Questions - AQA FP3 - A-Level Maths

1. Find the cartesian equation of $C _ { 1 }$.
2. Deduce that $C _ { 1 }$ is a circle and find its radius and the cartesian coordinates of its centre.
The diagram shows the curve $C _ { 2 }$ with polar equation $$r = 4 + \sin \theta , \quad 0 \leqslant \theta \leqslant 2 \pi$$ \includegraphics[max width=\textwidth, alt={}, center]{90a59b47-3799-46a2-b76b-ced5cc3e1aac-4_519_847_443_593}
1. Find the area of the region that is bounded by $C _ { 2 }$.
2. Prove that the curves $C _ { 1 }$ and $C _ { 2 }$ do not intersect.
3. Find the area of the region that is outside $C _ { 1 }$ but inside $C _ { 2 }$.

AQA FP3 2010 June Q7

Given that $x = t ^ { \frac { 1 } { 2 } } , x > 0 , t > 0$ and $y$ is a function of $x$, show that:
1. $\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 t ^ { \frac { 1 } { 2 } } \frac { \mathrm {~d} y } { \mathrm {~d} t }$;
  (2 marks)
2. $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 4 t \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} t }$.
  (3 marks)
Hence show that the substitution $x = t ^ { \frac { 1 } { 2 } }$ transforms the differential equation $$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - \left( 8 x ^ { 2 } + 1 \right) \frac { \mathrm { d } y } { \mathrm {~d} x } + 12 x ^ { 3 } y = 12 x ^ { 5 }$$ into $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 4 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 3 y = 3 t$$ (2 marks)
Hence find the general solution of the differential equation $$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - \left( 8 x ^ { 2 } + 1 \right) \frac { \mathrm { d } y } { \mathrm {~d} x } + 12 x ^ { 3 } y = 12 x ^ { 5 }$$ giving your answer in the form $y = \mathrm { f } ( x )$.

AQA FP3 2011 June Q1

1 The function $y ( x )$ satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x , y )$$ where $$\mathrm { f } ( x , y ) = x + \ln ( 1 + y )$$ and $$y ( 2 ) = 1$$ Use the improved Euler formula $$y _ { r + 1 } = y _ { r } + \frac { 1 } { 2 } \left( k _ { 1 } + k _ { 2 } \right)$$ where $k _ { 1 } = h \mathrm { f } \left( x _ { r } , y _ { r } \right)$ and $k _ { 2 } = h \mathrm { f } \left( x _ { r } + h , y _ { r } + k _ { 1 } \right)$ and $h = 0.2$, to obtain an approximation to $y ( 2.2 )$, giving your answer to four decimal places.

AQA FP3 2011 June Q2

Find the values of the constants $p$ and $q$ for which $p + q x \mathrm { e } ^ { - 2 x }$ is a particular integral of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + \frac { \mathrm { d } y } { \mathrm {~d} x } - 2 y = 4 - 9 \mathrm { e } ^ { - 2 x }$$
Hence find the general solution of this differential equation.
Hence express $y$ in terms of $x$, given that $y = 4$ when $x = 0$ and that $\frac { \mathrm { d } y } { \mathrm {~d} x } \rightarrow 0$ as $x \rightarrow \infty$.

AQA FP3 2011 June Q3

Find $\int x ^ { 2 } \ln x \mathrm {~d} x$.
Explain why $\int _ { 0 } ^ { \mathrm { e } } x ^ { 2 } \ln x \mathrm {~d} x$ is an improper integral.
Evaluate $\int _ { 0 } ^ { \mathrm { e } } x ^ { 2 } \ln x \mathrm {~d} x$, showing the limiting process used.

