A-LevelMathematicsNumerical methods9709/22 · Feb/Mar 2017 · Q58 Marks
It is given that a is a positive constant such that ∫[0 to a] (1 + 2x + 3e³ˣ) dx = 250.
- (i)Show that a = ⅓ ln(251 – a – a²).[5]
- (ii)Use an iterative formula based on the equation in part (i) to find the value of a correct to 4 significant figures. Give the result of each iteration to 6 significant figures.[3]
Mark scheme
Marking points from the official mark scheme (Feb/Mar 2017). The mark or mark type is shown in brackets.
- [M1] Integrate to obtain form k₁x+k₂x² + k₃e³ˣ for non-zero constants
- [A1] Obtain x + x² + e³ˣ
- [A1] Apply both limits to obtain a + a² + e³ᵃ −1= 250 or equivalent
- [M1] Apply correct process to reach form without e involved
- [A1] Confirm given a = ⅓ln(251−a − a²)
- [M1] Use iterative process correctly at least once
- [A1] Obtain final answer 1.835
- [A1] Show sufficient iterations to 6 sf to justify answer or show sign change in interval (1.8345, 1.8355)
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