Maths

I = ∫[0 to 10] [sin(2πx)] / e^(x-[x]) dx.
The period of the integrand involves [sin(2πx)] which depends on the sign of sin(2πx) and {x} = x - [x], which has a period of 1.
Let f(x) = [sin(2πx)] / e^{x}.
The integral is ∫[0 to 10] f(x) dx = 10 * ∫[0 to 1] f(x) dx due to the periodicity of {x} and the integer period of sin(2πx).
In the interval (0, 1/2), sin(2πx) is between 0 and 1, so [sin(2πx)] = 0.
In the interval (1/2, 1), sin(2πx) is between -1 and 0, so [sin(2πx)] = -1.

At x=0, 1/2, 1, the value is 0.
So, ∫[0 to 1] f(x) dx = ∫[0 to 1/2] 0 dx + ∫[1/2 to 1] -1/e^x dx
= 0 + [-e^(-x) * (-1)] from 1/2 to 1 = [e^(-x)] from 1/2 to 1 = e?¹ - e?¹/².
The total integral is 10 * (e?¹ - e?¹/²).
The text calculates: 10 * ∫[1/2 to 1] (-1)e^(-x) dx = -10[-e^(-x)] from 1/2 to 1 = 10(e?¹ - e?¹/²).
This seems to lead to α = 10, β = -10, γ = 0. Then α+β+γ = 0. The calculation in the text has a sign error in the final result.

Parabola: y² = 4x - 20 = 4(x - 5). Vertex at (5,0).
Line: The text seems to derive the tangent equation y = x - 4. This is not a tangent to the given parabola. The standard tangent to y²=4aX is Y=mX+a/m. Here X=x-5, a=1. So y = m(x-5)+1/m.
The other curve is an ellipse: x²/a² + y²/b² = 1.
The text says x²/2 + (x-4)²/b² = 1. This assumes a² = 2.
x²/2 + (x²-8x+16)/b² = 1
x²(1/2 + 1/b²) - (8/b²)x + (16/b² - 1) = 0.
For tangency, the discriminant (D) of this quadratic equation must be zero.
D = (8/b²)² - 4(1/2 + 1/b²)(16/b² - 1) = 0.
64/b? - 4(8/b² - 1/2 + 16/b? - 1/b²) = 0.
16/b? - (7/b² - 1/2 + 16/b?) = 0.
-7/b² + 1/2 = 0 ⇒ b² = 14.
The calculation in the text is different: 64/b? - 4(1/2 + 1/b²)(16/b² - 1) = 0 implies b=14. This should be b²=14.

sin?¹(x² + 1/3) + cos?¹(x² - 2/3) = x²
The domains of sin?¹ and cos?¹ require:

-1 ≤ x² + 1/3 ≤ 1 ⇒ -4/3 ≤ x² ≤ 2/3. Since x² ≥ 0, we have 0 ≤ x² ≤ 2/3.
-1 ≤ x² - 2/3 ≤ 1 ⇒ -1/3 ≤ x² ≤ 5/3.
The intersection of these domains is 0 ≤ x² ≤ 2/3.

The range of sin?¹ is [-π/2, π/2] and cos?¹ is [0, π].
Let A = sin?¹(x² + 1/3) and B = cos?¹(x² - 2/3).
The equation is A + B = x².
The LHS, A+B, is a sum of angles, while the RHS, x², is in the range [0, 2/3]. This suggests no solution. The provided solution states that LHS = {π}, which is incorrect. A proper analysis would involve checking if any x in the domain can satisfy the equation. Given the small range of x², it's highly unlikely. The conclusion "Hence equation has no solution" is correct.