Pure Maths ---75 pastpaper

(a)(i) F'(x) = sec x tan x - 1
F'(θ) = 0
sec θ tan θ - 1 = 0
sin θ/cos2 θ = 1
sin θ = cos2 θ
sin θ = 1 - sin2 θ
sin2 θ + sin θ - 1 = 0
sin θ = (-1 ± √5)/2
sin θ = (√5 - 1)/2 since 0 ≦ θ < π/2
θ = sin-1 [(√5 - 1)/2]

(ii) F'(x) = sec x tan x - 1
For 0 ≦ x ≦ θ
sin x ≦ (√5 - 1)/2
sin2 x ≦ (3 - √5)/2
cos2 x ≧ 1 - (3 - √5)/2
= (√5 - 1)/2
∴ sin x/cos2 x ≦ 1
sec x tan x - 1 ≦ 0
F'(x) ≦ 0 with equality holds with x = θ
So F is strictly decreasing on [0, θ] since F'(x) = 0 only holds for ONE POINT ONLY BUT NOT A RANGE OF VALUES OF x.
Conversely when θ ≦ x < π/2, by similar deductions, F'(x) ≧ 0 and hence F is strictly increasing on [θ, π/2).

(iii) Since F is strictly decreasing on [0, θ] and strictly increasing on [θ, π/2), by graphical inspection, we can see that F attains its absolute minimum at x = θ for F defined on [0, π/2).
For the value of F(θ) = sec θ - θ - 1, we have:
sin θ = (√5 - 1)/2
cos θ = √[(√5 - 1)/2]
sec θ = √[2/(√5 - 1)]
= √[(√5 + 1)/2]
sec θ - θ - 1 = √[(√5 + 1)/2] - sin-1 [(√5 - 1)/2] - 1
Now by Mean-Value theorem, consider a function g(x) = sin-1 x - x for x defined on [0, 1], we have:
g'(x) = 1/√(1 - x2) - 1
g(o) = 0
So we can find a value of c in (0, x) such that
g'(c) = [g(x) - g(0)]/(x - 0)
1/√(1 - c2) - 1 = g(x)/x
Hence g(x)/x > 0 since 0 < c < 1
g(x) > 0 for all x in (0, 1]
sin-1 x - x > 0
sin-1 x > x
sin-1 [(√5 - 1)/2] > (√5 - 1)/2
-sin-1 [(√5 - 1)/2] < - (√5 - 1)/2
-sin-1 [(√5 - 1)/2] - 1 < - (√5 - 1)/2 - 1
√[(√5 + 1)/2] - sin-1 [(√5 - 1)/2] - 1 < √[(√5 + 1)/2] - (√5 - 1)/2 - 1
F(θ) < √[(√5 + 1)/2] - (√5 + 1)/2 < 0 since (√5 + 1)/2 > 1 and for all k > 1, √k < k
So F(θ) < 0
Note, calculation of F(θ) by calculator is NOT acceptable.