| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2004 |
| Session | November |
| 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 standard techniques: (i) finding a normal line using the given derivative at a point (basic calculus), and (ii) integrating a simple function of the form k(ax+b)^{-1/2} using substitution or recognition. Both parts are routine A-level procedures with no conceptual challenges, making it slightly easier than average. |
| Spec | 1.07m Tangents and normals: gradient and equations1.08b Integrate x^n: where n != -1 and sums1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| \(x=3\), \(m=2\); perpendicular \(m = -\frac{1}{2}\) | M1 | Use of \(m_1 m_2 = -1\) even if algebraic |
| \(y - 3 = -\frac{1}{2}(x-3) \rightarrow x + 2y = 9\) | M1 A1 [3] | Correct form of line equation or \(y=mx+c\); needs putting as \(x+2y=9\) for A mark; tangent gets 0/3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int \rightarrow 6(4x-3)^{\frac{1}{2}} \div \frac{1}{2} \div 4\) | M1 A1 | M1 for \((4x-3)^k \div k\); A1 for \(k=\frac{1}{2}\) and \(\div 4\) |
| \(y = 3(4x-3) + c\) | M1 | Using \((3,3)\) to find \(c\) only after attempt at integration |
| Uses \((3,3) \rightarrow c = -6\) | A1 [4] | Allow full marks once \(-6\) obtained |
# Question 7:
## Part (i)
| $x=3$, $m=2$; perpendicular $m = -\frac{1}{2}$ | M1 | Use of $m_1 m_2 = -1$ even if algebraic |
| $y - 3 = -\frac{1}{2}(x-3) \rightarrow x + 2y = 9$ | M1 A1 [3] | Correct form of line equation or $y=mx+c$; needs putting as $x+2y=9$ for A mark; tangent gets 0/3 |
## Part (ii)
| $\int \rightarrow 6(4x-3)^{\frac{1}{2}} \div \frac{1}{2} \div 4$ | M1 A1 | M1 for $(4x-3)^k \div k$; A1 for $k=\frac{1}{2}$ and $\div 4$ |
| $y = 3(4x-3) + c$ | M1 | Using $(3,3)$ to find $c$ only after attempt at integration |
| Uses $(3,3) \rightarrow c = -6$ | A1 [4] | Allow full marks once $-6$ obtained |
---7 A curve is such that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 6 } { \sqrt { } ( 4 x - 3 ) }$ and $P ( 3,3 )$ is a point on the curve.\\
(i) Find the equation of the normal to the curve at $P$, giving your answer in the form $a x + b y = c$.\\
(ii) Find the equation of the curve.
\hfill \mbox{\textit{CAIE P1 2004 Q7 [7]}}