Date | May 2018 | Marks available | 7 | Reference code | 18M.2.hl.TZ2.6 |
Level | HL only | Paper | 2 | Time zone | TZ2 |
Command term | Use and Prove that | Question number | 6 | Adapted from | N/A |
Question
Use mathematical induction to prove that (1−a)n>1−na for {n:n∈Z+,n⩾ where 0 < a < 1.
Markscheme
Let {{\text{P}}_n} be the statement: {\left( {1 - a} \right)^n} > 1 - na for some {n \in {\mathbb{Z}^ + },\,n \geqslant 2} where 0 < a < 1 consider the case n = 2{\text{:}}\,\,\,{\left( {1 - a} \right)^2} = 1 - 2a + {a^2} M1
> 1 - 2a because {a^2} < 0. Therefore {{\text{P}}_2} is true R1
assume {{\text{P}}_n} is true for some n = k
{\left( {1 - a} \right)^k} > 1 - ka M1
Note: Assumption of truth must be present. Following marks are not dependent on this M1.
EITHER
consider {\left( {1 - a} \right)^{k + 1}} = \left( {1 - a} \right){\left( {1 - a} \right)^k} M1
> 1 - \left( {k + 1} \right)a + k{a^2} A1
> 1 - \left( {k + 1} \right)a \Rightarrow {{\text{P}}_{k + 1}} is true (as k{a^2} > 0) R1
OR
multiply both sides by \left( {1 - a} \right) (which is positive) M1
{\left( {1 - a} \right)^{k + 1}} > \left( {1 - ka} \right)\left( {1 - a} \right)
{\left( {1 - a} \right)^{k + 1}} > 1 - \left( {k + 1} \right)a + k{a^2} A1
{\left( {1 - a} \right)^{k + 1}} > 1 - \left( {k + 1} \right)a \Rightarrow {{\text{P}}_{k + 1}} is true (as k{a^2} > 0) R1
THEN
{{\text{P}}_2} is true {{\text{P}}_k} is true \Rightarrow {{\text{P}}_{k + 1}} is true so {{\text{P}}_n} true for all n > 2 (or equivalent) R1
Note: Only award the last R1 if at least four of the previous marks are gained including the A1.
[7 marks]