A Level Mathematics 9709 Paper 13, Oct/Nov 2023

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Question paper

9709/13 · Oct/Nov 2023

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A curve is such that its gradient at a point $(x, y)$ is given by $\frac{d y}{d x} = x - 3 x^{- \frac{1}{2}}$ . It is given that the curve passes through the point $(4, 1)$ . Find the equation of the curve. [4]

[4]

The circle with equation $(x - 3)^{2} + (y - 5)^{2} = 40$ intersects the $y$ -axis at points $A$ and $B$ .

Find the $y$ -coordinates of $A$ and $B$ , expressing your answers in terms of surds. [2]

[2]

Find the equation of the circle which has $A B$ as its diameter. [2]

[2]

Show that the equation $5 \cos θ - \sin θ \tan θ + 1 = 0$ may be expressed in the form $a \cos^{2} θ + b \cos θ + c = 0$ , where $a$ , $b$ and $c$ are constants to be found. [3]

[3]

Hence solve the equation $5 \cos θ - \sin θ \tan θ + 1 = 0$ for $0 < θ < 2 π$ . [4]

[4]

a(i)

Expand the following in ascending powers of $x$ up to and including the term in $x^{2}$ . $(1 + 2 x)^{5}$ . [1]

[1]

a(ii)

Expand the following in ascending powers of $x$ up to and including the term in $x^{2}$ . $(1 - a x)^{6}$ , where $a$ is a constant. [2]

[2]

In the expansion of $(1 + 2 x)^{5} (1 - a x)^{6}$ , the coefficient of $x^{2}$ is $- 5$ . Find the possible values of $a$ . [4]

[4]

The first, second and third terms of a geometric progression are $2 p + 6$ , $5 p$ and $8 p + 2$ respectively.

Find the possible values of the constant $p$ . [3]

[3]

One of the values of $p$ found in (a) is a negative fraction. Use this value of $p$ to find the sum to infinity of this progression. [4]

[4]

A line has equation $y = 6 x - c$ and a curve has equation $y = c x^{2} + 2 x - 3$ , where $c$ is a constant. The line is a tangent to the curve at point $P$ . Find the possible values of $c$ and the corresponding coordinates of $P$ . [7]

[7]

The function $f$ is defined by $f (x) = 1 + \frac{3}{x - 2}$ for $x > 2$ .

State the range of $f$ . [1]

[1]

Obtain an expression for $f^{- 1} (x)$ and state the domain of $f^{- 1}$ . [4]

[4]

The function $g$ is defined by $g (x) = 2 x - 2$ for $x > 0$ . Obtain a simplified expression for $g f (x)$ . [2]

[2]

The diagram shows part of the graph of $y = \sin (a (x + b))$ , where $a$ and $b$ are positive constants.

State the value of $a$ and one possible value of $b$ . [2]

[2]

Another curve, with equation $y = f (x)$ , has a single stationary point at the point $(p, q)$ , where $p$ and $q$ are constants. This curve is transformed to a curve with equation $y = - 3 f (\frac{1}{4} (x + 8)) .$ For the transformed curve, find the coordinates of the stationary point, giving your answer in terms of $p$ and $q$ . [3]

[3]

A curve has equation $y = 2 x^{\frac{1}{2}} - 1$ .

Find the equation of the normal to the curve at the point $A (4, 3)$ , giving your answer in the form $y = m x + c$ . [3]

[3]

A point is moving along the curve $y = 2 x^{\frac{1}{2}} - 1$ in such a way that at $A$ the rate of increase of the $x$ -coordinate is $3 cm s^{- 1}$ . Find the rate of increase of the $y$ -coordinate at $A$ . [2]

[2]

At $A$ the moving point suddenly changes direction and speed, and moves down the normal in such a way that the rate of decrease of the $y$ -coordinate is constant at $5 cm s^{- 1}$ . As the point moves down the normal, find the rate of change of its $x$ -coordinate. [3]

[3]

The diagram shows points $A$ , $B$ and $C$ lying on a circle with centre $O$ and radius $r$ . Angle $A O B$ is $2.8$ radians. The shaded region is bounded by two arcs. The upper arc is part of the circle with centre $O$ and radius $r$ . The lower arc is part of a circle with centre $C$ and radius $R$ .

State the size of angle $A C O$ in radians. [1]

[1]

Find $R$ in terms of $r$ . [1]

[1]

Find the area of the shaded region in terms of $r$ . [7]

[7]

The diagram shows part of the curve with equation $y = x + \frac{2}{(2 x - 1)^{2}}$ . The lines $x = 1$ and $x = 2$ intersect the curve at $P$ and $Q$ respectively and $R$ is the stationary point on the curve.

Verify that the $x$ -coordinate of $R$ is $\frac{3}{2}$ and find the $y$ -coordinate of $R$ . [4]

[4]

Find the exact value of the area of the shaded region. [6]

[6]

Mark scheme

9709/13 · Oct/Nov 2023

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1 · 4 marks

B1
${\frac{x^{2}}{2}}$
B1
${\frac{- 3 x^{\frac{1}{2}}}{\frac{1}{2}}}$
M1
$1 = 8 - 12 + c$
A1
$y = \frac{1}{2} x^{2} - 6 x^{\frac{1}{2}} + 5$ , allow $f (x) =$

2a · 2 marks

M1
$(0 - 3)^{2} + (y - 5)^{2} = 40$
A1
$y = 5 \pm \sqrt{31}$

2b · 2 marks

1 mark
${x^{2} + (y - 5)^{2}}$ Allow $(x - 0)^{2}$
1 mark
$= {31}$

3a · 3 marks

M1
$5 \cos^{2} θ - \sin^{2} θ + \cos θ [= 0]$
M1
$5 \cos^{2} θ - (1 - \cos^{2} θ) + \cos θ [= 0]$
A1
$6 \cos^{2} θ + \cos θ - 1 = 0$

