A Level Further Mathematics 9231 Paper 22, May/June 2024

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Question paper

9231/22 · May/June 2024

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1

Find the roots of the equation z3=−1083+108i, giving your answers in the form r(cos⁡θ+isin⁡θ), where r>0 and 0⩽θ<2π.

[5]
2

Find the Maclaurin's series for e1+x2+e1−x up to and including the term in x2.

[4]
3

It is given that x=sin⁡−1tandy=tcos⁡−1t,for 0⩽t<1.

a

Show that dydx=−t+1−t2cos⁡−1t.

[3]
b

Find d2ydx2 in terms of t.

[4]
4

It is given that, for n⩾0, In=∫0ln⁡3sechnx dx.

a

Show that, for n⩾2, (n−1)In=(35)n−2(45)+(n−2)In−2. [You may use the result that ddx(sech x)=−tanh⁡x sech x.]

[5]
b

Find the value of I4.

[3]
5

The diagram shows the curve with equation y=2x−x2 for 0⩽x⩽1, together with a set of n rectangles of width 1n.

Figure 1 of 1 for question 5
a

By considering the sum of the areas of these rectangles, show that ∫01(2x−x2) dx<Un, where Un=(1+1n)(23−16n).

[5]
b

Use a similar method to find, in terms of n, a lower bound Ln for ∫01(2x−x2) dx.

[4]
c

Show that lim⁡n→∞(Un−Ln)=0.

[2]
6
a

Show that (cosh⁡x+sinh⁡x)12=e12x.

[2]
b

Find the particular solution of the differential equation d2ydx2+dydx+3y=5(cosh⁡x+sinh⁡x)12, given that, when x=0, y=1 and dydx=43.

[10]
7
a

Use the substitution u=1+x2 to find ∫x1+x2 dx.

[2]
b

Find the solution of the differential equation xdydx−y=x2sinh⁡−1x, given that y=1 when x=1. Give your answer in the form y=f(x).

[10]
8
a

Find the set of values of a for which the system of equations 6x+ay=3,2x−y=1,x+5y+4z=2 has a unique solution.

[2]
b

Show that the system of equations in part (a) is consistent for all values of a.

[3]
c

The matrix A is given by A=(6002−10154). Find a matrix P and a diagonal matrix D such that (14A+24I)2=PDP−1.

[7]
d

Use the characteristic equation of A to show that (14A+24I)2=A4(A+bI)2, where b is an integer to be determined.

[4]

Mark scheme

9231/22 · May/June 2024

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1 · 5 marks

  • B1

    Finds modulus and argument of −1083+108i

  • M1

    Finds one root. For M1 must have divided their argument of −1083+108i by 3.

  • 1 mark

    Correct root. FT on their modulus.

  • M1

    Finds other two roots.

  • A1

    Correct other two roots.

2 · 4 marks

  • B1

    Changes e1+x2 or e1−x so that the formula given in the list of formulae (MF19) can be applied.

  • B1

    e1−x=e(1−x+12x2+… )

  • M1

    Sums power series.

  • A1

    e1+x2+e1−x=2e−ex+32ex2+…

3a · 3 marks

  • B1

    dydt=cos⁡−1t−t1−t2,dxdt=11−t2

  • M1

    Uses chain rule

  • A1

    dydx=−t+1−t2cos⁡−1t, AG

3b · 4 marks

  • M1

    Applies product rule.

  • A1

    ddt(dydx)=−2−tcos⁡−1t1−t2

  • M1

    Uses chain rule.

  • A1

    d2ydx2=−21−t2−tcos⁡−1t

4a · 5 marks

  • B1

    Separates into correct structure.

  • 1 mark

    Uses integration by parts correctly.

  • 1 mark

    Uses 1−sech⁡2x=tanh⁡2x.

  • A1

    (35)n−2(45)+(n−2)(In−2−In)

  • A1

    (n−1)In=(35)n−2(45)+(n−2)In−2, AG

4b · 3 marks

  • B1

    I2=45

  • M1

    Applies reduction formula with n=4.

