Questions FP2 - A-Level Maths

Using the definitions of $\cosh x$ and $\sinh x$ in terms of $\mathrm { e } ^ { x }$ and $\mathrm { e } ^ { - x }$, prove that $$\sinh 2 x = 2 \sinh x \cosh x$$
Show that the curve with equation $$y = \cosh 2 x - 6 \sinh x$$ has just one stationary point, and find its $x$-coordinate in logarithmic form. Determine the nature of the stationary point. \section*{June 2006} 1 Find the first three non-zero terms of the Maclaurin series for $$( 1 + x ) \sin x$$ simplifying the coefficients. 2
Given that $y = \tan ^ { - 1 } x$, prove that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 1 + x ^ { 2 } }$.
Verify that $y = \tan ^ { - 1 } x$ satisfies the equation $$\left( 1 + x ^ { 2 } \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 x \frac { \mathrm {~d} y } { \mathrm {~d} x } = 0$$ 3 The equation of a curve is $y = \frac { x + 1 } { x ^ { 2 } + 3 }$.
State the equation of the asymptote of the curve.
Show that $- \frac { 1 } { 6 } \leqslant y \leqslant \frac { 1 } { 2 }$. 4
Using the definition of $\cosh x$ in terms of $\mathrm { e } ^ { x }$ and $\mathrm { e } ^ { - x }$, prove that $$\cosh 2 x = 2 \cosh ^ { 2 } x - 1$$
Hence solve the equation $$\cosh 2 x - 7 \cosh x = 3$$ giving your answer in logarithmic form. 5
Express $t ^ { 2 } + t + 1$ in the form $( t + a ) ^ { 2 } + b$.
By using the substitution $\tan \frac { 1 } { 2 } x = t$, show that $$\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \frac { 1 } { 2 + \sin x } \mathrm {~d} x = \frac { \sqrt { 3 } } { 9 } \pi$$ 6
\includegraphics[max width=\textwidth, alt={}, center]{0ec9c4ff-8622-4dda-a000-6ffe36f38023-07_627_1356_260_392} The diagram shows the curve with equation $y = 3 ^ { x }$ for $0 \leqslant x \leqslant 1$. The area $A$ under the curve between these limits is divided into $n$ strips, each of width $h$ where $n h = 1$.
By using the set of rectangles indicated on the diagram, show that $A > \frac { 2 h } { 3 ^ { h } - 1 }$.
By considering another set of rectangles, show that $A < \frac { ( 2 h ) 3 ^ { h } } { 3 ^ { h } - 1 }$.
Given that $h = 0.001$, use these inequalities to find values between which $A$ lies. 7 The equation of a curve, in polar coordinates, is $$r = \sqrt { 3 } + \tan \theta , \quad \text { for } - \frac { 1 } { 3 } \pi \leqslant \theta \leqslant \frac { 1 } { 4 } \pi$$
Find the equation of the tangent at the pole.
State the greatest value of $r$ and the corresponding value of $\theta$.
Sketch the curve.
Find the exact area of the region enclosed by the curve and the lines $\theta = 0$ and $\theta = \frac { 1 } { 4 } \pi$. 8 The curve with equation $y = \frac { \sinh x } { x ^ { 2 } }$, for $x > 0$, has one turning point.
Show that the $x$-coordinate of the turning point satisfies the equation $x - 2 \tanh x = 0$.
Use the Newton-Raphson method, with a first approximation $x _ { 1 } = 2$, to find the next two approximations, $x _ { 2 }$ and $x _ { 3 }$, to the positive root of $x - 2 \tanh x = 0$.
By considering the approximate errors in $x _ { 1 }$ and $x _ { 2 }$, estimate the error in $x _ { 3 }$. (You are not expected to evaluate $x _ { 4 }$.) June 2006
9
Given that $y = \sinh ^ { - 1 } x$, prove that $y = \ln \left( x + \sqrt { x ^ { 2 } + 1 } \right)$.
