| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Combined region areas |
| Difficulty | Standard +0.3 This is a straightforward multi-part question requiring standard techniques: finding a normal line equation (differentiation, negative reciprocal gradient) and calculating area between a curve and line (definite integration). All steps are routine A-level procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Equation \(y - 1 = -\frac{1}{2}x\) | B1 B1 M1 M1 A1 | B1 Without "×4". B1 for "×4" even if first B mark lost. Use of \(m_1m_2=-1\). Correct method for eqn |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int\sqrt{1+4x}dx = \frac{(1+4x)^{\frac{3}{2}}}{3} \div 4\) | B1 B1 B1 | B1 Without the "÷4". For "÷4" even if first B mark lost |
| Limits \(-\frac{1}{4}\) to 0 → \(\frac{1}{6}\). Area \(BOC = \frac{1}{2} \times 2 \times 1 = 1\) → Shaded area \(= \frac{7}{6}\) | B1 | For \(1 +\) his "1/6" |
| [5] |
$y = \sqrt{1 + 4x}$
(i) $\frac{dy}{dx} = \frac{1}{2}(1+4x)^{-\frac{1}{2}} \times 4 = 2$ at $B(0, 1)$
Gradient of normal $= -\frac{1}{2}$
Equation $y - 1 = -\frac{1}{2}x$ | B1 B1 M1 M1 A1 | B1 Without "×4". B1 for "×4" even if first B mark lost. Use of $m_1m_2=-1$. Correct method for eqn
[5] |
(ii) At $x = -\frac{1}{4}$
$\int\sqrt{1+4x}dx = \frac{(1+4x)^{\frac{3}{2}}}{3} \div 4$ | B1 B1 B1 | B1 Without the "÷4". For "÷4" even if first B mark lost
Limits $-\frac{1}{4}$ to 0 → $\frac{1}{6}$. Area $BOC = \frac{1}{2} \times 2 \times 1 = 1$ → Shaded area $= \frac{7}{6}$ | B1 | For $1 +$ his "1/6"
[5] |11\\
\includegraphics[max width=\textwidth, alt={}, center]{fe4c3555-5736-48c4-b61a-9f6b9a1ee46e-4_598_789_255_678}
The diagram shows the curve $y = \sqrt { } ( 1 + 4 x )$, which intersects the $x$-axis at $A$ and the $y$-axis at $B$. The normal to the curve at $B$ meets the $x$-axis at $C$. Find\\
(i) the equation of $B C$,\\
(ii) the area of the shaded region.
\hfill \mbox{\textit{CAIE P1 2013 Q11 [10]}}