A Level Mathematics 9709 Paper 11, Oct/Nov 2023

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Question paper

9709/11 · Oct/Nov 2023

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Expand $(1 + 3 x)^{6}$ in ascending powers of $x$ up to, and including, the term in $x^{2}$ .

[2]

Hence find the coefficient of $x^{2}$ in the expansion of $(1 - 7 x + x^{2}) (1 + 3 x)^{6}$ .

[2]

A line has equation $y = 2 c x + 3$ and a curve has equation $y = c x^{2} + 3 x - c$ , where $c$ is a constant. Showing all necessary working, determine which of the following statements is correct. A The line and curve intersect only for a particular set of values of $c$ . B The line and curve intersect for all values of $c$ . C The line and curve do not intersect for any values of $c$ .

[4]

The diagram shows a cubical closed container made of a thin elastic material which is filled with water and frozen. During the freezing process the length, $x cm$ , of each edge of the container increases at the constant rate of $0.01 cm per minute$ . The volume of the container at time $t minutes$ is $V {cm}^{3}$ . Find the rate of increase of $V$ when $x = 20$ .

[3]

The transformation R denotes a reflection in the $x$ -axis and the transformation T denotes a translation of $(\begin{array}{c} 3 \\ - 1 \end{array})$ .

Find the equation, $y = g (x)$ , of the curve with equation $y = x^{2}$ after it has been transformed by the sequence of transformations R followed by T.

[2]

Find the equation, $y = h (x)$ , of the curve with equation $y = x^{2}$ after it has been transformed by the sequence of transformations T followed by R.

[2]

State fully the transformation that maps the curve $y = g (x)$ onto the curve $y = h (x)$ .

[2]

Show that the equation $4 \sin x + \frac{5}{\tan x} + \frac{2}{\sin x} = 0$ may be expressed in the form $a \cos^{2} x + b \cos x + c = 0$ , where $a$ , $b$ and $c$ are integers to be found.

[3]

Hence solve the equation $4 \sin x + \frac{5}{\tan x} + \frac{2}{\sin x} = 0$ for $0^{\circ} ⩽ x ⩽ 360^{\circ}$ .

[3]

The diagram shows a motif formed by the major arc $A B$ of a circle with radius $r$ and centre $O$ , and the minor arc $A O B$ of a circle, also with radius $r$ but with centre $C$ . The point $C$ lies on the circle with centre $O$ .

Given that angle $A C B = k π radians$ , state the value of the fraction $k$ .

[1]

State the perimeter of the shaded motif in terms of $π$ and $r$ .

[1]

Find the area of the shaded motif, giving your answer in terms of $π$ , $r$ and $\sqrt{3}$ .

[5]

The sum of the first two terms of a geometric progression is 15 and the sum to infinity is $\frac{125}{7}$ . The common ratio of the progression is negative. Find the third term of the progression.

[7]

The diagram shows the curves with equations $y = 2 (2 x - 3)^{4}$ and $y = (2 x - 3)^{2} + 1$ meeting at points $A$ and $B$ .

By using the substitution $u = 2 x - 3$ find, by calculation, the coordinates of $A$ and $B$ .

[4]

Find the exact area of the shaded region.

[5]

Express $4 x^{2} - 12 x + 13$ in the form $(2 x + a)^{2} + b$ , where $a$ and $b$ are constants.

[2]

The function $f$ is defined by $f (x) = 4 x^{2} - 12 x + 13$ for $p < x < q$ , where $p$ and $q$ are constants. The function $g$ is defined by $g (x) = 3 x + 1$ for $x < 8$ . Given that it is possible to form the composite function $g f$ , find the least possible value of $p$ and the greatest possible value of $q$ .

[3]

Find an expression for $g f (x)$ .

[1]

The function $h$ is defined by $h (x) = 4 x^{2} - 12 x + 13$ for $x < 0$ . Find an expression for $h^{- 1} (x)$ .

[3]

A curve has a stationary point at $(2, - 10)$ and is such that $\frac{d^{2} y}{d x^{2}} = 6 x$ .

