Solution: Step 1: Expressing the function in an alternate form. The given function is:
f(x) = sin x + cos x.
We rewrite it using the identity:
sin x + cos x = √2 sin(x + ^π⁄₄)
Thus, the function can be rewritten as:
f(x) = √2 sin(x + ^π⁄₄)

Step 2: Determining the maximum value. Since the maximum value of sin θ is 1, the maximum value of f(x) is:
√2 × 1 = √2.

This maximum is attained when:
sin(x + ^π⁄₄) = 1.

Step 3: Finding x in the given domain. Solving for x:
x + ^π⁄₄ = ^π⁄₂
x = ^π⁄₂ - ^π⁄₄
x = ^π⁄₄

Since x = ^π⁄₄ lies within the given domain [0, π], the global maximum value is confirmed as √2.

Conclusion: The global maximum value of f(x) in the given interval is √2.

Solution: Step 1: Differentiate x with respect to θ. Given:
x = a(1 − cos θ)
Differentiate both sides with respect to θ:
^dx⁄_dθ = a sin θ.

Step 2: Differentiate y with respect to θ. Given:
y = a(θ + sin θ)
Differentiate both sides with respect to θ:
^dy⁄_dθ = a(1 + cos θ).

Step 3: Compute ^dy⁄_dx. Using the chain rule:
^dy⁄_dx = ^dy⁄_dθ × ^dθ⁄_dx = ^{a(1 + cos θ)}⁄_{a sin θ}

Simplify:
^dy⁄_dx = ^{1 + cos θ}⁄_{sin θ}

Using the identity:
1 + cos θ = 2 cos²(^θ⁄₂)
and
sin θ = 2 sin(^θ⁄₂) cos(^θ⁄₂)

we get:
^dy⁄_dx = ^{2 cos²(θ⁄₂)}⁄_{2 sin(^θ⁄2) cos(^θ⁄2)}

Canceling common terms:
^dy⁄_dx = ^{cos(θ⁄₂)}⁄_sin(^θ⁄2) = cot(^θ⁄₂)

Conclusion: The correct answer is cot(^θ⁄₂).

Solution: Step 1: Substituting t = e^2x. Let:
t = e^2x ⇒ dt = 2e^2x dx = 2t dx.

Rewriting the integral:
I = ∫^e2x⁄_{e^2x + 1} dx.

Step 2: Splitting the fraction. Rewrite the numerator:
e^2x = e^2x + 1 − 1 = (e^2x + 1) − 2.

Thus, the integral becomes:
I = ∫^{(e2x + 1) − 2}⁄_{e^2x + 1} dx

Splitting:
I = ∫ (1 - ²⁄_{e^2x + 1}) dx.

Step 3: Solving the integral. The first term:
∫ 1 dx = x.

For the second term, let:
t = e^2x + 1 ⇒ dt = 2e^2x dx = 2(t − 1)dx.

Rearrange:
dx = ^dt⁄_2(t-1)

Thus:
∫²⁄_{e^2x + 1} dx = ∫ ²⁄_t ^dt⁄_2(t-1) = ∫^dt⁄_t = log|t|.

Substituting back t = e^2x + 1:
log|e^2x + 1|.

Step 4: Conclusion.
I = log(e^2x + 1) − x + C.

Thus, the correct answer is log(e^2x + 1) − x + C.

Solution: Step 1: Start by simplifying the given expression. We use the identity for ^{1 + sin x}⁄_{1 - sin x} to convert it into a more convenient form:
^{1 + sin x}⁄_{1 - sin x} = ²⁄_{cos²(^x⁄2)} = 2 tan²(^x⁄₂)

Step 2: Substitute this identity into the integral:
∫ e^{x(1 + sin x)}⁄_{1 - sin x} dx = ∫ e^x 2 tan²(^x⁄₂) dx.

Step 3: Now, perform substitution. Let:
u = ^x⁄₂ ⇒ du = ^dx⁄₂ ⇒ dx = 2du.

The integral becomes:
∫ e^2u 2 tan(u) 2du = 4∫ e^2u tan(u)du.

Step 4: Using standard integral techniques, the result of the integral is:
e^x tan(^x⁄₂) + C.

Solution: Step 1: Rewrite the quadratic expression inside the square root. We can complete the square for 4x – x²:
4x - x² = 4 − (x − 2)².

Thus, the integral becomes:
∫ ¹⁄_{√4 - (x-2)²} dx.

Step 2: Use the standard trigonometric substitution x − 2 = 2 sin θ. Then, dx = 2 cos θ dθ, and the expression inside the square root becomes:
4 − (x − 2)² = 4 − 4 sin² θ = 4 cos² θ.