AQA FP3 2011 June Q4

4 By using an integrating factor, find the solution of the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } + ( \cot x ) y = \sin 2 x , \quad 0 < x < \frac { \pi } { 2 }$$ given that $y = \frac { 1 } { 2 }$ when $x = \frac { \pi } { 6 }$.
(10 marks)

AQA FP3 2011 June Q5

Given that $y = \ln ( 1 + 2 \tan x )$, find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ and $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$.
(You may leave your expression for $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$ unsimplified.)
Hence, using Maclaurin's theorem, find the first two non-zero terms in the expansion, in ascending powers of $x$, of $\ln ( 1 + 2 \tan x )$.
(2 marks)
Find $$\lim _ { x \rightarrow 0 } \left[ \frac { \ln ( 1 + 2 \tan x ) } { \ln ( 1 - x ) } \right]$$ (4 marks)

AQA FP3 2011 June Q6

6 A differential equation is given by $$\left( x ^ { 3 } + 1 \right) \frac { d ^ { 2 } y } { d x ^ { 2 } } - 3 x ^ { 2 } \frac { d y } { d x } = 2 - 4 x ^ { 3 }$$

Show that the substitution $$u = \frac { \mathrm { d } y } { \mathrm {~d} x } - 2 x$$ transforms this differential equation into $$\left( x ^ { 3 } + 1 \right) \frac { \mathrm { d } u } { \mathrm {~d} x } = 3 x ^ { 2 } u$$ (4 marks)
Hence find the general solution of the differential equation $$\left( x ^ { 3 } + 1 \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 3 x ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x } = 2 - 4 x ^ { 3 }$$ giving your answer in the form $y = \mathrm { f } ( x )$.
$7 \quad$ The curve $C _ { 1 }$ is defined by $r = 2 \sin \theta , \quad 0 \leqslant \theta < \frac { \pi } { 2 }$. The curve $C _ { 2 }$ is defined by $r = \tan \theta , \quad 0 \leqslant \theta < \frac { \pi } { 2 }$.
Find a cartesian equation of $C _ { 1 }$.
1. Prove that the curves $C _ { 1 }$ and $C _ { 2 }$ meet at the pole $O$ and at one other point, $P$, in the given domain. State the polar coordinates of $P$.
2. The point $A$ is the point on the curve $C _ { 1 }$ at which $\theta = \frac { \pi } { 4 }$. The point $B$ is the point on the curve $C _ { 2 }$ at which $\theta = \frac { \pi } { 4 }$. Determine which of the points $A$ or $B$ is further away from the pole $O$, justifying your answer.
3. Show that the area of the region bounded by the arc $O P$ of $C _ { 1 }$ and the arc $O P$ of $C _ { 2 }$ is $a \pi + b \sqrt { 3 }$, where $a$ and $b$ are rational numbers.

AQA FP3 2012 June Q1

1 The function $y ( x )$ satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x , y )$$ where $$\mathrm { f } ( x , y ) = \sqrt { ( 2 x ) } + \sqrt { y }$$ and $$y ( 2 ) = 9$$ Use the improved Euler formula $$y _ { r + 1 } = y _ { r } + \frac { 1 } { 2 } \left( k _ { 1 } + k _ { 2 } \right)$$ where $k _ { 1 } = h \mathrm { f } \left( x _ { r } , y _ { r } \right)$ and $k _ { 2 } = h \mathrm { f } \left( x _ { r } + h , y _ { r } + k _ { 1 } \right)$ and $h = 0.25$, to obtain an approximation to $y ( 2.25 )$, giving your answer to two decimal places.

AQA FP3 2012 June Q2

Write down the expansion of $\sin 2 x$ in ascending powers of $x$ up to and including the term in $x ^ { 5 }$.
Show that, for some value of $k$, $$\lim _ { x \rightarrow 0 } \left[ \frac { 2 x - \sin 2 x } { x ^ { 2 } \ln ( 1 + k x ) } \right] = 16$$ and state this value of $k$.

AQA FP3 2012 June Q3

3 The diagram shows a sketch of a curve $C$, the pole $O$ and the initial line.
\includegraphics[max width=\textwidth, alt={}, center]{c4bce668-61f1-4be0-97ee-c635df7e1fc6-2_380_735_1827_648} The polar equation of $C$ is $$r = 2 \sqrt { 1 + \tan \theta } , \quad - \frac { \pi } { 4 } \leqslant \theta \leqslant \frac { \pi } { 4 }$$ Show that the area of the shaded region, bounded by the curve $C$ and the initial line, is $\frac { \pi } { 2 } - \ln 2$.
(4 marks)

AQA FP3 2012 June Q4

By using an integrating factor, find the general solution of the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } + \frac { 4 } { 2 x + 1 } y = 4 ( 2 x + 1 ) ^ { 5 }$$ giving your answer in the form $y = \mathrm { f } ( x )$.
The gradient of a curve at any point $( x , y )$ on the curve is given by the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = 4 ( 2 x + 1 ) ^ { 5 } - \frac { 4 } { 2 x + 1 } y$$ The point whose $x$-coordinate is zero is a stationary point of the curve. Using your answer to part (a), find the equation of the curve.

