Problem 5 - Entrance Test

The x-coordinate in polar coordinates is given by x = rcos(θ). Substitute r = 1 + 2cos(θ) into the equation for x: x(θ) = (1 + 2cos(θ))cos(θ) = cos(θ) + 2cos²(θ). To find the maximum x-coordinate, we need to find the critical points by taking the derivative dx/dθ and setting it to zero. dx/dθ = d/dθ [cos(θ) + 2cos²(θ)] = -sin(θ) + 2 * 2cos(θ) * (-sin(θ)) = -sin(θ) - 4sin(θ)cos(θ) = -sin(θ)(1 + 4cos(θ)) Set dx/dθ = 0: -sin(θ)(1 + 4cos(θ)) = 0 This gives two conditions: 1) sin(θ) = 0 θ = 0, π (for 0 ≤ θ < 2π) 2) 1 + 4cos(θ) = 0 ⇒ cos(θ) = -1/4 θ = arccos(-1/4) (let's call this θ_0) and θ = 2π - arccos(-1/4). Now, evaluate x(θ) at these critical points: For θ = 0: x(0) = cos(0) + 2cos²(0) = 1 + 2(1)² = 3. For θ = π: x(π) = cos(π) + 2cos²(π) = -1 + 2(-1)² = -1 + 2 = 1. For cos(θ) = -1/4: x = cos(θ) + 2cos²(θ) = (-1/4) + 2(-1/4)² = -1/4 + 2(1/16) = -1/4 + 1/8 = -2/8 + 1/8 = -1/8. Comparing the x-values: 3, 1, and -1/8. The maximum x-coordinate is 3. The final answer is C.