The equation eˣ - 3x = 0 has a root α in the interval $[0, 1]$.

Mark scheme

Marking points from the official mark scheme. The mark or mark type is shown in brackets.

1. (a) Let f(x) = eˣ - 3x. f(0) = 1 and f(1) = e-3 ≈ -0.281… (B1 for both values)
2. (a) Since f(0) > 0 and f(1) < 0, and f(x) is continuous, a root must lie between $0$ and $1$. (B1 for conclusion)
3. (b) f'(x) = eˣ - 3 (B1)
4. (b) f(0.5) = e^(0.5) - 3(0.5) = 0.1487… (M1)
5. (b) f'(0.5) = e^(0.5) - 3 = -1.3512… (M1)
6. (b) x₁ = 0.5 - (0.1487…)/(-1.3512…) (M1 for substitution into N-R formula)
7. (b) x₁ = 0.6099 (4 d.p.) (A1)
8. (c) The Newton-Raphson method can diverge if the starting value x₀ is too close to a point where f'(x) = 0. (B1)
9. (c) f'(x) = 0 when x = ln 3 ≈ 1.0986. If x₀ was chosen in the vicinity of x = ln 3, the tangent would be almost horizontal, causing the next approximation x₁ to be very far from the root, or even diverge, or converge to the other root α. (B1 for identifying x=ln 3 or derivative being close to zero as the issue)