A Level Further Mathematics 9231 Paper 23, May/June 2023

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Question paper

9231/23 · May/June 2023

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1
a

Find the Maclaurin series for sin⁡−1x up to and including the term in x3.

[5]
b

Deduce an approximation to ∫01511−u2 du, giving your answer as a fraction.

[1]
2

The variables x and y are related by the differential equation 6d2xdt2+5dxdt+x=t2+10t+13.

a

Find the general solution for x in terms of t.

[6]
b

State an approximate solution for large positive values of t.

[1]
3

By considering the binomial expansions of (z+1z)4 and (z−1z)4, where z=cos⁡θ+isin⁡θ, use de Moivre's theorem to show that cot⁡4θ=cos⁡4θ+acos⁡2θ+bcos⁡4θ−acos⁡2θ+b, where a and b are integers to be determined. [7]

[7]
4

The curve C has equation 4y3+(x+y)6=109.

a

Show that, at the point (−4,3) on C, dydx=117.

[3]
b

Find the value of d2ydx2 at the point (−4,3).

[5]
5
a

Starting from the definitions of cosh⁡ and sinh⁡ in terms of exponentials, prove that 2cosh⁡2x=cosh⁡2x+1.

[3]
b

Find the solution of the differential equation dydx+2ytanh⁡x=1 for which y=1 when x=0. Give your answer in the form y=f(x).

[8]
6

The diagram shows the curve with equation y=(1−x)2 for 0⩽x⩽1, together with a set of n rectangles of width 1n.

Figure 1 of 1 for question 6
a

By considering the sum of the areas of these rectangles, show that ∫01(1−x)2 dx<Un, where Un=2n2+3n+16n2.

[5]
b

Use a similar method to find, in terms of n, a lower bound Ln for ∫01(1−x)2 dx.

[4]
c

Show that lim⁡n→∞(Un−Ln)=0.

[2]
7

The integral In, where n is an integer, is defined by In=∫043(1+x2)12n dx.

a

Find the exact value of I−1 giving your answer in the form ln⁡a, where a is an integer to be determined.

[2]
b

By considering ddx(x(1+x2)12n), or otherwise, show that (n+1)In=nIn−2+43(53)n.

[5]
c

A curve has equation y=x2, for 0⩽x⩽23. The arc length of the curve is denoted by s. Use the substitution u=2x to show that s=12I1 and find the exact value of s.

[4]
8

The matrix A is given by A=(a−6a2a+201−a002−a−1) where a is a constant with a≠0 and a≠1.

a

Show that the equation A(xyz)=(123) has a unique solution and interpret this situation geometrically.

[3]
b

Show that the eigenvalues of A are a, 1−a and −1.

[2]
c

Find a matrix P and a diagonal matrix D such that A4=PDP−1.

[6]
d

Use the characteristic equation of A to find A4 in terms of A and a.

[3]

Mark scheme

9231/23 · May/June 2023

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1a · 5 marks

  • B1

    Correct first derivative.

  • B1

    Correct second derivative.

  • M1

    Differentiates their f′′(x) using product rule.

  • M1

    Evaluates their derivatives at x=0. Must have attempted all three derivatives.

  • A1

    CWO.

1b · 1 mark

  • B1

    151750

2a · 6 marks

  • M1

    Auxiliary equation.

  • A1

    Complementary function. Allow with 'x =' missing.

  • B1

    Particular integral and its derivatives.

  • M1

    Substitutes and equates coefficients.

  • A1

    r=1,q=0,p=1

  • A1

    Must have x=.

2b · 1 mark

  • B1

    Must have 'x ='. Accept 'x is approximately' but do not accept 'x→'. FT on their PI.

3 · 7 marks

  • B1

    Use of z−z−1=2isin⁡θ and z+z−1=2cos⁡θ

  • M1

    Expands and groups (z+z−1)4.

  • A1

    (z+z−1)4=(z4+z−4)+4(z2+z−2)+6

  • M1

    Expands and groups (z−z−1)4.

  • A1

    (z−z−1)4=(z4+z−4)−4(z2+z−2)+6

  • M1

    Substitutes zn+z−n=2cos⁡nθ once in LHS.

