Questions - AQA C4 - A-Level Maths

Use the Factor Theorem to show that ( $2 x + 3$ ) is a factor of $\mathrm { f } ( x )$.
Write $\mathrm { f } ( x )$ in the form $( 2 x + 3 ) \left( a x ^ { 2 } + b x + c \right)$, where $a , b$ and $c$ are integers.
1. Show that the equation $2 \cos 2 \theta \sin \theta + 9 \sin \theta + 3 = 0$ can be written as $4 x ^ { 3 } - 11 x - 3 = 0$, where $x = \sin \theta$.
2. Hence find all solutions of the equation $2 \cos 2 \theta \sin \theta + 9 \sin \theta + 3 = 0$ in the interval $0 ^ { \circ } < \theta < 360 ^ { \circ }$, giving your solutions to the nearest degree.

AQA C4 2013 June Q6

6 The points $A , B$ and $C$ have coordinates $( 3 , - 2,4 ) , ( 1 , - 5,6 )$ and $( - 4,5 , - 1 )$ respectively. The line $l$ passes through $A$ and has equation $\mathbf { r } = \left[ \begin{array} { r } 3
- 2
4 \end{array} \right] + \lambda \left[ \begin{array} { r } 7
- 7
5 \end{array} \right]$.

Show that the point $C$ lies on the line $l$.
Find a vector equation of the line that passes through points $A$ and $B$.
The point $D$ lies on the line through $A$ and $B$ such that the angle $C D A$ is a right angle. Find the coordinates of $D$.
The point $E$ lies on the line through $A$ and $B$ such that the area of triangle $A C E$ is three times the area of triangle $A C D$. Find the coordinates of the two possible positions of $E$.

AQA C4 2013 June Q7

7 The height of the tide in a certain harbour is $h$ metres at time $t$ hours. Successive high tides occur every 12 hours. The rate of change of the height of the tide can be modelled by a function of the form $a \cos ( k t )$, where $a$ and $k$ are constants. The largest value of this rate of change is 1.3 metres per hour. Write down a differential equation in the variables $h$ and $t$. State the values of the constants $a$ and $k$.

AQA C4 2013 June Q8

$\quad$ Find $\int t \cos \left( \frac { \pi } { 4 } t \right) \mathrm { d } t$.
The platform of a theme park ride oscillates vertically. For the first 75 seconds of the ride, $$\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { t \cos \left( \frac { \pi } { 4 } t \right) } { 32 x }$$ where $x$ metres is the height of the platform above the ground after time $t$ seconds.
At $t = 0$, the height of the platform above the ground is 4 metres.
Find the height of the platform after 45 seconds, giving your answer to the nearest centimetre.
(6 marks)

AQA C4 2014 June Q1

1 A curve is defined by the parametric equations $x = \frac { t ^ { 2 } } { 2 } + 1 , y = \frac { 4 } { t } - 1$.

Find the gradient at the point on the curve where $t = 2$.
Find a Cartesian equation of the curve.
\includegraphics[max width=\textwidth, alt={}]{9f03a5f3-7fea-4fb7-b3bd-b4c0cdf662a2-02_1730_1709_977_153}

AQA C4 2014 June Q2

4 marks

Given that $\frac { 4 x ^ { 3 } - 2 x ^ { 2 } + 16 x - 3 } { 2 x ^ { 2 } - x + 2 }$ can be expressed as $A x + \frac { B ( 4 x - 1 ) } { 2 x ^ { 2 } - x + 2 }$, find the values of the constants $A$ and $B$.
The gradient of a curve is given by $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 4 x ^ { 3 } - 2 x ^ { 2 } + 16 x - 3 } { 2 x ^ { 2 } - x + 2 }$$ The point $( - 1,2 )$ lies on the curve. Find the equation of the curve.
[0pt] [4 marks]

AQA C4 2014 June Q3

4 marks

Find the binomial expansion of $( 1 - 4 x ) ^ { \frac { 1 } { 4 } }$ up to and including the term in $x ^ { 2 }$.
[0pt] [2 marks]
Find the binomial expansion of $( 2 + 3 x ) ^ { - 3 }$ up to and including the term in $x ^ { 2 }$.
Hence find the binomial expansion of $\frac { ( 1 - 4 x ) ^ { \frac { 1 } { 4 } } } { ( 2 + 3 x ) ^ { 3 } }$ up to and including the term in $x ^ { 2 }$.
[0pt] [2 marks]

AQA C4 2014 June Q4

1 marks

4 A painting was valued on 1 April 2001 at $\pounds 5000$.
The value of this painting is modelled by $$V = A p ^ { t }$$ where $\pounds V$ is the value $t$ years after 1 April 2001, and $A$ and $p$ are constants.

