Pearson Edexcel IAL Mathematics Unit FP2: Further Pure Mathematics 2 Question Bank

Practise FP2 topics including inequalities, complex-number transformations, differential equations, series methods and polar-coordinate work.

Covered syllabus points

FP2.1 - Inequalities FP2.2 - Series FP2.3 - Further complex numbers FP2.4 - First order differential equations FP2.5 - Second order differential equations FP2.6 - Maclaurin and Taylor series FP2.7 - Polar coordinates

[Maximum number: 6]

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

Use algebra to determine the values of x for which

\frac{x+1}{(x-3)(x+2)}\le 1-\frac{2}{x-3}

Critical values from the denominator are x=-2 and x=3.

Equating the two sides:

\frac{x+1}{(x-3)(x+2)}=1-\frac{2}{x-3}

x+1=x^2-x-6-2(x+2)

x^2-4x-11=0.

x=\frac{4\pm\sqrt{16+44}}{2}=2\pm\sqrt{15}

Therefore

x<-2,\qquad 2-\sqrt{15}\le x<3,\qquad x\ge 2+\sqrt{15}

[Maximum number: 4]

Hence, using the method of differences, show that for all integer values of n,

\sum_{r=1}^{n} \frac{1}{(2 r-1)(2 r+1)(2 r+3)}=\frac{n(n+2)}{a(2 n+b)(2 n+c)}

where a, b and c are integers to be determined.

$\frac18\sum_{r=1}^{n}\left(\frac{1}{2r-1}-\frac{2}{2r+1}+\frac{1}{2r+3}\right)$ $=\frac18\left(1-\frac13-\frac{1}{2n+1}+\frac{1}{2n+3}\right)$ $=\frac18\left(\frac{2(2n+1)(2n+3)-3(2n+3)+3(2n+1)}{3(2n+1)(2n+3)}\right)$ $=\frac{n(n+2)}{3(2n+1)(2n+3)}$
M1
A1
dM1
A1

[Maximum number: 10]

(a)

Show that the transformation v=y-2 x transforms the differential equation

\frac{\mathrm{d} y}{\mathrm{~d} x}+2 y x(y-4 x)=2-8 x^{3}

into the differential equation

\frac{\mathrm{d} v}{\mathrm{~d} x}=-2 x v^{2}

[ 4 ]

$\frac{\mathrm{d} v}{\mathrm{~d} x}=\frac{\mathrm{d} y}{\mathrm{~d} x}-2$

B1

$\frac{\mathrm{d} y}{\mathrm{~d} x}+2 y x(y-4 x)=2-8 x^{3} \rightarrow \frac{\mathrm{~d} v}{\mathrm{~d} x}+2+2(v+2 x) x(v+2 x-4 x)=2-8 x^{3}$ }

$\rightarrow \frac{\mathrm{d} v}{\mathrm{~d} x}+2+2 x\left(v^{2}-4 x^{2}\right)=2-8 x^{3}$

М1

$\rightarrow \frac{\mathrm{d} v}{\mathrm{~d} x}=-2 x v^{2} *$

A1*

(4)

(b)

Hence obtain the general solution of the differential equation (I).

[ 1 ]

$y=2 x+\frac{1}{x^{2}+c}$

B1ft

(1)

(c)

Sketch the solution curve that passes through the point (-1,-1).

On your sketch show clearly the equation of any horizontal or vertical asymptotes.

You do not need to find the coordinates of any intercepts with the coordinate axes or the coordinates of any stationary points.

Table for Question (c) — Edexcel A-Level Mathematics A2

[ 5 ]

-1=2(-1)+\frac{1}{1+c}\Rightarrow c=0

y=2x+\frac{1}{x^2}

Sketch attempted for the correct curve.

Figure for Answer to Question (c) — Edexcel A-Level Mathematics A2

Two branches, both tending to infinity as x tends to ±infinity, with a minimum in the first quadrant.

B1ft

x=0\text{ is the vertical asymptote}

[Maximum number: 7]

(a)

Hence show that the transformation $t=\ln x$ , where x>0, transforms the differential equation

x^2\frac{d^2y}{dx^2}-2y=1+4\ln x-2(\ln x)^2

into the differential equation

\frac{d^2y}{dt^2}-\frac{dy}{dt}-2y=1+4t-2t^2

[ 1 ]

x^2\frac{d^2y}{dx^2}-2y=e^{2t}e^{-2t}\left(\frac{d^2y}{dt^2}-\frac{dy}{dt}\right)-2y

= =\frac{d^2y}{dt^2}-\frac{dy}{dt}-2y.

Also $\ln x=t$ , so

\frac{d^2y}{dt^2}-\frac{dy}{dt}-2y=1+4t-2t^2

(b)

Solve differential equation (II) to determine y in terms of t.

[ 5 ]

For the complementary function,

m^2-m-2=0

so m=2,-1 and

y=Ae^{2t}+Be^{-t}

For a particular integral, let $y=at^2+bt+c$ . Then

\frac{dy}{dt}=2at+b,\qquad \frac{d^2y}{dt^2}=2a

Substitution gives

2a-2at-b-2at^2-2bt-2c=1+4t-2t^2

Comparing coefficients:

a=1,\qquad b=-3,\qquad c=2

Therefore

y=Ae^{2t}+Be^{-t}+t^2-3t+2

(c)

Hence determine the general solution of differential equation (I).

[ 1 ]

Since $t=\ln x$ ,

e^{2t}=x^2,\qquad e^{-t}=x^{-1}

Therefore the general solution is

y=Ax^2+Bx^{-1}+(\ln x)^2-3\ln x+2

All question bank results loaded

Covered syllabus points

Unit FP2: Further Pure Mathematics 2 question 1

Unit FP2: Further Pure Mathematics 2 question 2

Unit FP2: Further Pure Mathematics 2 question 3

Question (a)

Question (b)

Question (c)

Unit FP2: Further Pure Mathematics 2 question 4

Question (a)

Question (b)

Question (c)