| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas Between Curves |
| Type | Tangent or Normal Bounded Area |
| Difficulty | Standard +0.3 This is a standard C2 integration question requiring differentiation to find the tangent equation (given as 'show that'), then integration to find area between curve and line. The algebra is straightforward with simple powers of x, and the setup is clearly defined. Slightly easier than average due to the 'show that' part providing the answer to verify. |
| Spec | 1.07m Tangents and normals: gradient and equations1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks |
|---|---|
| (i) \(x = 4 \therefore y = 12 - 8 + 2 = 6\) | B1 |
| \(\frac{dy}{dx} = 3 - 2x^{-1}\) | M1 A1 |
| grad \(= 3 - 1 = 2\) | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = 2x - 2\) | A1 | |
| (ii) area \(= \int_0^4 [(3x - 4\sqrt{x} + 2) - (2x - 2)] \, dx\) | M1 | |
| \(= \int_0^4 (x - 4\sqrt{x} + 4) \, dx\) | ||
| \(= [\frac{1}{2}x^2 - \frac{8}{3}x^{\frac{3}{2}} + 4x]_0^4\) | M1 A2 | |
| \(= (8 - \frac{64}{3} + 16) - (0) = \frac{8}{3}\) | M1 A1 | (12) |
(i) $x = 4 \therefore y = 12 - 8 + 2 = 6$ | B1 |
$\frac{dy}{dx} = 3 - 2x^{-1}$ | M1 A1 |
grad $= 3 - 1 = 2$ | M1 |
$\therefore y - 6 = 2(x - 4)$
$y - 6 = 2x - 8$
$y = 2x - 2$ | A1 |
(ii) area $= \int_0^4 [(3x - 4\sqrt{x} + 2) - (2x - 2)] \, dx$ | M1 |
$= \int_0^4 (x - 4\sqrt{x} + 4) \, dx$ | |
$= [\frac{1}{2}x^2 - \frac{8}{3}x^{\frac{3}{2}} + 4x]_0^4$ | M1 A2 |
$= (8 - \frac{64}{3} + 16) - (0) = \frac{8}{3}$ | M1 A1 | **(12)** |
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**Total (72)**9.
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{0744b3cf-2941-45cb-b6df-2aaf44588e5c-3_592_771_683_541}
\end{center}
The diagram shows the curve $C$ with equation $y = 3 x - 4 \sqrt { x } + 2$ and the tangent to $C$ at the point $A$.
Given that $A$ has $x$-coordinate 4,\\
(i) show that the tangent to $C$ at $A$ has the equation $y = 2 x - 2$.
The shaded region is bounded by $C$, the tangent to $C$ at $A$ and the $y$-axis.\\
(ii) Find the area of the shaded region.
\hfill \mbox{\textit{OCR C2 Q9 [12]}}