Find $\frac{d y}{d x}$ .

[3]

Find the equation of the curve.

[3]

Find the coordinates of the other stationary point and determine its nature.

[3]

Find the equation of the tangent to the curve at the point where the curve crosses the $y$ -axis.

[2]

The diagram shows the circle with equation $(x - 4)^{2} + (y + 1)^{2} = 40$ . Parallel tangents, each with gradient 1, touch the circle at points $A$ and $B$ .

Find the equation of the line $A B$ , giving the answer in the form $y = m x + c$ .

[3]

Find the coordinates of $A$ , giving each coordinate in surd form.

[4]

Find the equation of the tangent at $A$ , giving the answer in the form $y = m x + c$ , where $c$ is in surd form.

[2]

Mark scheme

9709/11 · Oct/Nov 2023

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1a · 2 marks

B2
$1 + 18 x + 135 x^{2}$

1b · 2 marks

M1
3 products, allow $10 x^{2}$
A1
10

2 · 4 marks

M1
$c x^{2} + 3 x - c = 2 c x + 3$ leading to $c x^{2} + (3 - 2 c) x - (c + 3) [= 0]$
M1
$b^{2} - 4 a c = (3 - 2 c)^{2} + 4 c (c + 3)$
A1
$= 8 c^{2} + 9$
A1
$> 0$ [for all values of $c$ ] leading to B [Intersects for all values of $c$ ]

3 · 3 marks

B1
$\frac{d V}{d x} = 3 x^{2}$
M1
$\frac{d V}{d t} = [\frac{d V}{d x} \times \frac{d x}{d t}] = 3 \times 20^{2} \times 0.01$
A1
$12$

4a · 2 marks

B1
$- (x - 3)^{2}$
B1
$- 1$

4b · 2 marks

B1
$- (x - 3)^{2}$
B1
$+ 1$

4c · 2 marks

B2
${Translation} {(\begin{array}{c} 0 \\ 2 \end{array})}$

5a · 3 marks

M1
$4 \sin^{2} x + 5 \cos x + 2 [= 0]$
1 mark
$4 (1 - \cos^{2} x) + 5 \cos x + 2 [= 0]$
A1
$4 \cos^{2} x - 5 \cos x - 6 = 0$

5b · 3 marks

M1
$(4 \cos x + 3) (\cos x - 2) [= 0]$
A1
${138.6}^{\circ}$
B1
${221.4}^{\circ}$

6a · 1 mark

B1
$k = \frac{2}{3}$

6b · 1 mark

B1
Perimeter of shaded area $= 2 π r$

6c · 5 marks

M1
Major sector $O A B = \frac{1}{2} r^{2} \times \frac{4 π}{3}$
M1
One or both segments $= [2] \times (\frac{1}{2} r^{2} \times \frac{π}{3} - \frac{1}{2} r^{2} \sin \frac{π}{3})$
A1
$= [2] (r^{2} \frac{π}{6} - r^{2} \frac{\sqrt{3}}{4})$
1 mark
Shaded area $= \frac{2}{3} π r^{2} - 2 (\frac{1}{6} π r^{2} - \frac{r^{2} \sqrt{3}}{4})$
A1
$= \frac{π r^{2}}{3} + \frac{r^{2} \sqrt{3}}{2}$

7 · 7 marks

B1
$a (1 + r) = 15$
B1
$\frac{a}{1 - r} = \frac{125}{7}$
M1
$\frac{125}{7} (1 - r) (1 + r) = 15$
M1
$1 - r^{2} = \frac{105}{125}$
A1
$r^{2} = \frac{4}{25}$ leading to $r = - \frac{2}{5}$
A1
$a = \frac{125}{7} \times \frac{7}{5} = 25$
A1
3rd term $= 25 \times \frac{4}{25} = 4$

