| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2020 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration using inverse trig and hyperbolic functions |
| Type | Completing square then standard inverse trig |
| Difficulty | Standard +0.3 This is a standard Further Maths question testing completing the square followed by direct application of inverse trig formulas (arctan and arcsin). While it requires knowledge of Further Pure content, the technique is routine and mechanical with no problem-solving insight needed—slightly easier than an average A-level question once the method is known. |
| Spec | 4.08h Integration: inverse trig/hyperbolic substitutions |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Notes |
| \(3x^2+12x+24 = 3(x^2+4x+8) = 3\left((x+2)^2+4\right)\) | M1 | Obtains \(3\left((x+2)^2+\ldots\right)\) or \(3(x+2)^2+\ldots\); must include 3 now or later |
| \(3\left((x+2)^2+4\right)\) or \(3(x+2)^2+12\) | A1 | Correct completed square form |
| \(\int \frac{1}{3x^2+12x+24}\,dx = \frac{1}{3}\int\frac{1}{(x+2)^2+4}\,dx = \frac{1}{6}\arctan\frac{x+2}{2}(+c)\) | M1A1 | M1: Use of arctan; A1: Fully correct expression (condone omission of \(+c\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Notes |
| \(27-6x-x^2 = -(x^2+6x-27) = -\left((x+3)^2-36\right)\) | M1 | Obtains \(-\left((x+3)^2+\ldots\right)\) or \(-(x+3)^2+\ldots\) |
| \(-\left((x+3)^2-36\right)\) or \(36-(x+3)^2\) | A1 | Correct completed square form |
| \(\int\frac{1}{\sqrt{27-6x-x^2}}\,dx = \int\frac{1}{\sqrt{36-(x+3)^2}}\,dx = \arcsin\!\left(\frac{x+3}{6}\right)(+c)\) (or \(-\arccos\!\left(\frac{x+3}{6}\right)(+c)\)) | M1A1 | M1: Use of arcsin (or \(-\)arccos); A1: Fully correct expression (condone omission of \(+c\)) |
# Question 2:
## Part (i):
| Working/Answer | Mark | Notes |
|---|---|---|
| $3x^2+12x+24 = 3(x^2+4x+8) = 3\left((x+2)^2+4\right)$ | M1 | Obtains $3\left((x+2)^2+\ldots\right)$ or $3(x+2)^2+\ldots$; must include 3 now or later |
| $3\left((x+2)^2+4\right)$ or $3(x+2)^2+12$ | A1 | Correct completed square form |
| $\int \frac{1}{3x^2+12x+24}\,dx = \frac{1}{3}\int\frac{1}{(x+2)^2+4}\,dx = \frac{1}{6}\arctan\frac{x+2}{2}(+c)$ | M1A1 | M1: Use of arctan; A1: Fully correct expression (condone omission of $+c$) |
## Part (ii):
| Working/Answer | Mark | Notes |
|---|---|---|
| $27-6x-x^2 = -(x^2+6x-27) = -\left((x+3)^2-36\right)$ | M1 | Obtains $-\left((x+3)^2+\ldots\right)$ or $-(x+3)^2+\ldots$ |
| $-\left((x+3)^2-36\right)$ or $36-(x+3)^2$ | A1 | Correct completed square form |
| $\int\frac{1}{\sqrt{27-6x-x^2}}\,dx = \int\frac{1}{\sqrt{36-(x+3)^2}}\,dx = \arcsin\!\left(\frac{x+3}{6}\right)(+c)$ (or $-\arccos\!\left(\frac{x+3}{6}\right)(+c)$) | M1A1 | M1: Use of arcsin (or $-$arccos); A1: Fully correct expression (condone omission of $+c$) |
---2. Determine\\
(i) $\int \frac { 1 } { 3 x ^ { 2 } + 12 x + 24 } \mathrm {~d} x$\\
(ii) $\int \frac { 1 } { \sqrt { 27 - 6 x - x ^ { 2 } } } \mathrm {~d} x$\\
\hfill \mbox{\textit{Edexcel F3 2020 Q2 [8]}}