So, the integral simplifies to:
∫ ^{2 cos θ dθ}⁄_{2 cos θ} = ∫ dθ.

Step 3: Integrating dθ, we get:
θ + C.

Step 4: Substitute θ = sin^-1(^{x - 2}⁄₂), yielding:
sin^-1(^{x - 2}⁄₂) + C.

Solution: Step 1: Begin by calculating the determinants of the two matrices. The determinant of a 2x2 matrix |^{a b}⁄_{c d}| is given by:
Determinant = ad − bc.

For the first matrix |^{2017 2018}⁄_{2019 2020}|, the determinant is:
|^{2017 2018}⁄_{2019 2020}| = (2017)(2020) – (2018)(2019).

Simplifying:
2017 × 2020 = 4064340, 2018 × 2019 = 4064342,
so the determinant is:
4064340 - 4064342 = -2.

For the second matrix |^{2021 2022}⁄_{2023 2024}| , the determinant is:
|^{2021 2022}⁄_{2023 2024}| = (2021)(2024) – (2022)(2023).

Simplifying:
2021 × 2024 = 4084644, 2022 × 2023 = 4084646,
so the determinant is:
4084644 - 4084646 = -2.

Step 2: Now, we add the two determinants:
−2 + (−2) = −4.

Step 3: According to the given equation |^{2017 2018}⁄_{2019 2020}| + |^{2021 2022}⁄_{2023 2024}| = 2k, we have:
−4 = 2k ⇒ k = -2.

Step 4: Finally, we calculate k³:
k³ = (−2)³ = -8.

Solution: Step 1: The area of a triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃) is given by the formula:
Area = ¹⁄₂|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|.

Substituting the coordinates P(k, 1), Q(2, 4), R(1, 1) into this formula:
Area = ¹⁄₂ |k(4 − 1) + 2(1 − 1) + 1(1 − 4)|.

Step 2: Simplifying the expression:
Area = ¹⁄₂ |k(3) + 0 + 1(−3)| = ¹⁄₂ |3k − 3| .

Step 3: We are given that the area is 3 square units, so:
¹⁄₂ |3k − 3| = 3 ⇒ |3k - 3| = 6.

Step 4: Solving the absolute value equation:
3k - 3 = 6 or 3k - 3 = -6.

For 3k − 3 = 6:
3k = 9 ⇒ k = 3.

For 3k − 3 = −6:
3k = -3 ⇒ k = -1.

Step 5: Therefore, the possible values of k are k = −1 and k = 3.

Solution: Step 1: Begin by calculating A². Multiply the matrix A by itself:
A = |^{0 0 -1}⁄_{0 -1 0}|.
|^{-1 0 0}|
Perform the matrix multiplication A × A:
A² = |^{0 0 -1}⁄_{0 -1 0}| × |^{0 0 -1}⁄_{0 -1 0}|.
|^{-1 0 0}| |^{-1 0 0}|

Step 2: Carry out the multiplication:
A² = |^{(0)(0) + (0)(0) + (-1)(-1) (0)(0) + (0)(-1) + (-1)(0) (0)(-1) + (0)(0) + (-1)(0)}⁄_{(0)(0) + (-1)(0) + (0)(-1) (0)(0) + (-1)(-1) + (0)(0) (0)(-1) + (-1)(0) + (0)(0)}|
|^{(-1)(0) + (0)(0) + (0)(-1) (-1)(0) + (0)(-1) + (0)(0) (-1)(-1) + (0)(0) + (0)(0)}|

Simplifying:
A² = |^{1 0 0}⁄_{0 1 0}|
|^{0 0 1}|
Step 3: Now, compute I + A², where I is the identity matrix:
I = |^{1 0 0}⁄_{0 1 0}| ; A² = |^{1 0 0}⁄_{0 1 0}|
|^{0 0 1}| |^{0 0 1}|
Thus:
I + A² = |^{1 0 0}⁄_{0 1 0}| + |^{1 0 0}⁄_{0 1 0}| = |^{2 0 0}⁄_{0 2 0}|.
|^{0 0 1}| |^{0 0 1}| |^{0 0 2}|

This is equal to 2I.

Solution: Step 1: Compute the transpose of matrix A. The transpose of A, denoted as A', is obtained by swapping the off-diagonal elements:
A' = |^{sin α cos α}⁄_{-cos α sin α}|

Step 2: Add A and A'.
A + A' = |^{sin α -cos α}⁄_{cos α sin α}| + |^{sin α cos α}⁄_{-cos α sin α}|

Adding corresponding elements:
A + A' = |^{sin α + sin α -cos α + cos α}⁄_{cos α - cos α sin α + sin α}|

Simplifying:
A + A' = |^{2 sin α 0}⁄_{0 2 sin α}|

Step 3: Equating to the identity matrix. The identity matrix is:
I = |^{1 0}⁄_{0 1}|

Thus, equating:
|^{2 sin α 0}⁄_{0 2 sin α}| = |^{1 0}⁄_{0 1}|

From which:
2 sin α = 1 ⇒ sin α = ¹⁄₂

Step 4: Finding cos α. Using the Pythagorean identity:
sin² α + cos² α = 1.

