Hint
For 0.98, choose x=0.01. The final two terms of the exact expansion determine the error, not a guessed convergence rule.
Worked solution
(a)
The first four terms are \(1+5(-2x)+10(-2x)^2+10(-2x)^3=1-10x+40x^2-80x^3\).
Answer: \(1-10x+40x^2-80x^3\).
(b)
Set x=0.01, so 1−2x=0.98. The retained polynomial gives \(1-0.1+0.004-0.00008=0.90392\).
Answer: 0.90392.
(c)
The exact omitted remainder is \(80x^4-32x^5\). At x=0.01 it equals \(0.0000008-0.0000000032=0.0000007968\). It is positive, so the estimate is low, and 7.968×10⁻⁷<10⁻⁶. The exact value is 0.9039207968.
Answer: The exact error is \(0.0000007968=7.968\times10^{-7}\), positive and less than \(10^{-6}\).
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official examination-board mark scheme. Award stated criteria once each, up to the part maximum. Accept valid equivalent methods. Follow-through applies only where explicitly stated; do not award an accuracy criterion for an incorrect final result.
(a) · 3 marks
- 1 mark: Give constant and linear terms 1−10x.
- 1 mark: Give the quadratic term 40x².
- 1 mark: Give the cubic term −80x³.
Alternative methods: Repeated multiplication with the correct retained terms is accepted.
Common error: Forgetting the factor 2 in higher powers.
(b) · 2 marks
- 1 mark: Use the correct substitution x=0.01.
- 1 mark: Evaluate the stated truncated estimate 0.90392.
Alternative methods: An equivalent exact decimal/fraction evaluation of the retained terms is accepted.
Common error: Substituting x=0.02 into 1−2x.
(c) · 3 marks
- 1 mark: Obtain the exact omitted remainder 80x⁴−32x⁵.
- 1 mark: Evaluate its positive value 7.968×10⁻⁷ and identify an underestimate.
- 1 mark: Verify that its magnitude is strictly below 10⁻⁶.
Alternative methods: Compare the retained result with the exact finite expansion or an exact rational power.
Common error: Assuming the leading omitted term alone equals the exact error.
Total: 8 marks.