| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2021 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof by induction |
| Type | Prove inequality: recurrence sequence |
| Difficulty | Challenging +1.8 This is a challenging Further Maths induction problem requiring students to work with logarithmic inequalities and a recursive sequence. The inductive step demands careful manipulation of the given hint and geometric growth factor 3^n, going beyond routine induction. However, the hint substantially guides part (a), and part (b) follows naturally from (a), preventing this from reaching the highest difficulty tier. |
| Spec | 4.01a Mathematical induction: construct proofs |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(a_2 = 2^3\) leading to \(\ln a_2 = 3\ln 2\) so true when \(n = 2\) | B1 | Checks base case. Must be \(n=2\) and exact |
| Assume that \(\ln a_k \geqslant 3^{k-1} \ln 2\) | B1 | States inductive hypothesis |
| \(\ln a_{k+1} = 3\ln\left(a_k + \frac{1}{a_k}\right) > 3\ln a_k\) | M1 | Applies result given in part (a) |
| \(\geqslant 3^k \ln 2\) | M1 A1 | Applies inductive hypothesis and writes in required form |
| So true when \(n = k+1\). By induction, true for all integers \(n \geqslant 2\) | A1 | States conclusion, all previous marks must be gained |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(3\ln\left(a_n + \frac{1}{a_n}\right) - \ln a_n > 2\ln a_n\) | M1 | Applies results given in parts (a) and (b) |
| \(> 2 \times 3^{n-1} \ln 2 = 3^{n-1} \ln 4\) | A1 | AG |
## Question 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $a_2 = 2^3$ leading to $\ln a_2 = 3\ln 2$ so true when $n = 2$ | B1 | Checks base case. Must be $n=2$ and exact |
| Assume that $\ln a_k \geqslant 3^{k-1} \ln 2$ | B1 | States inductive hypothesis |
| $\ln a_{k+1} = 3\ln\left(a_k + \frac{1}{a_k}\right) > 3\ln a_k$ | M1 | Applies result given in part (a) |
| $\geqslant 3^k \ln 2$ | M1 A1 | Applies inductive hypothesis and writes in required form |
| So true when $n = k+1$. By induction, true for all integers $n \geqslant 2$ | A1 | States conclusion, all previous marks must be gained |
## Question 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $3\ln\left(a_n + \frac{1}{a_n}\right) - \ln a_n > 2\ln a_n$ | M1 | Applies results given in parts (a) and (b) |
| $> 2 \times 3^{n-1} \ln 2 = 3^{n-1} \ln 4$ | A1 | AG |3 The sequence of real numbers $a _ { 1 } , a _ { 2 } , a _ { 3 } , \ldots$ is such that $a _ { 1 } = 1$ and
$$a _ { n + 1 } = \left( a _ { n } + \frac { 1 } { a _ { n } } \right) ^ { 3 }$$
\begin{enumerate}[label=(\alph*)]
\item Prove by mathematical induction that $\ln a _ { n } \geqslant 3 ^ { n - 1 } \ln 2$ for all integers $n \geqslant 2$.\\[0pt]
[You may use the fact that $\ln \left( x + \frac { 1 } { x } \right) > \ln x$ for $x > 0$.]
\item Show that $\ln \mathrm { a } _ { \mathrm { n } + 1 } - \ln \mathrm { a } _ { \mathrm { n } } > 3 ^ { \mathrm { n } - 1 } \ln 4$ for $n \geqslant 2$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2021 Q3 [8]}}