Questions FP2 - A-Level Maths

Show that, in the case $a = 1 , \mathrm { P }$ has coordinates $\left( 1 - \frac { 1 } { \sqrt { 2 } } , \frac { 1 } { \sqrt { 2 } } \right)$. Write down the coordinates of Q .
Show that, for all positive values of $a$, the coordinates of P are $$x = a \left( 1 - \frac { a } { \sqrt { a ^ { 2 } + 1 } } \right) , \quad y = \frac { a } { \sqrt { a ^ { 2 } + 1 } } .$$ Write down the coordinates of Q in a similar form. Now let the variable point P be defined by the parametric equations $( * )$ for all values of the parameter $a$, positive, zero and negative. Let Q be defined for all $a$ by your answer in part (ii).
Using your calculator, sketch the locus of P as $a$ varies. State what happens to P as $a \rightarrow \infty$ and as $a \rightarrow - \infty$. Show algebraically that this locus has an asymptote at $y = - 1$.
On the same axes, sketch, as a dotted line, the locus of Q as $a$ varies.
(The single curve made up of these two loci and including the point B is called a right strophoid.)
State, with a reason, the size of the angle POQ in Fig. 5. What does this indicate about the angle at which a right strophoid crosses itself? \section*{OCR
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Edexcel FP2 2008 June Q1

\begin{enumerate} \item Solve the differential equation $\frac { \mathrm { d } y } { \mathrm {~d} x } - 3 y = x$ to obtain $y$ as a function of $x$. \item (a) Simplify the expression $\frac { ( x + 3 ) ( x + 9 ) } { x - 1 } - ( 3 x - 5 )$, giving your answer in the form $\frac { a ( x + b ) ( x + c ) } { x - 1 }$, where $a , b$ and $c$ are integers.
(b) Hence, or otherwise, solve the inequality $\frac { ( x + 3 ) ( x + 9 ) } { x - 1 } > 3 x - 5 \quad$ (4)(Total $\mathbf { 8 }$ marks) \item (a) Find the general solution of the differential equation $3 \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - \frac { \mathrm { d } y } { \mathrm {~d} x } - 2 y = x ^ { 2 }$
(b) Find the particular solution for which, at $x = 0 , y = 2$ and $\frac { \mathrm { d } y } { \mathrm {~d} x } = 3$.(6)(Total 14 marks) \item The diagram above shows the curve $C _ { 1 }$ which has polar equation $\boldsymbol { r } = \boldsymbol { a } ( \mathbf { 3 } + \mathbf { 2 } \boldsymbol { \operatorname { c o s } } \boldsymbol { \theta } ) , 0 \leq \theta < 2 \pi$ and the circle $C _ { 2 }$ with equation $\boldsymbol { r } = \mathbf { 4 } \boldsymbol { a } , 0 \leq \theta < 2 \pi$, where $a$ is a positive constant.

Edexcel FP2 2008 June Q5

5. (a) Find, in terms of $k$, the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 4 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 3 x = k t + 5 , \text { where } k \text { is a constant and } t > 0 .$$ For large values of $t$, this general solution may be approximated by a linear function.
(b) Given that $k = 6$, find the equation of this linear function.(2)(Total 9 marks)

Edexcel FP2 2008 June Q6

6. (a) Find, in the simplest surd form where appropriate, the exact values of $x$ for which $$\frac { x } { 2 } + 3 = \left| \frac { 4 } { x } \right|$$ (b) Sketch, on the same axes, the line with equation $y = \frac { x } { 2 } + 3$ and the graph of $y = \left| \frac { 4 } { x } \right| , \quad x \neq 0$.
(c) Find the set of values of $x$ for which $\frac { x } { 2 } + 3 > \left| \frac { 4 } { x } \right|$.
(2)(Total 10 marks)

