A Level Mathematics 9709 Paper 12, Oct/Nov 2024

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9709/12 · Oct/Nov 2024

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The diagram shows the curve with equation $y = a \sin (b x) + c$ for $0 ⩽ x ⩽ 2 π$ , where $a$ , $b$ and $c$ are positive constants.

State the values of $a$ , $b$ and $c$ .

[3]

b(i)

For these values of $a$ , $b$ and $c$ , determine the number of solutions in the interval $0 ⩽ x ⩽ 2 π$ for each of the following equations: $a \sin (b x) + c = 7 - x$

[1]

b(ii)

$a \sin (b x) + c = 2 π (x - 1)$ .

[1]

The first term of an arithmetic progression is $- 20$ and the common difference is 5.

Find the sum of the first 20 terms of the progression.

[2]

It is given that the sum of the first $2 k$ terms is 10 times the sum of the first $k$ terms. Find the value of $k$ .

[3]

The equation of a curve is $y = 2 x^{2} - 3$ . Two points $A$ and $B$ with $x$ -coordinates 2 and $(2 + h)$ respectively lie on the curve.

Find and simplify an expression for the gradient of the chord $A B$ in terms of $h$ .

[3]

Explain how the gradient of the curve at the point $A$ can be deduced from the answer to part (a), and state the value of this gradient.

[2]

Find the term independent of $x$ in the expansion of each of the following:

${(x + \frac{3}{x^{2}})}^{6}$

[2]

$(4 x^{3} - 5) {(x + \frac{3}{x^{2}})}^{6}$ .

[4]

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

a(i)

State the value of $f (- 1)$ .

[1]

a(ii)

The diagram shows the graph of $y = f (x)$ . Sketch the graph of $y = f^{- 1} (x)$ on this diagram. Show any relevant mirror line.

[2]

a(iii)

Find an expression for $f^{- 1} (x)$ and state the domain of the function $f^{- 1}$ .

[4]

The function $g$ is defined by $g (x) = 3 x + 2$ for $x \in R$ . Solve the equation $f (x) = g f (\frac{1}{4})$ .

[3]

The diagram shows a metal plate $O A B C D E F$ consisting of sectors of two circles, each with centre $O$ . The radii of sectors $A O B$ and $E O F$ are $r cm$ and the radius of sector $C O D$ is $2 r cm$ . Angle $A O B = angle E O F = θ radians$ and angle $C O D = 2 θ radians$ . It is given that the perimeter of the plate is $14 cm$ and the area of the plate is $10 {cm}^{2}$ . Given that $r > \frac{3}{2}$ and $θ < \frac{3}{4}$ , find the values of $r$ and $θ$ .

[6]

By expressing $- 2 x^{2} + 8 x + 11$ in the form $- a (x - b)^{2} + c$ , where $a$ , $b$ and $c$ are positive integers, find the coordinates of the vertex of the graph with equation $y = - 2 x^{2} + 8 x + 11$ .

[3]

The diagram shows part of the curve with equation $y = - 2 x^{2} + 8 x + 11$ and the line with equation $y = 8 x + 9$ . Find the area of the shaded region.

[5]

The equation of a circle is $x^{2} + y^{2} + p x + 2 y + q = 0$ , where $p$ and $q$ are constants.

Express the equation in the form $(x - a)^{2} + (y - b)^{2} = r^{2}$ , where $a$ is to be given in terms of $p$ and $r^{2}$ is to be given in terms of $p$ and $q$ .

[2]

b(i)

The line with equation $x + 2 y = 10$ is the tangent to the circle at the point $A (4, 3)$ . Find the equation of the normal to the circle at the point $A$ .

[3]

b(ii)

Find the values of $p$ and $q$ .

[5]

The equation of a curve is $y = \frac{1}{2} k^{2} x^{2} - 2 k x + 2$ and the equation of a line is $y = k x + p$ , where $k$ and $p$ are constants with $0 < k < 1$ .

It is given that one of the points of intersection of the curve and the line has coordinates $(\frac{5}{2}, \frac{1}{2})$ . Find the values of $k$ and $p$ , and find the coordinates of the other point of intersection.

[7]

It is given instead that the line and the curve do not intersect. Find the set of possible values of $p$ .

[3]

A function $f$ with domain $x > 0$ is such that $f^{'} (x) = 8 (2 x - 3)^{\frac{1}{3}} - 10 x^{\frac{2}{3}}$ . It is given that the curve with equation $y = f (x)$ passes through the point $(1, 0)$ .

