rows of B (each row has length 3) by one of the two columns of A (each column has length 3).

So BA has two rows and two columns.

For Q.4 part (ii), find the determinant of C

Then the inverse of C is (1/determinant) times a matrix in which you swap the elements of C on the leading diagonal (top left to bottom right) and minus the other two elements (top right and bottom left).


QUESTION 5. The first step is to expand

Σ [(2r−1)2] = Σ [4r2 − 4r +1]

and then “take the sigma inside the bracket”

so it = 4 Σ(r2) − 4 Σ(r) + Σ 1

Σ(r2) is in the formula book.

You have to remember Σ(r), but it’s (1/2)n(n+1)

Σ 1 = 1+1+1+1+…+1 (n times) = n

You put in the formulas for Σ(r2) and Σ(r)

then take out the best common factor, which in this case is (1/3)n

collect terms inside the big bracket

and simplify.

(1/3)n goes into (4/6)n

2 times

so into

4/6n(n+1)(2n+1) it goes

2(n+1)(2n+1) times

(1/3)n goes into 2n

6 times

so into

2n(n+1)

it goes

6(n+1) times

And (1/3)n goes into n

3 times

So you end with

(1/3)n [ 2(n+1)(2n+1) − 6(n+1) + 3]

Simplify inside the big bracket, and you’re done.

The answer you got from part (a) tells us for every number n that the Σ up to n terms is (1/3)n (4n2−1) [α]

Now in part (b) we want the Σ from the (2n+1)th term to the 4n’th

The Σ from the first to the 4n’th is (1/3)(4n) [4×(4n)2 − 1] [β. I got this by substituting 4n for n in formula α]

And similarly the Σ from the first to the 2n’th is (1/3)(2n) [4×(2n)2 − 1] [γ]

What about the Σ from the (2n+1)th term to the 4n’th?

Just as the total marks you get for questions 5 to 8 equals your total for questions 1 to 8, minus your total for questions 1 to 4

the Σ from the (2n+1)th term to the 4n’th

equals the Σ from the first to the 4n’th

minus the Σ from the first to the 2n’th

So you can get the answer by subtracting the formula I’ve marked [γ] (for Σ from the first term to the 2n’th] from the formula I’ve marked [β] (for Σ from the first term to the 4n’th).

You’ll have

(1/3)(4n) [4×(4n)2 − 1] − (1/3)(2n) [4×(2n)2 − 1]

The best common factor is (2/3)n

so the expression equals

(2/3)n [ 2[4×(4n)2 − 1] − [4×(2n)2 − 1]]

which is

(2/3)n [128n2 − 2 − 16n2 + 1]

or

(2/3n) [112n2 − 1]


QUESTION 6. The slope of the tangent is dy/dx, which is (dy/dt) / (dx/dt)

y = ct−1 and x = ct, so

dy/dt = −ct−2 and dx/dt=c

dy/dx = −t−2 = slope of tangent = m

and then you find the equation of the tangent by using

y = mx + k
putting m = −t−2 and plugging in x=ct, y=ct−1 (because the tangent must go through that point) to find k

or (alternatively: whichever floats your boat) by using the formula for a line of slope m through a point (x1,y1)

y−y1 = m (x−x1)

Where does the normal meet the x−axis? Find that by putting y=0 in the equation of the normal

(t3)x = ct4−c

so x = ct−1−ct−3

Where does the tangent meet the x-axis? Find that by putting y=0 in the equation of the tangent

Then the area of the triangle APB is half base × height

And base = difference in the x−values for A and B

Height = y-coordinate of P


QUESTION 7. For each of (a) and (b) in Q.7 (i) (a) and (b), draw a diagram showing the real and imaginary axes

and the points (1,0) [i.e. the complex number 1, real part 1, imaginary part 0] and (0,1) [i.e. the complex number i, real part 0, imaginary part 1]

and where those points would go to

  • in first diagram, if they are reflected in the line y=−x (which goes across the diagram slantways, from top left to bottom right, through the origin)
  • in second diagram, if they are rotated (around the origin) 135 degrees anticlockwise

For (a), the matrix you want has first column = the point where (1,0) goes to when reflected (it’s (0,−1))

second column = the point where (0,1) goes to when reflected

For (b), the matrix has first column = the point where (1,0) goes to when rotated (it’s (−1/√2, 1/√2))

second column = the point where (0,1) goes to when rotated

For (c) in Q.7 (i) (a) and (b), multiply your answers for (a) and (b)

but remember: it’s rotation matrix × reflection matrix

because the reflection is done first

not reflection matrix × rotation matrix

Area of T is half base times height

which is half (11−3) × k

which is 4k

Area of T’ = area of T × (determinant of matrix)

so (determinant of matrix) × 4k = 364

and you can work out k from that


QUESTION 8. This is actually easier than most parabola/hyperbola questions, because you don’t have to get the equation of a tangent or a normal.

You just need the equation of the line joining the two points P and Q

and you can get that from

(y−yP) = [(yP − yQ)/(xP −xQ)] (x−xQ)

or by finding slope of the line m = (yP − yQ)/(xP −xQ)

and then substituting the (x,y) from either P or Q in the equation

y = mx + b

to find b

For the line l2

its slope is −1/m (if m is the slope you’ve just found)

so use

y = (−1/m)x + b

but this time the x and y you put in to find b are the (x,y) for the focus

and you find the focus from the formula sheet

the parabola is y2 = 16x

compared to the standard equation y2 = 4ax

so a=4

the formula sheet says the focus is (a,0)

so it’s (4,0)

Now you have the equation for the line l2

The directrix (from the formula sheet) is x=−a

so in this case x=−4

Plug x=−4 in your equation for the line l2

Work out y

and you’ve finished the question.


QUESTION 9. Step 2 is

6 | 8n − 2n

so 6 | 8 × 8n − 8 × 2n

(because if 6 divides into a number, it also divides into 8 times that number)

so 6 | 8 × 8n − 2 × 2n − 6 × 2n

so 6 | 8 × 8n − 2 × 2n

in other words 6 | 8n+1 − 2n+1

which is what you want.