Solution:

The integral is:

I_n = ∫₀¹ (1 - x^k)ⁿ · 1 dx

Using integration by parts:

I_n = (1 - x^k)ⁿ · x |₀¹ - nk ∫₀¹ (1 - x^k)^n-1 · x^k-1 dx

I_n = nk ∫₀¹ [(1 - x^k)ⁿ - (1 - x^k)^n-1] dx

I_n = nk I_n-1 - nk I_n

Rearranging terms:

I_n / I_n-1 = nk / (nk + 1)

Given:

I₂₁ / I₂₀ = 21k / (1 + 21k)

Equating:

147 / 148 = 21k / (1 + 21k)

Solve for k:

k = 7

Solution:

1. Substitution:

Let t = (8)^x. The equation becomes:
t² - 16t + 48 = 0.

2. Solve the Quadratic Equation:

Using the quadratic formula:
t = (-b ± √(b² - 4ac)) / 2a, where a = 1, b = -16, c = 48:
t = (16 ± √(16² - 4(1)(48))) / 2.
t = (16 ± √(256 - 192)) / 2.
t = (16 ± √64) / 2.
t = (16 ± 8) / 2.
t = 12 or t = 4.

3. Back Substitution:

Recall t = (8)^x:
- For t = 4:
(8)^x = 4 ⟹ x = log₈(4).
- For t = 12:
(8)^x = 12 ⟹ x = log₈(12).

4. Sum of Solutions:

The sum of solutions is:
log₈(4) + log₈(12).

Using the logarithmic property log_b(m) + log_b(n) = log_b(mn):
log₈(4) + log₈(12) = log₈(4 ⋅ 12) = log₈(48).

Simplify:

log₈(48) = log₈(8 ⋅ 6) = log₈(8) + log₈(6).
log₈(8) = 1, so the sum is:
1 + log₈(6).

Solution:

1. Determine the Coordinates of the Center of C₁:

The point (6, 6) divides the line joining (α, β) and (8, 15/2) in the ratio 2:1. Using the section formula:
6 = (2 × 8 + 1 × α) / 3, and 6 = (2 × 15/2 + 1 × β) / 3.

Solving for α and β:

16 + α = 18 ⟹ α = 2, and 15 + β = 18 ⟹ β = 3.

Thus, the center of C₁ is (α, β) = (2, 3).

2. Use the Touching Condition:

The distance between the centers of C₁ and C₂ equals the sum of the radii:
√[(2 - 8)² + (3 - 15/2)²] = r₁ + r₂.

Simplify:

√[(-6)² + (-9/2)²] = r₁ + r₂.
√[36 + 81/4] = r₁ + r₂.
√[(144 + 81) / 4] = r₁ + r₂.
√(225 / 4) = r₁ + r₂.
r₁ + r₂ = 15/2.

3. Determine r₁ and r₂:

From the external touching condition:
2r₂ = r₁ = 5.
r₂ = 5/2, and r₁ = 5.

4. Calculate α + β + 4(r₁² + r₂²):

Compute r₁² + r₂²:
r₁² = 5² = 25, and r₂² = (5/2)² = 25/4.
r₁² + r₂² = 25 + 25/4 = 100/4 + 25/4 = 125/4.

Now calculate:

α + β + 4(r₁² + r₂²) = 2 + 3 + 4 × 125/4.
α + β + 4(r₁² + r₂²) = 5 + 125 = 130.

Solution:

1. Coordinates of Points:

- P(x, y, z) is a point in 3D space.
- Q(x, y, 0) is the projection of P in the xy-plane.

2. Distance Relations:

- The distance OP = γ, so:
x² + y² + z² = γ².
- The distance OQ is the projection in the xy-plane:
OQ = √(x² + y²).

3. Using the Angles θ and φ:

- The angle θ is between OQ and the x-axis. From this, we have:
cosθ = x / √(x² + y²).
- The angle φ is between OP and the z-axis. From this, we derive:
cosφ = z / √(x² + y² + z²) = z / γ.

Using cos²φ + sin²φ = 1, we get:
sin²φ = 1 - cos²φ = (x² + y²) / γ².