Write down the value of $A$.
According to the model, the value of this painting on 1 April 2011 was $\pounds 25000$. Using this model:
1. show that $p ^ { 10 } = 5$;
2. use logarithms to find the year in which the painting will be valued at $\pounds 75000$.
A painting by another artist was valued at $\pounds 2500$ on 1 April 1991. The value of this painting is modelled by $$W = 2500 q ^ { t }$$ where $\pounds W$ is the value $t$ years after 1 April 1991, and $q$ is a constant.
1. Show that, according to the two models, the value of the two paintings will be the same $T$ years after 1 April 1991, $$\text { where } T = \frac { \ln \left( \frac { 5 } { 2 } \right) } { \ln \left( \frac { p } { q } \right) }$$
2. Given that $p = 1.029 q$, find the year in which the two paintings will have the same value.
  [0pt] [1 mark]

AQA C4 2014 June Q5

3 marks

1. Express $3 \sin x + 4 \cos x$ in the form $R \sin ( x + \alpha )$ where $R > 0$ and $0 ^ { \circ } < \alpha < 90 ^ { \circ }$, giving your value of $\alpha$ to the nearest $0.1 ^ { \circ }$.
2. Hence solve the equation $3 \sin 2 \theta + 4 \cos 2 \theta = 5$ in the interval $0 ^ { \circ } < \theta < 360 ^ { \circ }$, giving your solutions to the nearest $0.1 ^ { \circ }$.
1. Show that the equation $\tan 2 \theta \tan \theta = 2$ can be written as $2 \tan ^ { 2 } \theta = 1$.
2. Hence solve the equation $\tan 2 \theta \tan \theta = 2$ in the interval $0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }$, giving your solutions to the nearest $0.1 ^ { \circ }$.
1. Use the Factor Theorem to show that $2 x - 1$ is a factor of $8 x ^ { 3 } - 4 x + 1$.
2. Show that $4 \cos 2 \theta \cos \theta + 1$ can be written as $8 x ^ { 3 } - 4 x + 1$ where $x = \cos \theta$.
3. Given that $\theta = 72 ^ { \circ }$ is a solution of $4 \cos 2 \theta \cos \theta + 1 = 0$, use the results from parts (c)(i) and (c)(ii) to show that the exact value of $\cos 72 ^ { \circ }$ is $\frac { ( \sqrt { 5 } - 1 ) } { p }$ where $p$ is an integer.
  [0pt] [3 marks]

AQA C4 2014 June Q6

2 marks

6 The line $l _ { 1 }$ has equation $\mathbf { r } = \left[ \begin{array} { r } 4
- 5
3 \end{array} \right] + \lambda \left[ \begin{array} { r } - 1
3
1 \end{array} \right]$.
The line $l _ { 2 }$ has equation $\mathbf { r } = \left[ \begin{array} { r } 7
- 8
6 \end{array} \right] + \mu \left[ \begin{array} { r } 2
- 3
1 \end{array} \right]$.
The point $P$ lies on $l _ { 1 }$ where $\lambda = - 1$. The point $Q$ lies on $l _ { 2 }$ where $\mu = 2$.

Show that the vector $\overrightarrow { P Q }$ is parallel to $\left[ \begin{array} { r } 1
- 1
1 \end{array} \right]$.
The lines $l _ { 1 }$ and $l _ { 2 }$ intersect at the point $R ( 3 , b , c )$.
1. Show that $b = - 2$ and find the value of $c$.
2. The point $S$ lies on a line through $P$ that is parallel to $l _ { 2 }$. The line $R S$ is perpendicular to the line $P Q$.
  \includegraphics[max width=\textwidth, alt={}, center]{9f03a5f3-7fea-4fb7-b3bd-b4c0cdf662a2-16_887_1159_1320_443} Find the coordinates of $S$.
  $7 \quad$ A curve has equation $\cos 2 y + y \mathrm { e } ^ { 3 x } = 2 \pi$.
  The point $A \left( \ln 2 , \frac { \pi } { 4 } \right)$ lies on this curve.

