For f(x) to be differentiable at x = k, f'(k) must exist. Setting k = 1 makes f'(k) continuous.

The discriminant must be positive: k^2 - 4*1*9 > 0, which gives k < 6 or k > -6.

For no real roots, the discriminant must be less than zero: k^2 - 4*1*16 < 0 => k^2 < 64 => |k| < 8.

For no real roots, the discriminant must be less than zero: k^2 - 36 < 0, hence k < -6.

A · B = 1*2 + k*3 + 2*4 = 0. Thus, 2 + 3k + 8 = 0, so 3k = -10, k = -10/3.

The coefficient of x^4 is C(6,4) * k^2 = 15k^2. Setting 15k^2 = 240 gives k^2 = 16, so k = 4 or -4.

For a maximum, f'(x) = 2x + k = 0 at x = -2. Thus, k = 4.

Setting k(1) + 1 = 2(1) - 1 gives k + 1 = 1, so k = 0.

Setting k(1) + 1 = 3 gives k = 2 for continuity.

Setting k(2) + 1 = 2^2 - 3 gives 2k + 1 = 1, leading to k = 0.

Setting k(1) + 2 = 3 gives k = 1.

Setting k = 0 for continuity at x = 0 gives f(0) = 0.

Setting k(0) = 0^2 + 1 gives k = 1.

log2(8) = log2(2^3) = 3.

Setting 2(1) + m = m(1) + 3 gives m = 1.

Setting 2(1) + m = 1^2 + 1 gives m = 0.

Setting the two pieces equal at x = 3 gives us 6 + m = 6. Thus, m = 0.

Setting 3(1) + m = 2(1)^2 gives m = -1.

Setting the two pieces equal at x = 1: 1^2 + m = 4 - 1. Solving gives m = 2.

Setting x^2 + m = mx + 1 at x = 1 gives m = 1 for continuity.

sin(30°) = 1/2 and cos(60°) = 1/2, thus the sum is 1/2 + 1/2 = 1.

Let θ = cos^(-1)(1/2). Then cos(θ) = 1/2, which corresponds to θ = π/3. Therefore, sin(θ) = sin(π/3) = √3/2.

sin(tan^(-1)(1)) = 1/√2, using the triangle with opposite and adjacent sides equal.

Using the right triangle definition, sin(tan^(-1)(x)) = opposite/hypotenuse = x/√(1+x^2).

Since π/3 is in the range of sin^(-1), sin^(-1)(sin(π/3)) = π/3.

Since sin(π/4) = 1/√2, sin^(-1)(1/√2) = π/4.

Since sin^2(θ) = 1 - cos^2(θ), we have sin^(-1)(√(1 - cos^2(θ))) = sin^(-1)(sin(θ)) = θ.

sin^(-1)(√3/2) = π/3 and cos^(-1)(1/2) = π/3. Therefore, the sum is π/3 + π/3 = 2π/3.

sin^(-1)(√3/2) = π/3 and sin^(-1)(1/2) = π/6. Therefore, π/3 + π/6 = π/2.

sin^(-1)(√3/2) = π/3.