The line L intersects the axes at A(5,0) and B(0,9). The lines l₁ and l₂ trisect AB. The points trisecting AB are P(10/3, 3) and Q(5/3, 6). m₁ = 9/10 and m₂ = 18/10 = 9/5. m₁ + m₂ = 9/2 The equation of the line passing through the origin with slope 9/2 is y = (9/2)x or 9x - 2y = 0. Solving the system of equations 9x + 5y = 45 and 9x - 2y = 0, we get y=45/7 and x=10/7. This point satisfies y - x = 5.

For A, B, C, and D to be coplanar, the scalar triple product of vectors AB, AC, and AD must be zero. AB = -4i - 3j + (3-2λ)k, AC = -7i + (x-5)j + (4-2λ)k, and AD = -6i + (6-2λ)k. The scalar triple product gives the equation -4λ² + 20λ - 24 = 0, which simplifies to λ² - 5λ + 6 = 0. Solving for λ gives λ=2 or λ=3. Therefore S={2,3}. Σ_λ∈S(λ+2)² = (2+2)² + (3+2)² = 16 + 25 = 41.

Integrate by parts with u = x² and dv = ^{(xsec2x + tanx)}⁄_{(xtanx + 1)²}. Then v = -1/(xtanx + 1). I(x) = ^-x2⁄_(xtanx+1) + ∫ ^2x⁄_{(xtanx + 1)} dx = ^-x2⁄_(xtanx+1) + 2ln|xsinx+cosx|. Since I(0) = 0, the constant of integration is 0. Substitute x = π/4 and evaluate.

The differences between consecutive terms form an arithmetic progression: 6, 8, 10, 12, ... The nth term of this progression is 2n+4. The nth term of the original series is given by T_n = n² + 3n + 1. The sum of the first 20 terms is Σ²⁰_n=1(n²+3n+1) = Σn² + 3Σn + Σ1 = 2870 + 630 + 20 = 3520.

The probability of getting a sum of 5 is 4/36 = 1/9. Let p = 1/9 and q = 8/9. The probability of at least 4 successes in 5 trials is given by P(X=4) + P(X=5) = ⁵C₄p⁴q¹ + ⁵C₅p⁵ = 5(1/9)⁴(8/9) + (1/9)⁵ = ⁴¹⁄_9⁵ = ¹²³⁄₃₁₁. Thus, k=123.

Let A = [[α, β], [γ, δ]]. Since A² = I, |A²|=|I| => |A|² = 1 => |A| = b = ±1. Also, A² = [[α² + βγ, αβ + βδ],[αγ + γδ, βγ + δ²]] = I. This leads to α + δ = 0, or a = 0. Therefore, 3a² + 4b² = 3(0)² + 4(±1)² = 4.

Rationalize each term in the summation: Σ^n-1_k=1 ^{(√a_k+1 - √a_k)}⁄_d = ^{(√a_n - √a₁)}⁄_d. Substituting a_n = a₁ + (n-1)d and simplifying the limit yields 1.

Using the combination formula, we get ^4(2n-1)⁄_(n-2) = 10. Solving gives n=8. Substitute n=8 into the ratio (n² + 3n) : (n² - 3n + 4) to get 88:44, which simplifies to 2:1.

For set A, consider the cases x ≥ -3, -4 ≤ x < -3, and x < -4. Solving the inequality in each case gives -5 ≤ x ≤ -2, so A = [-5, -2]. For set B, the summation is a geometric series. Simplifying the inequality gives x+[x] < 3. Let x=n+ε, where n is an integer and 0 < ε < 1. This gives -5 ≤ x ≤ -2, so B=[-5, -2]. Therefore A=B.

Let the edge parallel to the z-axis be through (3,0,0). The direction vector of OP is <3,4,5>. The direction vector of the edge is <0,0,1>. The shortest distance is given by the formula |(a2-a1)·(b1 x b2)|/|b1 x b2|. Here, a1=(0,0,0), b1=<3,4,5>, a2=(3,0,0) and b2=<0,0,1>. b1 x b2 = <4,-3,0>. |b1 x b2| = 5. (a2-a1)·(b1 x b2) = 12. So, the shortest distance is 12/5. If we consider the edge parallel to z-axis through (0,4,0), a2 becomes (0,4,0), and the shortest distance becomes 12/√5.

The equation of the plane passing through the intersection of the two given planes is (2x - y + z - 3) + λ(4x - 3y + 5z + 9) = 0. This plane is parallel to the line with direction vector <-2, 4, 5>. The normal vector of the plane is <2+4λ, -1-3λ, 1+5λ>. Since the plane is parallel to the line, the normal vector is perpendicular to the direction vector of the line. Therefore, -2(2+4λ) + 4(-1-3λ) + 5(1+5λ) = 0, which gives λ=3/5. Substituting this value of λ into the equation of the plane and comparing it to ax + by + cz + 6 = 0 gives a=11, b=-7, and c=10. Thus, a+b+c = 14.

