A2 June 2024 Q5
5. A continuous random variable \(X\) has probability density function
\[\mathrm{f}(x) = \begin{cases} ax^{-2} - bx^{-3} & 2 \leqslant x \lt \infty \\ 0 & \text{otherwise} \end{cases}\]where \(a\) and \(b\) are constants.
Given that \(\mathrm{P}(X \leqslant 4) = \dfrac{3}{8}\)
Show your working clearly. (6)
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle\int ax^{-2} - bx^{-3}\,\mathrm{d}x = -\frac{a}{x} + \frac{b}{2x^2}\) | M1 | 1.1b |
| \(\begin{aligned} &\left[-\dfrac{a}{x} + \dfrac{b}{2x^2}\right]_2^\infty = 0 - \left(-\dfrac{a}{2} + \dfrac{b}{8}\right) && \left[= \dfrac{a}{2} - \dfrac{b}{8}\right] \\[8pt] \text{or } &\left[-\dfrac{a}{x} + \dfrac{b}{2x^2}\right]_2^4 = \left(-\dfrac{a}{4} + \dfrac{b}{32}\right) - \left(-\dfrac{a}{2} + \dfrac{b}{8}\right) && \left[= \dfrac{a}{4} - \dfrac{3b}{32}\right] \\[8pt] \text{or } &\left[-\dfrac{a}{x} + \dfrac{b}{2x^2}\right]_4^\infty = 0 - \left(-\dfrac{a}{4} + \dfrac{b}{32}\right) && \left[= \dfrac{a}{4} - \dfrac{b}{32}\right] \end{aligned}\) | M1 M1 | 1.1b 1.1b |
| \(\begin{aligned} &\dfrac{a}{2} - \dfrac{b}{8} = 1 \ \text{ or } \ 4a - b = 8 \ \ \text{oe} \\[6pt] &\dfrac{a}{4} - \dfrac{3b}{32} = \dfrac{3}{8} \ \text{ or } \ 8a - 3b = 12 \ \ \text{oe} \\[6pt] &\dfrac{a}{4} - \dfrac{b}{32} = \dfrac{5}{8} \ \text{ or } \ 8a - b = 20 \ \ \text{oe} \end{aligned}\) | dM1 A1 | 1.1b 1.1b |
| \(\therefore a = 3\)* | A1*cso | 2.1 |
| (6) |
Notes
M1: Attempt to integrate one term correct. Look for the power increasing by 1
M1: Integrating both terms and substitute limits the correct way round [any one of \((2, \infty)\) or \((2, 4)\) or \((4, \infty)\)] to form one expression where \(c\) is a non-zero constant.
Alternatively, allow this mark for:
Allow this mark for \(\mathrm{F}(x) = \left(-\dfrac{a}{x} + \dfrac{b}{2x^2}\right) - \left(-\dfrac{a}{2} + \dfrac{b}{8}\right)\) or one of \(\mathrm{F}(4)\) or \(\mathrm{F}(\infty)\)
M1: Integrating and substitute limits the correct way round [any one of \((2, \infty)\) or \((2, 4)\) or \((4, \infty)\)] to form a second expression. Alternatively, for \(\mathrm{F}(4)\) and \(\mathrm{F}(\infty)\)
dM1: Dependent on the 2nd M. For one of the expressions equal to correct value from 1, \(\dfrac{3}{8}\) or \(\dfrac{5}{8}\)
A1: Translating a problem in mathematical context into two correct equations with one \(a\) term, one \(b\) term and one number
A1*cso: Fully correct solution, achieving \(a = 3\)
Alternative
Alt (a) M1: As in main scheme
M1: substitutes two of 2, 4 or “\(\infty\)” into their integral which must have a constant of integration eg \(-\dfrac{a}{2} + \dfrac{b}{8} + c\ (= 0)\), \(-\dfrac{a}{4} + \dfrac{b}{32} + c\ \left(= \dfrac{3}{8}\right)\) or \(-\dfrac{a}{\text{“}\infty\text{”}} + \dfrac{b}{\text{“}\infty\text{”}} + c\ (= 1)\) oe
M1: substitutes all three of 2, 4 or “\(\infty\)”
dM1: forms at least two equations involving \(c\) eg \(-\dfrac{a}{2} + \dfrac{b}{8} + c = 0\), \(-\dfrac{a}{4} + \dfrac{b}{32} + c = \dfrac{3}{8}\) or \(-\dfrac{a}{\text{“}\infty\text{”}} + \dfrac{b}{\text{“}\infty\text{”}} + c = 1\) (may just state \(c = 1\))
A1A1: As in main scheme
| Scheme | Marks | AO |
|---|---|---|
| \(b = 4\) | B1 | 1.1b |
| \(\left[-\dfrac{3}{x} + \dfrac{\text{“}4\text{”}}{2x^2}\right]_2^m = 0.5\) or e.g. \(-3x^{-1} + \text{“}2\text{”}x^{-2} + 1 = 0.5\) | M1 | 1.2 |
| \(m^2 - 6m + 4 = 0\) oe | A1 | 1.1b |
| \((m =)\ \ 3 + \sqrt{5}\) | A1 | 2.2a |
| (4) | ||
| (10 marks) |
Notes
B1: Writing or using \(b = 4\) may be seen in (a)
M1: Equating their integral with \(b\), limits 2 and \(m\) substituted and equated to 0.5
Allow their \(\mathrm{F}(x) = 0.5\). It must be of the form \(\alpha x^{-1} + \beta x^{-2} + \gamma = 0.5\) oe
May be in terms of eg \(x\) instead of \(m\)
A1: A correct 3 term quadratic = 0 Terms do not need to be collected on the same side. May be implied by ans. May be in terms of eg \(x\)
A1: \(3 + \sqrt{5}\) and any other solutions should be eliminated