| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2018 |
| Session | Specimen |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration using inverse trig and hyperbolic functions |
| Type | Completing square then standard inverse trig |
| Difficulty | Challenging +1.2 This is a Further Maths F3 question testing standard techniques for inverse trig and hyperbolic integrals. Part (a) requires completing the square and recognizing arctan form (routine for FM students). Part (b) requires completing the square and recognizing arsinh form with slightly more algebraic manipulation. While the topic is advanced, these are textbook applications of memorized formulas with straightforward algebra, making it moderately above average difficulty but not requiring novel insight. |
| Spec | 1.08h Integration by substitution4.08h Integration: inverse trig/hyperbolic substitutions |
| Answer | Marks | Guidance |
|---|---|---|
| 3(a) | x2 (cid:14)4x(cid:14)13(cid:123)(cid:11)x(cid:14)2(cid:12)2 (cid:14) 9 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| (cid:11)x(cid:14)2(cid:12)2 (cid:14) 9 3 (cid:169) 3 (cid:185) | M1: karctanf(cid:11)x(cid:12). | M1 A1 |
| Answer | Marks |
|---|---|
| (cid:16)2 | Correct use of limits |
| arctan0 need not be shown | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 12 | cao | A1 |
| Answer | Marks |
|---|---|
| x2 (cid:14)4x(cid:14)13(cid:123)(cid:11)x(cid:14)2(cid:12)2 (cid:14) 9 | B1 |
| Answer | Marks |
|---|---|
| 9tan2t(cid:14)9 3 3 | M1 sub and integrate inc use of |
| Answer | Marks |
|---|---|
| limits | M1 A1 |
| Answer | Marks |
|---|---|
| 0 | Either change limits and |
| Answer | Marks |
|---|---|
| substitute original imits | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 12 | cao | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Scheme | Marks |
| Answer | Marks |
|---|---|
| 3(b) | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| or (cid:14)25 | M1: 4(cid:11)x(cid:114) p(cid:12)2 (cid:114)q, (cid:11)p,q(cid:122)0(cid:12) | M1 A1 |
| Answer | Marks |
|---|---|
| M1: karsinh f(cid:11)x(cid:12). A1: Correct expression | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| (cid:172) (cid:169) 2 (cid:185)(cid:188) (cid:16)1 | Correct use of limits | M1 |
| Answer | Marks |
|---|---|
| 2 | Uses the logarithmic |
| form of arsinh | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | cao | A1 |
| Answer | Marks |
|---|---|
| (cid:16)5 | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Scheme | Marks |
Question 3:
--- 3(a) ---
3(a) | x2 (cid:14)4x(cid:14)13(cid:123)(cid:11)x(cid:14)2(cid:12)2 (cid:14) 9 | B1
(cid:179) 1 1 (cid:167) x(cid:14)2(cid:183)
dx(cid:32) arctan
(cid:168) (cid:184)
(cid:11)x(cid:14)2(cid:12)2 (cid:14) 9 3 (cid:169) 3 (cid:185) | M1: karctanf(cid:11)x(cid:12). | M1 A1
A1: Correct expression
1
(cid:170)1 (cid:167)x(cid:14)2(cid:183)(cid:186) 1
arctan (cid:32) (cid:11)arctan1(cid:16)arctan0(cid:12)
(cid:171) (cid:168) (cid:184)(cid:187)
(cid:172)3 (cid:169) 3 (cid:185)(cid:188) 3
(cid:16)2 | Correct use of limits
arctan0 need not be shown | M1
(cid:83)
12 | cao | A1
(5)
Alternative
Sub x(cid:14)2(cid:32)3tant
x2 (cid:14)4x(cid:14)13(cid:123)(cid:11)x(cid:14)2(cid:12)2 (cid:14) 9 | B1
dx (cid:83)
(cid:32)3sec2t x(cid:32)(cid:16)2,tant (cid:32)0, t (cid:32)0; x(cid:32)1,tant (cid:32)1, t (cid:32)
dt 4
(cid:179) 3sec2t 1(cid:179) 1
dt (cid:32) dt (cid:32) t
9tan2t(cid:14)9 3 3 | M1 sub and integrate inc use of
tan2(cid:14)1(cid:32)sec2
A1 Correct expression Ignore
limits | M1 A1
(cid:83)
(cid:170)(cid:83)(cid:186)4
.
