| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2018 |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Reduction formulas with hyperbolic integrals |
| Difficulty | Challenging +1.8 This is a Further Maths question requiring reduction formula derivation using hyperbolic identities, followed by a non-trivial connection to tanh^(-1) via series summation. The multi-step nature, need for hyperbolic function manipulation (sech²x = 1 - tanh²x), integration by parts or substitution, and the conceptual leap in part (b) place it well above average difficulty but not at the extreme end for FM students. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.08a Maclaurin series: find series for function |
| Answer | Marks |
|---|---|
| 8(a) | ln2 |
| Answer | Marks |
|---|---|
| tanh2n x(cid:32)tanh2(cid:11)n(cid:16)1(cid:12) xtanh2x | B1 |
| tanh2n x(cid:32)(cid:114)tanh2(cid:11)n(cid:16)1(cid:12) x (cid:11) 1(cid:16)sech2x (cid:12) | M1 |
| Answer | Marks |
|---|---|
| 0 | M1: Correctly substitutes for I and |
| Answer | Marks |
|---|---|
| (cid:179) tanh2(cid:11)n(cid:16)1(cid:12) xsech2xdx(cid:32)ktanh2n(cid:16)1x | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| n(cid:16)1 2n(cid:16)1(cid:169)5(cid:185) | Correct completion with no errors | A1* |
| Answer | Marks |
|---|---|
| 0 | B1 |
| Answer | Marks |
|---|---|
| 0 | ln2 |
| Answer | Marks |
|---|---|
| 0 | M1 |
| Answer | Marks |
|---|---|
| 0 | M1: |
| Answer | Marks |
|---|---|
| A1: Correct expression | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| n(cid:16)1 2n(cid:16)1(cid:169)5(cid:185) | Correct completion with no errors | A1* |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Scheme | Marks |
| Answer | Marks | Guidance |
|---|---|---|
| 8(b) | I (cid:32)ln2 | |
| 0 | The integration must be seen. | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 1 3(cid:169)5(cid:185) | Applies the reduction formula once | M1 |
| Answer | Marks |
|---|---|
| 2 0 1(cid:169)5(cid:185) 3(cid:169)5(cid:185) | M1: Second application of the |
| reduction formula | M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 125 | cao | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 1 0 | Correct integration | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 1 3(cid:169)5(cid:185) | Applies the reduction formula once | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 1 5 | M1: Uses limits | M1A1 |
| Answer | Marks |
|---|---|
| 125 | A1 |
Question 8:
--- 8(a) ---
8(a) | ln2
(cid:179)
I (cid:32) tanh2nx dx, n(cid:116)0
n
0
tanh2n x(cid:32)tanh2(cid:11)n(cid:16)1(cid:12) xtanh2x | B1
tanh2n x(cid:32)(cid:114)tanh2(cid:11)n(cid:16)1(cid:12) x (cid:11) 1(cid:16)sech2x (cid:12) | M1
ln2 ln2
I (cid:32) (cid:179) tanh2(cid:11)n(cid:16)1(cid:12) xdx(cid:16) (cid:179) tanh2(cid:11)n(cid:16)1(cid:12) xsech2xdx
n
0 0
ln2
(cid:170) 1 (cid:186)
I (cid:32) I (cid:16) tanh2n(cid:16)1x
n n(cid:16)1 (cid:171) (cid:172)2n(cid:16)1 (cid:187) (cid:188)
0 | M1: Correctly substitutes for I and
n-1
obtains
