FP1 Further Pure 1

Example:Find the modulus and argument of z = –6 + 5i.

Solution:First sketch a diagram (it is easy to get the argument wrong if you don’t).

Example:Find the square roots of 5 + 12i, in the form a + bi,  a, b ∈ ℝ.

Solution:

the marker I above refers to a² – b² = 5, which Substitute in I re-uses.

Example:Show that 3 – 2i is a root of the equation z³ – 8z² + 25z – 26 = 0. Find the other two roots.

Solution:Put z = 3 – 2i in z³ – 8z² + 25z – 26

Factorising, by inspection,

Example:The roots of the quadratic equation 2x² − 5x − 4 = 0 are α and β. Without solving the equation, find the values of:

(a) α + β   (b) αβ   (c) 1α + 1β   (d) α² + β²

Solution:

Example:The roots of the quadratic equation ax² + bx + c = 0 are α = −32 and β = 54. Find integer values for a, b and c.

Solution:

ax² + bx + c = 0 can be written as x² + bax + ca = 0

So a = 8, b = 2, c = −15.

Example:The roots of the quadratic equation 3x² + x − 6 = 0 are α and β. Form a quadratic equation with integer coefficients which has roots:

(a) α² and β²    (b) 1α and 1β

Solution:(a)  3x² + x − 6 = 0 ⇒ a = 3, b = 1, c = −6

For the sum of roots we need α² + β²

For the product of roots we need α² × β²

The new equation with roots α² and β² can be written as:

(b)  Sum:  1α + 1β = α + βαβ

Product:  1α × 1β = 1αβ = −12

The new equation with roots 1α and 1β can be written as:

Example:The roots of the quadratic equation 2x² + 5x − 4 = 0 are α and β. Form a quadratic equation with integer coefficients which has roots:

(a) α³ and β³    (b) α + βα² and α + ββ²

Solution:(a)  2x² + 5x − 4 = 0 ⇒ a = 2, b = 5, c = −4

Sum of roots:  α³ + β³ = (α + β)³ − 3αβ(α + β)

Product of roots:  α³ × β³ = (αβ)³ = (−2)³ = −8

The new equation with roots α³ and β³ can be written as:

(b)  Sum of roots:

Product of roots:  α + βα² × α + ββ² = (α + β)²(αβ)²

The new equation can be written as:

Example:Show that α = 2⋅0946 is a root of the equation

Solution:

There is a change of sign and f is continuous

Example:(i)  Show that there is a root of the equation

(ii) Find an interval of width 0⋅25 which contains the root.

Solution:(i)  f(2) = 8 – 4 – 7  =  –3,   and   f(3) = 27 – 6 – 7 = 14

(ii)  Mid-point of [2, 3]  is  x = 2⋅5, and  f(2⋅5) = 15⋅625 – 5 – 7 = 3⋅625

Mid-point of [2, 2⋅5]  is  x = 2⋅25,

Example:(i)  Show that there is a root, α, of the equation

(ii) Use linear interpolation once to find an approximate value of α. Give your answer to 3 D.P.

Solution:(i)  f(2) = 8 – 4 – 9  =  –5,   and   f(3) = 27 – 6 – 9 = 12

(ii) From (i), curve passes through (2, –5)  and  (3, 12), and we assume that the curve is a straight line between these two points.

Let the line cross the x-axis at (α, 0)

Using similar triangles

Repeating the process will improve accuracy.

Example:(i)  Show that there is a root, α, of the equation

(ii) Starting with  x₀ = 2, use the Newton-Raphson formula to find x₁, x₂ and x₃, giving your answers to  3 D.P. where appropriate.

Solution:(i)  f(2) = 8 – 4 – 5  =  –1,   and   f(3) = 27 – 6 – 5 = 16

(ii)  f(x) = x³ – 2x – 5     ⇒   f′ (x) = 3x² – 2

Example:Find the equations of the tangents to y² = 8x at the points where x = 18, and show that the tangents meet on the x-axis.

Solution:x = 18⇒y² = 8 × 18⇒y = ±12

and

To find the intersection, add the equations to give

Example:Find the equation of the normal to the parabola given by x = 3t², y = 6t.

Solution:x = 3t², y = 6t⇒dxdt = 6t,dydt = 6,

Notice that this ‘general equation’ gives the equation of the normal for any particular value of t:–  when t = –3 the normal is y + 18 = 3(x – 27) ⇔ y = 3x – 99.

Example:Find the equation of the tangent to the hyperbola xy = 36 at the point where x = 3.

Solution:x = 3⇒3y = 36⇒y = 12

Example:Find the equation of the normal to the hyperbola given by x = 3t, y = 3t.

Solution:x = 3t, y = 3t⇒dxdt = 3,dydt = −3t²

Example:M = 4253  and  MN = −1221. Find N.

First find M^–1

Using M^–1 (MN) = IN = N

Example:The image of (2, 3) under T = 4512 is given by 451223 = 238

Example:Find the transformation represented by the matrix T = 2003

Solution:Find images of i, j and 11, and show on a sketch. Make sure that you letter the points

From sketch we can see that the transformation is a two-way stretch, of factor 2 parallel to the x-axis and of factor 3 parallel to the y-axis.

Example:Find the matrix for a shear with factor 2 and invariant line the x-axis.

Solution:Each point is moved in the x-direction by a distance of (2 × its y-coordinate).

i = 10 → 10 (does not move as it is on the invariant line).
This will be the first column of the matrix 1∗0∗

j = 01 → 21. This will be the second column of the matrix ∗2∗1

Example:Find the matrix for a reflection in y = –x.

Solution:First find the images of i and j. These will be the two columns of the matrix.

This will be the first column of the matrix 0∗−1∗

This will be the second column of the matrix ∗−1∗0

Example:Find n∑r=1 r(r² − 3).

Example:Find S_n = 2² + 4² + 6² + … + (2n)².

Example:Find n+2∑r=5 r²

Example:Use mathematical induction to prove that

Solution:When n = 1, S₁ = 1² = 1 and S₁ = 16× 1(1 + 1)(2 × 1 + 1) = 16× 1 × 2 × 3 = 1

Example:A sequence, 4, 9, 19, 39, … is defined by the recurrence relation

Solution:When n = 1, u₁ = 4, and u₁ = 5 × 2^{1 − 1} − 1 = 5 − 1 = 4, ⇒ formula true for n = 1.

Example:Prove that f(n) = 5ⁿ − 4n − 1 is divisible by 16 for all positive integers, n.

Solution:When n = 1, f(1) = 5¹ − 4 − 1 = 0, which is divisible by 16, and so f(n) is divisible by 16 when n = 1.

Example:Prove that f(n) = 2^{2n + 3} + 3^{2n − 1} is divisible by 5 for all positive integers n.

Solution:When n = 1, f(1) = 2^{2 + 3} + 3^{2 − 1} = 32 + 3 = 35 = 5 × 7, and so the result is true for n = 1.

Example:If M = 2−101, prove that Mⁿ = 2ⁿ1 − 2ⁿ01 for all positive integers n.

Solution:When n = 1, M¹ = 2¹1 − 2¹01 = 2−101 = M