A Level Mathematics 9709 Paper 23, May/June 2024

Question paper and mark scheme50 marks7 questions

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Question paper

9709/23 · May/June 2024

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1

Solve the inequality ∣5x+7∣>∣2x−3∣.

[4]
2

Use logarithms to solve the equation 62x−1=5e3x+2. Give your answer correct to 4 significant figures.

[4]
3

The diagram shows the curve with equation y=8e−x−e2x. The curve crosses the y-axis at the point A and the x-axis at the point B. The shaded region is bounded by the curve and the two axes.

Figure 1 of 1 for question 3
a

Find the gradient of the curve at A.

[3]
b

Show that the x-coordinate of B is ln⁡2 and hence find the area of the shaded region.

[5]
4

A curve is defined by the parametric equations x=4cos⁡2t,y=3sin⁡2t, for values of t such that 0<t<12π. Find the equation of the normal to the curve at the point for which t=16π. Give your answer in the form ax+by+c=0 where a, b and c are integers.

[7]
5

The polynomial p(x) is defined by p(x)=9x3+18x2+5x+4.

a

Find the quotient when p(x) is divided by (3x+2), and show that the remainder is 6.

[3]
b

Find the value of ∫02p(x)3x+2 dx, giving your answer in the form a+ln⁡b where a and b are integers.

[5]
6

The diagram shows the curve with equation y=ln⁡(2x+1)x+3. The curve has a maximum point M.

Figure 1 of 1 for question 6
a

Find an expression for dydx.

[2]
b

Show that the x-coordinate of M satisfies the equation x=x+3ln⁡(2x+1)−0.5.

[2]
c

Show by calculation that the x-coordinate of M lies between 2.5 and 3.0 .

[2]
d

Use an iterative formula based on the equation in part (b) to find the x-coordinate of M correct to 4 significant figures. Give the result of each iteration to 6 significant figures.

[3]
7
a

Prove that 2sin⁡θcosec⁡2θ≡sec⁡θ.

[2]
b

Solve the equation tan⁡2θ+7sin⁡θcosec⁡2θ=8 for −π<θ<π.

[5]
c

Find ∫8sin⁡212xcosec⁡2x dx.

[3]

Mark scheme

9709/23 · May/June 2024

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1 · 4 marks

  • B1

    Solve 5x+7=2x−3 to obtain −103

  • M1

    Attempt solution of linear equation where 5x and 2x have different signs

  • A1

    Obtain −47

  • A1

    State x<−103, x>−47

2 · 4 marks

  • M1

    (2x−1)ln⁡6

  • M1

    ln⁡5+3x+2

  • 1 mark

    Attempt solution of linear equation

  • A1

    Obtain 9.256

3a · 3 marks

  • M1

    Differentiate to obtain form k1e−x+k2e2x

  • A1

    Obtain −8e−x−2e2x

  • A1

    Substitute x=0 to obtain −10

3b · 5 marks

  • M1

    Attempt to find x-coordinate of B

  • A1

    Obtain e3x=8 and hence x=ln⁡2

  • B1

    Integrate to obtain −8e−x−12e2x

  • M1

    Use limits 0 and ln⁡2 correctly to find area

  • A1

    Obtain 52

4 · 7 marks

  • M1

    Obtain forms dxdt=k1cos⁡tsin⁡t or dydt=k2cos⁡2t

  • A1

    Obtain correct −8cos⁡tsin⁡t or −4sin⁡2t and 23cos⁡2t

  • 1 mark

    Attempt value of dydx when t=16π

  • A1

    Obtain dydx=−12

  • M1

    State or imply gradient of normal is 2

  • 1 mark

    Attempt equation of normal

  • A1

    Obtain 4x−2y−9=0

5a · 3 marks

  • M1

    Carry out division at least as far as 3x2+k1x

  • A1

    Obtain quotient 3x2+4x−1

  • A1

    Confirm remainder is 6

5b · 5 marks

  • B1

    Identify integrand as 3x2+4x−1+63x+2

  • M1

    Integrate to obtain at least x3 and k2ln⁡(3x+2) terms

  • A1

    Obtain x3+2x2−x+2ln⁡(3x+2)

  • 1 mark

    Apply limits and appropriate logarithm properties

  • A1

    Obtain 14+ln⁡16

6a · 2 marks

  • M1

    Attempt use of quotient rule

  • A1

    Obtain (x+3)22x+1−ln⁡(2x+1)(x+3)2

6b · 2 marks

  • M1

    Equate first derivative to zero and arrange as far as 2x+1=…

  • A1

    Confirm x=x+3ln⁡(2x+1)−0.5

6c · 2 marks

  • M1

    Consider sign of x−x+3ln⁡(2x+1)+0.5 or equivalent for 2.5 and 3.0

  • A1

    Obtain −0.07 (0.0696...) and 0.4 (0.4166...) and justify conclusion

6d · 3 marks

  • M1

    Use iterative process correctly at least once

  • A1

    Obtain final answer 2.569

  • A1

    Show sufficient iterations to 6 sf to justify answer or show sign change in interval [2.5685,2.5695]

7a · 2 marks

  • M1

    Express left-hand side in terms of sin⁡θ and cos⁡θ using cosec 2θ=1sin⁡2θ

  • A1

    Obtain 1cos⁡θ and confirm sec⁡θ

7b · 5 marks

  • M1

    Attempt to obtain quadratic equation in sec⁡θ or cos⁡θ only

  • A1

    Obtain sec⁡2θ−1+72sec⁡θ=8 involving one trigonometric ratio

  • 1 mark

    Attempt to solve 3-term quadratic equation for sec⁡θ, using a correct method, to find at least one value of θ

  • A1

    Obtain any two of the four correct solutions ±0.952, ±1.76

  • A1

    Obtain remaining two correct solutions

7c · 3 marks

  • B1

    Identify integrand as 2sec⁡212x

  • M1

    Integrate ksec⁡212x to obtain 2ktan⁡12x

  • A1

    Obtain correct 4tan⁡12x

Questions

  1. 1
    Question 14 marks
    Mark scheme ›
  2. 2
    Question 24 marks
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  3. 3
    Question 38 marks
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  4. 4
    Question 47 marks
    Mark scheme ›
  5. 5
    Question 58 marks
    Mark scheme ›
  6. 6
    Question 69 marks
    Mark scheme ›
  7. 7
    Question 710 marks
    Mark scheme ›

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