Edexcel FP2 2008 June Q7

7. (a) Show that the substitution $y = v x$ transforms the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x } { y } + \frac { 3 y } { x } , \quad x > 0 , \quad y > 0$$ into the differential equation $$x \frac { \mathrm {~d} v } { \mathrm {~d} x } = 2 v + \frac { 1 } { v } .$$ (b) By solving differential equation (II), find a general solution of differential equation (I) in the form $y = \mathrm { f } ( x )$. Given that $y = 3$ at $x = 1$,
(c)find the particular solution of differential equation (I).(2)

Edexcel FP2 2008 June Q8

8. The curve $C$ shown in the diagram above has polar equation $$r = 4 ( 1 - \cos \theta ) , 0 \leq \theta \leq \frac { \pi } { 2 }$$ At the point $P$ on $C$, the tangent to $C$ is parallel to the line $\theta = \frac { \pi } { 2 }$.

Show that $P$ has polar coordinates $\left( 2 , \frac { \pi } { 3 } \right)$. The curve $C$ meets the line $\theta = \frac { \pi } { 2 }$ at the point $A$. The tangent to $C$ at the initial line at the point $N$. The finite region $R$, shown shaded in
\includegraphics[max width=\textwidth, alt={}]{863ef52d-ae75-450c-9eab-8102804868f5-2_737_561_1395_1329} the diagram above, is bounded by the initial line, the line $\theta = \frac { \pi } { 2 }$, the arc $A P$ of $C$ and the line $P N$.
Calculate the exact area of $R$.

Edexcel FP2 2008 June Q9

9. $$\left( x ^ { 2 } + 1 \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 2 y ^ { 2 } + ( 1 - 2 x ) \frac { \mathrm { d } y } { \mathrm {~d} x }$$

By differentiating equation (I) with respect to $x$, show that $$\left( x ^ { 2 } + 1 \right) \frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } } = ( 1 - 4 x ) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + ( 4 y - 2 ) \frac { \mathrm { d } y } { \mathrm {~d} x }$$ Given that $y = 1$ and $\frac { \mathrm { d } y } { \mathrm {~d} x } = 1$ at $x = 0$,
find the series solution for $y$, in ascending powers of $x$, up to and including the term in $x _ { 3 }$.(4)
Use your series to estimate the value of $y$ at $x = - 0.5$, giving your answer to two decimal places.(1)

Edexcel FP2 2008 June Q10

10. The point $P$ represents a complex number $z$ on an Argand diagram such that $$| z - 3 | = 2 | z |$$

Show that, as $z$ varies, the locus of $P$ is a circle, and give the coordinates of the centre and the radius of the circle.(5) The point $Q$ represents a complex number $z$ on an Argand diagram such that $$| z + 3 | = | z - i \sqrt { } 3 |$$
Sketch, on the same Argand diagram, the locus of $P$ and the locus of $Q$ as $z$ varies.(5)
On your diagram shade the region which satisfies $$| z - 3 | \geq 2 | z | \text { and } | z + 3 | \geq | z - i \sqrt { } 3 |$$

Edexcel FP2 2008 June Q11

De Moivre's theorem states that $\quad ( \cos \theta + \mathrm { i } \sin \theta ) ^ { n } = \cos n \theta + \mathrm { i } \sin n \theta$ for $n \in \Re$
1. Use induction to prove de Moivre's theorem for $n \in \mathbb { Z } ^ { + }$.
2. Show that $\cos 5 \theta = 16 \cos ^ { 5 } \theta - 20 \cos ^ { 3 } \theta + 5 \cos \theta$
3. Hence show that $2 \cos \frac { \pi } { 10 }$ is a root of the equation
$$x ^ { 4 } - 5 x ^ { 2 } + 5 = 0$$

OCR MEI FP2 2011 June Q1

A curve has polar equation $r = a ( 1 - \sin \theta )$, where $a > 0$ and $0 \leqslant \theta < 2 \pi$.
1. Sketch the curve.
2. Find, in an exact form, the area of the region enclosed by the curve.
1. Find, in an exact form, the value of the integral $\int _ { - \frac { 1 } { 2 } } ^ { \frac { 1 } { 2 } } \frac { 1 } { 1 + 4 x ^ { 2 } } \mathrm {~d} x$.
2. Find, in an exact form, the value of the integral $\int _ { - \frac { 1 } { 2 } } ^ { \frac { 1 } { 2 } } \frac { 1 } { \left( 1 + 4 x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } } \mathrm {~d} x$.

