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.

tan(45°) = 1 and cot(45°) = 1, so 1 + 1 = 2.

tan(45°) = 1.

C(7, 4) = 7! / (4! * (7-4)!) = 7! / (4! * 3!) = (7*6*5)/(3*2*1) = 35.

The coefficient of x^2 is C(4,2) * (3)^2 * (-4)^2 = 6 * 9 * 16 = 864.

Using the binomial theorem, the coefficient of x^4 in (a + b)^n is given by nCk * a^(n-k) * b^k. Here, n=6, a=x, b=-2, and k=2. Thus, the coefficient is 6C2 * (1)^4 * (-2)^2 = 15 * 4 = 60.

∫(0 to 1) (x^2 + 2x) dx = [x^3/3 + x^2] from 0 to 1 = (1/3 + 1) - (0) = 4/3.

∫(0 to 2) (x^2 + 1) dx = [x^3/3 + x] from 0 to 2 = (8/3 + 2) - (0) = 4.

∫(0 to π) sin(x) dx = [-cos(x)] from 0 to π = [1 - (-1)] = 2.

∫(1 to 3) (x^2 - 2x + 1) dx = [(x^3/3 - x^2 + x)] from 1 to 3 = (9/3 - 9 + 3) - (1/3 - 1 + 1) = 0.

∫(1 to 4) (x^3) dx = [x^4/4] from 1 to 4 = (256/4 - 1/4) = 63.

f'(x) = 4x^3 - 12x^2 + 12x. Evaluating at x = 1 gives f'(1) = 4 - 12 + 12 = 4.

The determinant is \( 1*4 - 2*3 = 4 - 6 = -2 \).

The determinant is calculated as 1(1*0 - 4*6) - 2(0*0 - 4*5) + 3(0*6 - 1*5) = 1(0 - 24) - 2(0 - 20) + 3(0 - 5) = -24 + 40 - 15 = 1.

Using the determinant formula, we calculate it to be 1.

Using the determinant formula, we calculate it to be -10.

Using the determinant formula, we calculate it to be -20.

Substituting x = 1 gives the determinant | 1  1  2 |  | 3  1  4 |  | 5  6  1 | = 6.

The integral evaluates to [x - (1/3)x^3] from 0 to 1 = 1 - 1/3 = 2/3.

The integral ∫(3x^2)dx from 0 to 1 = [x^3] from 0 to 1 = 1.

The integral evaluates to [(1/3)x^3 + x^2] from 0 to 1 = (1/3 + 1) - 0 = 4/3.

The integral ∫(2x + 1)dx = x^2 + x. Evaluating from 0 to 2 gives (2^2 + 2) - (0 + 0) = 4.

The limit is 1.

To find the minimum, set f'(x) = e^x + 2x = 0. The minimum occurs at x = 0.

Set f'(x) = e^x - 2x = 0. The solution is approximately x = 1.