9231/22 — Mark Scheme

cosh^2 x = 1 + sinh^2 x = 1 + 9/16 = 25/16

cosh x = 5/4 (positive, as cosh x ≥ 1)

tanh x = (3/4)/(5/4) = 3/5

sinh 2x = 2(3/4)(5/4) = 15/8

det(A - λI) = (2-λ)(3-λ)^2 - 1(3-λ) + 1[-(3-λ)]

= (2-λ)(3-λ)^2 - 2(3-λ)

= (3-λ)[(2-λ)(3-λ) - 2]

= (3-λ)(6 - 5λ + λ^2 - 2)

= (3-λ)(λ^2 - 5λ + 4)

= (3-λ)(λ-1)(λ-4)

Setting to zero: eigenvalues are λ = 1, 3, 4

For λ = 1: (A - I)v = 0

Solve: v1+2v2=0, v1+2v3=0 ⇒ v2=v3=-v1/2

Eigenvector: (2, -1, -1)^T

For λ = 3: (A - 3I)v = 0

v1 = 0, v2+v3=0 ⇒ v3=-v2

Eigenvector: (0, 1, -1)^T

For λ = 4: (A - 4I)v = 0

v1=v2, v1=v3 ⇒ v1=v2=v3

Eigenvector: (1, 1, 1)^T

Q = [[2,0,1],[-1,1,1],[-1,-1,1]], D = diag(1,3,4)

Aux eq: m^2 + 2m + 5 = 0 ⇒ m = -1 ± 2i

CF: y_c = e^{-x}(A cos 2x + B sin 2x)

PI: try y_p = p cos x + q sin x

Substitute: cos: -p+2q+5p = 4p+2q = 0, sin: -q-2p+5q = -2p+4q = 10

Solving: p = -1, q = 2, so y_p = -cos x + 2 sin x

GS: y = e^{-x}(A cos 2x + B sin 2x) - cos x + 2 sin x

x=0,y=0 ⇒ A-1=0 ⇒ A=1

x=0,y'=2 ⇒ -A+2B+2=2 ⇒ B=1/2

PS: y = e^{-x}(cos 2x + 0.5 sin 2x) - cos x + 2 sin x

y = cosh x, dy/dx = sinh x

L = ∫_0^1 √(1 + sinh^2 x) dx = ∫_0^1 cosh x dx

= [sinh x]_0^1 = sinh 1

S = 2π ∫_0^1 y √(1+(y')^2) dx = 2π ∫_0^1 cosh^2 x dx

S = 2π ∫_0^1 (1+cosh 2x)/2 dx = π [x + (1/2) sinh 2x]_0^1

= π(1 + sinh 2 / 2)

I_n = ∫_0^{π/2} x^n cos x dx

IBP: u=x^n, dv=cos x dx ⇒ du=nx^{n-1}dx, v=sin x

I_n = [x^n sin x]_0^{π/2} - n ∫ x^{n-1} sin x dx

= (π/2)^n - n ∫ x^{n-1} sin x dx

IBP again: u=x^{n-1}, dv=sin x dx ⇒ du=(n-1)x^{n-2}dx, v=-cos x

I_n = (π/2)^n - n[(-x^{n-1} cos x)_0^{π/2} + (n-1) ∫ x^{n-2} cos x dx]

= (π/2)^n - n(n-1)I_{n-2}

I_0 = ∫ cos x dx = [sin x]_0^{π/2} = 1

I_2 = (π/2)^2 - 2(1)I_0 = π^2/4 - 2

I_4 = (π/2)^4 - 4(3)I_2 = π^4/16 - 12(π^2/4 - 2)

= π^4/16 - 3π^2 + 24

Let z = cos θ + i sin θ. Then z - z^{-1} = 2i sin θ

(2i sin θ)^5 = (z - z^{-1})^5 = z^5 - 5z^3 + 10z - 10z^{-1} + 5z^{-3} - z^{-5}

32i sin^5 θ = (z^5-z^{-5}) - 5(z^3-z^{-3}) + 10(z-z^{-1})

= 2i sin 5θ - 5(2i sin 3θ) + 10(2i sin θ)

sin^5 θ = (1/16)(sin 5θ - 5 sin 3θ + 10 sin θ)

∫ sin^5 θ dθ = (1/16) ∫ (sin 5θ - 5 sin 3θ + 10 sin θ) dθ

= (1/16)[-cos 5θ/5 + 5 cos 3θ/3 - 10 cos θ]

At θ=π/2: all cos terms = 0

At θ=0: -1/5 + 5/3 - 10 = -128/15

= (1/16)(0 + 128/15) = 8/15

det(B - λI) = (1-λ)[(1-λ)^2-4] - 2[2(1-λ)]

= (1-λ)(λ^2-2λ-3) - 4(1-λ)

= (1-λ)(λ^2-2λ-7)

= -λ^3 + 3λ^2 + 5λ - 7

Char eq: λ^3 - 3λ^2 - 5λ + 7 = 0

B^2 = [[5,4,4],[4,9,4],[4,4,5]]

B^3 = [[13,22,12],[22,25,22],[12,22,13]]

B^3 - 3B^2 - 5B + 7I =

[[13-15-5+7, 22-12-10, 12-12],
[22-12-10, 25-27-5+7, 22-12-10],
[12-12, 22-12-10, 13-15-5+7]]

= [[0,0,0],[0,0,0],[0,0,0]]

Hence CH verified.

Multiply CH by B^{-1}: B^2 - 3B - 5I + 7B^{-1} = 0

7B^{-1} = -B^2 + 3B + 5I

B^{-1} = (1/7)[[3,2,-4],[2,-1,2],[-4,2,3]]

From CH: B^3 = 3B^2 + 5B - 7I

B^4 = B B^3 = B(3B^2+5B-7I) = 3B^3 + 5B^2 - 7B

= 3(3B^2+5B-7I) + 5B^2 - 7B

= 9B^2+15B-21I+5B^2-7B = 14B^2 + 8B - 21I

Hence p=14, q=8, r=-21

sinh^{-1} x = ln(x + √(x^2+1))

f'(x) = 1/√(x^2+1) (standard derivative)

Maclaurin: f(x) = f(0) + f'(0)x + f''(0)x^2/2! + f'''(0)x^3/3! + ...

f(0) = sinh^{-1}0 = 0

f'(x) = (1+x^2)^{-1/2}, f'(0) = 1

f''(x) = -x(1+x^2)^{-3/2}, f''(0) = 0

f'''(x) = -(1+x^2)^{-3/2} + 3x^2(1+x^2)^{-5/2}, f'''(0) = -1

sinh^{-1}x = x - x^3/6 + ...

sinh^{-1}(0.4) ≈ 0.4 - 0.4^3/6 = 0.4 - 0.010667 = 0.389333...

≈ 0.3893 (4 d.p.)