OCR MEI FP2 2011 June Q2

Use de Moivre's theorem to find expressions for $\sin 5 \theta$ and $\cos 5 \theta$ in terms of $\sin \theta$ and $\cos \theta$.
Hence show that, if $t = \tan \theta$, then $$\tan 5 \theta = \frac { t \left( t ^ { 4 } - 10 t ^ { 2 } + 5 \right) } { 5 t ^ { 4 } - 10 t ^ { 2 } + 1 }$$
1. Find the 5th roots of $- 4 \sqrt { 2 }$ in the form $r \mathrm { e } ^ { \mathrm { j } \theta }$, where $r > 0$ and $0 \leqslant \theta < 2 \pi$. These 5th roots are represented in the Argand diagram, in order of increasing $\theta$, by the points A , $\mathrm { B } , \mathrm { C } , \mathrm { D } , \mathrm { E }$.
2. Draw the Argand diagram, making clear which point is which. The mid-point of AB is the point P which represents the complex number $w$.
3. Find, in exact form, the modulus and argument of $w$.
4. $w$ is an $n$th root of a real number $a$, where $n$ is a positive integer. State the least possible value of $n$ and find the corresponding value of $a$.

OCR MEI FP2 2011 June Q3

Find the value of $k$ for which the matrix $$\mathbf { M } = \left( \begin{array} { r r r } 1 & - 1 & k
5 & 4 & 6
3 & 2 & 4 \end{array} \right)$$ does not have an inverse.
Assuming that $k$ does not take this value, find the inverse of $\mathbf { M }$ in terms of $k$.
In the case $k = 3$, evaluate $$\mathbf { M } \left( \begin{array} { r } - 3
3
1 \end{array} \right)$$
State the significance of what you have found in part (ii).
Find the value of $t$ for which the system of equations $$\begin{array} { r } x - y + 3 z = t
5 x + 4 y + 6 z = 1
3 x + 2 y + 4 z = 0 \end{array}$$ has solutions. Find the general solution in this case and describe the solution geometrically.

OCR MEI FP2 2011 June Q4

Given that $\cosh y = x$, show that $y = \pm \ln \left( x + \sqrt { x ^ { 2 } - 1 } \right)$ and that $\operatorname { arcosh } x = \ln \left( x + \sqrt { x ^ { 2 } - 1 } \right)$.
Find $\int _ { \frac { 4 } { 5 } } ^ { 1 } \frac { 1 } { \sqrt { 25 x ^ { 2 } - 16 } } \mathrm {~d} x$, expressing your answer in an exact logarithmic form.
Solve the equation $$5 \cosh x - \cosh 2 x = 3$$ giving your answers in an exact logarithmic form.

OCR MEI FP2 2011 June Q5

5 In this question, you are required to investigate the curve with equation $$y = x ^ { m } ( 1 - x ) ^ { n } , \quad 0 \leqslant x \leqslant 1 ,$$ for various positive values of $m$ and $n$.

