Given that Re((z − 2i)/(z + 2i)) = 0, the argument of (z − 2i) is 90° away from (z + 2i), implying z lies on a circle centered at −2i with radius equal to its imaginary part.

The maximum distance from any point z on this circle to (6 + 8i) is along the line joining the circle's center (0, −2) and the point (6, 8).

Using the distance formula, the maximum value of |z − (6 + 8i)| is calculated as:

d = √((6 − 0)² + (8 − (−2))²) = √(36 + 100) = 12.

Hence, the correct answer is (1) 12.

The ellipse equation is rewritten as:

x²/18 + y²/6 = 1.

The line y = x intersects the ellipse in the first quadrant. Substitute y = x into the ellipse equation:

x²/18 + x²/6 = 1.

Simplify to find x²(1/18 + 1/6) = 1, giving x² = 18/3 = 6, so x = √6.

Now integrate y = ±√(6 − x²/3) between 0 and √6 to calculate the enclosed area, accounting for symmetry:

Area = 1/4 × π × √3 × √6.

The result is √3π.

For the ellipse E, the semi-major axis (a) is 10, and the semi-minor axis (b) is √75. The eccentricity is:

e = √(1 − b²/a²) = √(1 − 75/100) = √(25/100) = 1/2.

The foci are at (1 ± ae, 1) = (1 ± 5, 1).

The hyperbola H has the same foci. Its eccentricity is e_H = 2 (reciprocal of the ellipse eccentricity).

Using e = c/a for the hyperbola, where c = 5, we have:

2 = 5/a ⇒ a = 5/2.

The conjugate axis length is determined using b² = c² − a²:

b² = 5² − (5/2)² = 25 − 25/4 = 100/4 = 25/2 ⇒ b = √(25/2).

Finally, substituting α = 2a and β = 2b into 3α² + 2β²:

3(2a)² + 2(2b)² = 3(5)² + 2(√50)² = 3(25) + 2(50) = 75 + 100 = 225.

Thus, the answer is (2) 225.

The orthocenter P(1, 1) lies at the intersection of the altitudes of triangle ABC.

Using the centroid property of a triangle (G = (A + B + C)/3), the centroid is:

G = ((3 + (−2) + α)/3, (−1 + 3 + β)/3).

The centroid-to-orthocenter vector relation gives:

H = 3G − 2P.

Substituting known values, we calculate α and β.

Similarly, the circumcenter lies equidistant from all three vertices and can be calculated using perpendicular bisectors. Substituting coordinates of A, B, and C, we find h and k.

Summing (α + β) and 2(h + k), the result is:

(3 + 2) + 2(−2 + 1) = 5.

Hence, the correct answer is (3) 5.

The variance formula for a frequency distribution is:

σ² = Σf(x − μ)²/N.

Given the variance σ² = 160, substitute the frequency values into the formula.

Let c be the missing value, and calculate the mean μ using:

μ = Σ(fx)/Σf.

Using trial and error for c, calculate:

Variance = Σf(x − μ)²/N = 160.

For c = 7, the calculations match the given variance.

Thus, the correct answer is (3) 7.

The range of the function f(x) is determined by finding the extrema of g(x) = sin(3x) + cos(3x):

g′(x) = 3cos(3x) − 3sin(3x).

Setting g′(x) = 0, solve:

tan(3x) = 1 ⇒ 3x = π/4, 5π/4, ...

Substitute back into g(x):

Max = √2, Min = −√2. For f(x), this gives range (1/(2 + √2), 1/(2 − √2)).

Given the interval [a, b], calculate A.M. = (a + b)/2 and G.M. = √(ab).

The ratio α/β = √2.

Hence, the correct answer is (1) √2.

For Statement-I, calculate a×b:

a×b = |î ĵ k̂|
| 1 2 −3|
| 2 1 −1| = (−1)î − (−5)ĵ + (−3)k̂ = −î + 5ĵ − 3k̂.

The condition a·r = 0 implies r lies in the plane orthogonal to a. Solving for magnitude of r reveals it is not √10.

For Statement-II, use the identity cos 2A + cos 2B + cos 2C = 1 − 4sinA sinB sinC ≥ −3/2 (since −1 ≤ sinX ≤ 1).

