Split the numerator: 2n² + 3n + 4 = 2n(2n-1) + 8n + 8. The sum becomes ∑ ^2n(2n-1)⁄_(2n)! + 8∑ⁿ⁄_(2n)! + 4∑¹⁄_(2n)!. Use series expansions: ∑¹⁄_(2n-2)! = ^e+1⁄_e⁄₂, ∑¹⁄_(2n-1)! = ^e-1⁄_e⁄₂, and ∑¹⁄_(2n)! = ^e+1⁄_e-2⁄₂. Combining and simplifying gives ^13e⁄₄ + ⁵⁄₄.

Let tan^-1(^1-x⁄_1+x) = θ. Then 2θ = cos^-1(cos2θ). Since 0 < x < 1, 2θ ∈ (0, π). So 2θ = 2θ. This means x = tan(π/8) = √2 - 1 ≈ 0.414 < ¹⁄₂. Thus, n(S) = 1.

r x a = c x a implies (r - c) x a = 0, so r = c + λa. r · b = 0 implies (c + λa) · b = 0. Solving for λ gives λ = ²⁄₇. Then r = c + ²⁄₇a = ¹¹⁄₇i + ⁴⁄₇j - ¹¹⁄₇k. |r| = √(121+16+121)/49 = √2.

A can be written as a rotation matrix with angle 60°. Aⁿ corresponds to rotation by n*60°. A³⁰ is rotation by 1800°, which is equivalent to I. A²⁵ is rotation by 1500° which simplifies to A. Therefore, A³⁰ + A²⁵ - A = I + A - A = I.

n(A) = 15, n(B) = 9, n(C) = 9. (A ∪ B) ∩ C = (A ∩ C) ∪ (B ∩ C). A ∩ C: (1,2), (1,4), (1,6), (3,4), (3,6), (5,6) so n(A ∩ C) = 6. B and C are disjoint, so B ∩ C is empty. Therefore, n((A ∪ B) ∩ C) = 6.

Using logical equivalences, simplify each statement: (1) simplifies to ~p ∨ q. (2) simplifies to true (a tautology). (3) simplifies to ~p ∨ ~q ∨ ~q. (4) simplifies to p.

f(1)f(2) < 0 and f(2)f(3) < 0. f(x) = 2x² - 8x + k. f(1) = k-6, f(2) = k-8, f(3) = k-6. So, (k-6)(k-8)<0, which means k∈(6,8). Only integer is k=7.

f(2)+f(¹⁄₂)=3. f(-1) + f(-1) = 0 implies f(-1)=0. f(-1)+f(-1)=0. Adding these equations gives 2f(2) + 2f(¹⁄₂) + 2f(-1) = ⁹⁄₂ or f(2) = ⁹⁄₂.

Plane P: (2+λ)x + (3+2λ)y + (-1+3λ)z - (2+6λ) = 0. Perpendicular to 2x+y-z+1=0 means (2+λ)(2) + (3+2λ)(1) + (-1+3λ)(-1) = 0, so λ=-8. Then P: -6x-13y-25z+46 = 0. Distance d from (-7,1,1) is d = |42-13-25+46|/√(36+169+625) = 50/√830. d² = 2500/830 = ²⁵⁰⁄₈₃.

|^b+i⁄_1+ai| = 1 implies |b+i| = |1+ai| so b² + 1 = a² + 1. Since ab < 0, b = -a. |z-1| = |2z| gives |a+ib-1| = |2(a+ib)|, so (a-1)² + b² = 4(a²+b²). Substituting b = -a gives 6a²+2a-1 = 0. Solving for 'a', no option matches for 1 + [|a|].

f(x) = |x² - 5x + 6| - 3x + 2. Since x²-5x+6 = (x-2)(x-3), the absolute value changes sign at x=2 and x=3. In [-1,2], f(x) = x²-8x+8 and in [2,3], f(x) = -x²+2x-4. f(-1)=17, f(2)=-4, f(3)=-7. The maximum is 17 and minimum is -7, so the sum is 10.

R₁ implies A=B, which is clearly an equivalence relation (reflexive, symmetric, transitive). R₂ can be simplified to A=B using set algebra, which is also an equivalence relation.

The region is bounded by y=1, y=x², and y=⁸⁄_x. Split the integral into two parts: from x=1 to 2 (area under y=x² and above y=1), and from x=2 to 8 (area under y=⁸⁄_x and above y=1). The total area is ∫₁² (x²-1)dx + ∫₂⁸ (⁸⁄_x-1)dx = ⁷⁄₃ + (16ln2-6) = 16ln2 - ¹¹⁄₃.

Substitute y²=t. The equation becomes ^dt⁄_dx + ^t⁄_x = ¹⁄_x². Integrating factor is x. Solution is xy² = ln(2x). Comparing with αx = exp(x^βy^γ), we get α=2, β=1, γ=2. α+β-γ=1.

Using the substitution x → (^π⁄₂ - x) and adding the original integral, we get 2I = ∫_π/4^{π/2 (π/2)+8}⁄_2-cos2xdx = ∫₀^{π/4 π+16}⁄_2(2-cos2x)dx. Substitute tan x = t, so cos 2x = ^1-t2⁄_1+t² and dx = ^dt⁄_1+t². The integral becomes I = ^(π+16)⁄₄∫₀^1dt⁄_3t²+1. Substituting √3t=u gives I = ^(π+16)π⁄_12√3 - ^π⁄_√3 = ^π2⁄_6√3.

x_i = 9 + (i-1)d. Mean x̄ = 9+3d. Variance σ² = ^{∑(x_i - x̄)2}⁄_n = 4d². Standard deviation σ = 4 implies d=2. x̄ = 9+3(2) = 15. x₆ = 9 + 5(2) = 19. So x̄+x₆ = 34.

