| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2014 |
| Session | June |
| Marks | 7 |
| 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 straightforward Further Maths question requiring completing the square (routine algebraic manipulation) followed by direct application of standard inverse trig and hyperbolic integral formulas. While it's from FP3, the techniques are mechanical with no problem-solving insight needed—students simply recognize the forms and apply memorized results. Slightly above average difficulty due to being Further Maths content, but well below typical FP3 challenge level. |
| Spec | 1.02e Complete the square: quadratic polynomials and turning points4.08h Integration: inverse trig/hyperbolic substitutions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a=9,\quad b=\dfrac{1}{3},\quad c=4\) | B1, B1, B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\displaystyle\int \dfrac{1}{9(x+\frac{1}{3})^2+4}\,dx = \dfrac{1}{6}\arctan\!\left(\dfrac{3x+1}{2}\right)(+c)\) | M1A1 | M1: \(k\arctan\!\left(\dfrac{x+\text{"}\frac{1}{3}\text{"}}{\sqrt{\text{"}\frac{4}{9}\text{"}}} \right)\); A1: \(\dfrac{1}{6}\arctan\!\left(\dfrac{3x+1}{2}\right)\) oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\displaystyle\int \dfrac{1}{\sqrt{9(x+\frac{1}{3})^2+4}}\,dx = \dfrac{1}{3}\text{arsinh}\!\left(\dfrac{3x+1}{2}\right)(+c)\) | M1A1 | M1: \(k\,\text{arsinh}\!\left(\dfrac{x+\text{"}\frac{1}{3}\text{"}}{\sqrt{\text{"}\frac{4}{9}\text{"}}} \right)\); A1: \(\dfrac{1}{3}\text{arsinh}\!\left(\dfrac{3x+1}{2}\right)\) oe; Allow \(\dfrac{1}{\sqrt{9}}\) |
# Question 2: $9x^2+6x+5 \equiv a(x+b)^2+c$
**Part (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a=9,\quad b=\dfrac{1}{3},\quad c=4$ | B1, B1, B1 | |
**Part (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\displaystyle\int \dfrac{1}{9(x+\frac{1}{3})^2+4}\,dx = \dfrac{1}{6}\arctan\!\left(\dfrac{3x+1}{2}\right)(+c)$ | M1A1 | M1: $k\arctan\!\left(\dfrac{x+\text{"}\frac{1}{3}\text{"}}{\sqrt{\text{"}\frac{4}{9}\text{"}}} \right)$; A1: $\dfrac{1}{6}\arctan\!\left(\dfrac{3x+1}{2}\right)$ oe |
**Part (c):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\displaystyle\int \dfrac{1}{\sqrt{9(x+\frac{1}{3})^2+4}}\,dx = \dfrac{1}{3}\text{arsinh}\!\left(\dfrac{3x+1}{2}\right)(+c)$ | M1A1 | M1: $k\,\text{arsinh}\!\left(\dfrac{x+\text{"}\frac{1}{3}\text{"}}{\sqrt{\text{"}\frac{4}{9}\text{"}}} \right)$; A1: $\dfrac{1}{3}\text{arsinh}\!\left(\dfrac{3x+1}{2}\right)$ oe; Allow $\dfrac{1}{\sqrt{9}}$ |
---2.
$$9 x ^ { 2 } + 6 x + 5 \equiv a ( x + b ) ^ { 2 } + c$$
\begin{enumerate}[label=(\alph*)]
\item Find the values of the constants $a$, $b$ and $c$.
Hence, or otherwise, find
\item $\int \frac { 1 } { 9 x ^ { 2 } + 6 x + 5 } d x$
\item $\int \frac { 1 } { \sqrt { 9 x ^ { 2 } + 6 x + 5 } } \mathrm {~d} x$
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP3 2014 Q2 [7]}}