The maximum value of the expression 1+sin2θ+3sinθcosθ+5cos2θ is

Let the expression be denoted by E. Then

E = 1 + sin2θ + 3sinθcosθ + 5cos2θ

We can rewrite the expression using trigonometric identities:

sin2θ = 2sinθcosθ
cos2θ = cos²θ - sin²θ = 1 - 2sin²θ = 2cos²θ - 1

Substituting these into the expression for E, we get:

E = 1 + 2sinθcosθ + 3sinθcosθ + 5(1 - 2sin²θ)
E = 1 + 5sinθcosθ + 5 - 10sin²θ
E = 6 + 5sinθcosθ - 10sin²θ

Let x = sinθcosθ. Then 2x = sin2θ, and sin²θ + cos²θ = 1. Also, sin²θ = (1 - cos2θ)/2 and cos²θ = (1 + cos2θ)/2. Then

sin²2θ = 4sin²θcos²θ = 4x²

We also know that 2x = sin2θ, so x = sin2θ/2.

Since -1 ≤ sin2θ ≤ 1, we have -1/2 ≤ x ≤ 1/2

Using the identity sin²θ + cos²θ = 1, we have sin²θ = 1 - cos²θ. Thus,

x = sinθcosθ = sinθ√(1 - sin²θ)

This is a difficult function to maximize. Let's try another approach.

E = 1 + sin2θ + 3sinθcosθ + 5cos2θ

E = 1 + sin2θ + (3/2)sin2θ + 5cos2θ

E = 1 + (5/2)sin2θ + 5cos2θ

Let A = 5/2 and B = 5. Then

E = 1 + A sin2θ + B cos2θ

We can express this in the form Rsin(2θ + α) where R = √(A² + B²) and α = arctan(B/A)

R = √((5/2)² + 5²) = √(25/4 + 25) = √(125/4) = 5√5/2

Therefore, the maximum value of E is 1 + R = 1 + 5√5/2 ≈ 1 + 5(2.236)/2 ≈ 6.18

However, none of the options are close to 6.18. Let's re-examine the expression:

E = 6 + 5sinθcosθ - 10sin²θ = 6 + (5/2)sin2θ - 10(1-cos2θ)/2 = 6 + (5/2)sin2θ - 5 + 5cos2θ = 1 + (5/2)sin2θ + 5cos2θ

Let u = 2θ. Then E = 1 + (5/2)sinu + 5cosu. The maximum occurs when sinu = A/R and cosu = B/R where R = √(A² + B²) = 5√5/2

Maximum value is 1 + R = 1 + 5√5/2 ≈ 6.18. Let's check if there's a mistake in the problem statement.

If the expression was 1 + sin²θ + 3sinθcosθ + 5cos²θ, then

E = 1 + (1 - cos²θ) + (3/2)sin2θ + 5cos²θ = 2 + (3/2)sin2θ + 4cos²θ

This is still difficult to solve. There must be a mistake in the question or options.