| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2021 |
| Session | November |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Indefinite & Definite Integrals |
| Type | Integration with given constant |
| Difficulty | Moderate -0.8 This is a straightforward definite integration problem requiring only basic integration of a power function (x^{-1/2}), substitution of limits, and solving a simple equation for k. It's easier than average as it involves routine techniques with no conceptual challenges or multi-step reasoning. |
| Spec | 1.08d Evaluate definite integrals: between limits |
| VI4V SIHIL NI III HM IONOO | VIAV SIHI NI III M M O N OO | VIIIV SIHI NI IIIIM I I ON OC |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\int_k^9 \frac{6}{\sqrt{x}}\,dx = \left[ax^{\frac{1}{2}}\right]_k^9 = 20 \Rightarrow 36 - 12\sqrt{k} = 20\) | M1, A1 | Setting \(\left[ax^{\frac{1}{2}}\right]_k^9 = 20\) |
| Correct method of solving \(36 - 12\sqrt{k} = 20\) | dM1 | Use both limits, correct index work to find \(k\); cannot score if \(k^{\frac{1}{2}} < 0\) |
| \(\Rightarrow k = \frac{16}{9}\) | A1 | oe |
## Question 9:
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\int_k^9 \frac{6}{\sqrt{x}}\,dx = \left[ax^{\frac{1}{2}}\right]_k^9 = 20 \Rightarrow 36 - 12\sqrt{k} = 20$ | M1, A1 | Setting $\left[ax^{\frac{1}{2}}\right]_k^9 = 20$ |
| Correct method of solving $36 - 12\sqrt{k} = 20$ | dM1 | Use both limits, correct index work to find $k$; cannot score if $k^{\frac{1}{2}} < 0$ |
| $\Rightarrow k = \frac{16}{9}$ | A1 | oe |
---\begin{enumerate}
\item Find the value of the constant $k , 0 < k < 9$, such that
\end{enumerate}
$$\int _ { k } ^ { 9 } \frac { 6 } { \sqrt { x } } \mathrm {~d} x = 20$$
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VI4V SIHIL NI III HM IONOO & VIAV SIHI NI III M M O N OO & VIIIV SIHI NI IIIIM I I ON OC \\
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\hfill \mbox{\textit{Edexcel AS Paper 1 2021 Q9 [4]}}