On separate diagrams, sketch the curve in each of the following cases.
(A) $m = 1 , n = 1$,
(B) $m = 2 , n = 2$,
(C) $m = 2 , n = 4$,
(D) $m = 4 , n = 2$.
What feature does the curve have when $m = n$ ? What is the effect on the curve of interchanging $m$ and $n$ when $m \neq n$ ?
Describe how the $x$-coordinate of the maximum on the curve varies as $m$ and $n$ vary. Use calculus to determine the $x$-coordinate of the maximum.
Find the condition on $m$ for the gradient to be zero when $x = 0$. State a corresponding result for the gradient to be zero when $x = 1$.
Use your calculator to investigate the shape of the curve for large values of $m$ and $n$. Hence conjecture what happens to the value of the integral $\int _ { 0 } ^ { 1 } x ^ { m } ( 1 - x ) ^ { n } \mathrm {~d} x$ as $m$ and $n$ tend to infinity.
Use your calculator to investigate the shape of the curve for small values of $m$ and $n$. Hence conjecture what happens to the shape of the curve as $m$ and $n$ tend to zero. OCR is committed to seeking permission to reproduce all third-party content that it uses in its assessment materials. OCR has attempted to identify and contact all copyright holders whose work is used in this paper. To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced in the OCR Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download from our public website (\href{http://www.ocr.org.uk}{www.ocr.org.uk}) after the live examination series.
If OCR has unwittingly failed to correctly acknowledge or clear any third-party content in this assessment material, OCR will be happy to correct its mistake at the earliest possible opportunity. For queries or further information please contact the Copyright Team, First Floor, 9 Hills Road, Cambridge CB2 1 GE.
OCR is part of the Cambridge Assessment Group; Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge.

OCR FP2 2009 January Q1

Write down and simplify the first three terms of the Maclaurin series for $\mathrm { e } ^ { 2 x }$.
Hence show that the Maclaurin series for $$\ln \left( \mathrm { e } ^ { 2 x } + \mathrm { e } ^ { - 2 x } \right)$$ begins $\ln a + b x ^ { 2 }$, where $a$ and $b$ are constants to be found.

OCR FP2 2009 January Q2

2 It is given that $\alpha$ is the only real root of the equation $x ^ { 5 } + 2 x - 28 = 0$ and that $1.8 < \alpha < 2$.

The iteration $x _ { n + 1 } = \sqrt [ 5 ] { 28 - 2 x _ { n } }$, with $x _ { 1 } = 1.9$, is to be used to find $\alpha$. Find the values of $x _ { 2 } , x _ { 3 }$ and $x _ { 4 }$, giving the answers correct to 7 decimal places.
The error $e _ { n }$ is defined by $e _ { n } = \alpha - x _ { n }$. Given that $\alpha = 1.8915749$, correct to 7 decimal places, evaluate $\frac { e _ { 3 } } { e _ { 2 } }$ and $\frac { e _ { 4 } } { e _ { 3 } }$. Comment on these values in relation to the gradient of the curve with equation $y = \sqrt [ 5 ] { 28 - 2 x }$ at $x = \alpha$.
Prove that the derivative of $\sin ^ { - 1 } x$ is $\frac { 1 } { \sqrt { 1 - x ^ { 2 } } }$.
Given that $$\sin ^ { - 1 } 2 x + \sin ^ { - 1 } y = \frac { 1 } { 2 } \pi$$ find the exact value of $\frac { \mathrm { d } y } { \mathrm {~d} x }$ when $x = \frac { 1 } { 4 }$.
By means of a suitable substitution, show that $$\int \frac { x ^ { 2 } } { \sqrt { x ^ { 2 } - 1 } } \mathrm {~d} x$$ can be transformed to $\int \cosh ^ { 2 } \theta \mathrm {~d} \theta$.
Hence show that $\int \frac { x ^ { 2 } } { \sqrt { x ^ { 2 } - 1 } } \mathrm {~d} x = \frac { 1 } { 2 } x \sqrt { x ^ { 2 } - 1 } + \frac { 1 } { 2 } \cosh ^ { - 1 } x + c$.

OCR FP2 2009 January Q5

5
\includegraphics[max width=\textwidth, alt={}, center]{b9f29713-bc86-4869-9e54-195208e5e81d-3_661_734_267_703} The diagram shows the curve with equation $y = \mathrm { f } ( x )$, where $$f ( x ) = 2 x ^ { 3 } - 9 x ^ { 2 } + 12 x - 4.36$$ The curve has turning points at $x = 1$ and $x = 2$ and crosses the $x$-axis at $x = \alpha , x = \beta$ and $x = \gamma$, where $0 < \alpha < \beta < \gamma$.

