S2 June 2015 Q3
3. A random variable \(X\) has probability density function given by
\[\mathrm{f}(x) = \begin{cases} kx^2 & 0 \leqslant x \leqslant 2 \\ k\left(1 - \dfrac{x}{6}\right) & 2 \lt x \leqslant 6 \\ 0 & \text{otherwise} \end{cases}\]where \(k\) is a constant.
| Scheme | Marks |
|---|---|
| \(\displaystyle\int_0^2 kx^2\,\mathrm{d}x + \int_2^6 k\left(1 - \frac{x}{6}\right)\mathrm{d}x = 1\) \(k\left[\dfrac{x^3}{3}\right]_0^2 + k\left[x - \dfrac{x^2}{12}\right]_2^6 = 1\) | M1 A1 |
| \(k\left[\dfrac{8}{3}\right] + k\left[3 - \dfrac{5}{3}\right] = 1\) \(4k = 1\) \(k = \dfrac{1}{4}\ *\) | dM1 A1cso |
Notes
M1: for adding the two integrals, and attempting to integrate, at least one integral \(x^n \to x^{n+1}\), ignore limits and does not need to be put equal to 1. Do not award if they add before integrating
A1: correct integration, ignore limits and does not need to be put equal to 1
M1: dependent on first M being awarded, correct use of limits and putting equal to 1. This may be seen as \(\mathrm{F}(2) = \dfrac{8}{3}k\) and using \(\mathrm{F}(6) = 1\)
A1: cso answer given so need \(4k = 1\) leading to \(k = \dfrac{1}{4}\)
NB Validation – if they substitute in \(k = \frac{1}{4}\) you may award the 1st three marks as per scheme. For the Final A mark they must say “therefore \(k = \frac{1}{4}\)”
| Scheme | Marks |
|---|---|
| 2 | B1 |
Notes
B1: cao
| Scheme | Marks |
|---|---|
| \(\displaystyle\int_0^x kt^2\,\mathrm{d}t = \frac{kx^3}{3}\) | M1 |
| \(\displaystyle\int k\left(1 - \frac{t}{6}\right)\mathrm{d}t = k\left[t - \frac{t^2}{12}\right] + C\) \(= kt - k\dfrac{t^2}{12} + C\) \(\mathrm{F}(6) = 1\) \(6k - 3k + C = 1 \quad \therefore C = \dfrac{1}{4}\) | M1 |
| \(\mathrm{F}(x) = \begin{cases} 0 & x \lt 0 \\ \dfrac{x^3}{12} & 0 \leqslant x \leqslant 2 \\ \dfrac{x}{4} - \dfrac{x^2}{48} + \dfrac{1}{4} & 2 \lt x \leqslant 6 \\ 1 & x \gt 6 \end{cases}\) | A1 A1 B1 |
Notes
M1: attempting to find \(\displaystyle\int_0^x kt^2\,\mathrm{d}t\), \(t^2 \to t^3\), ignore limits, may leave in terms of \(k\)
M1: attempting to find \(\displaystyle\int k\left(1 - \frac{t}{6}\right)\mathrm{d}t\) at least one integral \(t^n \to t^{n+1}\) and either have \(+\,C\) \((C \ne 0)\) and use \(\mathrm{F}(6) = 1\) or have limits 2 and \(x\) and + “their \(\displaystyle\int_0^2 kt^2\,\mathrm{d}t\)” and attempt to integrate \(t^n \to t^{n+1}\)
NB: may use any letter, need not be \(t\), condone use of \(x\)
A1: second line correct
A1: third line correct
B1: first and fourth line correct they may use “otherwise” instead of \(x \lt 0\) or \(x \gt 6\) but not instead of both
NB: Condone use of < rather than \(\leqslant\) and vice versa
| Scheme | Marks |
|---|---|
| \(\dfrac{x}{4} - \dfrac{x^2}{48} + \dfrac{1}{4} = 0.75\) \(x^2 - 12x + 24 = 0\) oe | M1 A1 |
| \(x = \dfrac{12 \pm \sqrt{144 - 4 \times 24}}{2}\) \(= 2.54\) or \(6 - 2\sqrt{3}\) | dM1 A1 |
Notes
M1: putting their line 2 or their line 3 = 0.75
A1: The correct quadratic equation – like terms must be collected together
M1d: dep on previous M1 being awarded. A correct method for solving a 3 term quadratic equation = 0 leading to \(x = \ldots\) Use either the quadratic formula or completing the square - If they quote a correct formula and attempt to use it, award the method mark if there are small errors. Where the formula is not quoted, the method mark can be implied from correct working with values but is lost if there is a mistake. If they attempt to factorise award M1 if they have \(\left(x^2 + bx + c\right) = (x + p)(x + q)\), where \(|pq| = |c|\) leading to \(x = \ldots\) May be implied by a correct value for \(x\)
A1: awrt 2.54 or \(6 - 2\sqrt{3}\) or \(6 - \sqrt{12}\). If 2 values for \(x\) are given they must eliminate the incorrect one.