It is given that, for non-negative integers $n$, $$I _ { n } = \int _ { 0 } ^ { \alpha } \sinh ^ { n } \theta \mathrm {~d} \theta$$ where $\alpha = \sinh ^ { - 1 } 1$. Show that $$n I _ { n } = \sqrt { 2 } - ( n - 1 ) I _ { n - 2 } , \quad \text { for } n \geqslant 2$$
Evaluate $I _ { 4 }$, giving your answer in terms of $\sqrt { 2 }$ and logarithms. \section*{Jan 2007} 1 It is given that $\mathrm { f } ( x ) = \ln ( 3 + x )$.
Find the exact values of $\mathrm { f } ( 0 )$ and $\mathrm { f } ^ { \prime } ( 0 )$, and show that $\mathrm { f } ^ { \prime \prime } ( 0 ) = - \frac { 1 } { 9 }$.
Hence write down the first three terms of the Maclaurin series for $\mathrm { f } ( x )$, given that $- 3 < x \leqslant 3$. 2 It is given that $\mathrm { f } ( x ) = x ^ { 2 } - \tan ^ { - 1 } x$.
Show by calculation that the equation $\mathrm { f } ( x ) = 0$ has a root in the interval $0.8 < x < 0.9$.
Use the Newton-Raphson method, with a first approximation 0.8 , to find the next approximation to this root. Give your answer correct to 3 decimal places. 3
\includegraphics[max width=\textwidth, alt={}, center]{0ec9c4ff-8622-4dda-a000-6ffe36f38023-09_714_997_991_571} The diagram shows the curve with equation $y = \mathrm { e } ^ { x ^ { 2 } }$, for $0 \leqslant x \leqslant 1$. The region under the curve between these limits is divided into four strips of equal width. The area of this region under the curve is $A$.
By considering the set of rectangles indicated in the diagram, show that an upper bound for $A$ is 1.71 .
By considering an appropriate set of four rectangles, find a lower bound for $A$. 4
On separate diagrams, sketch the graphs of $y = \sinh x$ and $y = \operatorname { cosech } x$.
Show that $\operatorname { cosech } x = \frac { 2 \mathrm { e } ^ { x } } { \mathrm { e } ^ { 2 x } - 1 }$, and hence, using the substitution $u = \mathrm { e } ^ { x }$, find $\int \operatorname { cosech } x \mathrm {~d} x$. \section*{Jan 2007} 5 It is given that, for non-negative integers $n$, $$I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 2 } \pi } x ^ { n } \cos x \mathrm {~d} x$$
Prove that, for $n \geqslant 2$, $$I _ { n } = \left( \frac { 1 } { 2 } \pi \right) ^ { n } - n ( n - 1 ) I _ { n - 2 } .$$
Find $I _ { 4 }$ in terms of $\pi$. 6
\includegraphics[max width=\textwidth, alt={}, center]{0ec9c4ff-8622-4dda-a000-6ffe36f38023-10_719_1435_849_354} The diagram shows the curve with equation $y = \frac { 2 x ^ { 2 } - 3 a x } { x ^ { 2 } - a ^ { 2 } }$, where $a$ is a positive constant.
Find the equations of the asymptotes of the curve.
Sketch the curve with equation $$y ^ { 2 } = \frac { 2 x ^ { 2 } - 3 a x } { x ^ { 2 } - a ^ { 2 } }$$ State the coordinates of any points where the curve crosses the axes, and give the equations of any asymptotes. 7
Express $\frac { 1 - t ^ { 2 } } { t ^ { 2 } \left( 1 + t ^ { 2 } \right) }$ in partial fractions.
Use the substitution $t = \tan \frac { 1 } { 2 } x$ to show that $$\int _ { \frac { 1 } { 3 } \pi } ^ { \frac { 1 } { 2 } \pi } \frac { \cos x } { 1 - \cos x } \mathrm {~d} x = \sqrt { 3 } - 1 - \frac { 1 } { 6 } \pi$$ 8
Define tanh $y$ in terms of $\mathrm { e } ^ { y }$ and $\mathrm { e } ^ { - y }$.
Given that $y = \tanh ^ { - 1 } x$, where $- 1 < x < 1$, prove that $y = \frac { 1 } { 2 } \ln \left( \frac { 1 + x } { 1 - x } \right)$.
Find the exact solution of the equation $3 \cosh x = 4 \sinh x$, giving the answer in terms of a logarithm.
