Edexcel GCE Further Pure Mathematics FP2 June 2013 exam answers review

Find

1. |z|,(1)(1)
2. arg(z), in terms of π.(2)(2)
w = 2(cosπ/4 + isinπ/4)

Find

3. |w/z|

   (1)

4. arg(w/z), in terms of π(2)

Given that y = ½ and dy/dx = 1/8 at x=0,

find a series expansion for y in terms of x, up to and including the term in x3.

(5)

1. Given that

   z = r (cosθ + i sinθ),

   r ε ℜ

   prove, by induction, that zⁿ = rⁿ (cos nθ + isin nθ), nεΖ⁺(5)

w = 3(cos3π/4 + isin3π/4)
2. Find the exact value of w⁵, giving your answer in the form a + ib, where a, b ε ℜ

1. Find the general solution of the differential equation

   xdy/dx + 2y = 4x²

   (5)

2. Find the particular solution for which y = 5 at x = 1, giving your answer in the form y = f(x).(2)
3. 1. Find the exact values of the coordinates of the turning points of the curve with equation y = f(x), making your method clear.
   2. Sketch the curve with equation y = f(x), showing the coordinates of the turning points.

   (5)

1. Use algebra to find the exact solutions of the equation

   |2x² + 6x – 5| = 5 – 2x

   (6)

2. On the same diagram, sketch the curve with equation y = |2x² + 6x – 5| and the line with equation y = 5 – 2x, showing the x-coordinates of the points where the line crosses the curve.(3)

3. Find the set of values of x for which

   |2x² + 6x – 5| > 5 – 2x

   (3)

1. Show that the transformation y = xv transforms the equation
   4x²d²y/dx² – 8xdy/dx + (8 + 4x²)y = x^<4/sup>

   (II)

   into the equation

   4d²x/dx² + 4v = x

   (6)

2. Solve the differential equation (II) to find v as a function of x (6)
3. Hence state the general solution of the differential equation (I).(1)

Figure 1 shows a curve C with polar equation r = asin2θ, 0≤θ≤π/2, and a half-line l.

The half-line l meets C at the pole O and at the point P. The tangent to C at P is parallel to the initial line. The polar coordinates of P are (R, φ).

1. Show that cosφ = 1/√3(6)
2. Find the exact value of R.(2)

The region S, shown shaded in Figure 1, is bounded by C and l.

3. Use calculus to show that the exact area of S is

   1/36a²(9arccos(1/√3) + √2)

   (7)