A Level Further Mathematics 9231 Paper 12, May/June 2023

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Question paper

9231/12 · May/June 2023

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1

Let A=(3011).

a

Prove by mathematical induction that, for all positive integers n, 2An=(2×3n03n−12).

[5]
b

Find, in terms of n, the inverse of An.

[2]
2

The cubic equation x3+4x2+6x+1=0 has roots α,β,γ.

a

Find the value of α2+β2+γ2.

[2]
b

Use standard results from the list of formulae (MF19) to show that ∑r=1n((α+r)2+(β+r)2+(γ+r)2)=n(n2+an+b), where a and b are constants to be determined.

[6]
3
a

Use the method of differences to find ∑r=1n1(kr+1)(kr−k+1) in terms of n and k, where k is a positive constant.

[4]
b

Deduce the value of ∑r=1∞1(kr+1)(kr−k+1).

[1]
c

Find also ∑r=nn21(kr+1)(kr−k+1) in terms of n and k.

[2]
4

The matrix M is given by M=(ab2c2a), where a,b,c are real constants and b≠0.

a

Show that M does not represent a rotation about the origin.

[2]
b

Find the equations of the invariant lines, through the origin, of the transformation in the x–y plane represented by M.

[5]
c

It is given that M represents the sequence of two transformations in the x–y plane given by an enlargement, centre the origin, scale factor 5 followed by a shear, x-axis fixed, with (0,1) mapped to (5,1). Find M.

[3]
d

The triangle DEF in the x–y plane is transformed by M onto triangle PQR. Given that the area of triangle DEF is 12 cm2, find the area of triangle PQR.

[2]
5

The curve C has polar equation r2=1θ2+1, for 0⩽θ⩽π.

a

Sketch C and state the polar coordinates of the point of C furthest from the pole.

[3]
b

Find the area of the region enclosed by C, the initial line, and the half-line θ=π.

[4]
c

Show that, at the point of C furthest from the initial line, (θ+1θ)cot⁡θ−1=0 and verify that this equation has a root between 1.1 and 1.2.

[5]
6

The curve C has equation y=x2+2x−15x−2.

a

Find the equations of the asymptotes of C.

[3]
b

Show that C has no stationary points.

[3]
c

Sketch C, stating the coordinates of the intersections with the axes.

[3]
d

Sketch the curve with equation y=∣x2−2x−15x−2∣.

[2]
e

Find the set of values of x for which ∣2x2+4x−30x−2∣<15.

[4]
7

The plane Π1 has equation r=−4j−3k+λ(i−j+k)+μ(i+j−k).

a

Obtain an equation of Π1 in the form px+qy+rz=d.

[4]
b

The plane Π2 has equation r.(−5i+3j+5k)=4. Find a vector equation of the line of intersection of Π1 and Π2.

[4]
c

The line l passes through the point A with position vector ai+aj+(a−7)k and is parallel to (1−b)i+bj+bk, where a and b are positive constants. Given that the perpendicular distance from A to Π1 is 2, find the value of a.

[2]
d

Given that the obtuse angle between l and Π1 is 34π, find the exact value of b.

[4]

Mark scheme

9231/12 · May/June 2023

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1a · 5 marks

  • B1

    2A=(6022)=(2×31031−12) so true when n=1.

  • B1

    Assume that it is true for n=k, so 2Ak=(2×3k03k−12).

  • M1

    Then 2Ak+1=(2×3k03k−12)(3011)

  • A1

    =(2×3k+103k+1−3+22)

  • A1

    So, it is also true for n=k+1. Hence, by induction, true for all positive integers.

1b · 2 marks

  • B1

    det⁡An=det⁡(3n012(3n−1)1)=3n Or a multiple of A−n=2−13−n(201−3n2×3n) seen.

  • B1

    A−n=3−n(1012(1−3n)3n)

2a · 2 marks

  • M1

    (−4)2−2(6)

  • A1

    4

2b · 6 marks

  • B1

    (α+r)2=α2+2αr+r2

  • M1

    ∑r=1n((α+r)2+(β+r)2+(γ+r)2)

  • A1

    =∑r=1n(4+2(−4)r+3r2)

  • M1

    4n−4n(n+1)+12n(n+1)(2n+1)

  • M1

    −4n2+12n(n+1)(2n+1)=n(−4n+12(2n2+3n+1))

  • A1

    =n(n2−52n+12)

3a · 4 marks

  • M1

    1(kr+1)(kr−k+1)=1k(1k(r−1)+1−1kr+1) (method)