4. Distance of P from the x-Axis:

- The distance of P from the x-axis is calculated as:
Distance = √(y² + z²).

- Substituting y² + z² using x² + y² + z² = γ²:
y² + z² = γ² - x².

- Since sin²φ = (x² + y²) / γ², we have:
x² = γ² cos²θ sin²φ.

- Replacing x² in γ² - x²:
y² + z² = γ² (1 - cos²θ sin²φ).

- Taking the square root:
Distance = γ √(1 - sin²φ cos²θ).

Solution:

1. Given Function:

f(x) = (x - 2)^2/3(2x + 1)

2. Finding the First Derivative:

Using the product rule:

f'(x) = (2/3)(x - 2)^-1/3(2x + 1) + (x - 2)^2/3(2)

3. Combine Terms:

f'(x) = [2(2x + 1) + 2(x - 2)] / [3(x - 2)^1/3]

f'(x) = 2(3x - 1) / [3(x - 2)^1/3]

4. Critical Points:

f'(x) = 0 ⟹ 3x - 1 = 0 ⟹ x = 1/3

f'(x) is undefined when x = 2.

5. Conclusion:

The critical points are x = 1/3 and x = 2.

Solution:

1. Expression for Area:

∫₀^af(x) dx = e^-a + 4a² + a - 1

2. Differentiate:

f(a) = -e^-a + 8a + 1

f'(a) = e^-a + 8

f''(a) = -e^-a

3. Differential Equation:

(8e^x + 1)f''(x) + f'(x) = 0

Solution:

1. Find f'(x):

f'(x) = -sin(x)(12cos²(x) + 6√3cos(x))

2. Critical Points:

f'(x) = 0 ⟹ sin(x) = 0, cos(x) = 0, or 2cos(x) + √3 = 0

Solutions in (0, 2π):

x = π/2, 3π/2 (local maxima)

3. Conclusion:

The number of local maxima is 2.

Solution:

1. Trace and Determinant:

tr(A) = 2 + 3 + b, det(A) = 6b - ab

2. Given det(A) = -21:

6b - ab = -21

3. Solve for a and b:

a = -5, b = -1

4. Calculate:

2a + 3b = -13

Solution:

1. Direction Vectors:

p = i - 3j + 4k, q = 2i + 3j + k

2. Cross Product:

p × q = -15i + 7j + 9k

3. Dot Product:

AB • (p × q) = 32

4. Distance:

Distance = 32 / √355

m = 32, n = 355 ⟹ m + n = 387

Solution:

1. Maximum Product:

Max product = 144 when x = y = 12

2. Favorable Condition:

Product ≥ 108

3. Favorable Pairs:

13 pairs: (13, 11), ..., (6, 18)

4. Total Pairs:

23 pairs

5. Probability:

P = 13/23 ⟹ m = 13, n = 23

6. Find n - m:

n - m = 10

Solution:

1. Given Information:

sin x = -3/5, π < x < 3π/2.

Since x lies in the third quadrant:

• cos x < 0
• tan x > 0

2. Find cos x:

Using the Pythagorean identity:

sin²x + cos²x = 1.

Substitute sin x = -3/5:

(-3/5)² + cos²x = 1

9/25 + cos²x = 1

cos²x = 16/25

Since cos x < 0 (third quadrant):

cos x = -4/5

3. Find tan x:

Using the definition tan x = sin x / cos x:

tan x = (-3/5) / (-4/5) = 3/4

4. Calculate 80(tan²x - cos x):

Substitute tan x = 3/4 and cos x = -4/5:

tan²x = (3/4)² = 9/16

80(tan²x - cos x) = 80[(9/16) - (-4/5)]

80[(9/16) + (4/5)]

Find the common denominator for 9/16 and 4/5:

9/16 = 45/80, 4/5 = 64/80

Add the fractions:

45/80 + 64/80 = 109/80

Multiply by 80:

80 × 109/80 = 109

Conclusion:

The value of 80(tan²x - cos x) is 109.

