Step 1: Consider the scenario where the bag contains 4 white and 4 black balls (4W4B).

• Probability of drawing 2 white and 2 black balls: \( P(4W4B | 2W2B) = \frac{P(4W4B) \cdot P(2W2B|4W4B)}{\text{Total probability for all configurations}} \)

Step 2: Substitute probabilities for each case and calculate:

• Final Probability: \( \frac{1}{5} \cdot \frac{6 \cdot 6}{70} / (\text{Sum of all cases}) = 2/7 \).

Step 1: Rewrite the integral with substitution. Let 2x = t, so dx = 1/2 dt. Then, the limits become 0 to π/2.

Step 2: The integral becomes:

(1/4) ∫₀^π/2 t / (sin⁴(t) + cos⁴(t)) dt.

Step 3: Use symmetry to simplify:

The integral becomes:

(1/4) ∫₀^π/2 (π/2 - t) / (sin⁴(t) + cos⁴(t)) dt.

Step 4: Combine terms and simplify:

2I = π/8 ∫₀^π/2 1 / (sin⁴(t) + cos⁴(t)) dt.

Step 5: Solve using standard forms:

The result is I = √2π/32.

Step 1: Calculate det(A):

det(A) = (√2)(√2) - (1)(-1) = 2 + 1 = 3.

Step 2: Calculate det(C):

det(C) = (det(A))² * det(B) = 3² * 1 = 9.

Step 3: Calculate det(X):

det(X) = det(Aᵀ) * det(C²) * det(A).

Since det(Aᵀ) = det(A):

det(X) = 3 * 9² * 3 = 729.

Step 1: Use the identity tan(A + B) = (tan(A) + tan(B)) / (1 - tan(A) * tan(B)).

Step 2: Substitute the given values for tan(A) and tan(B):

tan(A + B) = [√(1 / (x² + x + 1)) + √(x / (x² + x + 1))] / [1 - √(1 / (x² + x + 1)) * √(x / (x² + x + 1))].

Step 3: Simplify the numerator:

tan(A + B) = [1 + √(x)] / √(x² + x + 1).

Step 4: Simplify the denominator:

1 - √(x / (x² + x + 1)) = (x² + x + 1 - x) / √(x² + x + 1) = √(x² + x + 1).

Step 5: Combine the terms:

tan(A + B) = √(x / (x² + x + 1)).

Step 6: Compare with tan(C):

tan(A + B) = tan(C), so A + B = C.

Step 1: Calculate the total number of ways to partition 5 into 4 parts:

• 5, 0, 0, 0 : 1 way
• 4, 1, 0, 0 : 5! / 4! = 5 ways
• 3, 2, 0, 0 : 5! / (3! * 2!) = 10 ways
• 2, 2, 0, 1 : 5! / (2! * 2! * 1!) = 15 ways
• 2, 1, 1, 1 : 5! / (2! * (1!)³) = 10 ways
• 3, 1, 1, 0 : 5! / (3! * 2!) = 10 ways

Step 2: Add all cases:

Total = 1 + 5 + 10 + 15 + 10 + 10 = 51 ways.

Step 1: Let z = x + iy, where x and y are real numbers. From |z - 1| = 1, we get:

(x - 1)² + y² = 1 ... (1)

From the second condition, (√2 - 1)(2x) + i(2y) = 2√2:

(√2 - 1)x + y = √2 ... (2)

Step 2: Solve equations (1) and (2):

x = 1 or x = -1/√2 ... (3)

For x = 1, y = 1, so z₁ = 1 + i.

For x = -1/√2, y = √2 - 1/√2, so z₂ = -1/√2 + i(√2 - 1/√2).

Step 3: Compute |z₁ - z₂|²:

|z₁ - z₂|² = [(1 - (-1/√2))² + (1 - (√2 - 1/√2))²].

After simplification, |z₁ - z₂|² = 2.

Step 1: Arrange the observations: 125, a, b, 170, 190, 210, 230. Median = 170.

