Mathematics — Calculus, Vectors, 3D Geometry, Probability & Linear Programming — Practice

Question 1

Prove that sin⁻¹(3/5) + sin⁻¹(4/5) = π/2.

Accepted Answer

Let A = sin⁻¹(3/5), so sin A = 3/5 → cos A = 4/5. Let B = sin⁻¹(4/5), so sin B = 4/5 → cos B = 3/5. Now: cos(A+B) = cos A cos B − sin A sin B = (4/5)(3/5) − (3/5)(4/5) = 12/25 − 12/25 = 0. Therefore A + B = cos⁻¹(0) = π/2. Hence sin⁻¹(3/5) + sin⁻¹(4/5) = π/2. ∎

Question 2

Evaluate ∫ x² · eˣ dx.

Accepted Answer

Use integration by parts twice. ILATE: u = x², dv = eˣdx. Then v = eˣ, du = 2x dx. ∫x²eˣ dx = x²eˣ − ∫eˣ · 2x dx = x²eˣ − 2∫xeˣ dx. Now ∫xeˣ dx by parts: u = x, dv = eˣdx → v = eˣ, du = dx. = xeˣ − ∫eˣ dx = xeˣ − eˣ. Therefore: ∫x²eˣ dx = x²eˣ − 2(xeˣ − eˣ) + C = eˣ(x² − 2x + 2) + C.

Question 3

Solve the differential equation: dy/dx + y/x = x², given that y = 1 when x = 1.

Accepted Answer

This is a LINEAR first-order ODE. dy/dx + Py = Q where P = 1/x, Q = x². IF = e^∫(1/x)dx = e^(ln x) = x. Multiplying: d/dx(xy) = x · x² = x³. Integrating: xy = ∫x³ dx = x⁴/4 + C. General solution: y = x³/4 + C/x. Applying IC: y(1) = 1 → 1 = 1/4 + C → C = 3/4. PARTICULAR SOLUTION: y = x³/4 + 3/(4x) = (x⁴ + 3)/(4x).

Practice — Mathematics — Calculus, Vectors, 3D Geometry, Probability & Linear Programming

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