| Exam Board | Edexcel |
|---|---|
| Module | AEA (Advanced Extension Award) |
| Year | 2020 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Approximation for small x |
| Difficulty | Challenging +1.8 This AEA question requires multiple sophisticated techniques: binomial expansion with algebraic manipulation, inequality proof requiring careful term analysis, and an intermediate value theorem application with given bounds. While each part uses standard methods, the combination of proving inequalities from series expansions and coordinating multiple results to establish a root's existence requires above-average mathematical maturity and multi-step reasoning beyond typical A-level questions. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps1.04c Extend binomial expansion: rational n, |x|<11.04d Binomial expansion validity: convergence conditions1.09a Sign change methods: locate roots |
3.(a)(i)Write down the binomial series expansion of
$$\left( 1 + \frac { 2 } { n } \right) ^ { n } \quad n \in \mathbb { N } , n > 2$$
in powers of $\left( \frac { 2 } { n } \right)$ up to and including the term in $\left( \frac { 2 } { n } \right) ^ { 3 }$\\
(ii)Hence prove that,for $n \in \mathbb { N } , n \geqslant 3$
$$\left( 1 + \frac { 2 } { n } \right) ^ { n } \geqslant \frac { 19 } { 3 } - \frac { 6 } { n }$$
(b)Use the binomial series expansion of $\left( 1 - \frac { x } { 4 } \right) ^ { \frac { 1 } { 2 } }$ to show that $\sqrt { 3 } < \frac { 7 } { 4 }$
$$\mathrm { f } ( x ) = \left( 1 + \frac { 2 } { x } \right) ^ { x } - 3 ^ { \frac { x } { 6 } } \quad x \in \mathbb { R } , x > 0$$
Given that the function $\mathrm { f } ( x )$ is continuous and that $\sqrt [ 6 ] { 3 } > \frac { 6 } { 5 }$\\
(c)prove that $\mathrm { f } ( x ) = 0$ has a root in the interval[9,10]
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\hfill \mbox{\textit{Edexcel AEA 2020 Q3 [13]}}