| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2012 |
| Session | November |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Geometric/graphical PDF with k |
| Difficulty | Standard +0.8 This Further Pure question requires multiple techniques: finding k using the integral property of PDFs, deriving a transformed PDF using the Jacobian method (Y=X²), calculating E(Y) with piecewise integration, and proving a median relationship. While each step is methodical, the transformation of random variables and handling piecewise functions across multiple parts requires solid understanding beyond routine A-level statistics. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03g Cdf of transformed variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(8mge/a = mg \Rightarrow e = a/8\) | B1 | Resolve vertically at equilibrium |
| \(m\,d^2x/dt^2 = mg - 8mg(e+x)/a\) | M1 A1 | Newton's Law at general point |
| \(d^2x/dt^2 = -(8g/a)x\), so \(\omega^2 = 8g/a\) | A1 | Simplify; allow stating result without derivation |
| OR: \(v^2 = \omega^2\{(\tfrac{1}{4}a)^2 - e^2\}\), \(3ga/8 = \omega^2(a^2/16 - a^2/64)\), \(\omega^2 = 8g/a\) | (M1)(A1)(A1) | Assume SHM and find \(\omega^2\) from speed \(v\) when first slack |
| \(\omega t = \cos^{-1}(-\tfrac{1}{2})\) or \(\tfrac{1}{2}\pi + \sin^{-1}(\tfrac{1}{2}) = 2\pi/3\) | M1 A1 A1 | Use \(x = \tfrac{1}{4}a\cos\omega t\) or \(\tfrac{1}{4}a\sin\omega t\) |
| \(t = (2\pi/3)\sqrt{(a/8g)}\) A.G. | A1 | Substitute \(\omega = \sqrt{(8g/a)}\) |
| \(v^2 = \omega^2(a^2/16 - e^2) = 3ga/8\) or \(\tfrac{1}{4}a\omega\sin 2\pi/3 = 3ga/8\) | M1 A1 | Find \(v^2\) when first slack |
| \(\tfrac{1}{2}mv^2 = \tfrac{1}{2}\cdot 8mg(e + \tfrac{1}{4}a)^2/a - mg(e + \tfrac{1}{4}a)\), \(v^2 = 9ga/8 - 3ga/4 = 3ga/8\) | (M1 A1) | OR: energy method |
| \(2gs_2 = v^2\), \(s_2 = 3a/16\) | M1 A1 | Find further distance \(s_2\) to rest |
| \(\tfrac{1}{4}a + e + s_2 = 9a/16\) or \(0 \cdot 562[5]a\) | M1 A1 | Find total distance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(k + 3k = 1\), \(k = \tfrac{1}{4}\) | M1 A1 | Find \(k\) by equating area under graph to 1 |
| \(\tfrac{1}{2}kx = x/8\) and \(k = \tfrac{1}{4}\) A.G. | B1 | \(f(x)\) for \(0 < x \leq 2\) and \(2 < x \leq 5\) |
| Part (i): \(F(x) = x^2/16\) \((0 \leq x \leq 2)\) | M1 A1 | Integrate to find \(F(x)\) |
| \(\tfrac{1}{4}x - \tfrac{1}{4}\) \((2 < x \leq 5)\) | M1 A1 | |
| \(G(y) = P(Y < y) = P(X^2 < y) = P(X < y^{1/2}) = F(y^{1/2})\) | M1 A1 | Relate dist. fn. \(G(y)\) of \(Y\) to \(X\) |
| \(= y/16\) and \(\tfrac{1}{4}y^{1/2} - \tfrac{1}{4}\) | M1 A1 | |
| \(g(y) = 1/16\) or \(0 \cdot 0625\) \((0 \leq y \leq 4)\) | M1 A1 | Differentiate to find \(g(y)\); both results required for M1 |
| \(\tfrac{1}{8}\sqrt{y}\) \((4 < y \leq 25)\); \([0\) otherwise\(]\) | M1 A1 | |
| Part (ii): \(E(Y) = (1/16)\int y\,dy + (1/8)\int y^{1/2}\,dy\) | M1 | *EITHER*: Find \(E(Y)\) using \(\int y\,g(y)\,dy\) |
| \(= [y^2/32]_0^4 + [y^{3/2}/12]_4^{25} = \tfrac{1}{2} + 117/12 = 10 \cdot 25\) A.G. | A1 A1 | |
| \(E(Y) = (1/8)\int x^3\,dx + \tfrac{1}{4}\int x^2\,dx = [x^4/32]_0^2 + [x^3/12]_2^5 = \tfrac{1}{2} + 117/12 = 10 \cdot 25\) A.G. | (M1)(A1)(A1) | *OR*: Find \(E(Y)\) using \(\int x^2 f(x)\,dx\) |
| Part (iii): \(F(m_x) = \tfrac{1}{4}m_x - \tfrac{1}{4} = \tfrac{1}{2}\), \(m_x = 3\) | M1 A1 | *EITHER*: Find median \(m_x\) of \(X\) |
| \(F(m_y) = \tfrac{1}{4}m_y^{1/2} - \tfrac{1}{4} = \tfrac{1}{2}\), \(m_y = 9\) | M1 A1 | median \(m_y\) of \(Y\) |
| \(P(Y < m_x^2) = P(X^2 < m_x^2) = P(X < m_x)\) | (M1 A1) | *OR*: Show \(m_y = m_x^2\) |
