Edexcel FP3 January 2011 exam answers review
- Express 2/(2r + 1)(2r + 3) in partial fractions.(2)
- Using your answer to (a), find, in terms of n,
[IMAGE]Give your answer as a single fraction in its simplest form.
(3)
Find
- |z|,(1)1
- arg(z), in terms of π.
w = 2(cosπ/4 + isinπ/4)
(2)
- [IMAGE](1)
- arg (w/z, in terms of π)(2)
Find
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)
-
Given that
z = r (cosθ + isinθ), r∈ℜprove, by induction, that zn = rn (cosnθ + i sin nθ), n ∈ Ζ+
w = 3(cos3π/4 + isin3π/4)(5)
- Find the exact value of w5, giving your answer in the form a + ib, where a, b∈ℜ(2)
- Find the general solution of the differential equation
xdy/dx + 2y = 4x²
(5)
- Find the particular solution for which y = 5 at x = 1, giving your answer in the form y = f(x).(2)
-
- Find the exact values of the coordinates of the turning points of the curve with equation y = f(x), making your method clear.
- Sketch the curve with equation y = f(x), showing the coordinates of the turning
points.
(5)
- Use algebra to find the exact solutions of the equation
|2x² + 6x – 5| = 5 – 2x
(6)
- 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)
- Find the set of values of x for which
|2x² + 6x – 5| > 5 – 2x
(3)
- Show that the transformation y = xv transforms the equation
4x²d²y/dx² – 8xdy/dx + (8 + 4x²)y = x² (I)
into the equation
4d²v/dx² + 4v = x (II)(6)
- Solve the differential equation (II) to find v as a function of x.(6)
- 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).
- Show that cosφ = 1/√3(6)
- Find the exact value of R.(2)
- Use calculus to show that the exact area of S is
1/36a² (9arccos(1/√3) + √2)
(7)
The region S, shown shaded in Figure 1, is bounded by C and l.