Solution:

1. Given Integral:

I(x) = ∫ [6 / sin²x(1 - cot x)²] dx

2. Simplify the Integrand:

Using the identity sin²x = 1 / csc²x, rewrite the integrand:

I(x) = ∫ [6 csc²x / (1 - cot x)²] dx

3. Substitution:

Let t = 1 - cot x, then:

csc²x dx = dt

Substitute into the integral:

I = ∫ [6 / t²] dt

4. Evaluate the Integral:

I = -6/t + c

Substituting back t = 1 - cot x, we get:

I(x) = -6 / (1 - cot x) + c

5. Given Condition:

At x = 0, I(0) = 3. Substituting x = 0, where cot 0 = ∞, we find:

c = 3

Therefore:

I(x) = -6 / (1 - cot x) + 3

6. Find I(π/12):

At x = π/12, we calculate:

cot(π/12) = 2 + √3

Substituting:

I(π/12) = 3 - 6 / [1 - (2 + √3)]

Simplify the denominator:

1 - (2 + √3) = -1 - √3

So:

I(π/12) = 3 + 6 / (1 + √3)

Rationalize the denominator:

6 / (1 + √3) = 6(1 - √3) / [(1 + √3)(1 - √3)]

= 6(1 - √3) / -2

6 / (1 + √3) = -3(1 - √3) = -3 + 3√3

7. Final Value:

Substituting back:

I(π/12) = 3 + (-3 + 3√3) = 3√3

Conclusion:

The value of I(π/12) is 3√3.

Solution:

1. Equations of the Lines AB and AC:

The equations of the sides are given as:

AB: 4x + y = 14, AC: 3x - 2y = 5.

2. Point Dividing the Line BC:

The point P = (2, -4/3) divides BC internally in the ratio 2:1.

3. Coordinates of B:

From the equation of line AB: 4x + y = 14,

y = 14 - 4x.

Let B = (x1, 14 - 4x1).

4. Coordinates of C:

From the equation of line AC: 3x - 2y = 5,

y = (3x - 5)/2.

Let C = (x2, (3x2 - 5)/2).

5. Section Formula for P:

The coordinates of P are given by the section formula:

x = (2x2 + x1)/3, y = (2y2 + y1)/3.

Substituting P = (2, -4/3):

2 = (2x2 + x1)/3, -4/3 = (2y2 + y1)/3.

6. Solve for x1 and x2:

From 2 = (2x2 + x1)/3:

6 = 2x2 + x1 → x1 = 6 - 2x2.

From -4/3 = (2y2 + y1)/3, substitute y1 = 14 - 4x1 and y2 = (3x2 - 5)/2:

-4 = 2y2 + y1 = 2((3x2 - 5)/2) + (14 - 4x1).

-4 = (3x2 - 5) + 14 - 4(6 - 2x2).

-4 = 3x2 - 5 + 14 - 24 + 8x2.

-4 = 11x2 - 15.

Solve for x2:

x2 = 1.

Substitute x2 = 1 into x1 = 6 - 2x2:

x1 = 6 - 2(1) = 4.

Therefore, B = (4, -2) and C = (1, -1).

7. Equation of Line BC:

The slope of BC is:

m = (y2 - y1)/(x2 - x1) = (-1 - (-2))/(1 - 4) = 1/-3 = -1/3.

The equation of BC is:

y - y1 = m(x - x1).

Substituting (x1, y1) = (4, -2) and m = -1/3:

y + 2 = -1/3(x - 4).

3(y + 2) = -(x - 4).

3y + 6 = -x + 4.

x + 3y + 2 = 0.

Conclusion:

The equation of BC is x + 3y + 2 = 0.

Solution:

1. Prime Factorization of 2310:

The prime factorization of 2310 is:

N = 2310 = 231 × 10 = 3 × 11 × 7 × 2 × 5.

Hence:

A = {2, 3, 5, 7, 11}.

2. Definition of f(x):

The function is given as:

f(x) = ⌊log₂(x² + ⌊x³/5⌋)⌋.

