OCR MEI FP2 2007 June — Question 4 18 marks

Exam BoardOCR MEI
ModuleFP2 (Further Pure Mathematics 2)
Year2007
SessionJune
Marks18
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHyperbolic functions
TypeFind stationary points of hyperbolic curves
DifficultyStandard +0.8 This is a substantial Further Maths question requiring multiple hyperbolic function techniques: inverse hyperbolic integration, proving a double-angle identity from exponential definitions, finding stationary points involving cosh 2x (requiring the identity from part b(i)), solving a hyperbolic equation to get x = ln 3, and evaluating a definite integral. While each component uses standard FM techniques, the multi-part structure, need to apply the proven identity, and algebraic manipulation to reach exact forms like ln 3 and 59/3 elevate this above routine FM exercises.
Spec4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 14.07d Differentiate/integrate: hyperbolic functions4.08h Integration: inverse trig/hyperbolic substitutions
4
  1. Find \(\int _ { 0 } ^ { 1 } \frac { 1 } { \sqrt { 9 x ^ { 2 } + 16 } } \mathrm {~d} x\), giving your answer in an exact logarithmic form.
    1. Starting from the definitions of \(\sinh x\) and \(\cosh x\) in terms of exponentials, prove that \(\sinh 2 x = 2 \sinh x \cosh x\).
    2. Show that one of the stationary points on the curve $$y = 20 \cosh x - 3 \cosh 2 x$$ has coordinates \(\left( \ln 3 , \frac { 59 } { 3 } \right)\), and find the coordinates of the other two stationary points.
    3. Show that \(\int _ { - \ln 3 } ^ { \ln 3 } ( 20 \cosh x - 3 \cosh 2 x ) \mathrm { d } x = 40\).
Question 4: Investigation of curves
This question requires the use of a graphical calculator.
The curve with equation \(y = \frac{x^2 - kx + 2k}{x - k}\) is to be investigated for different values of \(k\).
(i) [6]
Use your graphical calculator to obtain rough sketches of the curve in the cases \(k = -2\), \(k = -0.5\) and \(k = 1\).
(ii) [4]
Show that the equation of the curve may be written as \(y = x - 2k + \frac{2k(k - 1)}{x - k}\).
M1 Attempt polynomial division
A1 Correct quotient \(x - 2k\)
A1 Correct remainder \(2k(k - 1)\)
Hence find the two values of \(k\) for which the curve is a straight line.
A1 \(k = 0\) and \(k = 1\)
(iii) When the curve is not a straight line, it is a conic.
(A) [1]
Name the type of conic.
B1 Rectangular hyperbola
(B) [2]
Write down the equations of the asymptotes.
A1 \(x = k\)
A1 \(y = x - 2k\)
(iv) [5]
Draw a sketch to show the shape of the curve when \(1 < k < 8\). This sketch should show where the curve crosses the axes and how it approaches its asymptotes. Indicate the points A and B on the curve where \(x = 1\) and \(x = k\) respectively.
B1 Two branches with correct orientation relative to asymptotes
B1 Curve crosses \(y\)-axis at \((0, -2)\)
B1 Curve crosses \(x\)-axis at \(x = 2k\)
B1 Point A marked at \(x = 1\) (on left branch)
B1 Asymptotes clearly shown and labelled or evident
# Question 4: Investigation of curves

This question requires the use of a graphical calculator.

The curve with equation $y = \frac{x^2 - kx + 2k}{x - k}$ is to be investigated for different values of $k$.

## (i) [6]

Use your graphical calculator to obtain rough sketches of the curve in the cases $k = -2$, $k = -0.5$ and $k = 1$.

## (ii) [4]

Show that the equation of the curve may be written as $y = x - 2k + \frac{2k(k - 1)}{x - k}$.

M1 Attempt polynomial division

A1 Correct quotient $x - 2k$

A1 Correct remainder $2k(k - 1)$

Hence find the two values of $k$ for which the curve is a straight line.

A1 $k = 0$ and $k = 1$

## (iii) When the curve is not a straight line, it is a conic.

### (A) [1]

Name the type of conic.

B1 Rectangular hyperbola

### (B) [2]

Write down the equations of the asymptotes.

A1 $x = k$

A1 $y = x - 2k$

## (iv) [5]

Draw a sketch to show the shape of the curve when $1 < k < 8$. This sketch should show where the curve crosses the axes and how it approaches its asymptotes. Indicate the points A and B on the curve where $x = 1$ and $x = k$ respectively.

B1 Two branches with correct orientation relative to asymptotes

B1 Curve crosses $y$-axis at $(0, -2)$

B1 Curve crosses $x$-axis at $x = 2k$

B1 Point A marked at $x = 1$ (on left branch)

B1 Asymptotes clearly shown and labelled or evident
4
\begin{enumerate}[label=(\alph*)]
\item Find $\int _ { 0 } ^ { 1 } \frac { 1 } { \sqrt { 9 x ^ { 2 } + 16 } } \mathrm {~d} x$, giving your answer in an exact logarithmic form.
\item \begin{enumerate}[label=(\roman*)]
\item Starting from the definitions of $\sinh x$ and $\cosh x$ in terms of exponentials, prove that $\sinh 2 x = 2 \sinh x \cosh x$.
\item Show that one of the stationary points on the curve

$$y = 20 \cosh x - 3 \cosh 2 x$$

has coordinates $\left( \ln 3 , \frac { 59 } { 3 } \right)$, and find the coordinates of the other two stationary points.
\item Show that $\int _ { - \ln 3 } ^ { \ln 3 } ( 20 \cosh x - 3 \cosh 2 x ) \mathrm { d } x = 40$.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{OCR MEI FP2 2007 Q4 [18]}}
This paper (5 questions)
View full paper
Q1 18 Q2 18 Q3 18 Q4 18 Q5 18