Solve the equation $$\tanh ^ { - 1 } x + \ln ( 1 - x ) = \ln \left( \frac { 4 } { 5 } \right)$$ 9 The equation of a curve, in polar coordinates, is $$r = \sec \theta + \tan \theta , \quad \text { for } 0 \leqslant \theta \leqslant \frac { 1 } { 3 } \pi$$
Sketch the curve.
Find the exact area of the region bounded by the curve and the lines $\theta = 0$ and $\theta = \frac { 1 } { 3 } \pi$.
Find a cartesian equation of the curve. \footnotetext{Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. 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. }\section*{June 2007} 1 The equation of a curve, in polar coordinates, is $$r = 2 \sin 3 \theta , \quad \text { for } 0 \leqslant \theta \leqslant \frac { 1 } { 3 } \pi .$$ Find the exact area of the region enclosed by the curve between $\theta = 0$ and $\theta = \frac { 1 } { 3 } \pi$. 2
Given that $\mathrm { f } ( x ) = \sin \left( 2 x + \frac { 1 } { 4 } \pi \right)$, show that $\mathrm { f } ( x ) = \frac { 1 } { 2 } \sqrt { 2 } ( \sin 2 x + \cos 2 x )$.
Hence find the first four terms of the Maclaurin series for $\mathrm { f } ( x )$. [You may use appropriate results given in the List of Formulae.] 3 It is given that $\mathrm { f } ( x ) = \frac { x ^ { 2 } + 9 x } { ( x - 1 ) \left( x ^ { 2 } + 9 \right) }$.
Express $\mathrm { f } ( x )$ in partial fractions.
Hence find $\int f ( x ) \mathrm { d } x$. 4
Given that $$y = x \sqrt { 1 - x ^ { 2 } } - \cos ^ { - 1 } x$$ find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in a simplified form.
Hence, or otherwise, find the exact value of $\int _ { 0 } ^ { 1 } 2 \sqrt { 1 - x ^ { 2 } } \mathrm {~d} x$. 5 It is given that, for non-negative integers $n$, $$I _ { n } = \int _ { 1 } ^ { \mathrm { e } } ( \ln x ) ^ { n } \mathrm {~d} x$$
Show that, for $n \geqslant 1$, $$I _ { n } = \mathrm { e } - n I _ { n - 1 } .$$
Find $I _ { 3 }$ in terms of e. 6
\includegraphics[max width=\textwidth, alt={}, center]{0ec9c4ff-8622-4dda-a000-6ffe36f38023-13_816_1369_267_388} The diagram shows the curve with equation $y = \frac { 1 } { x ^ { 2 } }$ for $x > 0$, together with a set of $n$ rectangles of unit width, starting at $x = 1$.
By considering the areas of these rectangles, explain why $$\frac { 1 } { 1 ^ { 2 } } + \frac { 1 } { 2 ^ { 2 } } + \frac { 1 } { 3 ^ { 2 } } + \ldots + \frac { 1 } { n ^ { 2 } } > \int _ { 1 } ^ { n + 1 } \frac { 1 } { x ^ { 2 } } \mathrm {~d} x$$
By considering the areas of another set of rectangles, explain why $$\frac { 1 } { 2 ^ { 2 } } + \frac { 1 } { 3 ^ { 2 } } + \frac { 1 } { 4 ^ { 2 } } + \ldots + \frac { 1 } { n ^ { 2 } } < \int _ { 1 } ^ { n } \frac { 1 } { x ^ { 2 } } \mathrm {~d} x$$
Hence show that $$1 - \frac { 1 } { n + 1 } < \sum _ { r = 1 } ^ { n } \frac { 1 } { r ^ { 2 } } < 2 - \frac { 1 } { n }$$
Hence give bounds between which $\sum _ { r = 1 } ^ { \infty } \frac { 1 } { r ^ { 2 } }$ lies. 7
Using the definitions of hyperbolic functions in terms of exponentials, prove that $$\cosh x \cosh y - \sinh x \sinh y = \cosh ( x - y )$$
Given that $\cosh x \cosh y = 9$ and $\sinh x \sinh y = 8$, show that $x = y$.
Hence find the values of $x$ and $y$ which satisfy the equations given in part (ii), giving the answers in logarithmic form. \section*{June 2007} 8 The iteration $x _ { n + 1 } = \frac { 1 } { \left( x _ { n } + 2 \right) ^ { 2 } }$, with $x _ { 1 } = 0.3$, is to be used to find the real root, $\alpha$, of the equation $x ( x + 2 ) ^ { 2 } = 1$.