8a · 4 marks

B1
$u = 2 x - 3$ leading to $2 u^{4} = u^{2} + 1$ leading to $2 u^{4} - u^{2} - 1 [= 0]$
M1
$(2 u^{2} + 1) (u^{2} - 1) [= 0]$
A1
$u = \pm 1$ leading to $2 x - 3 = \pm 1$ leading to $x = 1$ or $2$
A1
$(1, 2), (2, 2)$

8b · 5 marks

B1
$\frac{(2 x - 3)^{3}}{3 \times 2} + x$
B1
$\frac{2 (2 x - 3)^{5}}{5 \times 2}$
M1
$(\frac{1}{6} + 2) - (- \frac{1}{6} + 1) - {\frac{1}{5} - (- \frac{1}{5})}$
M1
$\frac{4}{3} - \frac{2}{5}$
A1
$\frac{14}{15}$

9a · 2 marks

B1
$(2 x - 3)^{2}$
B1
$+ 4$

9b · 3 marks

M1
$t h e i r (2 x - 3)^{2} + 4 [<] 8$ OR $4 x^{2} - 12 x + 13 [<] 8$
1 mark
$(2 x - 3)^{2} < 4$ leading to $- 2 < 2 x - 3 < 2$ OR $4 x^{2} - 12 x + 5 < 0$ leading to $(2 x - 1) (2 x - 5) < 0$
A1
$\frac{1}{2} < x < 2 \frac{1}{2}$ leading to [LEAST] $p = \frac{1}{2}$ , [GREATEST] $q = 2 \frac{1}{2}$

9c · 1 mark

B1
$g f (x) = 12 x^{2} - 36 x + 40$

9d · 3 marks

M1
$y = (2 x - 3)^{2} + 4$ leading to $(2 x - 3)^{2} = y - 4$ leading to $2 x - 3 = [\pm] \sqrt{y - 4}$
1 mark
$2 x = 3 [\pm] \sqrt{y - 4}$ leading to $x = \frac{3}{2} [\pm] \frac{\sqrt{y - 4}}{2}$
A1
$h^{- 1} (x) = \frac{3}{2} - \frac{\sqrt{x - 4}}{2}$

10a · 3 marks

B1
$\frac{d y}{d x} = 3 x^{2} [+ c]$
M1
$3 \times 2^{2} + c = 0$
A1
$\frac{d y}{d x} = 3 x^{2} - 12$

10b · 3 marks

B1
$y = x^{3} - 12 x [+ k]$
M1
$- 10 = 2^{3} - 12 \times 2 + k$
A1
$y = x^{3} - 12 x + 6$

10c · 3 marks

M1
$3 x^{2} - 12 = 0$ [leading to $x = - 2$ ]
A1
$y = (- 2)^{3} - 12 (- 2) + 6 = 22$ leading to $(- 2, 22)$
A1
When $x = - 2, \frac{d^{2} y}{d x^{2}} < 0$ (or $- 12$ ) hence Maximum

10d · 2 marks

M1
At $x = 0, \frac{d y}{d x} = - 12, y = 6$
A1
$y - 6 = - 12 x$

11a · 3 marks

B1
Gradient of $A B = - 1$
B1
Centre of circle $= (4, - 1)$
B1
Equation of $A B$ is $y + 1 = - 1 (x - 4)$ leading to $y = - x + 3$

11b · 4 marks

M1
$(x - 4)^{2} + (- x + 3 + 1)^{2} = 40$
1 mark
$2 (x - 4)^{2} = 40$ OR $[2] (x^{2} - 8 x - 4)$ leading to $\frac{8 \pm \sqrt{64 + 16}}{2}$ or $\frac{16 \pm \sqrt{256 + 64}}{4}$
A1
$x = 4 [\pm] \sqrt{20}$
A1
$(4 - \sqrt{20}, - 1 + \sqrt{20})$

11c · 2 marks

M1
$y - their (- 1 + \sqrt{20}) = 1 {x - their (4 - \sqrt{20})}$
A1
$y = x - 5 + 2 \sqrt{20}$ or $y = x - 5 + \sqrt{80}$ or $y = x - 5 + 4 \sqrt{5}$

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