AQA FP3 2012 June Q5

Find $\int x ^ { 2 } \mathrm { e } ^ { - x } \mathrm {~d} x$.
Hence evaluate $\int _ { 0 } ^ { \infty } x ^ { 2 } \mathrm { e } ^ { - x } \mathrm {~d} x$, showing the limiting process used.

AQA FP3 2012 June Q6

6 It is given that $y = \ln ( 1 + \sin x )$.

Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
Show that $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = - \mathrm { e } ^ { - y }$.
Express $\frac { \mathrm { d } ^ { 4 } y } { \mathrm {~d} x ^ { 4 } }$ in terms of $\frac { \mathrm { d } y } { \mathrm {~d} x }$ and $\mathrm { e } ^ { - y }$.
Hence, by using Maclaurin's theorem, find the first four non-zero terms in the expansion, in ascending powers of $x$, of $\ln ( 1 + \sin x )$.

AQA FP3 2012 June Q7

Show that the substitution $x = \mathrm { e } ^ { t }$ transforms the differential equation $$\text { into } \quad \begin{aligned} x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 4 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 6 y & = 3 + 20 \sin ( \ln x )
\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 5 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 6 y & = 3 + 20 \sin t \end{aligned}$$ (7 marks)
Find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 5 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 6 y = 3 + 20 \sin t$$ (11 marks)
Write down the general solution of the differential equation $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 4 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 6 y = 3 + 20 \sin ( \ln x )$$

AQA FP3 2012 June Q8

A curve has cartesian equation $x y = 8$. Show that the polar equation of the curve is $r ^ { 2 } = 16 \operatorname { cosec } 2 \theta$.
The diagram shows a sketch of the curve, $C$, whose polar equation is $$r ^ { 2 } = 16 \operatorname { cosec } 2 \theta , \quad 0 < \theta < \frac { \pi } { 2 }$$ \includegraphics[max width=\textwidth, alt={}, center]{c4bce668-61f1-4be0-97ee-c635df7e1fc6-4_364_567_1635_726}
1. Find the polar coordinates of the point $N$ which lies on the curve $C$ and is closest to the pole $O$.
2. The circle whose polar equation is $r = 4 \sqrt { 2 }$ intersects the curve $C$ at the points $P$ and $Q$. Find, in an exact form, the polar coordinates of $P$ and $Q$.
3. The obtuse angle $P N Q$ is $\alpha$ radians. Find the value of $\alpha$, giving your answer to three significant figures.
  (5 marks)

AQA FP3 2013 June Q1

1 It is given that $y ( x )$ satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x , y )$$ where $$\mathrm { f } ( x , y ) = ( x - y ) \sqrt { x + y }$$ and $$y ( 2 ) = 1$$ Use the improved Euler formula $$y _ { r + 1 } = y _ { r } + \frac { 1 } { 2 } \left( k _ { 1 } + k _ { 2 } \right)$$ where $k _ { 1 } = h \mathrm { f } \left( x _ { r } , y _ { r } \right)$ and $k _ { 2 } = h \mathrm { f } \left( x _ { r } + h , y _ { r } + k _ { 1 } \right)$ and $h = 0.2$, to obtain an approximation to $y ( 2.2 )$, giving your answer to three decimal places.

AQA FP3 2013 June Q2

2 The Cartesian equation of a circle is $( x + 8 ) ^ { 2 } + ( y - 6 ) ^ { 2 } = 100$.
Using the origin $O$ as the pole and the positive $x$-axis as the initial line, find the polar equation of this circle, giving your answer in the form $r = p \sin \theta + q \cos \theta$.
(4 marks)

AQA FP3 2013 June Q3

Find the values of the constants $a , b$ and $c$ for which $a + b x + c x \mathrm { e } ^ { - 3 x }$ is a particular integral of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } - 3 y = 3 x - 8 \mathrm { e } ^ { - 3 x }$$
Hence find the general solution of this differential equation.
Hence express $y$ in terms of $x$, given that $y = 1$ when $x = 0$ and that $\frac { \mathrm { d } y } { \mathrm {~d} x } \rightarrow - 1$ as $x \rightarrow \infty$.

