| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure (Further Additional Pure) |
| Year | 2020 |
| Session | November |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Surface area of revolution with hyperbolics |
| Difficulty | Challenging +1.8 This is a challenging Further Maths question requiring arc length and surface of revolution formulas with hyperbolic functions. Part (a) involves computing arc length of y cosh x using the identity 1 + sinh²x = cosh²x. Part (b) requires deriving a complex surface area integral from z = 1 (so y = sech x), involving implicit differentiation and algebraic manipulation to reach the given form. While the techniques are standard for Further Maths, the algebraic complexity with hyperbolics and the multi-step derivation make this substantially harder than typical A-level questions. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 18.06b Arc length and surface area: of revolution, cartesian or parametric |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (a) | (i) |
| [1] | 3.4 | Penalise line extending outside x ∈[–ln20, ln20] |
| Answer | Marks |
|---|---|
| (ii) | 2 |
| Answer | Marks |
|---|---|
| < 20 so YES (design requirement met) | M1 |
| Answer | Marks |
|---|---|
| [4] | 3.1b |
| Answer | Marks |
|---|---|
| 3.5a | Condone use of y instead of z |
| Answer | Marks | Guidance |
|---|---|---|
| (b) | (i) | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Axes must be x- and y-axes and labelled thus | B1 | |
| [1] | 3.4 | Allow ft for reciprocal of previous function |
| Answer | Marks |
|---|---|
| (ii) | dy −sinhx |
| Answer | Marks |
|---|---|
| 0 | B1 |
| Answer | Marks |
|---|---|
| [5] | 1.1 |
| Answer | Marks |
|---|---|
| 2.3 | Correct derivative (with correct sign) |
| Answer | Marks |
|---|---|
| (iii) | ∫ above (given) is 4π × 1.564 710 33… = 19.7 |
| so YES (design requirement met) | B1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 3.5a | BC correct to ≥ 3 s.f. (19.662 73 to 5 d.p.) |
Question 5:
5 | (a) | (i) | z = cosh x (catenary curve) through (0, 1), symmetrical about z-axis | B1
[1] | 3.4 | Penalise line extending outside x ∈[–ln20, ln20]
Axes must be x- and z-axes and labelled thus
(ii) | 2
dz dz
L = ∫ 1+ dx used with = sinh x
dx dx
= ∫ 1+sinh2 xdx or ∫coshxdx with correct limits
= 19.95 or via 2sinh(ln20)
< 20 so YES (design requirement met) | M1
A1
A1
A1
[4] | 3.1b
1.1
1.1
3.5a | Condone use of y instead of z
May have a factor of 2 if limits (0, ln20) used
BC or via correct integration
From cao with stated conclusion
(b) | (i) | 1
y = sketched
coshx
(must be through (0, 1) and symmetrical about y-axis)
Axes must be x- and y-axes and labelled thus | B1
[1] | 3.4 | Allow ft for reciprocal of previous function
(provided all positive)
Condone line extending outside x ∈[–ln20, ln20]
only if already penalised in (a) (i)
(ii) | dy −sinhx
=
dx cosh2 x
2
dy dy
A = 2π∫y 1+ dx used with y and = …
dx dx
ln20 1 sinh2x
= 2π ∫ 1+ dx
coshx cosh4 x
−ln20
Use of sinh2x ≡ cosh2x – 1
ln20 1
= 4π ∫ cosh4x+cosh2x−1dx
cosh3x
0 | B1
M1
A1
M1
A1
[5] | 1.1
1.1
1.1
3.1a
2.3 | Correct derivative (with correct sign)
Condone (incorrect sign)-squared
AG Must show clearly how the limits give k = 4
and how the integrand is as shown
(iii) | ∫ above (given) is 4π × 1.564 710 33… = 19.7
so YES (design requirement met) | B1
B1
[2] | 1.1
3.5a | BC correct to ≥ 3 s.f. (19.662 73 to 5 d.p.)
Cao Correct conclusion must be stated
5 A designer intends to manufacture a product using a 3-D printer. The product will take the form of a surface $S$ which must meet a number of design specifications. The designer chooses to model $S$ with the equation $\mathrm { Z } = \mathrm { y } \cosh \mathrm { x }$ for $- \ln 20 \leqslant x \leqslant \ln 20 , - 2 \leqslant y \leqslant 2$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item In the Printed Answer Booklet, on the axes provided, sketch the section of $S$ given by $y = 1$.
\item One of the design specifications of the product is that this section should have a length no greater than 20 units.
Determine whether the product meets this requirement according to the model.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item In the Printed Answer Booklet, on the axes provided, sketch the contour of $S$ given by $z = 1$.
\item When this contour is rotated through $2 \pi$ radians about the $x$-axis, the surface $T$ is generated. The surface area of $T$ is denoted by $A$.
Show that $A$ can be written in the form $\mathrm { k } \pi \int _ { 0 } ^ { \ln 20 } \frac { 1 } { \cosh ^ { 3 } \mathrm { x } } \sqrt { \cosh ^ { 4 } \mathrm { x } + \cosh ^ { 2 } \mathrm { x } - 1 } \mathrm { dx }$ for some\\
integer $k$ to be determined. integer $k$ to be determined.
\item A second design specification is that the surface area of $T$ must not be greater than 20 square units.
Use your calculator to decide whether the product meets this requirement according to the model.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure 2020 Q5 [13]}}