3. Calculate f(x) for Each Element of A:

For each x ∈ A:

f(2) = ⌊log₂(2² + ⌊2³/5⌋)⌋ = ⌊log₂(4 + ⌊8/5⌋)⌋ = ⌊log₂(4 + 1)⌋ = ⌊log₂(5)⌋ = 2.

f(3) = ⌊log₂(3² + ⌊3³/5⌋)⌋ = ⌊log₂(9 + ⌊27/5⌋)⌋ = ⌊log₂(9 + 5)⌋ = ⌊log₂(14)⌋ = 3.

f(5) = ⌊log₂(5² + ⌊5³/5⌋)⌋ = ⌊log₂(25 + ⌊125/5⌋)⌋ = ⌊log₂(25 + 25)⌋ = ⌊log₂(50)⌋ = 5.

f(7) = ⌊log₂(7² + ⌊7³/5⌋)⌋ = ⌊log₂(49 + ⌊343/5⌋)⌋ = ⌊log₂(49 + 68)⌋ = ⌊log₂(117)⌋ = 6.

f(11) = ⌊log₂(11² + ⌊11³/5⌋)⌋ = ⌊log₂(121 + ⌊1331/5⌋)⌋ = ⌊log₂(121 + 266)⌋ = ⌊log₂(387)⌋ = 8.

• f(2):
• f(3):
• f(5):
• f(7):
• f(11):

4. Range of f:

From the above calculations, the range of f is:

B = {2, 3, 5, 6, 8}.

5. Number of One-to-One Functions:

The number of one-to-one functions from A to B is given by:

|A| = |B| = 5.

Therefore, the number of one-to-one functions is:

5! = 120.

Conclusion:

The number of one-to-one functions is 120.

Solution:

We are given:

|z + 2| = 1, and Im((z + 1) / (z + 2)) = 1/5.

Let z + 2 = cosθ + i sinθ.

The reciprocal becomes:

1 / (z + 2) = cosθ - i sinθ.

Now:

(z + 1) / (z + 2) = 1 - (1 / (z + 2)) = 1 - (cosθ - i sinθ).

Simplifying:

(z + 1) / (z + 2) = (1 - cosθ) + i sinθ.

From the given condition:

Im((z + 1) / (z + 2)) = sinθ, sinθ = 1/5.

Using the Pythagorean identity:

cos²θ = 1 - sin²θ = 1 - (1/25) = 24/25.

Therefore:

cosθ = ±√(24/25) = ±(2√6)/5.

Finally, the real part of z + 2 is:

|Re(z + 2)| = 2√6/5.

Solution:

We are given:

a + 5b = 42, a, b ∈ ℕ.

Rewrite a = 42 - 5b:

• For b = 1, a = 37.
• For b = 2, a = 32.
• For b = 3, a = 27.
• For b = 4, a = 22.
• For b = 5, a = 17.
• For b = 6, a = 12.
• For b = 7, a = 7.
• For b = 8, a = 2.

The set R has 8 elements, so m = 8.

The summation is:

∑(n=1 to 8) [1 + i^(n!)] = x + iy.

For n ≥ 4, n! is a multiple of 4, and i^(n!) = 1.

The terms alternate cyclically as i, -1, -i, 1, and repeat.

Therefore:

∑ = (1 + i) + (1 - 1) + (1 - i) + (1 + 1) + ...

Simplifying:

x = 5, y = -1.

m + x + y = 8 + 5 - 1 = 12.

Solution:

The function f(x) = cos x - x + 1 is given.

Step 1: Analyze the derivative f′(x):

Differentiate f(x): f′(x) = -sin x - 1.

Since sin x ∈ [-1, 1], f′(x) = -sin x - 1 ∈ [-2, 0], which means f′(x) < 0 for all x ∈ ℜ. Hence, f(x) is strictly decreasing.

Step 2: Analyze statement (S1):

Since f(x) is strictly decreasing, it is one-to-one. Check values at endpoints of [0, π]:

• f(0) = 2 and f(π) = -π.

By the Intermediate Value Theorem, f(x) = 0 has exactly one solution in [0, π]. Thus, (S1) is correct.