AQA C4 2014 June Q8

7 marks

Express $\frac { 16 x } { ( 1 - 3 x ) ( 1 + x ) ^ { 2 } }$ in the form $\frac { A } { 1 - 3 x } + \frac { B } { 1 + x } + \frac { C } { ( 1 + x ) ^ { 2 } }$.
Solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 16 x \mathrm { e } ^ { 2 y } } { ( 1 - 3 x ) ( 1 + x ) ^ { 2 } }$$ where $y = 0$ when $x = 0$.
Give your answer in the form $\mathrm { f } ( y ) = \mathrm { g } ( x )$.
[0pt] [7 marks]

AQA C4 2015 June Q1

6 marks

1 It is given that $\mathrm { f } ( x ) = \frac { 19 x - 2 } { ( 5 - x ) ( 1 + 6 x ) }$ can be expressed as $\frac { A } { 5 - x } + \frac { B } { 1 + 6 x }$, where $A$ and $B$ are integers.

Find the values of $A$ and $B$.
Hence show that $\int _ { 0 } ^ { 4 } \mathrm { f } ( x ) \mathrm { d } x = k \ln 5$, where $k$ is a rational number.
[0pt] [6 marks]

AQA C4 2015 June Q2

3 marks

Express $2 \cos x - 5 \sin x$ in the form $R \cos ( x + \alpha )$, where $R > 0$ and $0 < \alpha < \frac { \pi } { 2 }$, giving your value of $\alpha$, in radians, to three significant figures.
1. Hence find the value of $x$ in the interval $0 < x < 2 \pi$ for which $2 \cos x - 5 \sin x$ has its maximum value. Give your value of $x$ to three significant figures.
2. Use your answer to part (a) to solve the equation $2 \cos x - 5 \sin x + 1 = 0$ in the interval $0 < x < 2 \pi$, giving your solutions to three significant figures.
  [0pt] [3 marks]

AQA C4 2015 June Q3

4 marks

The polynomial $\mathrm { f } ( x )$ is defined by $\mathrm { f } ( x ) = 8 x ^ { 3 } - 12 x ^ { 2 } - 2 x + d$, where $d$ is a constant. When $\mathrm { f } ( x )$ is divided by ( $2 x + 1$ ), the remainder is - 2 . Use the Remainder Theorem to find the value of $d$.
The polynomial $\mathrm { g } ( x )$ is defined by $\mathrm { g } ( x ) = 8 x ^ { 3 } - 12 x ^ { 2 } - 2 x + 3$.
1. Given that $x = - \frac { 1 } { 2 }$ is a solution of the equation $\mathrm { g } ( x ) = 0$, write $\mathrm { g } ( x )$ as a product of three linear factors.
2. The function h is defined by $\mathrm { h } ( x ) = \frac { 4 x ^ { 2 } - 1 } { \mathrm {~g} ( x ) }$ for $x > 2$. Simplify $\mathrm { h } ( x )$, and hence show that h is a decreasing function.
  [0pt] [4 marks]

AQA C4 2015 June Q4

2 marks

Find the binomial expansion of $( 1 + 5 x ) ^ { \frac { 1 } { 5 } }$ up to and including the term in $x ^ { 2 }$.
1. Find the binomial expansion of $( 8 + 3 x ) ^ { - \frac { 2 } { 3 } }$ up to and including the term in $x ^ { 2 }$.
2. Use your expansion from part (b)(i) to find an estimate for $\sqrt [ 3 ] { \frac { 1 } { 81 } }$, giving your answer to four decimal places.
  [0pt] [2 marks]

AQA C4 2015 June Q5

5 marks

5 A curve is defined by the parametric equations $x = \cos 2 t , y = \sin t$.
The point $P$ on the curve is where $t = \frac { \pi } { 6 }$.

