| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2018 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Separable variables |
| Difficulty | Standard +0.3 This question is misclassified - it's actually a probability/statistics transformation problem, not a differential equations question. The transformation Y = √X requires the Jacobian method (standard A-level Further Maths statistics), and finding E(Y) is routine integration. Both parts are textbook exercises with no novel insight required, making this slightly easier than average. |
| Spec | 5.03c Calculate mean/variance: by integration5.03g Cdf of transformed variables |
| Answer | Marks | Guidance |
|---|---|---|
| \(F(x) = \int f(x)\ dx = (1/20)(3x - 2\sqrt{x} + c)\), \(c = -1\) so \(F(x) = (1/20)(3x - 2\sqrt{x} - 1)\) | M1 | Find or state distribution function \(F(x)\) for \(1 \leq x \leq 9\) |
| or \((3/20)x - (1/10)\sqrt{x} - 1/20\) | A1 | Find or state \(G(y)\) from \(Y = \sqrt{X}\) for \(1 \leq x \leq 9\) or \(1 \leq y \leq 3\) |
| \(G(y) = F(y^2) = (1/20)(3y^2 - 2y - 1)\) | M1 | Allow \(A1\sqrt{}\) as FT on expression found for \(F(x)\) |
| or \((3/20)y^2 - (1/10)y - 1/20\) | A2 | Verify \(g(y)\) (differentiation may be implied) |
| \(g(y) = G'(y) = (1/10)(3y-1)\) [for \(1 \leq y \leq 3\), \(g(y) = 0\) otherwise] | M1 A1 AG | SC Missing/incorrect \(c\) can earn M1 M1 \(A1\sqrt{}\) M1 (max 4/7) |
| Answer | Marks | Guidance |
|---|---|---|
| OR: Use of \(g(y) = f(x) \times | dx/dy | \) |
| \(f(x) = (1/20)(3 - 1/y)\) | (M1 A1) | Find \(f(x)\) using \(x = y^2\) |
| \(dx/dy = 2y\) | (M1 A1) | Find \(dx/dy\) using \(x = y^2\) |
| \(g(y) = f(x) \times dx/dy = (1/10)(3y-1)\) [for \(1 \leq y \leq 3\), \(g(y) = 0\) otherwise] | (M1 A1) AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(Y) = (1/10)\int_1^3 (3y^2 - y)\ dy\) | M1 | Find mean of \(Y\) from \(\int y\ g(y)\ dy\) |
| \(= (1/10)[y^3 - \frac{1}{2}y^2]_1^3 = 11/5\) or \(2.2\) | A1 |
## Question 9:
### Part 9(i):
| $F(x) = \int f(x)\ dx = (1/20)(3x - 2\sqrt{x} + c)$, $c = -1$ so $F(x) = (1/20)(3x - 2\sqrt{x} - 1)$ | M1 | Find or state distribution function $F(x)$ for $1 \leq x \leq 9$ |
| or $(3/20)x - (1/10)\sqrt{x} - 1/20$ | A1 | Find or state $G(y)$ from $Y = \sqrt{X}$ for $1 \leq x \leq 9$ or $1 \leq y \leq 3$ |
| $G(y) = F(y^2) = (1/20)(3y^2 - 2y - 1)$ | M1 | Allow $A1\sqrt{}$ as FT on expression found for $F(x)$ |
| or $(3/20)y^2 - (1/10)y - 1/20$ | A2 | Verify $g(y)$ (differentiation may be implied) |
| $g(y) = G'(y) = (1/10)(3y-1)$ [for $1 \leq y \leq 3$, $g(y) = 0$ otherwise] | M1 A1 AG | SC Missing/incorrect $c$ can earn M1 M1 $A1\sqrt{}$ M1 (max 4/7) |
**Total: 7 marks**
| OR: Use of $g(y) = f(x) \times |dx/dy|$ | (*M1) | Reference to standard result required (not in syllabus) |
| $f(x) = (1/20)(3 - 1/y)$ | (M1 A1) | Find $f(x)$ using $x = y^2$ |
| $dx/dy = 2y$ | (M1 A1) | Find $dx/dy$ using $x = y^2$ |
| $g(y) = f(x) \times dx/dy = (1/10)(3y-1)$ [for $1 \leq y \leq 3$, $g(y) = 0$ otherwise] | (M1 A1) AG | |
### Part 9(ii):
| $E(Y) = (1/10)\int_1^3 (3y^2 - y)\ dy$ | M1 | Find mean of $Y$ from $\int y\ g(y)\ dy$ |
| $= (1/10)[y^3 - \frac{1}{2}y^2]_1^3 = 11/5$ or $2.2$ | A1 | |
**Total: 2 marks**
---9 The continuous random variable $X$ has probability density function given by
$$f ( x ) = \begin{cases} \frac { 1 } { 20 } \left( 3 - \frac { 1 } { \sqrt { } x } \right) & 1 \leqslant x \leqslant 9 \\ 0 & \text { otherwise } \end{cases}$$
The random variable $Y$ is defined by $Y = \sqrt { } X$.\\
(i) Show that the probability density function of $Y$ is given by
$$\operatorname { g } ( y ) = \begin{cases} \frac { 1 } { 10 } ( 3 y - 1 ) & 1 \leqslant y \leqslant 3 \\ 0 & \text { otherwise } \end{cases}$$
(ii) Find the mean value of $Y$.\\
\hfill \mbox{\textit{CAIE FP2 2018 Q9 [9]}}