Edexcel A-Level Mathematics AS Unit P2 Pure Mathematics 2 Questions

Practise P2 Pure methods across proof, algebra, coordinate geometry, sequences, binomial expansion, calculus and functions.

Covered syllabus points

P2.1 - Proof P2.2 - Algebra and functions P2.3 - Coordinate geometry in the (x, y) plane P2.4 - Sequences and series P2.5 - Exponentials and logarithms P2.6 - Trigonometry P2.7 - Differentiation P2.8 - Integration

[Maximum number: 6]

In this question you must show detailed reasoning.

(a)

Given that x and y are positive numbers such that

(x-y)^{3}>x^{3}-y^{3}

prove that

y>x

[ 4 ]

$(x-y)^3=x^3-3x^2y+3xy^2-y^3$ $(x-y)^3>x^3-y^3\Rightarrow -3x^2y+3xy^2>0$ 3xy(y-x)>0
As x and y are positive, (y-x)>0, so y>x.
A1*

M1: Attempts expansion of $(x-y)^3$ .
A1: Correct simplified inequality with cubed terms cancelled.
dM1: Takes out or divides by a common factor of kxy.
A1*: Correct working and reasoning using positivity of x and y.

(b)

Using a counter example, show that the result in part (a) is not true for all real numbers.

[ 2 ]

M1: Chooses a suitable counterexample, for example x=3, y=-1. Any positive x and negative y will work;
substitution is not needed for this mark.
A1: Shows that the result is not true for their values. For example, $(3-(-1))^3=64$ and $3^3-(-1)^3=28$ , so 64>28 but -1<3.

[Maximum number: 2]

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

\mathrm{f}(z)=4 z^{3}+p z^{2}-24 z+108

where p is a constant.
Given that -3 is a root of the equation f(z)=0

determine the value of p

$\mathrm{f}(-3)=0 \Rightarrow 4(-3)^3+p(-3)^2-24(-3)+108=0 \Rightarrow p=\ldots$ p=-8
(2)

[Maximum number: 3]

Figure 2

Figure 2 shows a sketch of
- the circle C with centre X(4,-3)
- the line l with equation $y=\frac{5}{2} x-\frac{55}{2}$

Given that l is the tangent to C at the point N,

Hence find

an equation for C.

M1: Attempts radius or radius ${ }^{2}$ . Condone slips on lhs, for example r or $d=(\text { " } 9 \text { " }-4)^{2}+(\text { " }-5 \text { " }+3)^{2}$ You must see an attempt to take the difference for both components before each are squared.

M1: Correct form of equation E.g. $(x-4)^{2}+(y+3)^{2}=" r "{ }^{2}$
Look for $(x-4)^{2}+(y+3)^{2}=(\text { " } 9 \text { " }-4)^{2}+(\text { " }-5 \text { " }+3)^{2}$
Condone for this mark a slip on one of the components when finding their $r^{2}$
A1: $(x-4)^{2}+(y+3)^{2}=29$ or exact equivalent. The constant terms must be collected and not 25+4 ISW after a correct answer.

Alt (b)(ii)
Note that an alternative version is $x^{2}+y^{2}-8 x+6 y-4=0$ . It is possible to proceed using this alternative version of the formula for a circle

M1: States equation of circle is $x^{2}+y^{2}-8 x+6 y+c=0$
M1: Substitutes their N=(9,-5) into $x^{2}+y^{2}-8 x+6 y+c=0$ o.e. and finds " c "

A1: $x^{2}+y^{2}-8 x+6 y-4=0$ o.e.

Figure for Answer to Question 3 — Edexcel A-Level Mathematics AS

[Maximum number: 5]

A sequence is defined by

\begin{aligned} u_{1} & =6 \\ u_{n+1} & =k u_{n}+3 \end{aligned}

where k is a positive constant.

(a)

Find, in terms of k, an expression for $u_{3}$

Given that $\sum_{n=1}^{3} u_{n}=117$

[ 2 ]

Substitute $u_2=6k+3$ into $u_3=ku_2+3$ :
$u_3=k(6k+3)+3=6k^2+3k+3.$

(b)

find the value of k.

[ 3 ]

M1: Sets their $u_{1}+u_{2}+u_{3}=117$ to produce an equation in just k.
dM1: Solves a 3TQ by any valid method to find at least one value for k.
A1: $\quad k=\frac{7}{2}$ ONLY

Figure for Answer to Question (b) — Edexcel A-Level Mathematics AS

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Covered syllabus points

Question 1

Question (a)

Question (b)

Question 2

Question 3

Question 4

Question (a)

Question (b)