Substituting sin α = ¹⁄₂:
(¹⁄₂)² + cos² α = 1.
¹⁄₄ + cos² α = 1.
cos² α = ³⁄₄
cos α = ^√3⁄₂.

Conclusion: The correct answer is ^√3⁄₂.

Solution: Step 1: Given that A² = A, this means that A is a idempotent matrix.

Step 2: Let's first expand (I – A)³ using the binomial expansion:
(I – A)³ = I³ − 3I²A + 3IA² − A³ = I − 3A + 3A – A = I – A.

Thus:
(I – A)³ = I – A.

Step 3: Now, expand (I + A)²:
(I + A)² = I² + 2IA + A² = I + 2A + A = I + 3A.

Step 4: Subtract (I + A)² from (I – A)³:
(I – A)³ – (I + A)² = (I – A) − (I + 3A) = I – A – I – 3A = -4A.

Step 5: Now, factorize the result:
−4A = 2(I – 2A).

Thus, the final expression is:
(I – A)³ – (I + A)² = 2(I – 2A).

Solution: Step 1: We are given sin^-1(sin ^23π⁄₆), and we need to simplify the expression.

Step 2: First, simplify ^23π⁄₆ by reducing it within the range of the sine inverse function, which is -^π⁄₂ ≤ θ ≤ ^π⁄₂.

To do this, observe that:
^23π⁄₆ = 2π + ^11π⁄₆

Since sin(θ) is periodic with period 2π, we have:
sin(^23π⁄₆) = sin(^11π⁄₆)

Step 3: Now, ^11π⁄₆ lies in the fourth quadrant. In the fourth quadrant,
sin(^11π⁄₆) = - sin(^π⁄₆) = -¹⁄₂.

Step 4: Therefore, we have:
sin^-1(sin ^23π⁄₆) = sin^-1(-¹⁄₂).

Step 5: The angle whose sine is −¹⁄₂ in the range -^π⁄₂ ≤ θ ≤ ^π⁄₂ is - ^π⁄₆.

Thus,
sin^-1(sin ^23π⁄₆) = -^π⁄₆

Solution: Step 1: Begin by evaluating each term separately.

Step 2: For tan^-1(−1), we know that:
tan^-1(-1) = -^π⁄₄
since the tangent of -^π⁄₄ is -1.

Step 3: Next, evaluate sec^-1(−2). The value of sec^-1(−2) corresponds to the angle θ such that sec θ = -2. Since sec θ = ¹⁄_{cos θ}, we have cos θ = -¹⁄₂, and the angle that satisfies this condition is θ = ^2π⁄₃ (since cos(^2π⁄₃) = -¹⁄₂).

Step 4: Now, evaluate sin^-1(¹⁄_√2). The angle whose sine is ¹⁄_√2 is ^π⁄₄ , since sin(^π⁄₄) = ¹⁄_√2.

Step 5: Now, add all the results together:
tan^-1(-1) + sec^-1(-2) + sin^-1(¹⁄_√2) = -^π⁄₄ + ^2π⁄₃ + ^π⁄₄.

Step 6: Simplify the expression:
-^π⁄₄ + ^π⁄₄ = 0,
so the expression becomes: ^2π⁄₃

Thus, the final value is:
tan^-1(-1) + sec^-1(-2) + sin^-1(¹⁄_√2) = ^2π⁄₃.

Solution: Step 1: The function y = tan^-1 x is the inverse of the tangent function. By the definition of the inverse tangent (or arctangent), it returns the angle y such that:
tan y = x.

Step 2: The range of the arctangent function tan^-1 x is restricted to -^π⁄₂ < y < ^π⁄₂. This is because the tangent function is periodic, and the principal value of the inverse tangent is chosen to lie within this interval to ensure it is a one-to-one function.

Thus, we conclude:
-^π⁄₂ < y < ^π⁄₂.

Solution: Step 1: Understanding one-to-one (injective) functions. A function f is one-to-one (injective) if:
f(a) = f(b) ⇒ a = b.

Step 2: Checking injectivity of f(x). Given:
f(x) = x³ + 2.

Let f(a) = f(b):
a³ + 2 = b³ + 2.

Canceling 2 from both sides:
a³ = b³.

Taking the cube root:
a = b.

Since the function satisfies the condition for injectivity, it is one-to-one.