Find the equation of the normal to the curve at the point $(1, 0)$ .

[3]

Find $f (x)$ .

[4]

It is given that the equation $f^{'} (x) = 0$ can be expressed in the form $125 x^{2} - 128 x + 192 = 0$ . Determine, making your reasoning clear, whether $f$ is an increasing function, a decreasing function or neither.

[3]

Mark scheme

9709/12 · Oct/Nov 2024

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1a · 3 marks

B1
$a = 4$
B1
$b = 2$
B1
$c = 3$

1b(i) · 1 mark

B1
5

1b(ii) · 1 mark

B1
1

2a · 2 marks

M1
$\frac{20}{2} (2 \times - 20 + (20 - 1) \times 5)$ or $\frac{20}{2} (- 20 + 75)$
A1
550

2b · 3 marks

M1
$\frac{2 k}{2} (- 40 + (2 k - 1) \times 5)$ or $\frac{k}{2} (- 40 + (k - 1) \times 5)$
1 mark
$[- 40 k + 10 k^{2} - 5 k = - 200 k + 25 k^{2} - 25 k \Rightarrow] 15 k^{2} - 180 k = 0$
A1
$k = 12$

3a · 3 marks

B1
$[f (2 + h) =] 2 (2 + h)^{2} - 3$
M1
$\frac{(2 (2 + h)^{2} - 3) - 5}{(2 + h) - 2} [= \frac{2 h^{2} + 8 h}{h}]$
A1
$2 h + 8$ or $2 (h + 4)$

3b · 2 marks

B1
$h \to 0$ , or chord [AB] $\to$ tangent [at A]
B1
8

4a · 2 marks

B1
15 or $(\binom{6}{2}) \times x^{4} {(\frac{3}{x^{2}})}^{2}$ or $x^{6} \times \frac{6 \times 5}{2} {(\frac{3}{x^{3}})}^{2}$
B1
135

4b · 4 marks

B1
20 or $(\binom{6}{3}) \times x^{3} {(\frac{3}{x^{2}})}^{3}$ or $x^{6} \times \frac{6.5.4}{3!} {(\frac{3}{x^{3}})}^{3}$
B1
$= \frac{540}{x^{3}}$
M1
$4 \times 540 - 5 \times 135$
A1
1485

5a(i) · 1 mark

B1
$[f (- 1) =] \frac{1}{3}$

5a(ii) · 2 marks

B1
For showing the correct mirror line.
B1
For correct shape: the curves should intersect in the first square in the third quadrant. To the left of the point of intersection, the reflection is below the original and crosses the $x$ -axis. To the right of the point of intersection, the reflection is to the right the original.

5a(iii) · 4 marks

M1
$\frac{2 x + 1}{2 x - 1} = y \Rightarrow 2 x + 1 = y (2 x - 1)$
1 mark
$2 x y - 2 x = y + 1$
A1
$\frac{x + 1}{2 (x - 1)}, \frac{- x - 1}{2 - 2 x}$
B1
$[Domain of f^{- 1} is] x < 1$

5b · 3 marks

B1
$gf (\frac{1}{4}) = - 7$
M1
$\frac{2 x + 1}{2 x - 1} = - 7$
A1
$[x =] \frac{3}{8}$

6 · 6 marks

B1
$[Perimeter =] r + r θ + r + 2 r \times 2 θ + r + r θ + r [= 4 r + 6 r θ]$
B1
$[Area =] \frac{1}{2} r^{2} θ + \frac{1}{2} (2 r)^{2} \times 2 θ + \frac{1}{2} r^{2} θ [= 5 r^{2} θ]$
M1
$4 r + 6 r θ = 14$ and $5 r^{2} θ = 10$
1 mark
$5 r^{2} \frac{14 - 4 r}{6 r} = 10$ or $4 r + 6 r (\frac{10}{5 r^{2}}) = 14$ OR $5 {(\frac{14}{4 + 6 θ})}^{2} θ = 10$ or $4 \sqrt{\frac{10}{5 θ}} + 6 θ \sqrt{\frac{10}{5 θ}} = 14$
1 mark
$[\Rightarrow 2 r^{2} - 7 r + 6 = 0 \Rightarrow] (r - 2) (2 r - 3) = 0$ OR $[\Rightarrow 18 θ^{2} - 25 θ + 8 = 0 \Rightarrow] (9 θ - 8) (2 θ - 1) = 0$
B1
Then $r = 2$ and $θ = 0.5$

