| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 1 (Further Pure Core 1) |
| Year | 2020 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Integrate using hyperbolic substitution |
| Difficulty | Challenging +1.2 This is a structured Further Maths question on hyperbolic functions with clear scaffolding. Parts (a) and (b) are routine proofs using standard techniques. Part (c) requires executing a given substitution (the hardest part, but methodical), and part (d) is straightforward evaluation. While Further Maths content is inherently harder, the extensive guidance and step-by-step structure make this more accessible than typical FM questions, placing it moderately above average difficulty. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 14.07d Differentiate/integrate: hyperbolic functions4.08e Mean value of function: using integral4.08h Integration: inverse trig/hyperbolic substitutions |
| Answer | Marks | Guidance |
|---|---|---|
| 8 | (a) | eu −e−u 2 |
| Answer | Marks |
|---|---|
| AG | M1 |
| A1 | 2.1 |
| 2.1 | Use of exponential form for sinh u |
| Answer | Marks |
|---|---|
| (b) | cosh2u ≡2sinh2u+1 |
| Answer | Marks | Guidance |
|---|---|---|
| AG | B1 | 2.1 |
| Answer | Marks | Guidance |
|---|---|---|
| M1 | 1.1a | Use exponentials and attempt to integrate. |
| Answer | Marks |
|---|---|
| So f(x)= x(1+x) +c, a =−1, b=1 | = 1e2u −1e−2u −u+c= 1sinh2u−u+c |
| Answer | Marks | Guidance |
|---|---|---|
| = x(1+x)−sinh−1 x +c | A1 | |
| A1 | A1 | 1.1 |
| 1.1 | 1.1 | Ignore c. |
| So f(x)= | x(1+x) +c, a =−1, b=1 | A1 |
| Answer | Marks |
|---|---|
| (d) | 2 |
| Answer | Marks |
|---|---|
| 2+ 3 | M1 |
| A1 | 1.1 |
| 1.1 | Correct limits substituted and subtracted into |
Question 8:
8 | (a) | eu −e−u 2
2sinh2u+1≡2 +1
2
e2u −2+e−2u e2u +e−2u
= +1≡ ≡cosh2u
2 2
AG | M1
A1 | 2.1
2.1 | Use of exponential form for sinh u
[2]
(b) | cosh2u ≡2sinh2u+1
⇒2sinh2u ≡4sinhucoshu
⇒sinh2u ≡2sinhucoshu
AG | B1 | 2.1
[1]
Alternative method
eu −e−u 2
=2∫ du = 1∫e2u −2+e−2udu
2 2
M1 | 1.1a | Use exponentials and attempt to integrate.
= 1e2u −1e−2u −u+c= 1sinh2u−u+c
4 4 2
= x(1+x)−sinh−1 x +c
So f(x)= x(1+x) +c, a =−1, b=1 | = 1e2u −1e−2u −u+c= 1sinh2u−u+c
4 4 2
= x(1+x)−sinh−1 x +c | A1
A1 | A1 | 1.1
1.1 | 1.1 | Ignore c.
So f(x)= | x(1+x) +c, a =−1, b=1 | A1 | 1.1 | c must be included here as part of f(x) – allow a
and b not being stated explicitly but f(x) must be
[5]
(d) | 2
Area = x(1+x)−sinh−1 x
1
( ) ( )
= 6−ln( 2+ 3) − 2−ln(1+ 2)
1+ 2
= 6− 2+ln
2+ 3
1+ 2
So p = 6− 2, q =1, r =
2+ 3 | M1
A1 | 1.1
1.1 | Correct limits substituted and subtracted into
their answer to (c) soi
p, q, r must be stated
[2]8
\begin{enumerate}[label=(\alph*)]
\item Using exponentials, show that $\cosh 2 u \equiv 2 \sinh ^ { 2 } u + 1$.
\item By differentiating both sides of the identity in part (a) with respect to $u$, show that $\sinh 2 u \equiv 2 \sinh u \cosh u$.
\item Use the substitution $\mathrm { x } = \sinh ^ { 2 } \mathrm { u }$ to find $\int \sqrt { \frac { x } { x + 1 } } \mathrm {~d} x$. Give your answer in the form asinh $^ { - 1 } \mathrm {~b} \sqrt { \mathrm { x } } + \mathrm { f } ( \mathrm { x } )$ where $a$ and $b$ are integers and $\mathrm { f } ( x )$ is a function to be determined.
\item Hence determine the exact area of the region between the curve $\mathrm { y } = \sqrt { \frac { \mathrm { x } } { \mathrm { x } + 1 } }$, the $x$-axis, the line $x = 1$ and the line $x = 2$. Give your answer in the form $\mathrm { p } + \mathrm { q } \mid \mathrm { nr }$ where $p , q$ and $r$ are numbers to be determined.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core 1 2020 Q8 [10]}}