Thus, Statement-I is incorrect, and Statement-II is correct.

Start by evaluating the integral and numerator. For small t near π/2:

Approximate sin(2t^(1/3)) + cos(t^(1/3)) using Taylor series expansions:

sin(2t^(1/3)) ≈ 2t^(1/3), cos(t^(1/3)) ≈ 1 − (t^(1/3))²/2.

The integral becomes:

∫[(2t^(1/3) + 1 − (t^(1/3))²/2)] dt.

Substitute back into the numerator and calculate the derivative of the integral with respect to x:

Using L’Hopital’s rule for the limit:

lim x→π/2 ∫/ (x − π/2)² = (d²/dx²)[∫] at x = π/2.

After calculations, the result simplifies to 9π²/8.

The binomial expansion of (x^2/3 + 1/2x^−2/5)⁹ is:

(x^(2/3) + 1/2x^(−2/5))⁹ = Σ(k=0 to 9) C(9, k) * (x^(2/3))^(9−k) * (1/2x^(−2/5))^k.

For x^2/3, solve:

(9−k)(2/3) − k(2/5) = 2/3. Solve for k = 7. Substitute into the term to find coefficient.

For x^−2/5, solve:

(9−k)(2/3) − k(2/5) = −2/5. Solve for k = 6. Substitute to find coefficient.

Sum both coefficients to get 21/4.

Start with the condition AB⁻¹ = A⁻¹. From matrix algebra:

AB⁻¹ = A⁻¹ ⇒ A = BCB⁻¹.

The characteristic equation of C is:

C⁴ + αC² + βI = O.

Substitute B and find eigenvalues of C using determinant properties.

Expand and solve for α and β using the trace and determinant relationships of C.

The result is 2β − α = 10.

Differentiating logₑ y = 3 sin⁻¹ x gives:

y′/y = 3/(√(1−x²)).

Differentiating again:

y″/y − (y′/y)² = −3x/(1−x²)^(3/2).

Substituting y′ and y″ into the given equation:

(1−x)²y″ − xy′ = (1−1/2)²[−3(1/2)/(1−(1/4))^(3/2)] − (1/2)y′.

Simplify and calculate at x = 1/2 to get:

9eπ/2.

Use the substitution for cot⁻¹(√(1−x)/(1+x)):

t = cot⁻¹(√(1−x)/(1+x)), dt = −dx/(1+x²).

The integral becomes:

∫ cos(2t) (−1/(1+x²)) dt.

Expand cos(2t) as cos²(t) − sin²(t).

Simplify and integrate using trigonometric identities to get:

−1/4.

The sum of an infinite geometric progression is given by:

Σ arⁿ = a / (1−r) and Σ a³r³ⁿ = a³ / (1−r³).

From the first equation:

a / (1−r) = 57 ⇒ a = 57(1−r).

From the second equation:

a³ / (1−r³) = 9747.

Substitute a = 57(1−r) into the second equation:

[57(1−r)]³ / (1−r³) = 9747.

Simplify and solve for r and then a. Finally:

a + 18r = 31.

Each roll of a die is independent, and there are 6 outcomes per roll. For the i-th roll to be greater than the (i−1)-th roll:

The valid pairs are (1,2), (1,3), ..., (1,6); (2,3), ..., (2,6), and so on.

For two rolls, the number of valid outcomes is:

1 + 2 + 3 + 4 + 5 = 15.

For three rolls, calculate combinations for consecutive increases:

(1/6) × (15/36) = 5/54.

The probability is 5/54.

Use the substitution:

t = x + √(x² + 1), dt = (1/√(x² + 1)) dx.

The limits transform as:

x = −1 ⇒ t = 1 + √2; x = 2 ⇒ t = 2 + √5.

The integral becomes:

∫ logₑ(t) dt from t = (1 + √2) to t = (2 + √5).

Evaluate the integral as:

[t logₑ(t) − t] from (1 + √2) to (2 + √5).

After simplifications, the result is:

√2 − √5 + logₑ(9 + 4√5)/(1 + √2).

Using the recurrence relation for Pₙ = αⁿ − βⁿ:

Pₙ = √2Pₙ₋₁ + √3Pₙ₋₂.