The fifth term from the beginning is ⁿC₄(√2)^n-4(1/√3)⁴. The fifth term from the end is ⁿC_n-4(√2)⁴(1/√3)^n-4. The ratio of these terms is given as √6. Simplifying and solving for n, we get n=10. The third term from the beginning is ¹⁰C₂(√2)⁸(1/√3)² = 60√3.

We consider two cases: x² - 8x + 15 ≥ 0 and x² - 8x + 15 < 0. In the first case, x² - 10x + 22 = 0, which gives x = 5±√3. Only 5+√3 satisfies the condition. In the second case, -x² + 6x - 8 = 0, which gives x=2 or x=4. Only x=4 satisfies the condition. The sum of the roots is 4 + (5+√3) = 9 + √3.

Let BQ = y. tan60° = AB/BQ => y = 10√3. Let AP = x. tan15° = x/y => x = 15(2-√3). PQ = 30-x = 10(3√3-3). The area of the trapezium BCPQ is (1/2)(BC+PQ) * BQ = (1/2)(15 + 10(3√3-3)) * 10√3 = 600(√3-1).

Since d is perpendicular to both b and c, d is parallel to b x c. b x c = 2i - j + 2k. Let d = λ(2i - j + 2k). Since a·d = 18, we have 2(2λ) + 3(-λ) + 4(2λ) = 18 => λ=2. So, d = 4i - 2j + 4k. |a|² = 29, |d|² = 36. |a x d|² = |a|²|d|² - (a·d)² = 29*36 - 18² = 720.

Differentiate both sides with respect to x using logarithmic differentiation: 2^xy(ylnx + x(1/y)^dy⁄_dx) + 3^yx(lny*y^x+y^x*xlny(1/y)^dy⁄_dx) = 0. Substituting x=2 and y=2 and simplifying gives ^dy⁄_dx = ^{-(8 + 12ln2)}⁄_{(12 + 8ln2)} = ^-(2+3ln2)⁄_(3+2ln2) = ^-(2+ln8)⁄_(3+ln4).

For infinitely many solutions, the determinant of the coefficient matrix and the determinants of the matrices obtained by replacing each column with the constants must be zero. The coefficient matrix determinant is 7-a. Setting this to zero gives a=7. The determinant Δ_x (replacing the x column with the constants) is 11b-33. Setting this to zero gives b=3. Therefore, 2a+3b = 2(7)+3(3) = 23.

(P⇒Q)∧(R⇒Q) is equivalent to (¬P∨Q)∧(¬R∨Q), which can be written as (¬P∧¬R)∨Q. Using De Morgan's Law, this becomes ¬(P∨R)∨Q, which is equivalent to (P∨R)⇒Q.

Substitute 1/x for x to get another equation: 5f(¹⁄_x) + 4f(x) = x + 3. Solve this system of equations to get f(x) = ^{(5 - 4x)}⁄_(9x) + ¹⁄₃. Integrate 18f(x) from 1 to 2 to get 10loge2 - 6.

The combined variance is given by σ²_c = ^{(n₁σ2₁ + n₂σ2₂)}⁄_(n1+n2) + ^{(n₁n₂(m₁-m₂)2)}⁄_(n1+n2)². Substituting n₁=n₂=15, m₁=12, m₂=14, σ²₁=14, and σ²_c = 13, and solving for σ²₂ = σ², we get σ²=10.

The given curve can be rewritten as (x+1)² = 4(y-2). The equation of the tangent at P(1,3) is x - y + 2 = 0, so A is (0,2). The line through P parallel to x - 3y = 6 is x - 3y = -8. Substituting x = (3y-8) into y²=4x, we get y² - 12y + 32 = 0, so y=4 or y=8. Thus the intersection points are (4,4) and (16,8). Since B lies on 2x-3y=8, B must be (16,8). Then (AB)² = (16-0)² + (8-2)² = 292.

For the point (p,p+1) to lie inside E, we must have 3-p ≤ p+1 ≤ √(9-p²). The first inequality gives p ≥ 1. The second inequality gives p² + p - 4 ≤ 0, which means ^(-1-√17)⁄₂ ≤ p ≤ ^(-1+√17)⁄₂. Combining these gives 1 ≤ p < ^(-1+√17)⁄₂. Therefore, a=1 and b= ^(-1+√17)⁄₂. b² + b - a² = ^{(1-17+2√17+4)}⁄₄ + ^(-1+√17)⁄₂ - 1 = 3.