Step 3: Analyze statement (S2):

Since f′(x) < 0, f(x) is strictly decreasing on [0, π], and it cannot be increasing in [π/2, π]. Hence, (S2) is incorrect.

Conclusion: (S1) is correct, and (S2) is incorrect.

Solution:

To find the set of α:

• Step 1: Calculate the dot product of a and b:
• a ∙ b = αt² - 12 + 6αt.
• Step 2: For obtuse angle, a ∙ b < 0 for all t:
• The inequality becomes αt² + 6αt - 12 < 0.
• Step 3: For this inequality to hold, the discriminant must be negative:
• D = 36α² + 48α < 0 → 12α(α + 4) < 0.

Solving gives α ∈ (-4/3, 0].

Solution:

The given differential equation can be rewritten and integrated to give:

tan^-1(e^{tan x}) + tan^-1(y) = C.

Applying initial conditions x = 0, y = 1, we find C = π/2.

Substituting x = π/4:

tan^-1(e) + tan^-1(y) = π/2.

Solving gives y = 1/e.

Solution:

The hyperbola is given by:

y²/b² - x²/a² = 1.

With e = √3, a² = 6, b² = 3, we find α² = 66.

Calculating the focal distances and their product β:

β = 105.

Thus, α² + β = 171.

Solution:

Step 1: Matrix Definition:

The matrix is given as:

A = 
| 2  -1 |
| 1   1 |

Step 2: Finding Powers of A:

Calculate A²:

A2 = 
| 3  -3 |
| 3   0 |

Calculate A³:

A3 = 
| 3  -6 |
| 6  -3 |

Calculate A⁴:

A4 = 
| 0  -9 |
| 9  -9 |

Calculate A⁵:

A5 = 
| -9  -9 |
|  9 -18 |

Calculate A⁶:

A6 = 
| -27   0 |
|   0 -27 |

Step 3: Sum of Diagonal Elements:

From A⁶, observe the diagonal elements pattern:

• For Aⁿ, the sum of the diagonal elements is 3ⁿ.

For A¹³, the sum of the diagonal elements is:

3ⁿ = 3⁷.

Thus, n = 7.

Solution:

Step 1: Find the centroid (G) of the second triangle:

The centroid G of a triangle is calculated using the formula:

G = [(x₁ + x₂ + x₃) / 3, (y₁ + y₂ + y₃) / 3]

Using the given circumcentre (3, 4) and orthocentre (-6, -8), the centroid G is:

G = [(3 + (-6)) / 3, (4 + (-8)) / 3]

G = [-3 / 3, -4 / 3] = (-1, -4/3).

Step 2: Orthocentre of the first triangle:

The orthocentre of the first triangle lies at G = (-1, -4/3).

Step 3: Solving for |a - b|:

The lines 2x + 3y - 1 = 0 and x + 2y - 1 = 0 form two sides of the triangle. The third side ax + by - 1 = 0 must satisfy the condition that the orthocentre lies at (-1, -4/3).

By substituting (-1, -4/3) into the equations and solving for a and b, we find:

a = 2, b = -14.

Step 4: Calculate |a - b|:

|a - b| = |2 - (-14)| = |2 + 14| = 16.

Solution:

Step 1: Total Number of Ways to Draw 3 Balls:

The total number of ways to select 3 balls from 9 (5 blue and 4 yellow) is:

????(9, 3) = 84.

Step 2: Probability Distribution for Blue Balls:

Calculate the probabilities for each possible number of blue balls X (0, 1, 2, 3):

• Pr(X = 0) = ????(5, 0) ????(4, 3) / ????(9, 3) = 4/84
• Pr(X = 1) = ????(5, 1) ????(4, 2) / ????(9, 3) = 30/84
• Pr(X = 2) = ????(5, 2) ????(4, 1) / ????(9, 3) = 30/84
• Pr(X = 3) = ????(5, 3) ????(4, 0) / ????(9, 3) = 20/84

Calculate 7????X:

7????X = 7 × ∑k × Pr(X = k).

Substitute values:

7????X = 7 × (0 × 4 + 1 × 30 + 2 × 30 + 3 × 20) / 84.