3b · 4 marks

M1
$(3 \cos θ - 1) (2 \cos θ + 1) = 0$
A1
$θ = {1.23}$
A1
$θ = {2.09 or \frac{2 π}{3}}$
1 mark
$θ = {5.05 and 4.19 (allow \frac{4 π}{3})}$

4a(i) · 1 mark

B1
$1 + 10 x + 40 x^{2}$

4a(ii) · 2 marks

B2
${1} {- 6 a x} {+ 15 a^{2} x^{2}}$

4b · 4 marks

M1
$15 a^{2} - 60 a + 40 = - 5$ (method)
A1
$15 a^{2} - 60 a + 40 = - 5$ (accuracy)
1 mark
$[15] (a - 1) (a - 3) [= 0] OE$
A1
$a = 1 and 3$

5a · 3 marks

M1
$\frac{5 p}{2 p + 6} = \frac{8 p + 2}{5 p}$
1 mark
$9 p^{2} - 52 p - 12 [= 0]$
A1
$[(9 p + 2) (p - 6) = 0] leading to p = - \frac{2}{9} and 6$

5b · 4 marks

M1
$a = 2 (- \frac{2}{9}) + 6 [= \frac{50}{9}]$
M1
$r = - \frac{10}{9} \div \frac{50}{9} [= - \frac{1}{5}]$
1 mark
$S_{\infty} = \frac{50}{9} \div (1 - (- \frac{1}{5}))$
A1
$= \frac{125}{27}$

6 · 7 marks

B1
$c x^{2} + 2 x - 3 = 6 x - c leading to c x^{2} - 4 x + (c - 3) [= 0]$
M1
$16 - 4 c (c - 3) = 0$
A1
$4 c^{2} - 12 c - 16 [= 0] leading to [4] (c - 4) (c + 1) [= 0] leading to c = 4 and - 1$
1 mark
$When c = 4, 4 x^{2} - 4 x + 1 [= 0] [(2 x - 1)^{2} = 0]$
A1
$x = \frac{1}{2}, y = - 1$
1 mark
$When c = - 1, x^{2} + 4 x + 4 [= 0] [(x + 2)^{2} = 0]$
A1
$x = - 2, y = - 11$

7a · 1 mark

B1
Range is [y] > 1

7b · 4 marks

M1
$y = \frac{3}{x - 2} + 1 leading to y - 1 = \frac{3}{x - 2} leading to (x - 2) (y - 1) = 3$
1 mark
$x - 2 = \frac{3}{y - 1}$
A1
$x = \frac{3}{y - 1} + 2 leading to f^{- 1} (x) = \frac{3}{x - 1} + 2$
1 mark
[Domain is] $x > 1$

7c · 2 marks

M1
$gf (x) = 2 (\frac{3}{x - 2} + 1) - 2 or 2 (\frac{x + 1}{x - 2}) - 2$
A1
$\frac{6}{x - 2}$

8a · 2 marks

B1
$a = \frac{1}{2}$
B1
$b = \frac{π}{3}$

8b · 3 marks

B1
$x -coordinate = {4 p}$
B1
${- 8}$
B1
$y -coordinate = - 3 q$

9a · 3 marks

B1
$\frac{d y}{d x} = x^{- \frac{1}{2}} \to m = \frac{1}{2} at x = 4$
M1
Equation of normal is $y - 3 = - 2 (x - 4)$
A1
$y = - 2 x + 11$

9b · 2 marks

M1
$\frac{d y}{d t} = their \frac{1}{2} \times 3$
A1
$\frac{3}{2}$

9c · 3 marks

1 mark
$Required gradient [= \frac{d y}{d x}] = - 2$
M1
$\frac{d x}{d t} = \frac{1}{their normal gradient} \times - 5$
A1
$\frac{5}{2}$

10a · 1 mark

B1
Angle $A C O = 0.7$

10b · 1 mark

B1
$[R =] 1.53 r$

10c · 7 marks

B1
$Sector O A B = \frac{1}{2} r^{2} \times 2.8 [= 1.4 r^{2}]$
M1
$Sector C A B = \frac{1}{2} (their R)^{2} \times 2 \times their 0.7$
A1
$1.638 r^{2}$
M1
$[2] \times \frac{1}{2} r^{2} \sin (π - 1.4) OR [2] \times \frac{1}{2} r \times their R \sin 0.7$
A1
$2 \times 0.4927 r^{2}$
1 mark
$1.4 r^{2} - (their 1.638 r^{2} - their 0.985 r^{2})$
A1
$0.747 r^{2} to 0.748 r^{2}$

11a · 4 marks

B1
${- 2 \times 2 \times (2 x - 1)^{- 3} \times 2}$
B1
${1}$
1 mark
$Substitute x = \frac{3}{2} leading to \frac{d y}{d x} = \frac{- 8}{8} + 1 = 0. Hence x -coordinate of R is \frac{3}{2}$
B1
$When x = \frac{3}{2}, y = \frac{2}{4} + \frac{3}{2} = 2$

11b · 6 marks

B1
$y -coordinate of P = 3, y -coordinate of Q = \frac{20}{9}$
B1
${\frac{2 (2 x - 1)^{- 1}}{- 1 \times 2}}$
B1
${\frac{1}{2} x^{2}}$
M1
$[(- \frac{1}{3} + 2) - (- 1 + \frac{1}{2})] = \frac{5}{3} - (- \frac{1}{2})$
M1
$\frac{1}{2} (3 + \frac{20}{9}) = \frac{47}{18}$
A1
$\frac{47}{18} - \frac{13}{6} = \frac{4}{9}$

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