(cid:179) tanh2(cid:11)n(cid:16)1(cid:12) xsech2xdx(cid:32)ktanh2n(cid:16)1x | M1 A1
A1: Correct expression
2n(cid:16)1
1 (cid:167)3(cid:183)
(cid:32) I (cid:16) *
(cid:168) (cid:184)
n(cid:16)1 2n(cid:16)1(cid:169)5(cid:185) | Correct completion with no errors | A1*
(5)
Alternative
ln2
I (cid:16)I (cid:32) (cid:179) (cid:11) tanh2n x(cid:16)tanh2(cid:11)n(cid:16)1(cid:12) x (cid:12) dx
n n(cid:16)1
0
ln2
(cid:32) (cid:179) tanh2(cid:11)n(cid:16)1(cid:12) x (cid:11) tanh2 x(cid:16)1 (cid:12) dx
0 | B1
ln2
(cid:32) (cid:179) tanh2(cid:11)n(cid:16)1(cid:12) x (cid:11) (cid:16)sech2 x (cid:12) dx
0 | ln2
(cid:32) (cid:179) tanh2(cid:11)n(cid:16)1(cid:12) x (cid:11) (cid:114)sech2 x (cid:12) dx
0 | M1
ln2
(cid:170) 1 (cid:186)
I (cid:16)I (cid:32)(cid:16) tanh2n(cid:16)1x
n n(cid:16)1 (cid:171) (cid:172)2n(cid:16)1 (cid:187) (cid:188)
0 | M1:
Obtains
(cid:179) tanh2(cid:11)n(cid:16)1(cid:12) xsech2xdx(cid:32)ktanh2n(cid:16)1x
A1: Correct expression | M1 A1
2n(cid:16)1
1 (cid:167)3(cid:183)
(cid:32) I (cid:16) *
(cid:168) (cid:184)
n(cid:16)1 2n(cid:16)1(cid:169)5(cid:185) | Correct completion with no errors | A1*
(5)
Question | Scheme | Marks
--- 8(b) ---
8(b) | I (cid:32)ln2
0 | The integration must be seen. | B1
3
1(cid:167)3(cid:183)
I (cid:32) I (cid:16)
(cid:168) (cid:184)
2 1 3(cid:169)5(cid:185) | Applies the reduction formula once | M1
1 3
1(cid:167)3(cid:183) 1(cid:167)3(cid:183)
I (cid:32) I (cid:16) (cid:16)
(cid:168) (cid:184) (cid:168) (cid:184)
2 0 1(cid:169)5(cid:185) 3(cid:169)5(cid:185) | M1: Second application of the
reduction formula | M1A1
A1: Correct expression
84
I (cid:32)ln2(cid:16)
2 125 | cao | A1
Special Case: If I is found award B1 for I or I and M1M0A0A0
4 o 1
(5)
Alternative
ln2 ln2
I (cid:32) (cid:179) tanh2x dx(cid:32) (cid:179) (cid:11) 1(cid:16)sech2x (cid:12) dx
1
0 0
(cid:32)(cid:62)x(cid:16)tanhx(cid:64)ln2
I
1 0 | Correct integration | B1
3
1(cid:167)3(cid:183)
I (cid:32) I (cid:16)
(cid:168) (cid:184)
2 1 3(cid:169)5(cid:185) | Applies the reduction formula once | M1
3
I (cid:32)ln2(cid:16)tanh(cid:11)ln2(cid:12)(cid:32)ln2(cid:16)
1 5 | M1: Uses limits | M1A1
A1: Correct expression
3
3 1(cid:167)3(cid:183)
I (cid:32)ln2(cid:16) (cid:16)
(cid:168) (cid:184)
2 5 3(cid:169)5(cid:185)
84
(cid:32)ln2(cid:16)
125 | A1
(5)
(10 marks)
PPPMMMTTT
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Sample Assessment Materials for first teaching September 2018
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1/1/1/1/1/1/
322 Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and
Pure Mathematics – Sample Assessment Materials (SAMs) – Issue 3 – June 2018 © Pearson Education Limited 2018$$I_n = \int_{0}^{\ln 2} \tanh^{2n} x \, dx, \quad n \geq 0$$
\begin{enumerate}[label=(\alph*)]
\item Show that, for $n \geq 1$
$$I_n = I_{n-1} - \frac{1}{2n-1}\left(\frac{3}{5}\right)^{2n-1}$$
[5]
\item Hence show that
$$\int_{0}^{\ln 2} \tanh^{-1} x \, dx = p + \ln 2$$
where $p$ is a rational number to be found.
[5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel F3 2018 Q8 [10]}}