  • A1

    1(kr+1)(kr−k+1)=1k(1k(r−1)+1−1kr+1) (correct fractions)

  • M1

    ∑r=1n1(kr+1)(kr−k+1)=1k(1−1k+1+1k+1−12k+1+⋯+1k(n−1)+1−1kn+1)

  • A1

    1k(1−1kn+1)

3b · 1 mark

  • B1

    1k

3c · 2 marks

  • M1

    ∑r=nn21(kr+1)(kr−k+1)=∑r=1n21(kr+1)(kr−k+1)−∑r=1n−11(kr+1)(kr−k+1)

  • A1

    1k(1−1kn2+1−(1−1k(n−1)+1))=1k(1k(n−1)+1−1kn2+1)

4a · 2 marks

  • M1

    (ab2c2a)=(cos⁡θ−sin⁡θsin⁡θcos⁡θ)

  • A1

    b2=−c2 which is impossible since b and c are real and b≠0.

4b · 5 marks

  • B1

    (ab2c2a)(xy)=(ax+b2yc2x+ay)

  • M1

    Uses y=mx and Y=mX

  • A1

    c2x+amx=m(ax+b2mx)

  • A1

    c2+am=ma+b2m2  ⟹  c2=b2m2

  • A1

    y=cbx and y=−cbx

4c · 3 marks

  • M1

    M=(1501)(5005)

  • A1

    M=(1501)(5005) (correct order)

  • A1

    (55205)

4d · 2 marks

  • M1

    12×det⁡M

  • A1

    300 cm2

5a · 3 marks

  • B1

    Correct shape and domain, polar graph with r strictly decreasing but condone if not strictly decreasing close to θ=π.

  • B1

    Fully correct including shape at θ=0 correct, shape at θ=π correct

  • B1

    (1,0)

5b · 4 marks

  • M1

    12∫0π1θ2+1 dθ

  • M1

    Integrates 1θ2+1 to tan⁡−1θ

  • A1

    12[tan⁡−1θ]0π

  • A1

    12tan⁡−1π=0.631

5c · 5 marks

  • B1

    y=sin⁡θθ2+1

  • M1

    dydθ=(θ2+1)12cos⁡θ−θ(θ2+1)−12sin⁡θθ2+1=0

  • A1

    Derivative correctly found and set to zero

  • A1

    θ≠0  ⟹  (θ+1θ)cot⁡θ−1=0

  • B1

    (1.1+11.1)cot⁡1.1−1=0.02… and (1.2+11.2)cot⁡1.2−1=−0.209…

6a · 3 marks

  • B1

    x=2

  • M1

    x2+2x−15=(x−2)(x+4)−7

  • A1

    y=x+4

6b · 3 marks

  • M1

    dydx=(x−2)(2x+2)−(x2+2x−15)(x−2)2

  • A1

    x2−4x+11=0(or dydx=1+7(x−2)2)

  • A1

    42−4(1)(11)=−28<0 (or y′>0) There are no turning points.

6c · 3 marks

  • B1

    Axes and asymptotes.

  • B1

    Branches correct.

  • B1

    (0,7.5),(−5,0),(3,0)

6d · 2 marks

  • B1

    FT from sketch in (c).

  • B1

    Correct shape at infinity and on x axis.

6e · 4 marks

  • M1

    x2+2x−15x−2=152  ⟹  x2−112x=0

  • M1

    x2+2x−15x−2=−152  ⟹  x2+192x−30=0

  • A1

    x=0,112 or x=−12,52

  • A1

    −12<x<0 or 52<x<112

7a · 4 marks

  • M1

    ∣ijk1−1111−1∣

  • A1

    =(022)∼(011)

  • M1

    (−4)+(−3)=−7

  • A1

    y+z=−7

7b · 4 marks

  • B1

    States point common to both planes e.g. (−7−2−5).

  • M1

    ∣ijk011−535∣

  • A1

    =(2−55)

  • A1

    r=(−7−2−5)+λ(2−55)

7c · 2 marks

  • M1

    ∣a+a−7+72∣=2

  • A1

    a=1

7d · 4 marks

  • M1

    ∣b+b2(1−b)2+2b2∣=122 (method)

  • A1

    ∣b+b2(1−b)2+2b2∣=122 (correct equation)

  • M1

    2b=(1−b)2+2b2  ⟹  b2+2b−1=0

  • A1

    b=−1+2

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