Rewrite the differential equation as ^dy⁄_dx + (^{(xsinx + cosx)}⁄_(xcosx))y = ¹⁄_(xcosx). The integrating factor is xsecx. Multiplying the equation by the integrating factor gives d(xysecx)/dx = sec²x, so xysecx = tanx + C. Using y(π/3) = √3, we get C = √3. Thus y = ^{(sinx + √3cosx)}⁄_x. Differentiating twice and substituting x=π/6 gives |y"(π/6) + 2y'(π/6)| = 2.

Since 0 < x < π, we have 0 < sinx ≤ 1, and thus 0 < 13sinx ≤ 13. The greatest integer function is discontinuous (and hence not differentiable) at integer values. Thus, f(x) is not differentiable when 13sinx is an integer from 1 to 13. The equation 13sinx = k has two solutions for k=1,2,...,12 and one solution for k=13 in (0,π). Thus there are 2*12 + 1 = 25 points.

The region S is defined by the inequalities 2y-y² ≤ x ≤ 2y and x ≥ y. This region lies between the parabola x = 2y-y² and the lines x=2y and x=y. The area of the required region can be calculated by integrating from y=0 to y=1 (between the parabola and x=2y) and subtracting the area from y=0 to y=1 between x=y and x=2y. This gives ∫¹₀ (2y - (2y-y²))dy = ¹⁄₃. The area of the triangle formed by x=y, x=2y, and y=1 is 1/2. Therefore the required area is 1/2 - 1/3 = 1/6. Equating this to ^(1-1⁄_n)π⁄₄ gives n=5.

The total number of ways to distribute 20 oranges among 3 children is 3²⁰. The number of ways where at least one child gets no oranges is ³C₁ * 2²⁰ - ³C₂ * 1²⁰. The number of ways where each child gets at least one orange is 3²⁰ - (3*2²⁰ - 3) = 3483638676.

Let Q be (α, β, γ). The midpoint of PQ, M(^(α+1)⁄₂, ^(β+2)⁄₂, ^(γ+3)⁄₂), lies on the plane. The direction ratios of PQ are proportional to the normal of the plane, so ^(α-1)⁄₂ = ^(β-2)⁄_-1 = ^(γ-3)⁄₁ = k. This gives α=2k+1, β=-k+2, γ=k+3. Substituting in the equation of the plane gives k=2, so Q is (5,0,5). Now, PQ = <4,-2,2> and PR = <5,8,4>. The area of triangle PQR is (1/2)|PQ x PR|. PQ x PR = <-24, -6, 42>. The magnitude of this cross product is 2√594. Thus, the square of the area is 594.

The equation of a circle touching both axes is (x-a)² + (y-a)² = a². Since P(α,β) lies on the circle, (α-a)² + (β-a)² = a², which simplifies to α² + β² - 2a(α+β) + a² = 0. The equation of line AB is x+y=a. The distance PQ = |α + β - a|/√2 = 11. Thus, (α + β - a)² = 242. Substituting this into the simplified circle equation gives 2αβ=242, so αβ=121.

The general term in the expansion of (1+x²+2x)¹⁵ is given by ¹⁵C_r1,r2,r3 (1)^r₁(x²)^r₂(2x)^r₃, where r₁ + r₂ + r₃ = 15. To find the coefficient of x¹⁸, we need 2r₂ + r₃ = 18. Since r₁, r₂, r₃ are non-negative integers, we can have r₂=9, r₃=0, r₁=6; r₂=8, r₃=2, r₁=5; ... ; r₂=0, r₃=18, r₁=-3 (not possible). Since r₁+r₂+r₃ = 15, when r₃=18, r₂ must be negative or zero since r₁ is nonnegative. If r₂=0, r₁ = -3 which is not allowed. Therefore, the maximum value of r₃ is 2r₂+r₃=18 along with r₁+r₂+r₃ = 15, from which r₁ = 15-18 = -3, so not allowed. Only one possible solution is r₁ = 6, r₂=6, and r₃=3. If r₃=0, then 2r₂=18 => r₂=9 and r₁=6. Term = ¹⁵C₆*⁹C₀ = 5005. This is the only case.

Let k = a-b. Then we need to find the number of pairs (a,b) such that 2k²+3k ∈ B. Since a,b ∈ A, -9 ≤ k ≤ 9. If k=0, 2k²+3k = 0 ∈ B. This gives a=b. Since there are 10 possible values for a (and hence b), there are 10 pairs. If k=-2, 2k²+3k = 2 ∈ B. This gives a=b-2. Since 1 ≤ a ≤ 10 and 1 ≤ b ≤ 10, we have 3 ≤ b ≤ 10, which gives 8 possible values for b, and hence 8 pairs. No other values of k give a result in B. Therefore, the total number of elements in R is 10 + 8 = 18.