| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2007 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Indefinite & Definite Integrals |
| Type | Area under curve |
| Difficulty | Standard +0.3 Part (i) is trivial substitution (1 mark). Part (ii) requires setting up and evaluating a definite integral with a negative power, then solving a resulting equation - standard C2 integration technique with straightforward algebra. The 9 marks reflect working steps rather than conceptual difficulty. |
| Spec | 1.08b Integrate x^n: where n != -1 and sums1.08d Evaluate definite integrals: between limits1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(0 = 1 - \frac{1}{\sqrt{9}}\) | B1 (1 mark) | Verification of \((9, 0)\), with at least one step shown |
| (ii) \(\int_9^1 1 - 3x^{-\frac{1}{2}} dx = \left[x - 6\sqrt{x}\right]_9^1\) | M1, A1 (2 marks) | Attempt integration – increase in power for at least 1 term; For second term of form \(kx^{1/a}\); For correct integral |
| \(= (a - 6\sqrt{a}) - (b - 6\sqrt{b})\) \(= a - 6\sqrt{a} + 9\) | M1, A1 (2 marks) | Attempt \(F(a) - F(9)\); Obtain \(a - 6\sqrt{a} + 9\) |
| \(a - 6\sqrt{a} + 9 = 4\) \(a - 6\sqrt{a} + 5 = 0\) \((\sqrt{a} - 1)(\sqrt{a} - 5) = 0\) \(\sqrt{a} = 1, \sqrt{a} = 5\) \(a = 1, a = 25\) but \(a > 9\), so \(a = 25\) | M1, M1, A1, A1 (9 marks) | Equate expression for area to 4; Attempt to solve 'disguised' quadratic; Obtain at least \(\sqrt{a} = 5\); Obtain \(a = 25\) only |
**(i)** $0 = 1 - \frac{1}{\sqrt{9}}$ | B1 (1 mark) | Verification of $(9, 0)$, with at least one step shown
**(ii)** $\int_9^1 1 - 3x^{-\frac{1}{2}} dx = \left[x - 6\sqrt{x}\right]_9^1$ | M1, A1 (2 marks) | Attempt integration – increase in power for at least 1 term; For second term of form $kx^{1/a}$; For correct integral
$= (a - 6\sqrt{a}) - (b - 6\sqrt{b})$ $= a - 6\sqrt{a} + 9$ | M1, A1 (2 marks) | Attempt $F(a) - F(9)$; Obtain $a - 6\sqrt{a} + 9$
$a - 6\sqrt{a} + 9 = 4$ $a - 6\sqrt{a} + 5 = 0$ $(\sqrt{a} - 1)(\sqrt{a} - 5) = 0$ $\sqrt{a} = 1, \sqrt{a} = 5$ $a = 1, a = 25$ but $a > 9$, so $a = 25$ | M1, M1, A1, A1 (9 marks) | Equate expression for area to 4; Attempt to solve 'disguised' quadratic; Obtain at least $\sqrt{a} = 5$; Obtain $a = 25$ only
**Total: 10 marks**\includegraphics{figure_10}
The diagram shows the graph of $y = 1 - 3x^{-\frac{1}{2}}$.
\begin{enumerate}[label=(\roman*)]
\item Verify that the curve intersects the $x$-axis at $(9, 0)$. [1]
\item The shaded region is enclosed by the curve, the $x$-axis and the line $x = a$ (where $a > 9$). Given that the area of the shaded region is 4 square units, find the value of $a$. [9]
\end{enumerate}
\hfill \mbox{\textit{OCR C2 2007 Q10 [10]}}