A-LevelFurther MathematicsPolar coordinates9231/12 · Oct/Nov 2023 · Q613 Marks

The curve C has polar equation r = e^(-θ) - e^(-1/2π), where 0 ≤ θ ≤ 1/2π.

(a)Sketch C and state, in exact form, the greatest distance of a point on C from the pole.[3]
(b)Find the exact value of the area of the region bounded by C and the initial line.[5]
(c)Show that, at the point on C furthest from the initial line, 1 - e^(θ - 1/2π) - tanθ = 0 and verify that this equation has a root between 0.56 and 0.57.[5]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2023). The mark or mark type is shown in brackets.

[B1] Initial line drawn. Correct shape, r strictly decreasing.
[B1] Correct shape at extremities.
[B1] `1 - e^(-1/2π)`. May be seen on their diagram.
[M1] Uses correct formula with correct limits. `A = 1/2 ∫₀^(1/2π) r² dθ`.
[A1] `1/2 ∫₀^(1/2π) (e^(-2θ) - 2e^(-θ)e^(-1/2π) + e^(-π)) dθ`.
[M1 A1] Integrates to `1/2 [-1/2e^(-2θ) + 2e^((-θ-1/2π)) + θ e^(-π)]₀^(1/2π)`.
[A1] Exact value: `3/4e^(-π) + 1/4π e^(-π) + 1/4 - e^(-1/2π)`.
[B1] Uses `y = r sin θ`. `y = (e^(-θ) - e^(-1/2π)) sinθ`.
[M1 A1] Sets derivative equal to zero. `dy/(dθ) = -e^(-θ) sinθ + (e^(-θ) - e^(-1/2π)) cosθ = 0` leads to `1 - e^(θ - 1/2π) - tanθ = 0`. AG.
[B1] Shows sign change for f(0.56) and f(0.57): `f(0.56) = 0.0156 > 0` and `f(0.57) = -0.0044 < 0`.

On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.