Question 7 - A-Level Maths

Edexcel P2 2022 June — Question 7 7 marks

Exam Board	Edexcel
Module	P2 (Pure Mathematics 2)
Year	2022
Session	June
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Factor & Remainder Theorem
Type	Single polynomial, two remainder/factor conditions
Difficulty	Standard +0.3 This is a straightforward application of the factor theorem and integration. Students substitute x=-2 to get one equation, evaluate the definite integral to get another, then solve the simultaneous equations. While it requires multiple steps and combining two different techniques, both are standard P2 methods with no conceptual difficulty or novel insight required.
Spec	1.02j Manipulate polynomials: expanding, factorising, division, factor theorem 1.08d Evaluate definite integrals: between limits

$$f(x) = Ax^3 + 6x^2 - 4x + B$$ where $A$ and $B$ are constants. Given that

$(x + 2)$ is a factor of $f(x)$
$\int_{-3}^{5} f(x)dx = 176$

Find the value of $A$ and the value of $B$. [7]

Show mark scheme Show mark scheme source

Question 7:

Answer	Marks	Guidance
7	(x+2) a factor f(−2)=0−8A+24+8+B=0	M1A1

 A

f(x)dx= x4 +2x3−2x2 +Bx

Answer	Marks
4	M1

A1

5 5

  A 

f ( x ) d x = 1 7 6  x 4 + 2 x 3 − 2 x 2 + B x = 1 7 6

4 3 3

 A   A 

 54 +2(53)−2(52)+5B − 34 +2(33)−2(32)+3B =176

 

Answer	Marks
 4   4 	dM1

8 A − B = 3 2 

 A = ..., B = ...

Answer	Marks
1 3 6 A + 2 B = 1 2	dM1

1 A = , B = − 2 8

Answer	Marks
2	A1

(7)

(7 marks)

Notes:

M1: Uses the factor theorem or long division (see below) to find one linear equation relating A and

B. The “= 0” must be seen or implied by later work.

A1: Correct equation. Award once a correct equation is seen unsimplified or simplified.

So allow e.g. A ( − 2 ) 3 + 6 ( − 2 ) 2 − 4 ( − 2 ) + B = 0 but not e.g. A  − 2 3 + 6  − 2 2 − 4  − 2 + B = 0

unless the indices are processed correctly subsequently to e.g. − 8 A + 2 4 + 8 + B = 0

The “= 0” must be seen or implied by later work.

M1: Attempts to integrate the given expression, look for at least one power increased by 1.

This can also be evidenced by B → Bx

6x2+1

A1: Correct integration. Allow unsimplified e.g. for 2x3

2+1

  A

f(x)dx= x4 +2x3−2x2 +Bx

Condone poor notation e.g. and ignore any “+ c”

4 dM1: Applies limits 5 and 3 either way round and subtracts and equates to 176 to obtain an equation

in A and B. Depends on previous M.

Condone poor use of brackets when subtracting e.g. allow

A A

54 +2(53)−2(52)+5B− 34 +2(33)−2(32)+3B=176

4 4

dM1: Solves their two equations in A and B to find values for both constants.

Depends on all previous M marks but allow this mark if the only error was to attempt

f (2) = 0 at the beginning.

A1: Correct values.

Note that some candidates eliminate A or B by rearranging the first equation and substituting into the integral.

This essentially follows the main scheme:

Answer	Marks
f ( − 2 ) = 0  − 8 A + 2 4 + 8 + B = 0  B = 8 A − 3 2	M1A1

 A

f ( x ) d x = x 4 + 2 x 3 − 2 x 2 + 8 A x − 3 2 x

Answer	Marks
4	M1A1

 A   A 

 5 4 + 2 ( 5 3 ) − 2 ( 5 2 ) + 4 0 A − 1 6 0 − 3 4 + 2 ( 3 3 ) − 2 ( 3 2 ) + 2 4 A − 9 6 = 1 7 6

Answer	Marks
4 4	dM1
 A = . . . , B = . . .	dM1

1 A = , B = − 2 8

Answer	Marks
2	A1

Long Division for reference:

Score the first M1 for a complete method to obtain a remainder in terms of A and B that is set = 0

and A1 for a correct equation e.g. B – 8A + 32 = 0

Answer	Marks	Guidance
Question	Scheme	Marks

Question 7:
7 | (x+2) a factor f(−2)=0−8A+24+8+B=0 | M1A1
 A
f(x)dx= x4 +2x3−2x2 +Bx
4 | M1
A1
5 5
  A 
f ( x ) d x = 1 7 6  x 4 + 2 x 3 − 2 x 2 + B x = 1 7 6
4
3 3
 A   A 
 54 +2(53)−2(52)+5B − 34 +2(33)−2(32)+3B =176
 
 4   4  | dM1
8 A − B = 3 2 
 A = ..., B = ...
1 3 6 A + 2 B = 1 2 | dM1
1
A = , B = − 2 8
2 | A1
(7)
(7 marks)
Notes:
M1: Uses the factor theorem or long division (see below) to find one linear equation relating A and
B. The “= 0” must be seen or implied by later work.
A1: Correct equation. Award once a correct equation is seen unsimplified or simplified.
So allow e.g. A ( − 2 ) 3 + 6 ( − 2 ) 2 − 4 ( − 2 ) + B = 0 but not e.g. A  − 2 3 + 6  − 2 2 − 4  − 2 + B = 0
unless the indices are processed correctly subsequently to e.g. − 8 A + 2 4 + 8 + B = 0
The “= 0” must be seen or implied by later work.
M1: Attempts to integrate the given expression, look for at least one power increased by 1.
This can also be evidenced by B → Bx
6x2+1
A1: Correct integration. Allow unsimplified e.g. for 2x3
2+1
  A
f(x)dx= x4 +2x3−2x2 +Bx
Condone poor notation e.g. and ignore any “+ c”
4
dM1: Applies limits 5 and 3 either way round and subtracts and equates to 176 to obtain an equation
in A and B. Depends on previous M.
Condone poor use of brackets when subtracting e.g. allow
A A
54 +2(53)−2(52)+5B− 34 +2(33)−2(32)+3B=176
4 4
dM1: Solves their two equations in A and B to find values for both constants.
Depends on all previous M marks but allow this mark if the only error was to attempt
f (2) = 0 at the beginning.
A1: Correct values.
Note that some candidates eliminate A or B by rearranging the first equation and substituting into the integral.
This essentially follows the main scheme:
f ( − 2 ) = 0  − 8 A + 2 4 + 8 + B = 0  B = 8 A − 3 2 | M1A1
 A
f ( x ) d x = x 4 + 2 x 3 − 2 x 2 + 8 A x − 3 2 x
4 | M1A1
 A   A 
 5 4 + 2 ( 5 3 ) − 2 ( 5 2 ) + 4 0 A − 1 6 0 − 3 4 + 2 ( 3 3 ) − 2 ( 3 2 ) + 2 4 A − 9 6 = 1 7 6
4 4 | dM1
 A = . . . , B = . . . | dM1
1
A = , B = − 2 8
2 | A1
Long Division for reference:
Score the first M1 for a complete method to obtain a remainder in terms of A and B that is set = 0
and A1 for a correct equation e.g. B – 8A + 32 = 0
Question | Scheme | Marks

Show LaTeX source

$$f(x) = Ax^3 + 6x^2 - 4x + B$$
where $A$ and $B$ are constants.

Given that
\begin{itemize}
\item $(x + 2)$ is a factor of $f(x)$
\item $\int_{-3}^{5} f(x)dx = 176$
\end{itemize}

Find the value of $A$ and the value of $B$.
[7]

\hfill \mbox{\textit{Edexcel P2 2022 Q7 [7]}}

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