| Exam Board | WJEC |
|---|---|
| Module | Unit 1 (Unit 1) |
| Session | Specimen |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Area between curve and line |
| Difficulty | Standard +0.3 This is a straightforward AS-level calculus question requiring basic differentiation to find a tangent equation (standard procedure) and integration to find an area between curve and line. Both parts follow textbook methods with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07m Tangents and normals: gradient and equations1.08e Area between curve and x-axis: using definite integrals1.08f Area between two curves: using integration |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\frac{dy}{dx} = 3 – 2x\) | M1 | AO1 (At least one non-zero term correct) |
| An attempt to find the value of \(\frac{dy}{dx}\) at \(x = 2\) | m1 | AO1 |
| At \(x = 2\), \(\frac{dy}{dx} = -1\) | A1 | AO1 (c.a.o.) |
| Equation of tangent at \(B\) is: \(y – 2 = -1(x – 2)\) | A1 | AO1 (f.t. candidate's value for \(\frac{dy}{dx}\) at \(x = 2\)) |
| (b) x-coordinate of \(A = 3\) | B1 | AO1 (derived) |
| x-coordinate of \(C = 4\) | B1 | AO1 (derived) |
| If \(D\) is the foot of the perpendicular from \(B\) to the x-axis, area of triangle \(BDC = 2\) | B1 | AO1 (f.t. candidate's derived x-coordinate of \(C\)) |
| Area under curve \(= \int_2^3 (3x – x^2) dx\) | M1 | AO3 (use of integration) |
| \(= \left[\frac{3x^2}{2} – \frac{x^3}{3}\right]_2^3\) | A1 | AO3 (correct integration) |
| Area under curve \(= (27/2 – 9) – (6 – 8/3)\) | m1 | AO3 (an attempt to substitute limits, f.t. candidate's derived x-coordinate of \(A\)) |
| Shaded area = Area of triangle \(BDC\) – Area under curve | m1 | AO3 (f.t. candidate's derived x-coordinates of \(A\) and \(C\)) |
| Shaded area \(= 5/6\) | A1 | AO3 (c.a.o.) |
**(a)** $\frac{dy}{dx} = 3 – 2x$ | M1 | AO1 (At least one non-zero term correct)
An attempt to find the value of $\frac{dy}{dx}$ at $x = 2$ | m1 | AO1
At $x = 2$, $\frac{dy}{dx} = -1$ | A1 | AO1 (c.a.o.)
Equation of tangent at $B$ is: $y – 2 = -1(x – 2)$ | A1 | AO1 (f.t. candidate's value for $\frac{dy}{dx}$ at $x = 2$)
**(b)** x-coordinate of $A = 3$ | B1 | AO1 (derived)
x-coordinate of $C = 4$ | B1 | AO1 (derived)
If $D$ is the foot of the perpendicular from $B$ to the x-axis, area of triangle $BDC = 2$ | B1 | AO1 (f.t. candidate's derived x-coordinate of $C$)
Area under curve $= \int_2^3 (3x – x^2) dx$ | M1 | AO3 (use of integration)
$= \left[\frac{3x^2}{2} – \frac{x^3}{3}\right]_2^3$ | A1 | AO3 (correct integration)
Area under curve $= (27/2 – 9) – (6 – 8/3)$ | m1 | AO3 (an attempt to substitute limits, f.t. candidate's derived x-coordinate of $A$)
Shaded area = Area of triangle $BDC$ – Area under curve | m1 | AO3 (f.t. candidate's derived x-coordinates of $A$ and $C$)
Shaded area $= 5/6$ | A1 | AO3 (c.a.o.)
**Total: [12]**
---\includegraphics{figure_17}
The diagram above shows a sketch of the curve $y = 3x - x^2$. The curve intersects the $x$-axis at the origin and at the point $A$. The tangent to the curve at the point $B(2, 2)$ intersects the $x$-axis at the point $C$.
\begin{enumerate}[label=(\alph*)]
\item Find the equation of the tangent to the curve at $B$. [4]
\item Find the area of the shaded region. [8]
\end{enumerate}
\hfill \mbox{\textit{WJEC Unit 1 Q17 [12]}}