| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2013 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Compound expressions with binomial expansion |
| Difficulty | Challenging +1.8 This is a Further Maths FP3 reduction formula question requiring integration by parts to derive the recurrence relation, then applying it iteratively. While the integration by parts setup is standard for this topic, students must carefully handle the fractional power, work through algebraic manipulation to reach the given form, compute Iā from first principles, then apply the formula twice. The multi-step nature and need for precision in algebraic manipulation elevate this above average difficulty, though it follows a well-established template for FP3 reduction formula questions. |
| Spec | 8.06a Reduction formulae: establish, use, and evaluate recursively |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I_n = \left[x^n(2x-1)^{\frac{1}{2}}\right]_1^5 - \int_1^5 nx^{n-1}(2x-1)^{\frac{1}{2}}\,dx\) | M1 | Parts in the correct direction including a valid attempt to integrate \((2x-1)^{-\frac{1}{2}}\) |
| A1 | Fully correct application ā may be unsimplified. (Ignore limits.) | |
| \(I_n = 5^n\times3 - 1 - \int_1^5 nx^{n-1}(2x-1)(2x-1)^{-\frac{1}{2}}\,dx\) | B1 | Obtains correct (possibly unsimplified) expression using limits 5 and 1 and writes \((2x-1)^{\frac{1}{2}}\) as \((2x-1)(2x-1)^{-\frac{1}{2}}\) |
| \(I_n = 5^n\times3 - 1 - 2nI_n + nI_{n-1}\) | dM1 | Replaces \(\int x^n(2x-1)^{-\frac{1}{2}}\,dx\) with \(I_n\) and \(\int x^{n-1}(2x-1)^{-\frac{1}{2}}\,dx\) with \(I_{n-1}\) |
| \((2n+1)I_n = nI_{n-1} + 3\times5^n - 1\) * | A1cso | Correct completion to printed answer with no errors seen |
| (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I_0 = \int_1^5(2x-1)^{-\frac{1}{2}}\,dx = \left[(2x-1)^{\frac{1}{2}}\right]_1^5 = 2\) | B1 | \(I_0 = 2\) |
| \(5I_2 = 2I_1 + 74\) and \(3I_1 = I_0 + 14\) | M1 | Correctly applies the given reduction formula twice |
| A1 | Correct equations for \(I_2\) and \(I_1\) (may be implied) | |
| \(I_1 = \frac{16}{3}\) and \(I_2 = \ldots\) or \(5I_2 = 2\cdot\frac{I_0+14}{3}+74\) and \(I_2 = \ldots\) | dM1 | Completes to obtain a numerical expression for \(I_2\) |
| \(I_2 = \frac{254}{15}\) | B1 | |
| (5) | ||
| Total 10 |
# Question 5:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_n = \left[x^n(2x-1)^{\frac{1}{2}}\right]_1^5 - \int_1^5 nx^{n-1}(2x-1)^{\frac{1}{2}}\,dx$ | M1 | Parts in the correct direction including a valid attempt to integrate $(2x-1)^{-\frac{1}{2}}$ |
| | A1 | Fully correct application ā may be unsimplified. (Ignore limits.) |
| $I_n = 5^n\times3 - 1 - \int_1^5 nx^{n-1}(2x-1)(2x-1)^{-\frac{1}{2}}\,dx$ | B1 | Obtains correct (possibly unsimplified) expression using limits 5 and 1 **and** writes $(2x-1)^{\frac{1}{2}}$ as $(2x-1)(2x-1)^{-\frac{1}{2}}$ |
| $I_n = 5^n\times3 - 1 - 2nI_n + nI_{n-1}$ | dM1 | Replaces $\int x^n(2x-1)^{-\frac{1}{2}}\,dx$ with $I_n$ and $\int x^{n-1}(2x-1)^{-\frac{1}{2}}\,dx$ with $I_{n-1}$ |
| $(2n+1)I_n = nI_{n-1} + 3\times5^n - 1$ * | A1cso | Correct completion to printed answer with no errors seen |
| | **(5)** | |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_0 = \int_1^5(2x-1)^{-\frac{1}{2}}\,dx = \left[(2x-1)^{\frac{1}{2}}\right]_1^5 = 2$ | B1 | $I_0 = 2$ |
| $5I_2 = 2I_1 + 74$ and $3I_1 = I_0 + 14$ | M1 | Correctly applies the given reduction formula twice |
| | A1 | Correct equations for $I_2$ and $I_1$ (may be implied) |
| $I_1 = \frac{16}{3}$ and $I_2 = \ldots$ or $5I_2 = 2\cdot\frac{I_0+14}{3}+74$ and $I_2 = \ldots$ | dM1 | Completes to obtain a numerical expression for $I_2$ |
| $I_2 = \frac{254}{15}$ | B1 | |
| | **(5)** | |
| | **Total 10** | |5.
$$I _ { n } = \int _ { 1 } ^ { 5 } x ^ { n } ( 2 x - 1 ) ^ { - \frac { 1 } { 2 } } \mathrm {~d} x , \quad n \geqslant 0$$
\begin{enumerate}[label=(\alph*)]
\item Prove that, for $n \geqslant 1$,
$$( 2 n + 1 ) I _ { n } = n I _ { n - 1 } + 3 \times 5 ^ { n } - 1$$
\item Using the reduction formula given in part (a), find the exact value of $I _ { 2 }$
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP3 2013 Q5 [10]}}