# Question 11(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $8mge/a = mg \Rightarrow e = a/8$ | B1 | Resolve vertically at equilibrium |
| $m\,d^2x/dt^2 = mg - 8mg(e+x)/a$ | M1 A1 | Newton's Law at general point |
| $d^2x/dt^2 = -(8g/a)x$, so $\omega^2 = 8g/a$ | A1 | Simplify; allow stating result without derivation |
| **OR:** $v^2 = \omega^2\{(\tfrac{1}{4}a)^2 - e^2\}$, $3ga/8 = \omega^2(a^2/16 - a^2/64)$, $\omega^2 = 8g/a$ | (M1)(A1)(A1) | Assume SHM and find $\omega^2$ from speed $v$ when first slack |
| $\omega t = \cos^{-1}(-\tfrac{1}{2})$ or $\tfrac{1}{2}\pi + \sin^{-1}(\tfrac{1}{2}) = 2\pi/3$ | M1 A1 A1 | Use $x = \tfrac{1}{4}a\cos\omega t$ or $\tfrac{1}{4}a\sin\omega t$ |
| $t = (2\pi/3)\sqrt{(a/8g)}$ **A.G.** | A1 | Substitute $\omega = \sqrt{(8g/a)}$ |
| $v^2 = \omega^2(a^2/16 - e^2) = 3ga/8$ or $\tfrac{1}{4}a\omega\sin 2\pi/3 = 3ga/8$ | M1 A1 | Find $v^2$ when first slack |
| $\tfrac{1}{2}mv^2 = \tfrac{1}{2}\cdot 8mg(e + \tfrac{1}{4}a)^2/a - mg(e + \tfrac{1}{4}a)$, $v^2 = 9ga/8 - 3ga/4 = 3ga/8$ | (M1 A1) | OR: energy method |
| $2gs_2 = v^2$, $s_2 = 3a/16$ | M1 A1 | Find further distance $s_2$ to rest |
| $\tfrac{1}{4}a + e + s_2 = 9a/16$ or $0 \cdot 562[5]a$ | M1 A1 | Find total distance |
---
# Question 11(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $k + 3k = 1$, $k = \tfrac{1}{4}$ | M1 A1 | Find $k$ by equating area under graph to 1 |
| $\tfrac{1}{2}kx = x/8$ and $k = \tfrac{1}{4}$ **A.G.** | B1 | $f(x)$ for $0 < x \leq 2$ and $2 < x \leq 5$ |
| **Part (i):** $F(x) = x^2/16$ $(0 \leq x \leq 2)$ | M1 A1 | Integrate to find $F(x)$ |
| $\tfrac{1}{4}x - \tfrac{1}{4}$ $(2 < x \leq 5)$ | M1 A1 | |
| $G(y) = P(Y < y) = P(X^2 < y) = P(X < y^{1/2}) = F(y^{1/2})$ | M1 A1 | Relate dist. fn. $G(y)$ of $Y$ to $X$ |
| $= y/16$ and $\tfrac{1}{4}y^{1/2} - \tfrac{1}{4}$ | M1 A1 | |
| $g(y) = 1/16$ or $0 \cdot 0625$ $(0 \leq y \leq 4)$ | M1 A1 | Differentiate to find $g(y)$; both results required for M1 |
| $\tfrac{1}{8}\sqrt{y}$ $(4 < y \leq 25)$; $[0$ otherwise$]$ | M1 A1 | |
| **Part (ii):** $E(Y) = (1/16)\int y\,dy + (1/8)\int y^{1/2}\,dy$ | M1 | *EITHER*: Find $E(Y)$ using $\int y\,g(y)\,dy$ |
| $= [y^2/32]_0^4 + [y^{3/2}/12]_4^{25} = \tfrac{1}{2} + 117/12 = 10 \cdot 25$ **A.G.** | A1 A1 | |
| $E(Y) = (1/8)\int x^3\,dx + \tfrac{1}{4}\int x^2\,dx = [x^4/32]_0^2 + [x^3/12]_2^5 = \tfrac{1}{2} + 117/12 = 10 \cdot 25$ **A.G.** | (M1)(A1)(A1) | *OR*: Find $E(Y)$ using $\int x^2 f(x)\,dx$ |
| **Part (iii):** $F(m_x) = \tfrac{1}{4}m_x - \tfrac{1}{4} = \tfrac{1}{2}$, $m_x = 3$ | M1 A1 | *EITHER*: Find median $m_x$ of $X$ |
| $F(m_y) = \tfrac{1}{4}m_y^{1/2} - \tfrac{1}{4} = \tfrac{1}{2}$, $m_y = 9$ | M1 A1 | median $m_y$ of $Y$ |
| $P(Y < m_x^2) = P(X^2 < m_x^2) = P(X < m_x)$ | (M1 A1) | *OR*: Show $m_y = m_x^2$ |\begin{center}
\includegraphics[max width=\textwidth, alt={}]{bcd7ee99-e382-4cb6-aa39-d8b385b01319-5_453_807_1041_667}
\end{center}
The continuous random variable $X$ takes values in the interval $0 \leqslant x \leqslant 5$ only. For $0 \leqslant x \leqslant 5$ the graph of its probability density function f consists of two straight line segments, as shown in the diagram. Find $k$ and show that f is given by
$$f ( x ) = \begin{cases} \frac { 1 } { 8 } x & 0 \leqslant x \leqslant 2 \\ \frac { 1 } { 4 } & 2 < x \leqslant 5 \\ 0 & \text { otherwise } \end{cases}$$
The random variable $Y$ is given by $Y = X ^ { 2 }$.\\
(i) Find the probability density function of $Y$.\\
(ii) Show that $\mathrm { E } ( Y ) = 10.25$.\\
(iii) Show that the median of $Y$ is the square of the median of $X$.
\hfill \mbox{\textit{CAIE FP2 2012 Q11 OR}}