| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2010 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Normal or tangent line problems |
| Difficulty | Moderate -0.3 This is a straightforward two-part question requiring basic integration (power rule after simple substitution or recognition) and finding a normal line equation. Part (i) involves evaluating the derivative at x=2 and using point-slope form with negative reciprocal slope. Part (ii) requires integrating (3x-2)^(-1/2) and applying the initial condition. Both are standard A-level techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations1.08a Fundamental theorem of calculus: integration as reverse of differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(x = 2\), tangent has gradient \(3\) | M1 | Use of \(m_1m_2 = -1\) with \(dy/dx\) |
| \(\to\) normal has gradient \(-\frac{1}{3}\) | M1 A1 | Correct form of line eqn. for normal |
| \(\to y - 11 = -\frac{1}{3}(x-2)\) | [3] | |
| (ii) Integrate \(\to 6\frac{\sqrt{3x-2}}{\frac{3}{2}} + 3\) | B1 | Without the \(+3\) |
| B1 | For \(+3\), even if B0 above | |
| \(\to y = 4\sqrt{3x-2} + c\) through \((2,11)\) | M1 | Using \((2, 11)\) for \(c\) |
| \(\to y = 4\sqrt{3x-2} + 3\) | A1 | co |
| [4] |
$\frac{dy}{dx} = \frac{6}{\sqrt{3x-2}}$
(i) $x = 2$, tangent has gradient $3$ | M1 | Use of $m_1m_2 = -1$ with $dy/dx$
$\to$ normal has gradient $-\frac{1}{3}$ | M1 A1 | Correct form of line eqn. for normal
$\to y - 11 = -\frac{1}{3}(x-2)$ | [3] |
(ii) Integrate $\to 6\frac{\sqrt{3x-2}}{\frac{3}{2}} + 3$ | B1 | Without the $+3$
| B1 | For $+3$, even if B0 above
$\to y = 4\sqrt{3x-2} + c$ through $(2,11)$ | M1 | Using $(2, 11)$ for $c$
$\to y = 4\sqrt{3x-2} + 3$ | A1 | co
| [4] |
---5 The equation of a curve is such that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 6 } { \sqrt { } ( 3 x - 2 ) }$. Given that the curve passes through the point $P ( 2,11 )$, find\\
(i) the equation of the normal to the curve at $P$,\\
(ii) the equation of the curve.
\hfill \mbox{\textit{CAIE P1 2010 Q5 [7]}}