  • A1

    cot⁡4θ=cos⁡4θ+4cos⁡2θ+3cos⁡4θ−4cos⁡2θ+3

4a · 3 marks

  • B1

    Differentiates y3 correctly.

  • B1

    Differentiates (x+y)6 correctly.

  • B1

    Substitutes (−4,3), AG.

4b · 5 marks

  • B1

    Differentiates y2y′.

  • B1

    Differentiates first part of (x+y)5(1+y′).

  • B1

    Differentiates second part of (x+y)5(1+y′).

  • M1

    Substitutes (−4,3).

  • A1

    y′′=−96289

5a · 3 marks

  • B1

    cosh⁡x=12(ex+e−x)

  • M1

    Expands.

  • A1

    cosh⁡2x+1, AG.

5b · 8 marks

  • M1

    Finds integrating factor.

  • A1

    e2ln⁡cosh⁡x=cosh⁡2x

  • M1

    Correct form on LHS, ddx(yI) for their integrating factor I, and attempt to integrate their RHS.

  • M1

    Uses cosh⁡2x=12(cosh⁡2x+1).

  • A1

    12(12sinh⁡2x+x)+C (M1 A1 is for RHS).

  • M1

    Finds C.

  • M1

    Divides through by their coefficient of y.

  • A1

    y=sech2x(14sinh⁡2x+12x+1)

6a · 5 marks

  • M1

    Forms the sum of the areas of the rectangles.

  • A1

    Must have first two terms and last term for A1. Accept (1n)(1−nn)2 for the last term. Accept written in summation form with correct limits.

  • M1

    Setting up correct series so that formulae from MF19 can be applied. Can also recognise sum as 1n3∑r=1nr2

  • A1

    Correct intermediate expanded summation expression.

  • A1

    AG. Must have gained previous accuracy mark.

6b · 4 marks

  • M1

    Forms the sum of the areas of appropriate rectangles.

  • A1

    Must have first two and last terms for A1. Accept (1n)(1−nn)2 for the last term. Accept written in summation form with correct limits.

  • M1

    Setting up correct series so that formulae from MF19 can be applied. Recognising sum as 1n3∑r=1n−1r2=1n3∑r=1nr2−1n without wrong working scores M1 A1 M1.

  • A1

    2n2−3n+16n2

6c · 2 marks

  • M1

    Simplifies Un−Ln to cn.

  • A1

    1n→0 as n→∞, AG.

7a · 2 marks

  • M1

    Recognises integral.

  • A1

    ln⁡3

7b · 5 marks

  • M1

    Uses the product rule to differentiate.

  • A1

    nx2(1+x2)12n−1+(1+x2)12n

  • M1

    Uses x2=1+x2−1.

  • M1

    Integrates both sides using the limits given. Requires previous method mark.

  • A1

    Substitutes limits and rearranges. AG.

7c · 4 marks

  • M1

    Forms correct integral with correct limits.

  • A1

    12∫0431+u2 du=12I1, AG.

  • M1

    Applies reduction formula with n=1.

  • A1

    s=14ln⁡3+59

8a · 3 marks

  • M1

    Shows that determinant is non-zero or solves equations.

  • A1

    a(a−1)≠0 or correct solution for x,y,z.

  • B1

    Must be clear that there is just one point where all 3 planes intersect.

8b · 2 marks

  • M1

    Equates determinant to zero or working to find characteristic equation.

  • A1

    AG. Factorisation must be clear.

8c · 6 marks

  • M1

    Uses vector product (or equations) to find corresponding eigenvectors.

  • A1

    Eigenvector for λ=a.

  • A1

    Eigenvector for λ=−1.

  • A1

    Eigenvector for λ=1−a.

  • M1

    Correctly matched permutations of columns. Their eigenvectors must be non-zero for M1.

  • A1

    P and D matrices correct.

8d · 3 marks

  • B1

    Finds characteristic equation multiplied out.

  • M1

    Multiplies through by A.

  • A1

    A4=(a2−a+1)A2+(a2−a)A

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  3. 3
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  4. 4
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  5. 5
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  6. 6
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  7. 7
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