Find the value of $\alpha$, correct to 4 decimal places. You should show the result of each step of the iteration process.
Given that $\mathrm { f } ( x ) = \frac { 1 } { ( x + 2 ) ^ { 2 } }$, show that $\mathrm { f } ^ { \prime } ( \alpha ) \neq 0$.
The difference, $\delta _ { r }$, between successive approximations is given by $\delta _ { r } = x _ { r + 1 } - x _ { r }$. Find $\delta _ { 3 }$.
Given that $\delta _ { r + 1 } \approx \mathrm { f } ^ { \prime } ( \alpha ) \delta _ { r }$, find an estimate for $\delta _ { 10 }$. 9 It is given that the equation of a curve is $$y = \frac { x ^ { 2 } - 2 a x } { x - a }$$ where $a$ is a positive constant.
Find the equations of the asymptotes of the curve.
Show that $y$ takes all real values.
Sketch the curve $y = \frac { x ^ { 2 } - 2 a x } { x - a }$. \footnotetext{Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. 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. }1 It is given that $\mathrm { f } ( x ) = \ln ( 1 + \cos x )$.
Find the exact values of $f ( 0 ) , f ^ { \prime } ( 0 )$ and $f ^ { \prime \prime } ( 0 )$.
Hence find the first two non-zero terms of the Maclaurin series for $\mathrm { f } ( x )$. 2
\includegraphics[max width=\textwidth, alt={}, center]{0ec9c4ff-8622-4dda-a000-6ffe36f38023-15_584_707_575_721} The diagram shows parts of the curves with equations $y = \cos ^ { - 1 } x$ and $y = \frac { 1 } { 2 } \sin ^ { - 1 } x$, and their point of intersection $P$.
Verify that the coordinates of $P$ are $\left( \frac { 1 } { 2 } \sqrt { 3 } , \frac { 1 } { 6 } \pi \right)$.
Find the gradient of each curve at $P$. 3
\includegraphics[max width=\textwidth, alt={}, center]{0ec9c4ff-8622-4dda-a000-6ffe36f38023-15_650_791_1619_678} The diagram shows the curve with equation $y = \sqrt { 1 + x ^ { 3 } }$, for $2 \leqslant x \leqslant 3$. The region under the curve between these limits has area $A$.
Explain why $3 < A < \sqrt { 28 }$.
The region is divided into 5 strips, each of width 0.2 . By using suitable rectangles, find improved lower and upper bounds between which $A$ lies. Give your answers correct to 3 significant figures. 4 The equation of a curve, in polar coordinates, is $$r = 1 + 2 \sec \theta , \quad \text { for } - \frac { 1 } { 2 } \pi < \theta < \frac { 1 } { 2 } \pi .$$
Find the exact area of the region bounded by the curve and the lines $\theta = 0$ and $\theta = \frac { 1 } { 6 } \pi$. [The result $\int \sec \theta \mathrm { d } \theta = \ln | \sec \theta + \tan \theta |$ may be assumed.]
Show that a cartesian equation of the curve is $( x - 2 ) \sqrt { x ^ { 2 } + y ^ { 2 } } = x$. 5
\includegraphics[max width=\textwidth, alt={}, center]{0ec9c4ff-8622-4dda-a000-6ffe36f38023-16_609_892_815_628} The diagram shows the curve with equation $y = x \mathrm { e } ^ { - x } + 1$. The curve crosses the $x$-axis at $x = \alpha$.
Use differentiation to show that the $x$-coordinate of the stationary point is 1 . $\alpha$ is to be found using the Newton-Raphson method, with $\mathrm { f } ( x ) = x \mathrm { e } ^ { - x } + 1$.
Explain why this method will not converge to $\alpha$ if an initial approximation $x _ { 1 }$ is chosen such that $x _ { 1 } > 1$.
Use this method, with a first approximation $x _ { 1 } = 0$, to find the next three approximations $x _ { 2 } , x _ { 3 }$ and $x _ { 4 }$. Find $\alpha$, correct to 3 decimal places. 6 The equation of a curve is $y = \frac { 2 x ^ { 2 } - 11 x - 6 } { x - 1 }$.
