| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2013 |
| Session | November |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Invariant lines and eigenvalues and vectors |
| Type | Find eigenvectors given eigenvalue |
| Difficulty | Standard +0.3 The question text is corrupted/unreadable, but based on the topic (eigenvectors given eigenvalue) and exam context (CAIE FP2), this is typically a straightforward application: substitute the given eigenvalue into (A - λI)v = 0 and solve the resulting system. This is a standard textbook exercise requiring routine algebraic manipulation with minimal problem-solving insight, making it slightly easier than average. |
| Spec | 4.10d Second order homogeneous: auxiliary equation method4.10e Second order non-homogeneous: complementary + particular integral |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int_0^{\infty} Ae^{-\lambda t}\,dt = \left[-(A/\lambda)e^{-\lambda t}\right]_0^{\infty}\) | Show that \(A = \lambda\) | |
| \(= A/\lambda = 1\) if \(A = \lambda\) | M1 A1 | A.G. |
| Part total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int_0^1 \lambda e^{-\lambda t}\,dt\) or \(\left[-e^{-\lambda t}\right]_0^1 = 16/100\) | M1 | Find estimate for \(\lambda\) |
| \(1 - e^{-\lambda} = 0.16\) | ||
| \(\lambda = -\ln 0.84 = 0.174\) | M1 A1 | |
| \(\left[-e^{-\lambda t}\right]_0^m = \frac{1}{2}\) (A.E.F.) | M1 | State or use eqn for median \(m\) of \(T\) |
| \(e^{-\lambda m} = \frac{1}{2}\), \(m = \lambda^{-1}\ln 2 = 3.98\) | M1 A1 | Find value of \(m\) |
| Part total: 6 | [Total: 8] |
## Question 8(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int_0^{\infty} Ae^{-\lambda t}\,dt = \left[-(A/\lambda)e^{-\lambda t}\right]_0^{\infty}$ | | Show that $A = \lambda$ |
| $= A/\lambda = 1$ if $A = \lambda$ | M1 A1 | **A.G.** |
| **Part total: 2** | | |
## Question 8(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int_0^1 \lambda e^{-\lambda t}\,dt$ or $\left[-e^{-\lambda t}\right]_0^1 = 16/100$ | M1 | Find estimate for $\lambda$ |
| $1 - e^{-\lambda} = 0.16$ | | |
| $\lambda = -\ln 0.84 = 0.174$ | M1 A1 | |
| $\left[-e^{-\lambda t}\right]_0^m = \frac{1}{2}$ (A.E.F.) | M1 | State or use eqn for median $m$ of $T$ |
| $e^{-\lambda m} = \frac{1}{2}$, $m = \lambda^{-1}\ln 2 = 3.98$ | M1 A1 | Find value of $m$ |
| **Part total: 6** | **[Total: 8]** | |
---8 The lifetime, in years, of an electrical component is the random variable $T$, with probability density function f given by
$$\mathrm { f } ( t ) = \begin{cases} A \mathrm { e } ^ { - \lambda t } & t \geqslant 0 \\ 0 & \text { otherwise } \end{cases}$$
where $A$ and $\lambda$ are positive constants.\\
(i) Show that $A = \lambda$.
It is known that out of 100 randomly chosen components, 16 failed within the first year.\\
(ii) Find an estimate for the value of $\lambda$, and hence find an estimate for the median value of $T$.
\hfill \mbox{\textit{CAIE FP2 2013 Q8}}