Calculate P₁₀, P₁₁, and P₁₂ in terms of P₉ using this relation.

Substitute into the expression:

(11√3 − 10√2)P₁₀ + (11√2 + 10)P₁₁ − 11P₁₂.

Simplify using the relation, and the result becomes:

10√3P₉.

Calculate the vectors of the diagonals:

a + b = (1, α, 1), b + c = (−1, β, 0).

The cross product of these diagonals gives the area of the parallelogram:

|(α × β)| = √21/2.

Substitute the condition αβ = −6 and solve for integer pairs (α, β).

Possible values are (α₁, β₁) = (−3, 2), (α₂, β₂) = (2, −3).

Substitute into the final expression:

α₂₁ + β₂₁ − α₂β₂ = 19.

The equation of the circle is x² + y² = 4, and the equation of the parabola is y² = 8x.

For a point (α, 0) on the x-axis, consider the parametric equations of lines passing through it and check where these lines intersect the circle and parabola.

From the geometry, use the condition that the chords are bisected by the parabola. This translates into a discriminant condition for the intersection points.

Solve for α to find the interval (p, q). Calculate (2q − p)², which gives the result 80.

The critical points of f(x) occur where f'(x) = 0. Differentiate f(x):

f'(x) = (p² − 6p + 8)(2sin 4x) + 2(2 − p).

For no critical points, f'(x) ≠ 0 for all x. Analyze the discriminant condition for p² − 6p + 8.

This gives the interval (a, b) for p. Multiply 16ab to find the result 252.

The given equation is a first-order linear differential equation:

f'(x) − 3f(x) = α.

The integrating factor is e^−3x. Multiply through by the integrating factor and integrate:

f(x) = Ce^(3x) + α/3.

Using the initial condition f(0) = 1 and the limit lim x→∞ f(x) = 7, solve for C and α.

Substitute x = −log 3 to find f(−log 3). The result is 61.

Let the three-digit number be 100a + 10b + c, where a, b, c are digits.

The sum of the digits is a + b + c = 14, and 1 ≤ a ≤ 9, 0 ≤ b, c ≤ 9.

Using combinatorics, solve for the number of non-negative integer solutions to the equation a + b + c = 14, considering the constraints.

After counting valid cases, the total number is 70.

Substitute values for x, y in 2x + 3y = 23 to find all integer pairs in A.

For each valid x, there is exactly one y such that the equation holds. The number of pairs in A is the size of B.

The number of one-to-one functions from A to B is the factorial of the size of A:

n! = 24 for n = 4.

The parabola y² = 6x is parameterized as (3t², 6t) for a point on it.

Let A, B, and C correspond to parameters t₁, t₂, and t₃ respectively. The line segment AB has equation determined by its endpoints.

The line L through C, parallel to the x-axis, has equation y = y₃. Find the intersection point D of AB and L.

Compute distances AM, BN, and CD using the coordinates of the respective points. Simplify the ratio (AM · BN / CD)² to get 36.

The equation of the line is parameterized as (x, y, z) = (1+3t, 2t, 2+4t).

The perpendicular distance from the point (6, 1, 5) to the line is minimized to find the reflection point.

Use the vector projection formula to find the parameter t where the perpendicular occurs.

Substitute back into the line equation to get the image point. Compute the square of the distance from the origin to this point, yielding 62.

The first series is simplified using the harmonic number formula Hₙ = Σ (1/k).

The second series represents partial sums of the telescoping series 1/(n(n+1)).

Equate the two series after subtraction and solve for α. Simplifying the harmonic series difference gives α = 1011.

The domain of sin⁻¹x and cos⁻¹x is x ∈ [-1, 1].

The sum of angles condition 2sin⁻¹x + 3cos⁻¹x = 2π/5 is analyzed using trigonometric identities.

Check if the range of sin⁻¹x and cos⁻¹x allows the given sum. No real solutions satisfy the equation.

Write the system of equations AX = B and solve for X using the inverse of A.

Find the determinant |A| = 2m + 15, and ensure it is non-zero for invertibility.

Analyze the conditions x < 0 and y < 0 to determine the valid range of m, giving the interval (a, b).

Integrate |A| over this range and multiply by 8 to find the result 450.