Find the equations of the asymptotes of the curve.
Show that $y$ takes all real values. \section*{Jan 2008} 7 It is given that, for integers $n \geqslant 1$, $$I _ { n } = \int _ { 0 } ^ { 1 } \frac { 1 } { \left( 1 + x ^ { 2 } \right) ^ { n } } \mathrm {~d} x$$
Use integration by parts to show that $I _ { n } = 2 ^ { - n } + 2 n \int _ { 0 } ^ { 1 } \frac { x ^ { 2 } } { \left( 1 + x ^ { 2 } \right) ^ { n + 1 } } \mathrm {~d} x$.
Show that $2 n I _ { n + 1 } = 2 ^ { - n } + ( 2 n - 1 ) I _ { n }$.
Find $I _ { 2 }$ in terms of $\pi$. 8
By using the definition of $\sinh x$ in terms of $\mathrm { e } ^ { x }$ and $\mathrm { e } ^ { - x }$, show that $$\sinh ^ { 3 } x = \frac { 1 } { 4 } \sinh 3 x - \frac { 3 } { 4 } \sinh x$$
Find the range of values of the constant $k$ for which the equation $$\sinh 3 x = k \sinh x$$ has real solutions other than $x = 0$.
Given that $k = 4$, solve the equation in part (ii), giving the non-zero answers in logarithmic form. 9
Prove that $\frac { \mathrm { d } } { \mathrm { d } x } \left( \cosh ^ { - 1 } x \right) = \frac { 1 } { \sqrt { x ^ { 2 } - 1 } }$.
Hence, or otherwise, find $\int \frac { 1 } { \sqrt { 4 x ^ { 2 } - 1 } } \mathrm {~d} x$.
By means of a suitable substitution, find $\int \sqrt { 4 x ^ { 2 } - 1 } \mathrm {~d} x$. \footnotetext{Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (OCR) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. 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. }June 2008
$1 \frac { 2 a x } { \text { It is given that } \mathrm { f } ( x ) = \frac { \text { where } a \text { is a non-zero constant. Express } \mathrm { f } ( x ) \text { in partial } } { ( x - 2 a ) \left( x ^ { 2 } + a ^ { 2 } \right) } \text {, whent } }$ fractions. 2
\includegraphics[max width=\textwidth, alt={}, center]{0ec9c4ff-8622-4dda-a000-6ffe36f38023-18_341_1043_466_552} The diagram shows the curve $y = \mathrm { f } ( x )$. The curve has a maximum point at ( 0,5 ) and crosses the $x$-axis at $( - 2,0 ) , ( 3,0 )$ and $( 4,0 )$. Sketch the curve $y ^ { 2 } = \mathrm { f } ( x )$, showing clearly the coordinates of any turning points and of any points where this curve crosses the axes. 3 By using the substitution $t = \tan \frac { 1 } { 2 } x$, find the exact value of $$\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \frac { 1 } { 2 - \cos x } \mathrm {~d} x$$ giving the answer in terms of $\pi$. 4
Sketch, on the same diagram, the curves with equations $y = \operatorname { sech } x$ and $y = x ^ { 2 }$.
By using the definition of $\operatorname { sech } x$ in terms of $\mathrm { e } ^ { x }$ and $\mathrm { e } ^ { - x }$, show that the $x$-coordinates of the points at which these curves meet are solutions of the equation $$x ^ { 2 } = \frac { 2 \mathrm { e } ^ { x } } { \mathrm { e } ^ { 2 x } + 1 } .$$
The iteration $$x _ { n + 1 } = \sqrt { \frac { 2 \mathrm { e } ^ { x _ { n } } } { \mathrm { e } ^ { 2 x _ { n } } + 1 } }$$ can be used to find the positive root of the equation in part (ii). With initial value $x _ { 1 } = 1$, the approximations $x _ { 2 } = 0.8050 , x _ { 3 } = 0.8633 , x _ { 4 } = 0.8463$ and $x _ { 5 } = 0.8513$ are obtained, correct to 4 decimal places. State with a reason whether, in this case, the iteration produces a 'staircase' or a ‘cobweb’ diagram. 5 It is given that, for $n \geqslant 0$, $$I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \tan ^ { n } x \mathrm {~d} x$$
By considering $I _ { n } + I _ { n - 2 }$, or otherwise, show that, for $n \geqslant 2$, $$( n - 1 ) \left( I _ { n } + I _ { n - 2 } \right) = 1 .$$
Find $I _ { 4 }$ in terms of $\pi$. \section*{June 2008} 6 It is given that $\mathrm { f } ( x ) = 1 - \frac { 7 } { x ^ { 2 } }$.