Step 3: Checking onto (surjective) property. For surjectivity, f(x) must cover all integers. Since cube functions do not necessarily map to all integers, f(x) is not necessarily onto over Z.

Conclusion: The function f(x) is one-to-one.

Solution: Step 1: To determine the type of relation, we need to analyze the properties of the given relation R on the set {a, b, c}.
The relation R is defined as:
R = {(a, a), (b, b), (c, c), (a, c)}.

Step 2: Check for the following properties:
- Reflexive: A relation is reflexive if for every element x ∈ S, (x, x) belongs to the relation. In this case, (a, a), (b, b), (c, c) are in the relation, so the relation is reflexive.
- Symmetric: A relation is symmetric if for every pair (x, y) in the relation, the pair (y, x) is also in the relation. Since (a, c) is in the relation but (c, a) is not, the relation is not symmetric.
- Transitive: A relation is transitive if whenever (x, y) and (y, z) are in the relation, (x, z) must also be in the relation. Since (a, c) is in the relation but there is no corresponding (c, a), the relation is not transitive.

Step 3: The relation is reflexive but not symmetric or transitive, which makes it spontaneous, traditional, but not conformist.

Thus, the relation R is spontaneous, traditional, but not conformist.

Solution: Step 1: We are given that P(B) ≠ 0 and P(A|B) = 1. The conditional probability P(A|B) is defined as:
P(A|B) = ^{P(A ∩ B)}⁄_P(B).

Since P(A|B) = 1, this implies:
^{P(A ∩ B)}⁄_P(B) = 1.

Multiplying both sides by P(B), we get:
P(A ∩ B) = P(B).

Step 2: The equation P(A ∩ B) = P(B) means that the probability of the intersection of A and B is equal to the probability of B. This implies that every outcome in B is also in A, i.e., B ⊂ A.

Thus, the correct conclusion is:
B ⊂ A.

Solution: Step 1: An even prime number is a number that is both even and prime. The only even prime number is 2.

Step 2: Each die has the numbers 1, 2, 3, 4, 5, 6. For the condition to be satisfied (getting an even prime number on each die), both dice must show the number 2.

Step 3: The total number of possible outcomes when two dice are thrown is:
6 × 6 = 36.

Step 4: The only favorable outcome is when both dice show 2, so there is exactly one favorable outcome.

Step 5: Therefore, the probability of getting an even prime number on each die is:
¹⁄₃₆

Thus, the correct answer is ¹⁄₃₆.

Solution: Step 1: We are given the following probabilities:
P(A) = p, P(B) = ¹⁄₂ , P(A ∪ B) = ³⁄₅.

The formula for the probability of the union of two events is:
P(A ∪ B) = P(A) + P(B) − P(A ∩ B).

Step 2: Substitute the given values into the formula:
³⁄₅ = p + ¹⁄₂ − P(A ∩ B).

Step 3: Since events A and B are absolute, P(A ∩ B) = 0, as they do not overlap. Thus, the equation becomes:
³⁄₅ = p + ¹⁄₂.

Step 4: Solving for p:
p = ³⁄₅ - ¹⁄₂
To subtract the fractions, find a common denominator:
p = ⁶⁄₁₀ - ⁵⁄₁₀ = ¹⁄₁₀.

Thus, the value of p is ¹⁄₁₀.

Solution: Step 1: Understanding the constraints. The given constraints are:
x + y ≤ 55,
x + y ≥ 10,
x ≥ 0, y ≥ 0.

These constraints define a feasible region in the first quadrant where the values of x and y lie.

Step 2: Identifying the feasible region. The inequalities define a strip in the first quadrant between the lines: x + y = 10 (lower boundary). - x + y = 55 (upper boundary). However, for a solution to exist, the feasible region should be a bounded region where an optimal solution can be determined.

Step 3: Checking feasibility for optimization. Since the given constraints do not form a closed bounded region (it extends infinitely), there is no minimum bound for the function z = 7x + 3y. Thus, the solution region is not feasible for determining the minimum value.

Conclusion: Since the feasible region does not bound the function properly, the minimum value cannot be determined.

Solution: Step 1: The objective function is z = 6x + 12y.

Step 2: To find the maximum value of the objective function, we evaluate z at each of the given vertices:
- At (0,6):
z = 6(0) + 12(6) = 0 + 72 = 72.

- At (3, 3):
z = 6(3) + 12(3) = 18 + 36 = 54.

- At (9,9):
z = 6(9) + 12(9) = 54 + 108 = 162.

- At (0, 12):
z = 6(0) + 12(12) = 0 + 144 = 144.

Step 3: The maximum value of z is 162, which occurs at the vertex (9,9).

Thus, the maximum value of the objective function is 162.