7????X = 7 × 150 / 84 = 1050 / 84 = 35 / 3.

Step 3: Probability Distribution for Yellow Balls:

Similarly, calculate probabilities for Y (0, 1, 2, 3):

• Pr(Y = 0) = ????(5, 3) ????(4, 0) / ????(9, 3) = 20/84
• Pr(Y = 1) = ????(5, 2) ????(4, 1) / ????(9, 3) = 30/84
• Pr(Y = 2) = ????(5, 1) ????(4, 2) / ????(9, 3) = 30/84
• Pr(Y = 3) = ????(5, 0) ????(4, 3) / ????(9, 3) = 4/84

Calculate 4????Y:

4????Y = 4 × ∑k × Pr(Y = k).

Substitute values:

4????Y = 4 × (0 × 20 + 1 × 30 + 2 × 30 + 3 × 4) / 84.

4????Y = 4 × 102 / 84 = 408 / 84 = 16 / 3.

Step 4: Final Calculation of 7????X + 4????Y:

Combine the results:

7????X + 4????Y = 35 / 3 + 16 / 3 = 51 / 3 = 17.

Solution:

Step 1: Total Number of 3-Digit Numbers:

The digits available are 2, 3, 4, 5, and 7, and repetition is not allowed. The total number of 3-digit numbers that can be formed is:

Total = 5 × 4 × 3 = 60.

Step 2: Numbers Divisible by 3:

A number is divisible by 3 if the sum of its digits is divisible by 3. The available digits are:

2, 3, 4, 5, and 7. The sum of all the digits is:

2 + 3 + 4 + 5 + 7 = 21 (divisible by 3).

To determine how many 3-digit numbers are divisible by 3, calculate the combinations of digits whose sum is divisible by 3:

• {2, 4, 7}: Sum = 13 (not divisible by 3)
• {2, 5, 7}: Sum = 14 (not divisible by 3)
• {3, 4, 5}: Sum = 12 (divisible by 3)
• {3, 5, 7}: Sum = 15 (divisible by 3)

Using these combinations, the total numbers divisible by 3 are:

Total divisible = 24.

Step 3: Numbers Not Divisible by 3:

Total numbers not divisible by 3 = Total numbers - Numbers divisible by 3:

Total not divisible = 60 - 24 = 36.

Final Answer: The number of 3-digit numbers not divisible by 3 is 36.

Solution:

Step 1: Understanding the Pattern:

The arrangement of numbers is such that the n-th row contains n numbers. The cumulative sum of numbers up to the n-th row is given by the formula for triangular numbers:

T_n = 1 + 2 + 3 + ... + n = n(n + 1) / 2.

Step 2: Finding the Row Containing 5310:

We need to find n such that:

T_n-1 < 5310 ≤ T_n.

Start with the formula for T_n:

T_n = n(n + 1) / 2.

Multiply through by 2:

n(n + 1) = 2 × 5310 = 10620.

Rearrange into a quadratic equation:

n² + n - 10620 = 0.

Use the quadratic formula:

n = [-1 ± √(1 + 4 × 10620)] / 2 = [-1 ± √(42481)] / 2.

Simplify:

n = (-1 + 206) / 2 = 103 (since n must be positive).

Step 3: Verification:

Calculate T₁₀₃ and T₁₀₂:

T₁₀₃ = (103 × 104) / 2 = 5356,
T₁₀₂ = (102 × 103) / 2 = 5253.

Since 5253 < 5310 ≤ 5356, the number 5310 lies in the 103rd row.

Solution:

1. Given Function:
f(θ) = ∕frac;∕sin;⁴θ + 3∕cos;²θ∕sin;⁴θ + ∕cos;²θ

2. Simplification of f(θ):
Factorize the numerator and denominator:
f(θ) = 1 + ∕frac;2∕cos;²θ∕sin;⁴θ + ∕cos;²θ.
Substitute ∕cos;²θ = x, and since ∕sin;²θ + ∕cos;²θ = 1, ∕sin;⁴θ = (1 - x)². Rewrite:
f(θ) = 1 + ∕frac;2x∕(1 - x)² + x.
Further simplify:
f(θ) = 1 + ∕frac;2x∕x² - x + 1.