The Newton-Raphson method is to be used to find the roots of the equation $\mathrm { f } ( x ) = 0$, with $x _ { 1 } = k$.
(a) To which root, if any, would successive approximations converge in each of the cases $k < 0$ and $k = 1$ ?
(b) What happens if $1 < k < 2$ ?
Sketch the curve with equation $y ^ { 2 } = \mathrm { f } ( x )$. State the coordinates of the points where the curve crosses the $x$-axis and the coordinates of any turning points.
Using the definitions of $\cosh x$ and $\sinh x$ in terms of $\mathrm { e } ^ { x }$ and $\mathrm { e } ^ { - x }$, show that $$1 + 2 \sinh ^ { 2 } x \equiv \cosh 2 x .$$
Solve the equation $$\cosh 2 x - 5 \sinh x = 4$$ giving your answers in logarithmic form.

OCR FP2 2009 January Q7

7
\includegraphics[max width=\textwidth, alt={}, center]{b9f29713-bc86-4869-9e54-195208e5e81d-4_511_609_264_769} The diagram shows the curve with equation, in polar coordinates, $$r = 3 + 2 \cos \theta , \quad \text { for } 0 \leqslant \theta < 2 \pi .$$ The points $P , Q , R$ and $S$ on the curve are such that the straight lines $P O R$ and $Q O S$ are perpendicular, where $O$ is the pole. The point $P$ has polar coordinates ( $r , \alpha$ ).

Show that $O P + O Q + O R + O S = k$, where $k$ is a constant to be found.
Given that $\alpha = \frac { 1 } { 4 } \pi$, find the exact area bounded by the curve and the lines $O P$ and $O Q$ (shaded in the diagram).

OCR FP2 2009 January Q8

8
\includegraphics[max width=\textwidth, alt={}, center]{b9f29713-bc86-4869-9e54-195208e5e81d-5_579_1363_267_390} The diagram shows the curve with equation $y = \frac { 1 } { x + 1 }$. A set of $n$ rectangles of unit width is drawn, starting at $x = 0$ and ending at $x = n$, where $n$ is an integer.

By considering the areas of these rectangles, explain why $$\frac { 1 } { 2 } + \frac { 1 } { 3 } + \ldots + \frac { 1 } { n + 1 } < \ln ( n + 1 ) .$$
By considering the areas of another set of rectangles, show that $$1 + \frac { 1 } { 2 } + \frac { 1 } { 3 } + \ldots + \frac { 1 } { n } > \ln ( n + 1 ) .$$
Hence show that $$\ln ( n + 1 ) + \frac { 1 } { n + 1 } < \sum _ { r = 1 } ^ { n + 1 } \frac { 1 } { r } < \ln ( n + 1 ) + 1$$
State, with a reason, whether $\sum _ { r = 1 } ^ { \infty } \frac { 1 } { r }$ is convergent.

OCR FP2 2009 January Q9

9 A curve has equation $$y = \frac { 4 x - 3 a } { 2 \left( x ^ { 2 } + a ^ { 2 } \right) }$$ where $a$ is a positive constant.

Explain why the curve has no asymptotes parallel to the $y$-axis.
Find, in terms of $a$, the set of values of $y$ for which there are no points on the curve.
Find the exact value of $\int _ { a } ^ { 2 a } \frac { 4 x - 3 a } { 2 \left( x ^ { 2 } + a ^ { 2 } \right) } \mathrm { d } x$, showing that it is independent of $a$.

OCR FP2 2010 January Q1

1 It is given that $\mathrm { f } ( x ) = x ^ { 2 } - \sin x$.

The iteration $x _ { n + 1 } = \sqrt { \sin x _ { n } }$, with $x _ { 1 } = 0.875$, is to be used to find a real root, $\alpha$, of the equation $\mathrm { f } ( x ) = 0$. Find $x _ { 2 } , x _ { 3 }$ and $x _ { 4 }$, giving the answers correct to 6 decimal places.
The error $e _ { n }$ is defined by $e _ { n } = \alpha - x _ { n }$. Given that $\alpha = 0.876726$, correct to 6 decimal places, find $e _ { 3 }$ and $e _ { 4 }$. Given that $\mathrm { g } ( x ) = \sqrt { \sin x }$, use $e _ { 3 }$ and $e _ { 4 }$ to estimate $\mathrm { g } ^ { \prime } ( \alpha )$.