Use the Newton-Raphson method, with a first approximation $x _ { 1 } = 2.5$, to find the next approximations $x _ { 2 }$ and $x _ { 3 }$ to a root of $\mathrm { f } ( x ) = 0$. Give the answers correct to 6 decimal places. [3]
The root of $\mathrm { f } ( x ) = 0$ for which $x _ { 1 } , x _ { 2 }$ and $x _ { 3 }$ are approximations is denoted by $\alpha$. Write down the exact value of $\alpha$.
The error $e _ { n }$ is defined by $e _ { n } = \alpha - x _ { n }$. Find $e _ { 1 } , e _ { 2 }$ and $e _ { 3 }$, giving your answers correct to 5 decimal places. Verify that $e _ { 3 } \approx \frac { e _ { 2 } ^ { 3 } } { e _ { 1 } ^ { 2 } }$. 7 It is given that $\mathrm { f } ( x ) = \tanh ^ { - 1 } \left( \frac { 1 - x } { 2 + x } \right)$, for $x > - \frac { 1 } { 2 }$.
Show that $\mathrm { f } ^ { \prime } ( x ) = - \frac { 1 } { 1 + 2 x }$, and find $\mathrm { f } ^ { \prime \prime } ( x )$.
Show that the first three terms of the Maclaurin series for $\mathrm { f } ( x )$ can be written as $\ln a + b x + c x ^ { 2 }$, for constants $a , b$ and $c$ to be found. 8 The equation of a curve, in polar coordinates, is $$r = 1 - \sin 2 \theta , \quad \text { for } 0 \leqslant \theta < 2 \pi$$
\includegraphics[max width=\textwidth, alt={}, center]{0ec9c4ff-8622-4dda-a000-6ffe36f38023-19_268_793_1567_717} The diagram shows the part of the curve for which $0 \leqslant \theta \leqslant \alpha$, where $\theta = \alpha$ is the equation of the tangent to the curve at $O$. Find $\alpha$ in terms of $\pi$.
(a) If $\mathrm { f } ( \theta ) = 1 - \sin 2 \theta$, show that $\mathrm { f } \left( \frac { 1 } { 2 } ( 2 k + 1 ) \pi - \theta \right) = \mathrm { f } ( \theta )$ for all $\theta$, where $k$ is an integer.
(b) Hence state the equations of the lines of symmetry of the curve $$r = 1 - \sin 2 \theta , \quad \text { for } 0 \leqslant \theta < 2 \pi$$
Sketch the curve with equation $$r = 1 - \sin 2 \theta , \quad \text { for } 0 \leqslant \theta < 2 \pi$$ State the maximum value of $r$ and the corresponding values of $\theta$. \section*{June 2008} 9
Prove that $\int _ { 0 } ^ { N } \ln ( 1 + x ) \mathrm { d } x = ( N + 1 ) \ln ( N + 1 ) - N$, where $N$ is a positive constant.
\includegraphics[max width=\textwidth, alt={}, center]{0ec9c4ff-8622-4dda-a000-6ffe36f38023-20_616_1261_406_482} The diagram shows the curve $y = \ln ( 1 + x )$, for $0 \leqslant x \leqslant 70$, together with a set of rectangles of unit width.
(a) By considering the areas of these rectangles, explain why $$\ln 2 + \ln 3 + \ln 4 + \ldots + \ln 70 < \int _ { 0 } ^ { 70 } \ln ( 1 + x ) d x$$ (b) By considering the areas of another set of rectangles, show that $$\ln 2 + \ln 3 + \ln 4 + \ldots + \ln 70 > \int _ { 0 } ^ { 69 } \ln ( 1 + x ) d x$$ (c) Hence find bounds between which $\ln ( 70 ! )$ lies. Give the answers correct to 1 decimal place.

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