| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2002 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Finding curve equation from derivative |
| Difficulty | Moderate -0.8 This is a straightforward integration problem requiring a simple substitution (u = 1 + 2x) to find y from dy/dx = √(1+2x), then using the given point to find the constant. The y-intercept calculation is trivial substitution. Standard textbook exercise with routine techniques and no problem-solving insight required. |
| Spec | 1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.08a Fundamental theorem of calculus: integration as reverse of differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(y = (1+2x)^{3/2} ÷ (3/2) ÷ 2\) (+C) | M1, A1 | Attempt at \(\int n\). Needs \((\phantom{})^{3/2} + k\); A1 for ÷ 2 and \(k = \frac{3}{4}\) |
| Use of (4,11) to find C = 2. | M1, A1 | Attempt to use (4,11); Correct only |
| (ii) If \(x = 0\), \(y = 7/3\) | M1 A1∨ | Use of \(x = 0\) providing there is some integration |
**(i)** $y = (1+2x)^{3/2} ÷ (3/2) ÷ 2$ (+C) | M1, A1 | Attempt at $\int n$. Needs $(\phantom{})^{3/2} + k$; A1 for ÷ 2 and $k = \frac{3}{4}$
Use of (4,11) to find C = 2. | M1, A1 | Attempt to use (4,11); Correct only
**(ii)** If $x = 0$, $y = 7/3$ | M1 A1∨ | Use of $x = 0$ providing there is some integration4 The gradient at any point $( x , y )$ on a curve is $\sqrt { } ( 1 + 2 x )$. The curve passes through the point $( 4,11 )$. Find\\
(i) the equation of the curve,\\
(ii) the point at which the curve intersects the $y$-axis.
\hfill \mbox{\textit{CAIE P1 2002 Q4 [6]}}