(cid:171) (cid:187)
(cid:172)12(cid:188)
0 | Either change limits and
substitute
Or reverse substitution and
substitute original imits | M1
(cid:83)
12 | cao | A1
(5)
Question | Scheme | Marks
--- 3(b) ---
3(b) | 2
(cid:167) 3(cid:183)
4x2 (cid:16)12x(cid:14)34(cid:32)4 x(cid:16) (cid:14)25
(cid:168) (cid:184)
(cid:169) 2(cid:185)
(cid:11)2x(cid:16)3(cid:12)2
or (cid:14)25 | M1: 4(cid:11)x(cid:114) p(cid:12)2 (cid:114)q, (cid:11)p,q(cid:122)0(cid:12) | M1 A1
2
(cid:167) 3(cid:183)
A1: 4 x(cid:16) (cid:14)25
(cid:168) (cid:184)
(cid:169) 2(cid:185)
(cid:179) 1 1(cid:179) 1 1 (cid:167) x(cid:16) 3 (cid:183)
dx(cid:32) dx (cid:32) arsinh(cid:168) 2 (cid:184)
4(cid:11)x(cid:16) 3(cid:12)2 (cid:14) 25 2 (cid:11)x(cid:16) 3(cid:12)2 (cid:14) 25 2 (cid:169) 5 2 (cid:185)
2 2 4
M1: karsinh f(cid:11)x(cid:12). A1: Correct expression | M1 A1
4
(cid:170)1 (cid:167) x(cid:16)3 (cid:183)(cid:186) 1
(cid:171) arsinh(cid:168) 2 (cid:184)(cid:187) (cid:32) (cid:11) arsinh(cid:11)1(cid:12)(cid:16)arsinh(cid:11)(cid:16)1(cid:12)(cid:12)
2 5 2
(cid:172) (cid:169) 2 (cid:185)(cid:188) (cid:16)1 | Correct use of limits | M1
1(cid:11) (cid:11) (cid:12) (cid:11) (cid:12)(cid:12)
(cid:32) ln 1(cid:14) 2 (cid:16)ln (cid:16)1(cid:14) 2
2 | Uses the logarithmic
form of arsinh | M1
1 (cid:11) (cid:12) (cid:11) (cid:12)
(cid:32) ln 3(cid:14)2 2 or ln 1(cid:14) 2
2 | cao | A1
(7)
Alternative: Second M1 A1
Sub 2x(cid:16)3(cid:32)u or 2x(cid:16)3(cid:32)5sinhu
5
arsinh1 1 (cid:170)1 (cid:167)u(cid:183)(cid:186)
(cid:179) 5coshudu (cid:32) arsinh
(cid:171) (cid:168) (cid:184)(cid:187)
arsinh(cid:16)1 25sinh2u(cid:14)25 (cid:172)2 (cid:169)5(cid:185)(cid:188)
(cid:16)5 | M1 A1
5
5 1 (cid:170)1 (cid:167)u(cid:183)(cid:186)
(cid:179) du (cid:32) arsinh
(cid:171) (cid:168) (cid:184)(cid:187)
(cid:16)52 u2 (cid:14)25 (cid:172)2 (cid:169)5(cid:185)(cid:188)
(cid:16)5
(12 marks)
Question | Scheme | MarksWithout using a calculator, find
\begin{enumerate}[label=(\alph*)]
\item $\int_{-2}^{1} \frac{1}{x^2 + 4x + 13} \, dx$, giving your answer as a multiple of $\pi$,
[5]
\item $\int_{-1}^{4} \frac{1}{\sqrt{4x^2 - 12x + 34}} \, dx$, giving your answer in the form $p \ln\left(q + r\sqrt{2}\right)$,
[7]
\end{enumerate}
where $p$, $q$ and $r$ are rational numbers to be found.
\hfill \mbox{\textit{Edexcel F3 2018 Q3 [12]}}