The determinant of the coefficient matrix is (α-1)²(α+2). If α=1, the system becomes x+y+z=1, x+y+z=1, x+y+z=β which has infinite solutions if β=1, and no solution if β ≠ 1. If α=-2, the system becomes -2x+y+z=1, x-2y+z=1, x+y-2z=β. This system is inconsistent if β=1. If α=2, the determinant is non-zero. If β=1, x=y=z=¹⁄₃. If β=-1, it is inconsistent. Thus (1) is incorrect.

Projection of a on b is (a·b)/|b| = (5-3-15)/√(1+9+25) = -13/√35. The direction is opposite to b. None of the given options match the correct projection value.

The hyperbola is ^x2⁄₁₂ - ^y2⁄₉ = 1. The line's slope is -³⁄₂. For the nearest point, the tangent to the hyperbola should be parallel to the line. Let the point be (√12 secθ, 3tanθ). Slope of tangent is 3√3 secθ/(4√4 tanθ) = ³⁄_2√3(¹⁄_sinθ). Equating slopes gives sinθ=¹⁄_√3. The point becomes (-6/√2, -3), so √2(y₀-x₀) = √2(-3+⁶⁄_√2) = -9.

y=x^x. y' = x^x(1+ln x). y″ = x^x(1+ln x)² + x^x-1. y'(2) = 4(1+ln 2). y″(2) = 4(1+ln 2)² + 2. y″(2) - 2y'(2) = 4(1+ln 2)² + 2 - 8(1+ln 2) = 4(ln 2)² - 2.

For divisibility by 6, the number must be divisible by 2 and 3. The last digit must be 4. The sum of digits must be divisible by 3. Cases: all digits same (444444 - 1 number); two distinct digits (using 4,5 or 4,9 - 10 each); three distinct digits (4,5,9 with last digit 4 - 5!/(2!2!) arrangements with 4 last = 30, and permutations like 444554, 444994, etc.). Total = 1 + 10 + 10 + 20 + 5 + 5 + 30 = 81.

Let x' = x-1, y' = y-3, z' = z-4. Then x'+y'+z'=13 where x',y',z' ≥ 0. Number of solutions is ^13+3-1C_3-1 = ¹⁵C₂ = 105.

a/e=8, ae=2 gives a=4, e=¹⁄₂, b²=12. Tangent equation: ^{x cosθ}⁄₄ + ^{y sinθ}⁄_2√3 = 1. Passes through (0, 4√3) means sinθ = ¹⁄₂, so θ=30°. P is (2√3, √3). Q is (4/√3, 0). PQ = √[(4/√3-2√3)² + 3] = √13/3. (3PQ)² = 39.

Rewrite the parabola equation as (y-2)² = 8(x+1). Focus is (1,2). x-intercept is 3 means the point is (3,0). Focal chord equation is y-2 = m(x-1). Substituting (3,0) gives m=-1, so y=-x+3. Length of focal chord is 4a where a=2 (from the rewritten equation), so the length is 16.

Using property ∫₀^a f(x) dx = ∫₀^a f(a-x) dx and simplifying the numerator using trigonometric identities gives 2I = ∫₀^π(1 + cos x cos 3x + cos² x + cos 3x cos³x ) dx. Using further trigonometric identities and integrating gives I = ^13π⁄₁₆. Therefore, k=13.

T₆ = ^mC₅ (2^1/5 log₂(10-3^x)/5)^m-5 (√2^(x-2)log₂3 / (√10-3^1/2))⁵ = 21. The binomial coefficients condition implies 2(^mC₁) = ^mC₀ + ^mC₂, so m=2 (rejected) or m=7. With m=7, the T₆ equation simplifies to (10-3^x)3^x = 9. Let y=3^x, then y²-10y+9=0 gives y=1 or y=9, so x=0 or x=2. Sum of squares is 0+4=4.

General term is T_r+1 = ²²C_r (x^1/2/a)^22-r (-α/x³)^r. For the term independent of x, (22-r)/2 - 3r = 0, so r=4. T₅ = ²²C₄ a^-18 (-α)⁴ = 7315. Solving for α gives α=1.

Common differences are 4, 3, and 5. LCM is 60. Common terms must be of the form 2 + 60n. Find the common terms in the given ranges: 47, 107, and 167. Sum is 321.

Direction vector of the line is <-3,1,2>. The plane's normal vector is parallel to this, so the plane is 3x-y-2z=k. Since it passes through (3,-2,5), k=1. α=3, β=-1, γ=-2. αβγ = 6.

Line equation is ^x+3⁄₅ = ^y+6⁄₁₀ = ^z-1⁄_-4 = λ. A point on the line is (5λ-3, 10λ-6, -4λ+1). Intersection with plane gives λ=-¹⁄₃, so C is (^-14⁄₃, ^-28⁄₃, ⁷⁄₃). The ratio k:1 is -λ:1, which is ¹⁄₃ : 1. Let a point D on the given line be (-μ+1, 2μ-4, 3μ-2). CD is perpendicular to <-1, 2, 3>, which gives μ=¹¹⁄₁₄. Direction ratios of CD are <-1, -26, 17>. |a+b+c| = |-1-26+17| = 10.