Find the gradient at $P$.
Find the equation of the normal to the curve at $P$ in the form $y = m x + c$.
The normal at $P$ intersects the curve again at the point $Q ( \cos 2 q , \sin q )$. Use the equation of the normal to form a quadratic equation in $\sin q$ and hence find the $x$-coordinate of $Q$.
[0pt] [5 marks]

AQA C4 2015 June Q6

4 marks

6 The points $A$ and $B$ have coordinates $( 3,2,10 )$ and $( 5 , - 2,4 )$ respectively.
The line $l$ passes through $A$ and has equation $\mathbf { r } = \left[ \begin{array} { r } 3
2
10 \end{array} \right] + \lambda \left[ \begin{array} { r } 3
1
- 2 \end{array} \right]$.

Find the acute angle between $l$ and the line $A B$.
The point $C$ lies on $l$ such that angle $A B C$ is $90 ^ { \circ }$.
\includegraphics[max width=\textwidth, alt={}, center]{fdd3905e-11f7-4b20-adfe-4c686018a221-12_360_339_762_852} Find the coordinates of $C$.
The point $D$ is such that $B D$ is parallel to $A C$ and angle $B C D$ is $90 ^ { \circ }$. The point $E$ lies on the line through $B$ and $D$ and is such that the length of $D E$ is half that of $A C$. Find the coordinates of the two possible positions of $E$.
[0pt] [4 marks]

AQA C4 2015 June Q7

7 A curve has equation $y ^ { 3 } + 2 \mathrm { e } ^ { - 3 x } y - x = k$, where $k$ is a constant.
The point $P \left( \ln 2 , \frac { 1 } { 2 } \right)$ lies on this curve.

Show that the exact value of $k$ is $q - \ln 2$, where $q$ is a rational number.
Find the gradient of the curve at $P$.

AQA C4 2015 June Q8

2 marks

A pond is initially empty and is then filled gradually with water. After $t$ minutes, the depth of the water, $x$ metres, satisfies the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { \sqrt { 4 + 5 x } } { 5 ( 1 + t ) ^ { 2 } }$$ Solve this differential equation to find $x$ in terms of $t$.
Another pond is gradually filling with water. After $t$ minutes, the surface of the water forms a circle of radius $r$ metres. The rate of change of the radius is inversely proportional to the area of the surface of the water.
1. Write down a differential equation, in the variables $r$ and $t$ and a constant of proportionality, which represents how the radius of the surface of the water is changing with time.
  (You are not required to solve your differential equation.)
2. When the radius of the pond is 1 metre, the radius is increasing at a rate of 4.5 metres per second. Find the radius of the pond when the radius is increasing at a rate of 0.5 metres per second.
  [0pt] [2 marks]
  \includegraphics[max width=\textwidth, alt={}]{fdd3905e-11f7-4b20-adfe-4c686018a221-18_1277_1709_1430_153}
  
  \includegraphics[max width=\textwidth, alt={}]{fdd3905e-11f7-4b20-adfe-4c686018a221-20_2288_1707_221_153}

AQA C4 2016 June Q1

1 marks

Express $\frac { 19 x - 3 } { ( 1 + 2 x ) ( 3 - 4 x ) }$ in the form $\frac { A } { 1 + 2 x } + \frac { B } { 3 - 4 x }$.
1. Find the binomial expansion of $\frac { 19 x - 3 } { ( 1 + 2 x ) ( 3 - 4 x ) }$ up to and including the term in $x ^ { 2 }$.
2. State the range of values of $x$ for which this expansion is valid.
  [0pt] [1 mark]

AQA C4 2016 June Q3

4 marks

Express $\frac { 3 + 13 x - 6 x ^ { 2 } } { 2 x - 3 }$ in the form $A x + B + \frac { C } { 2 x - 3 }$.
Show that $\int _ { 3 } ^ { 6 } \frac { 3 + 13 x - 6 x ^ { 2 } } { 2 x - 3 } \mathrm {~d} x = p + q \ln 3$, where $p$ and $q$ are rational numbers.
[0pt] [4 marks]

AQA C4 2016 June Q4

4 marks

4 The mass of radioactive atoms in a substance can be modelled by the equation $$m = m _ { 0 } k ^ { t }$$ where $m _ { 0 }$ grams is the initial mass, $m$ grams is the mass after $t$ days and $k$ is a constant. The value of $k$ differs from one substance to another.