7a · 3 marks

M1
$- 2 ((x \pm p)^{2} \pm q)$ or $- 2 (x \pm p)^{2} \pm q$
1 mark
$- 2 ((x - 2)^{2} \pm q)$ or $- 2 (x - 2)^{2} \pm q$
A1
$- 2 (x - 2)^{2} + 19$ and $(2, 19)$

7b · 5 marks

B1
$[x =] \pm 1$
M1
Subtract and attempt to integrate / Attempt to integrate and subtract
B1
Both terms / All terms correct integrated expression
M1
Apply their limits, one positive and one negative, obtained from equating the line and the curve to their integrated expression.
1 mark
$= \frac{8}{3}, 2 \frac{2}{3}$

8a · 2 marks

B1
${(x - (- \frac{1}{2} p))}^{2} + (y - (- 1))^{2}$ OE
1 mark
${(x - (- \frac{1}{2} p))}^{2} + (y - (- 1))^{2} = - q + 1 + {(- \frac{1}{2} p)}^{2}$ OE

8b(i) · 3 marks

B1
$[Gradient of tangent =] - \frac{1}{2}$
M1
$[Gradient of normal =] 2$
A1
$\frac{y - 3}{x - 4} = 2 [y = 2 x - 5]$

8b(ii) · 5 marks

M1
Using their stated centre or $(\frac{\pm p}{2}, \pm 1)$ in their equation of the normal.
A1
$p = - 4$
M1
$r^{2} = (4 - 2)^{2} + (3 - (- 1))^{2} [= 20]$ or distance from point to line or substituting $(4, 3)$ into circle equation
1 mark
Using their expression for $r^{2}$ (from (a)) equated to their 20 (or substituting their $p = - 4$ ).
A1
$q = - 15$

9a · 7 marks

M1
$[\frac{1}{2} k^{2} \times \frac{25}{4} - 2 k \times \frac{5}{2} + 2 = \frac{1}{2}]$ OR $[\frac{1}{2} k^{2} \times \frac{25}{4} - 2 k \times \frac{5}{2} + 2 = k \times \frac{5}{2} + (\frac{1}{2} - \frac{5}{2} k)]$ $25 k^{2} - 40 k + 12 [= 0]$
A1
$k = \frac{2}{5}$
1 mark
$\frac{1}{2} = (their \frac{2}{5}) (\frac{5}{2}) + p \Rightarrow p =$
A1
$p = - \frac{1}{2}$
1 mark
$\frac{2}{25} x^{2} - \frac{6}{5} x + \frac{5}{2} [= 0] [4 x^{2} - 60 x + 125 [= 0]]$
A1
$x = \frac{25}{2}$
A1
$y = \frac{9}{2}$

9b · 3 marks

M1
$[\frac{1}{2} k^{2} x^{2} - 2 k x + 2 = k x + p \Rightarrow] \frac{1}{2} k^{2} x^{2} - 3 k x + 2 - p$
1 mark
$9 k^{2} - 4 \times \frac{1}{2} k^{2} (2 - p)$
A1
$p < - \frac{5}{2}$

10a · 3 marks

B1
$- 18$
M1
$\frac{1}{18}$
A1
$\frac{y [- 0]}{x - 1} = \frac{1}{18}$

10b · 4 marks

B1
$8 (2 x - 3)^{\frac{4}{3}} \cdot \frac{1}{2} \cdot \frac{1}{\frac{4}{3}}$
B1
$- 10 x^{\frac{5}{3}} \cdot \frac{1}{\frac{5}{3}} [+ c]$
M1
$0 = 3 (2 (1) - 3)^{\frac{4}{3}} - 6 (1)^{\frac{5}{3}} + c [0 = 3 - 6 + c]$
A1
$[f (x) or y =] 3 (2 x - 3)^{\frac{4}{3}} - 6 x^{\frac{5}{3}} + 3$

10c · 3 marks

M1
$b^{2} - 4 a c = 128^{2} - 4 \times 125 \times 192$ and stating " $< 0$ " OR use of the quadratic formula and stating "No solutions" OR completing the square for the given quadratic and stating positive or $> 0$ . OR sketch of the given quadratic and stating positive.
1 mark
No turning points [in the original function.]
A1
Decreasing because $f^{'} (any positive x value) < 0$

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