AQA FP3 2013 June Q4

4 Evaluate the improper integral $$\int _ { 0 } ^ { \infty } \left( \frac { 2 x } { x ^ { 2 } + 4 } - \frac { 4 } { 2 x + 3 } \right) \mathrm { d } x$$ showing the limiting process used and giving your answer in the form $\ln k$, where $k$ is a constant.

AQA FP3 2013 June Q5

Differentiate $\ln ( \ln x )$ with respect to $x$.
1. Show that $\ln x$ is an integrating factor for the first-order differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } + \frac { 1 } { x \ln x } y = 9 x ^ { 2 } , \quad x > 1$$
2. Hence find the solution of this differential equation, given that $y = 4 \mathrm { e } ^ { 3 }$ when $x = \mathrm { e }$.
  (6 marks)

AQA FP3 2013 June Q6

6 It is given that $y = ( 4 + \sin x ) ^ { \frac { 1 } { 2 } }$.

Express $y \frac { \mathrm {~d} y } { \mathrm {~d} x }$ in terms of $\cos x$.
Find the value of $\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } }$ when $x = 0$.
Hence, by using Maclaurin's theorem, find the first four terms in the expansion, in ascending powers of $x$, of $( 4 + \sin x ) ^ { \frac { 1 } { 2 } }$.
(2 marks)

AQA FP3 2013 June Q7

7 A differential equation is given by $$\sin ^ { 2 } x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 \sin x \cos x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y = 2 \sin ^ { 4 } x \cos x , \quad 0 < x < \pi$$

Show that the substitution $$y = u \sin x$$ where $u$ is a function of $x$, transforms this differential equation into $$\frac { \mathrm { d } ^ { 2 } u } { \mathrm {~d} x ^ { 2 } } + u = \sin 2 x$$
Hence find the general solution of the differential equation $$\sin ^ { 2 } x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 \sin x \cos x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y = 2 \sin ^ { 4 } x \cos x$$ giving your answer in the form $y = \mathrm { f } ( x )$.
(6 marks)

AQA FP3 2013 June Q8

8 The diagram shows a sketch of a curve and a circle.
\includegraphics[max width=\textwidth, alt={}, center]{a2bc95fe-5588-4ff7-a8a3-0cd07df412c9-4_460_693_370_680} The polar equation of the curve is $$r = 3 + 2 \sin \theta , \quad 0 \leqslant \theta \leqslant 2 \pi$$ The circle, whose polar equation is $r = 2$, intersects the curve at the points $P$ and $Q$, as shown in the diagram.

Find the polar coordinates of $P$ and the polar coordinates of $Q$.
A straight line, drawn from the point $P$ through the pole $O$, intersects the curve again at the point $A$.
1. Find the polar coordinates of $A$.
2. Find, in surd form, the length of $A Q$.
3. Hence, or otherwise, explain why the line $A Q$ is a tangent to the circle $r = 2$.
Find the area of the shaded region which lies inside the circle $r = 2$ but outside the curve $r = 3 + 2 \sin \theta$. Give your answer in the form $\frac { 1 } { 6 } ( m \sqrt { 3 } + n \pi )$, where $m$ and $n$ are integers.

AQA FP3 2014 June Q1

5 marks

1 It is given that $y ( x )$ satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x , y )$$ where $$\mathrm { f } ( x , y ) = \frac { \ln ( x + y ) } { \ln y }$$ and $$y ( 6 ) = 3$$ Use the improved Euler formula $$y _ { r + 1 } = y _ { r } + \frac { 1 } { 2 } \left( k _ { 1 } + k _ { 2 } \right)$$ where $k _ { 1 } = h \mathrm { f } \left( x _ { r } , y _ { r } \right)$ and $k _ { 2 } = h \mathrm { f } \left( x _ { r } + h , y _ { r } + k _ { 1 } \right)$ and $h = 0.4$, to obtain an approximation to $y ( 6.4 )$, giving your answer to three decimal places.
[0pt] [5 marks]

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