A Level Mathematics 9709 Paper 23, May/June 2023

Question paper and mark scheme50 marks7 questions

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Question paper

9709/23 · May/June 2023

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1

Solve the equation sec⁡2θ+5tan⁡2θ=9+17sec⁡θ for 0∘<θ<360∘.

[5]
2

The variables x and y satisfy the equation y=Ae(A−B)x, where A and B are constants. The graph of ln⁡y against x is a straight line passing through the points (0.4,3.6) and (2.9,14.1), as shown in the diagram. Find the values of A and B correct to 3 significant figures.

[5]
Figure 1 of 1 for question 2
3

The diagram shows part of the curve y=62x+3. The shaded region is bounded by the curve and the lines x=6 and y=2. Find the exact area of the shaded region, giving your answer in the form a−ln⁡b, where a and b are integers.

[5]
Figure 1 of 1 for question 3
4
Figure 1 of 1 for question 4
a

The diagram shows the graph of y=3−e−12x. On the diagram, sketch the graph of y=∣5x−4∣, and show that the equation 3−e−12x=∣5x−4∣ has exactly two real roots.

[2]
b

It is given that the two roots of 3−e−12x=∣5x−4∣ are denoted by α and β, where α<β. Show by calculation that α lies between 0.36 and 0.37.

[2]
c

Use the iterative formula xn+1=15(7−e−12xn) to find β correct to 4 significant figures. Give the result of each iteration to 6 significant figures.

[3]
5

The diagram shows the curve with equation y=e−12x(x2−5x+4). The curve crosses the x-axis at the points A and B, and has a maximum at the point C.

Figure 1 of 1 for question 5
a

Find the exact gradient of the curve at B.

[5]
b

Find the exact coordinates of C.

[4]
6
a

Show that 4sin⁡(θ+13π)cos⁡(θ−13π)≡3+2sin⁡2θ.

[4]
b

Find the exact value of 4sin⁡1724πcos⁡124π.

[2]
c

Find the exact value of ∫018π4sin⁡(2x+13π)cos⁡(2x−13π) dx.

[4]
7

A curve has parametric equations x=2t+3t+2,y=t2+at+1, where a is a constant. It is given that, at the point P on the curve, the gradient is 1.

a

Show that the value of t at P satisfies the equation 2t3+(a+8)t2+(4a+8)t+4a−1=0.

[4]
b

It is given that (t+1) is a factor of 2t3+(a+8)t2+(4a+8)t+4a−1. Find the value of a.

[2]
c

Hence show that P is the only point on the curve at which the gradient is 1.

[3]

Mark scheme

9709/23 · May/June 2023

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1 · 5 marks

  • M1

    Attempt to express equation in terms of sec⁡θ only

  • A1

    Obtain 6sec⁡2θ−17sec⁡θ−14(=0)

  • M1

    Attempt solution of 3-term quadratic equation to find one value of θ, from cos⁡θ=...

  • A1

    Obtain 73.4

  • A1

    Obtain 286.6

2 · 5 marks

  • B1

    State or imply equation ln⁡y=ln⁡A+(A−B)x

  • M1

    Equate A−B to gradient of line

  • A1

    Obtain A−B=4.2

  • M1

    Substitute appropriate values to find value of ln⁡A

  • A1

    Obtain ln⁡A=1.92 and hence A=6.82 and B=2.62

3 · 5 marks

  • 1 mark

    Integrate to obtain the form $k\ln(2x + 3)

  • A1

    Obtain correct 3ln⁡(2x+3)

  • 1 mark

    Apply limits 0 and 6 correctly to obtain kln⁡15−kln⁡3

  • 1 mark

    Apply relevant logarithm properties correctly to obtain form ln⁡b

  • A1

    Obtain 12−ln⁡125

4a · 2 marks

  • 1 mark

    Draw (more or less) correct sketch with vertex on positive x-axis

  • 1 mark

    Indicate in some way the two roots

4b · 2 marks

  • M1

    Consider sign of 3−e−12x+5x−4 or of 3−e−12x−∣5x−4∣ for 0.36 and 0.37

  • A1

    Obtain −0.035... and 0.018..., or equivalents, and justify conclusion

4c · 3 marks

  • M1

    Use iteration process correctly at least once

  • A1

    Obtain final answer 1.295

  • A1

    Show sufficient iterations to 6 sf to justify answer or show sign change in interval [1.2945, 1.2955]

5a · 5 marks

  • 1 mark

    Attempt use of product rule to find first derivative

  • A1

    Obtain −12e−12x(x2−5x+4)+e−12x(2x−5)

  • B1

    Obtain x=4 for point B

  • 1 mark

    Substitute x=4 to find the value of the derivative

  • A1

    Obtain 3e−2

5b · 4 marks

  • 1 mark

    Equate their first derivative to zero and simplify as far as quadratic equation

  • A1

    Obtain at least x2−9x+14=0

  • 1 mark

    Solve to find relevant x value and substitute to find the value of y

  • A1

    Obtain x=7 and y=18e−72

6a · 4 marks

  • B1

    Obtain at least either (12sin⁡θ+123cos⁡θ) or (12cos⁡θ+123sin⁡θ)

  • M1

    Expand and simplify with correct use of sin⁡2θ+cos⁡2θ=1

  • M1

    Use sin⁡θcos⁡θ=12sin⁡2θ

  • A1

    Confirm given result 3+2sin⁡2θ

6b · 2 marks

  • 1 mark

    Identify value of θ is 38π

  • 1 mark

    Obtain 3+2sin⁡34π and conclude 3+2

6c · 4 marks

  • B1

    Identify integrand as 3+2sin⁡4x

  • M1

    Integrate to obtain form k1x+k2cos⁡4x

  • A1

    Obtain correct 3x−12cos⁡4x

  • A1

    Obtain 18π3+12

7a · 4 marks

  • 1 mark

    Attempt use of quotient rule (or equivalent) to find dxdt

  • A1

    Obtain 1(t+2)2

  • 1 mark

    Equate dydx to 1

  • A1

    Obtain (2t+a)(t+2)2=1 and expand to confirm given result

7b · 2 marks

  • M1

    Substitute t=−1, equate to zero and attempt solution for a

  • A1

    Obtain a=3

7c · 3 marks

  • M1

    Divide their cubic by t+1 at least as far as the x term

  • A1

    Obtain 2t2+9t+11

  • A1

    Calculate discriminant, obtain −7 and conclude no further value of t

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