1. A sample of radioactive iodine reduced in mass from 24 grams to 12 grams in 8 days. Show that the value of the constant $k$ for this substance is 0.917004 , correct to six decimal places.
2. A similar sample of radioactive iodine reduced in mass to 1 gram after 60 days. Calculate the initial mass of this sample, giving your answer to the nearest gram.
The half-life of a radioactive substance is the time it takes for a mass of $m _ { 0 }$ to reduce to a mass of $\frac { 1 } { 2 } m _ { 0 }$. A sample of radioactive vanadium reduced in mass from exactly 10 grams to 8.106 grams in 100 days. Find the half-life of radioactive vanadium, giving your answer to the nearest day. [4 marks]
\includegraphics[max width=\textwidth, alt={}]{c42685e9-bfa4-48d4-8abb-13e88a4b765e-08_1182_1707_1525_153}

AQA C4 2016 June Q5

5 It is given that $\sin A = \frac { \sqrt { 5 } } { 3 }$ and $\sin B = \frac { 1 } { \sqrt { 5 } }$, where the angles $A$ and $B$ are both acute.

1. Show that the exact value of $\cos B = \frac { 2 } { \sqrt { 5 } }$.
2. Hence show that the exact value of $\sin 2 B$ is $\frac { 4 } { 5 }$.
1. Show that the exact value of $\sin ( A - B )$ can be written as $p ( 5 - \sqrt { 5 } )$, where $p$ is a rational number.
2. Find the exact value of $\cos ( A - B )$ in the form $r + s \sqrt { 5 }$, where $r$ and $s$ are rational numbers.

AQA C4 2016 June Q6

5 marks

6 The line $l _ { 1 }$ passes through the point $A ( 0,6,9 )$ and the point $B ( 4 , - 6 , - 11 )$.
The line $l _ { 2 }$ has equation $\mathbf { r } = \left[ \begin{array} { r } - 1
5
- 2 \end{array} \right] + \lambda \left[ \begin{array} { r } 3
- 5
1 \end{array} \right]$.

The acute angle between the lines $l _ { 1 }$ and $l _ { 2 }$ is $\theta$. Find the value of $\cos \theta$ as a fraction in its lowest terms.
Show that the lines $l _ { 1 }$ and $l _ { 2 }$ intersect and find the coordinates of the point of intersection.
The points $C$ and $D$ lie on line $l _ { 2 }$ such that $A C B D$ is a parallelogram.
\includegraphics[max width=\textwidth, alt={}, center]{c42685e9-bfa4-48d4-8abb-13e88a4b765e-12_392_949_1018_548} The length of $A B$ is three times the length of $C D$.
Find the coordinates of the points $C$ and $D$.
[0pt] [5 marks] $7 \quad$ A curve $C$ is defined by the parametric equations $$x = \frac { 4 - \mathrm { e } ^ { 2 - 6 t } } { 4 } , \quad y = \frac { \mathrm { e } ^ { 3 t } } { 3 t } , \quad t \neq 0$$

AQA C4 2016 June Q8

10 marks

8 It is given that $\theta = \tan ^ { - 1 } \left( \frac { 3 x } { 2 } \right)$.

By writing $\theta = \tan ^ { - 1 } \left( \frac { 3 x } { 2 } \right)$ as $2 \tan \theta = 3 x$, use implicit differentiation to show that $\frac { \mathrm { d } \theta } { \mathrm { d } x } = \frac { k } { 4 + 9 x ^ { 2 } }$, where $k$ is an integer.
[0pt] [3 marks]
Hence solve the differential equation $$9 y \left( 4 + 9 x ^ { 2 } \right) \frac { \mathrm { d } y } { \mathrm {~d} x } = \operatorname { cosec } 3 y$$ given that $x = 0$ when $y = \frac { \pi } { 3 }$. Give your answer in the form $\mathrm { g } ( y ) = \mathrm { h } ( x )$.
[0pt] [7 marks]

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