  • A1

    I4=236375=0.629

5a · 5 marks

  • M1

    Forms the sum of the areas of the rectangles, may be in summation form. M1 is for correct number of rectangles.

  • A1

    Correct sum of the areas of the rectangles.

  • M1

    Applies formulae from MF19. Substituted in correctly once for M1.

  • A1

    Summations must have (correct) limits for A1.

  • A1

    AG. This A1 requires the previous A1.

5b · 4 marks

  • M1

    Forms the sum of the areas of appropriate rectangles. M1 is for seeing correct height of last rectangle.

  • A1

    Correct sum of the areas of appropriate rectangles.

  • M1

    Applies formulae from MF19, substitutes correct limit. Substituted in correctly once for M1.

  • A1

    (1−1n)(23+16n)=23−12n−16n2

5c · 2 marks

  • M1

    Expresses their Un−Ln in terms of 1n. For, M1 they must taking the limit of an expression that tends to a constant. (Must be fully expanded if they do not take the limit at any stage.)

  • A1

    1n→0 as n→∞ or 23−23=0, AG, CWO. Their Ln must be correct.

6a · 2 marks

  • M1

    Substitutes sinh and cosh in terms of exponentials.

  • A1

    (ex)12=e12x, AG

6b · 10 marks

  • M1

    Auxiliary equation.

  • A1

    Complimentary function. Allow 'y=' missing.

  • B1

    Particular integral and its derivatives.

  • M1

    Substitutes and equates coefficients. PI must be correct form (aebx).

  • A1

    k=43

  • 1 mark

    Must have '$y = $'. FT on their CF.

  • B1

    dydx=e−12x(−112Asin⁡112x+112Bcos⁡112x)−12e−12x(Acos⁡112x+Bsin⁡112x)+23e12x

  • M1

    Substitutes initial conditions and forms simultaneous equations. For M1, CF must be correct form

  • A1

    A=−13,B=111

  • A1

    Must have '$y = $'.

7a · 2 marks

  • M1

    Applies substitution.

  • A1

    1+x2 [+C]

7b · 10 marks

  • B1

    Divides through by x.

  • M1

    Finds integrating factor.

  • A1

    x−1

  • M1

    Correct form on LHS. ddx(Iy) for their integrating factor I.

  • M1

    Integrates RHS. RHS must be of the form csinh⁡−1x.

  • A1

    x−1y=xsinh⁡−1x−∫x1+x2 dx

  • A1

    sinh⁡−1x=ln⁡(x+x2+1)

  • 1 mark

    Substitutes initial conditions.

  • 1 mark

    Divides through by their integrating factor.

  • A1

    y=x2sinh⁡−1x−x1+x2−xln⁡(1+2)+x(1+2)

8a · 2 marks

  • M1

    Finds determinant or solves equations.

  • A1

    a≠−3

8b · 3 marks

  • B1

    If a≠−3 then system has unique solution (so consistent). (FT)

  • M1

    Eliminates variable using all 3 equations. Or states there are two distinct equations and three unknowns. Or gives accurate geometrical description with a=−3.

  • A1

    So the system has infinitely many solutions (so consistent).

8c · 7 marks

  • B1

    Eigenvalues of A are 6, −1 and 4. Lower diagonal matrix or characteristic equation.

  • M1

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

  • A1

    Correct eigenvector for λ=6.

  • A1

    Correct eigenvector for λ=−1.

  • A1

    Correct eigenvector for λ=4.

  • M1

    Forms matrix P and matrix D. Column of zeros in P gets M0.

  • A1

    Correct P and D (or correctly matched permutations of columns).

8d · 4 marks

  • B1

    Characteristic equation.

  • M1

    Substitutes for A and makes 14A+24I the subject.

  • M1

    Squares and factorises.

  • A1

    (14A+24I)2=A4(A−9I)2. CWO.

Questions

  1. 1
    Question 15 marks
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  2. 2
    Question 24 marks
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  3. 3
    Question 37 marks
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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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  8. 8
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