OCR FP2 2010 January Q2

2 It is given that $\mathrm { f } ( x ) = \tan ^ { - 1 } ( 1 + x )$.

Find $\mathrm { f } ( 0 )$ and $\mathrm { f } ^ { \prime } ( 0 )$, and show that $\mathrm { f } ^ { \prime \prime } ( 0 ) = - \frac { 1 } { 2 }$.
Hence find the Maclaurin series for $\mathrm { f } ( x )$ up to and including the term in $x ^ { 2 }$.

OCR FP2 2010 January Q3

3
\includegraphics[max width=\textwidth, alt={}, center]{63afce50-e15f-4634-b2f1-ad5d78ab8bf5-2_597_1006_973_571} A curve with no stationary points has equation $y = \mathrm { f } ( x )$. The equation $\mathrm { f } ( x ) = 0$ has one real root $\alpha$, and the Newton-Raphson method is to be used to find $\alpha$. The tangent to the curve at the point $\left( x _ { 1 } , \mathrm { f } \left( x _ { 1 } \right) \right)$ meets the $x$-axis where $x = x _ { 2 }$ (see diagram).

Show that $x _ { 2 } = x _ { 1 } - \frac { \mathrm { f } \left( x _ { 1 } \right) } { \mathrm { f } ^ { \prime } \left( x _ { 1 } \right) }$.
Describe briefly, with the help of a sketch, how the Newton-Raphson method, using an initial approximation $x = x _ { 1 }$, gives a sequence of approximations approaching $\alpha$.
Use the Newton-Raphson method, with a first approximation of 1 , to find a second approximation to the root of $x ^ { 2 } - 2 \sinh x + 2 = 0$.

OCR FP2 2010 January Q4

4 The equation of a curve, in polar coordinates, is $$r = \mathrm { e } ^ { - 2 \theta } , \quad \text { for } 0 \leqslant \theta \leqslant \pi .$$

Sketch the curve, stating the polar coordinates of the point at which $r$ takes its greatest value.
The pole is $O$ and points $P$ and $Q$, with polar coordinates ( $r _ { 1 } , \theta _ { 1 }$ ) and ( $r _ { 2 } , \theta _ { 2 }$ ) respectively, lie on the curve. Given that $\theta _ { 2 } > \theta _ { 1 }$, show that the area of the region enclosed by the curve and the lines $O P$ and $O Q$ can be expressed as $k \left( r _ { 1 } ^ { 2 } - r _ { 2 } ^ { 2 } \right)$, where $k$ is a constant to be found.

OCR FP2 2010 January Q7

7
\includegraphics[max width=\textwidth, alt={}, center]{63afce50-e15f-4634-b2f1-ad5d78ab8bf5-3_591_1131_986_507} The diagram shows the curve with equation $y = \sqrt [ 3 ] { x }$, together with a set of $n$ rectangles of unit width.

By considering the areas of these rectangles, explain why $$\sqrt [ 3 ] { 1 } + \sqrt [ 3 ] { 2 } + \sqrt [ 3 ] { 3 } + \ldots + \sqrt [ 3 ] { n } > \int _ { 0 } ^ { n } \sqrt [ 3 ] { x } \mathrm {~d} x$$
By drawing another set of rectangles and considering their areas, show that $$\sqrt [ 3 ] { 1 } + \sqrt [ 3 ] { 2 } + \sqrt [ 3 ] { 3 } + \ldots + \sqrt [ 3 ] { n } < \int _ { 1 } ^ { n + 1 } \sqrt [ 3 ] { x } \mathrm {~d} x$$
Hence find an approximation to $\sum _ { n = 1 } ^ { 100 } \sqrt [ 3 ] { n }$, giving your answer correct to 2 significant figures.

Questions FP2 (1157 questions)

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