| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2009 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Rotation about x-axis: rational or reciprocal function |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question testing standard techniques: differentiation of a rational function with chain rule, finding an inverse function by swapping and rearranging, and applying the volume of revolution formula. All parts are routine A-level procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks | Guidance |
|---|---|---|
| \(f'(x)\) is negative \(\rightarrow\) decreasing | B1 B1, B1√ [3] | B1 for \(-3(2x + 5)^{-2}\). B1 for \(\times2\). √ providing bracket is squared (using value or values only B0). |
| (ii) \(y = \frac{3}{2x + 5} \rightarrow 2x + 5 = \frac{3}{y} \rightarrow f^{-1}(x) = \frac{1}{2}\left(\frac{3}{x} - 5\right)\) or \(\frac{3-5x}{2x}\) | M1, A1 [2] | Attempt at making \(x\) the subject. co including f(x) not f(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Limits 0 to \(2 → π(-\frac{1}{2} - (-0.9) → = 0.4π\) (or 1.26) | B1, B1, M1, A1 [4] | For \(-9(2x + 5)^{-1}\). For ÷ 2 in \([\quad]\) of \(y^2\). Use of correct limits with \(\int\) of \(y^2\). co |
$x \mapsto \frac{3}{2x + 5}$
**(i)** $f'(x) = -3(2x + 5)^{-2} \times 2$
$f'(x)$ is negative $\rightarrow$ decreasing | B1 B1, B1√ [3] | B1 for $-3(2x + 5)^{-2}$. B1 for $\times2$. √ providing bracket is squared (using value or values only B0).
**(ii)** $y = \frac{3}{2x + 5} \rightarrow 2x + 5 = \frac{3}{y} \rightarrow f^{-1}(x) = \frac{1}{2}\left(\frac{3}{x} - 5\right)$ or $\frac{3-5x}{2x}$ | M1, A1 [2] | Attempt at making $x$ the subject. co including f(x) not f(y)
**(iii)** $\int \frac{9}{(2x+5)^2}dx = (-9(2x+5)^{-1} \cdot 2)$
$= (-9π(2x + 5)^{-1} · 2)$
Limits 0 to $2 → π(-\frac{1}{2} - (-0.9) → = 0.4π$ (or 1.26) | B1, B1, M1, A1 [4] | For $-9(2x + 5)^{-1}$. For ÷ 2 in $[\quad]$ of $y^2$. Use of correct limits with $\int$ of $y^2$. co
**Total: [9]**
---8 The function f is such that $\mathrm { f } ( x ) = \frac { 3 } { 2 x + 5 }$ for $x \in \mathbb { R } , x \neq - 2.5$.\\
(i) Obtain an expression for $\mathrm { f } ^ { \prime } ( x )$ and explain why f is a decreasing function.\\
(ii) Obtain an expression for $\mathrm { f } ^ { - 1 } ( x )$.\\
(iii) A curve has the equation $y = \mathrm { f } ( x )$. Find the volume obtained when the region bounded by the curve, the coordinate axes and the line $x = 2$ is rotated through $360 ^ { \circ }$ about the $x$-axis.
\hfill \mbox{\textit{CAIE P1 2009 Q8 [9]}}