3. Finding the Range of f(θ):
The minimum and maximum values of f(θ) occur at:
f(θ)_min = 1, f(θ)_max = 3.
Hence:
[α, β] = [1, 3].

4. Sum of the Infinite G.P.:
The first term a = 64 and the common ratio r = ∕frac;αβ = ∕frac;1∕3. The sum of the infinite G.P. is:
S = ∕frac;a∕1 - r.
Substitute the values:
S = ∕frac;64∕1 - ∕frac;1∕3 = ∕frac;64∕∕frac;2∕3 = 64 × ∕frac;3∕2 = 96.

Solution:

Step 1: Simplify α:

α = ∑_r=0ⁿ (4r² + 2r + 1) ⋅ C(n, r)

Expand terms:

α = 4 ∑_r=0ⁿ r² ⋅ C(n, r) + 2 ∑_r=0ⁿ r ⋅ C(n, r) + ∑_r=0ⁿ C(n, r).

Using binomial identities:

• ∑_r=0ⁿ C(n, r) = 2ⁿ
• ∑_r=0ⁿ r ⋅ C(n, r) = n ⋅ 2^n-1
• ∑_r=0ⁿ r² ⋅ C(n, r) = n(n+1) ⋅ 2^n-2

Substitute back:

α = 4n(n+1) ⋅ 2^n-2 + 2n ⋅ 2^n-1 + 2ⁿ

α = 2^n-2(4n² + 8n + 4)

α = 2ⁿ ⋅ (n+1)²

Step 2: Simplify β:

β = ∑_r=0ⁿ C(n, r) ⋅ 1/(r+1) + 1/(n+1)

Using properties of binomial coefficients:

β = 1/(n+1) ⋅ ∑_r=0ⁿ⁺¹ C(n+1, r)

β = 1/(n+1) ⋅ 2ⁿ⁺¹

Step 3: Compute 2α/β:

2α/β = (2 ⋅ 2ⁿ ⋅ (n+1)²) / (2ⁿ⁺¹ / (n+1))

2α/β = (n+1)³

Step 4: Solve the inequality:

140 < (n+1)³ < 281

Take cube roots:

∛140 < n+1 < ∛281

Approximate:

5 < n+1 < 6

n = 5

Solution:

1. Given Vectors:
⃗a = 9⁢ı - 13⁢ᵑ + 25⁢ᵒ
⃗b = 3⁢ı + 7⁢ᵑ - 13⁢ᵒ
⃗c = 17⁢ı - 2⁢ᵑ + ᵒ

2. Calculate ⃗b + ⃗c:
⃗b + ⃗c = (3 + 17)⁢ı + (7 - 2)⁢ᵑ + (-13 + 1)⁢ᵒ = 20⁢ı + 5⁢ᵑ - 12⁢ᵒ

3. Calculate ⃗b - ⃗c:
⃗b - ⃗c = (3 - 17)⁢ı + (7 - (-2))⁢ᵑ + (-13 - 1)⁢ᵒ = -14⁢ı + 9⁢ᵑ - 14⁢ᵒ

4. Condition ⃗r × ⃗a = (⃗b + ⃗c) × ⃗a:
From the cross product condition:
⃗r - (⃗b + ⃗c) = λ⃗a
Therefore:
⃗r = λ⃗a + (⃗b + ⃗c)

5. Condition ⃗r · (⃗b - ⃗c) = 0:
Substituting ⃗r = λ⃗a + (⃗b + ⃗c):
(λ⃗a + (⃗b + ⃗c)) · (⃗b - ⃗c) = 0
Expanding:
λ(⃗a · (⃗b - ⃗c)) + ((⃗b + ⃗c) · (⃗b - ⃗c)) = 0
Simplify:
λ(⃗a · ⃗b - ⃗a · ⃗c) + (⃗b · ⃗b - ⃗c · ⃗c) = 0

