A-LevelFurther MathematicsProof by induction9231/12 · Oct/Nov 2015 · Q36 Marks
Given that a is a constant, prove by mathematical induction that, for every positive integer n, dⁿ/dxⁿ(xeᵃˣ) = naⁿ⁻¹eᵃˣ + aⁿxeᵃˣ.
- (a)Given that a is a constant, prove by mathematical induction that, for every positive integer n, dⁿ/dxⁿ(xeᵃˣ) = naⁿ⁻¹eᵃˣ + aⁿxeᵃˣ.[6]
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Marking points from the official mark scheme (Oct/Nov 2015). The mark or mark type is shown in brackets.
- [B1] n = 1 in formula gives a⁰eᵃˣ + axeᵃˣ = eᵃˣ + axeᵃˣ
- [B1] d/dx(xeᵃˣ) = eᵃˣ x1 + x.aeᵃˣ = eᵃˣ + axeᵃˣ ⇒ H₁ is true oe
- [B1] Assume Hₖ is true, i.e. dᵏ/dxᵏ(xeᵃˣ) = kaᵏ⁻¹eᵃˣ + aᵏxeᵃˣ.
- [M1] dᵏ⁺¹/dxᵏ⁺¹(xeᵃˣ) = kaᵏeᵃˣ + aᵏeᵃˣ + aᵏ⁺¹xeᵃˣ
- [A1] = (k+1)aᵏeᵃˣ + aᵏ⁺¹xeᵃˣ
- [A1] ⇒ Hₖ₊₁ is true, hence by PMI Hₙ is true for all positive integers n.
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