| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2006 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Normal or tangent line problems |
| Difficulty | Standard +0.3 This is a straightforward two-part question combining basic calculus concepts. Part (i) requires finding the gradient at a point, then the normal line equation, then intersection points and a midpoint—all routine A-level techniques. Part (ii) is standard integration with substitution u=6-2x to find the curve equation using the given point. No novel insight required, just methodical application of learned procedures. |
| Spec | 1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\frac{dy}{dx} = \frac{4}{\sqrt{6-2x}}\) | ||
| If \(x = 1, m=2\) and perp \(m = -\frac{1}{2}\) | M1 A1 | Use of \(m_1m_2 = -1\). A1 co for \(-\frac{1}{2}\) |
| → \(y - 8 = -\frac{1}{2}(x-1)\) (2y + x = 17) | DM1 | Any correct form of perpendicular. |
| → \((0, 8\frac{1}{2})\) and \((17, 0)\) | A1 | co. For his answers. |
| → \(M(8\frac{1}{2}, 4\frac{1}{4})\) | B1 | co. |
| (ii) \(y = \frac{4(6-2x)^{\frac{1}{2}}}{\frac{1}{2}} + c\) | B1 | For \(4, (6-2x)^{\frac{1}{2}}\) and \(÷ \frac{1}{2}\) and no other f(x) |
| For \(+ -2\) (only if no other f(x)) | M1 | |
| → subs \((1,8) \to c = 16\) | M1A1 | Substituting into any integrated expression to find c. |
**(i)** $\frac{dy}{dx} = \frac{4}{\sqrt{6-2x}}$ | |
If $x = 1, m=2$ and perp $m = -\frac{1}{2}$ | M1 A1 | Use of $m_1m_2 = -1$. A1 co for $-\frac{1}{2}$
→ $y - 8 = -\frac{1}{2}(x-1)$ (2y + x = 17) | DM1 | Any correct form of perpendicular.
→ $(0, 8\frac{1}{2})$ and $(17, 0)$ | A1 | co. For his answers.
→ $M(8\frac{1}{2}, 4\frac{1}{4})$ | B1 | co.
**(ii)** $y = \frac{4(6-2x)^{\frac{1}{2}}}{\frac{1}{2}} + c$ | B1 | For $4, (6-2x)^{\frac{1}{2}}$ and $÷ \frac{1}{2}$ and no other f(x)
For $+ -2$ (only if no other f(x)) | M1
→ subs $(1,8) \to c = 16$ | M1A1 | Substituting into any integrated expression to find c.
**Total: [5] [4]**
---9 A curve is such that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 4 } { \sqrt { } ( 6 - 2 x ) }$, and $P ( 1,8 )$ is a point on the curve.\\
(i) The normal to the curve at the point $P$ meets the coordinate axes at $Q$ and at $R$. Find the coordinates of the mid-point of $Q R$.\\
(ii) Find the equation of the curve.
\hfill \mbox{\textit{CAIE P1 2006 Q9 [9]}}