6. Dot Products:
⃗a · ⃗b = 9(3) + (-13)(7) + 25(-13) = 27 - 91 - 325 = -389
⃗a · ⃗c = 9(17) + (-13)(-2) + 25(1) = 153 + 26 + 25 = 204
⃗b · ⃗b = 3² + 7² + (-13)² = 9 + 49 + 169 = 227
⃗c · ⃗c = 17² + (-2)² + 1² = 289 + 4 + 1 = 294

7. Solve for λ:
Substituting:
λ(-389 - 204) + (227 - 294) = 0
Simplify:
λ(-593) - 67 = 0
λ = -67 / 593

8. Calculate ⃗r:
Substituting λ back:
⃗r = (-67 / 593)⃗a + (⃗b + ⃗c)

9. Expression 593⃗r + 67⃗a:
593⃗r + 67⃗a = 593(⃗b + ⃗c)
|593⃗r + 67⃗a|² = 593²|⃗b + ⃗c|²

10. Magnitude of ⃗b + ⃗c:
⃗b + ⃗c = 20⁢ı + 5⁢ᵑ - 12⁢ᵒ
|⃗b + ⃗c|² = 20² + 5² + (-12)² = 400 + 25 + 144 = 569

11. Final Result:
(|593⃗r + 67⃗a|²) / (593²) = |⃗b + ⃗c|² = 569

Correct Answer: (569)

Solution:

Step 1: Understanding the Function:

The function y = min{sin(x), cos(x)} represents the smaller value between sin(x) and cos(x) at any given x. The intersection points of sin(x) and cos(x) occur at:

x = ±π/4.

The intervals for y are as follows:

• From x = -π to x = -π/4, y = sin(x).
• From x = -π/4 to x = π/4, y = cos(x).
• From x = π/4 to x = π, y = sin(x).

Step 2: Total Area A:

The total area A is the sum of the absolute values of the integrals of min{sin(x), cos(x)} over the respective intervals:

A = ∫_-π^-π/4 sin(x) dx + ∫_-π/4^π/4 cos(x) dx + ∫_π/4^π sin(x) dx.

Step 3: Calculate Each Integral:

∫_-π^-π/4 sin(x) dx = [-cos(x)]_-π^-π/4 = -cos(-π/4) + cos(-π).

Result: 1 + 1/√2.

∫_-π/4^π/4 cos(x) dx = [sin(x)]_-π/4^π/4 = sin(π/4) - sin(-π/4).

Result: √2.

∫_π/4^π sin(x) dx = [-cos(x)]_π/4^π = -cos(π) + cos(π/4).

Result: 1 + 1/√2.

• For ∫_-π^-π/4 sin(x) dx:
• For ∫_-π/4^π/4 cos(x) dx:
• For ∫_π/4^π sin(x) dx:

Step 4: Combine the Results:

Add the absolute values of the integrals:

A = (1 + 1/√2) + √2 + (1 + 1/√2).

Combine terms:

A = 4.

Step 5: Final Result:

The square of the area is:

A² = 4² = 16.

Solution:

We are given:

&lim;_{x → 0} 2 × ^{1 - ∑_k=110 (1 - k2x2/2)}/_x²

Step 1: Expand the product:

Each term in the product is of the form (1 - k²x²/2). For small x, the product simplifies as:

(1 - x²/2)(1 - 4x²/2)(1 - 9x²/2)...(1 - 100x²/2).

Expanding to the first-order term:

1 - (x²/2 + 4x²/2 + 9x²/2 + ... + 100x²/2).

Step 2: Simplify the numerator:

The numerator becomes:

1 - (1 - x² × ¹/₂ ∑_k=1¹⁰ k²).

Simplify further:

x² × ¹/₂ ∑_k=1¹⁰ k².

Step 3: Sum of squares:

The sum of squares of the first 10 natural numbers is:

∑_k=1¹⁰ k² = ^n(n+1)(2n+1)/₆ = ^10(11)(21)/₆ = 385.

Step 4: Simplify the limit:

Substitute the sum into the limit:

&lim;_{x → 0} 2 × ^{(x2/2) × 385}/_x² = 2 × ³⁸⁵/₂ = 55.

Final Answer: 55