| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 1 (Further Pure Core 1) |
| Year | 2019 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Express hyperbolic in exponential form |
| Difficulty | Standard +0.8 This is a Further Maths question requiring students to work with hyperbolic functions in exponential form and perform a non-standard integration. Part (a) is routine algebraic manipulation, but part (b) requires recognizing a clever substitution (likely u = e^x) to transform the integral into a standard form, which is beyond typical A-level and requires problem-solving insight characteristic of Further Maths content. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.08h Integration: inverse trig/hyperbolic substitutions |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (a) | ex +e−x e2x +1 |
| Answer | Marks |
|---|---|
| e2x +1 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| [2] | 1.1 | |
| 2.1 | Use of coshx in exponentials | |
| 7 | (b) | u =ex ⇒du =exdx |
| Answer | Marks |
|---|---|
| +c A1 | M1 |
| Answer | Marks |
|---|---|
| [4] | 3.1a |
| Answer | Marks |
|---|---|
| 1.1 | Substitute and use (a) |
| Answer | Marks |
|---|---|
| Must include c | Allow absence of du |
Question 7:
7 | (a) | ex +e−x e2x +1
coshx= =
2 2ex
2ex
⇒sechx= AG
e2x +1 | M1
A1
[2] | 1.1
2.1 | Use of coshx in exponentials
7 | (b) | u =ex ⇒du =exdx
du
⇒dx=
u
2ex
⇒∫sechxdx=∫ dx
e2x +1
2u du
=∫ .
u2 +1 u
=2tan−1(u)+c=2tan−1(ex)+c
Alternatively:
u =sinhx⇒du =coshxdx M1
1 du du
⇒∫sech xdx=∫ . =∫
coshx coshx cosh2 x
du du
=∫ ==∫ A1
1+sinh2 x 1+u2
=tan−1u+c M1
=tan−1( )+c
sinhx A1
Alternatively:
2ex
∫sech xdx=∫ dx
e2x +1
sec2u
Let ex =tanu⇒exdx=sec2udu⇒dx= du M1
tanu
2tanu sec2u
⇒∫sech xdx=∫ . du =2∫du A1
tan2u+1 tanu
=2u+c M1
=2tan−1( ex)
+c A1 | M1
A1
M1
A1
[4] | 3.1a
1.1
3.1a
1.1 | Substitute and use (a)
any form entirely in terms
of u
Use standard form for
integral and substitute back
Must include c
Substitute
In correct form
Integrate and substitute back
Must include c | Allow absence of du7 The function sech $x$ is defined by $\operatorname { sech } x = \frac { 1 } { \cosh x }$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\operatorname { sech } x = \frac { 2 \mathrm { e } ^ { x } } { \mathrm { e } ^ { 2 x } + 1 }$.
\item Using a suitable substitution, find $\int \operatorname { sech } x \mathrm {~d} x$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core 1 2019 Q7 [6]}}