Step 2: Mean deviation about the median is given:

|125 - 170| + |a - 170| + |b - 170| + 0 + |190 - 170| + |210 - 170| + |230 - 170| / 7 = 205/7.

Simplify:

45 + |a - 170| + |b - 170| + 0 + 20 + 40 + 60 = 205.

|a - 170| + |b - 170| = 40 ... (1).

Step 3: Calculate mean:

Mean = (125 + a + b + 170 + 190 + 210 + 230) / 7 = 175.

Step 4: Mean deviation about the mean:

|125 - 175| + |a - 175| + |b - 175| + |170 - 175| + |190 - 175| + |210 - 175| + |230 - 175| / 7.

After substituting a + b = 300:

Total deviation = 50 + 50 + 5 + 15 + 35 + 55 = 210.

Mean deviation = 210 / 7 = 30.

Step 1: Compute the cross product ⃗a × ⃗b:

Using the determinant method:

⃗a × ⃗b = |i j k| |-5 3 -3| | 1 2 -4|

Expand the determinant:

⃗a × ⃗b = i[(3)(-4) - (-3)(2)] - j[(-5)(-4) - (-3)(1)] + k[(-5)(2) - (3)(1)]

⃗a × ⃗b = -6⃗i - 17⃗j - 13⃗k

Step 2: Compute ⃗c · (-⃗i + ⃗j + ⃗k):

⃗c · (-⃗i + ⃗j + ⃗k) = (-6)(-1) + (-17)(1) + (-13)(1)

⃗c · (-⃗i + ⃗j + ⃗k) = 6 - 17 - 13 = -12

Step 1: Let t = (√3 + √2)^x. Then its reciprocal is:

(√3 - √2)^x = 1/t

Substitute these into the equation:

t + 1/t = 10

Step 2: Multiply through by t to form a quadratic equation:

t² - 10t + 1 = 0

Use the quadratic formula:

t = [10 ± √(100 - 4)] / 2 = (10 ± √96) / 2 = 5 ± 2√6

Step 3: Solve for x:

If (√3 + √2)^x = 5 + 2√6, then x = log_{√3 + √2}(5 + 2√6)

If (√3 + √2)^x = 5 - 2√6, then x = log_{√3 + √2}(5 - 2√6)

Step 4: Since there are two valid values of x, the total number of elements in S is 2.

Step 1: Solve \( y = 6 - x \), substitute into \( xy + 4y = 16 \).

• Obtain quadratic \( x^2 - 2x - 8 = 0 \), roots are \( x = 4, x = -2 \).

Step 2: Integrate between \( x = -2 \) to \( x = 4 \).

• Area = \( \int_{-2}^4 \frac{16}{x + 4} - (6 - x) \, dx = 30 - 32 \log 2 \).

Case 1: x > 0

Here, f(x) = log(x). Then g(f(x)) = g(log(x)) = log(x), since g(x) = x for x ≥ 0. However, log(x) > 0 for x > 1, so g(f(x)) = log(x) maps to (0, ∞).

Case 2: x ≤ 0

Here, f(x) = e^-x. Then g(f(x)) = g(e^-x) = e^-x, since g(x) = x for x ≥ 0. The range is again (0, ∞).

Thus, g ◦ f is neither one-one nor onto, as it does not cover the entire set of real numbers.

Express the first equation as a linear combination of the other two:

2x + 3y - z = k₁(x + αy + 3z) + k₂(3x - y + βz)

Expand and equate coefficients:

2 = k₁ + 3k₂

3 = k₁α - k₂

-1 = 3k₁ + βk₂

5 = -4k₁ + 7k₂

Solve these equations to find α = -70, β = -16/13, and 13αβ = 1120.

The eccentricities of the hyperbola and ellipse are:

For the hyperbola: e_h = √(1 + sin²θ)

For the ellipse: e_e = √(1 - sin²θ)

Given e_h = √7 × e_e, substitute the values:

√(1 + sin²θ) = √7 × √(1 - sin²θ)

Squaring both sides:

1 + sin²θ = 7(1 - sin²θ)

8 sin²θ = 6, so sin²θ = 3/4

Taking the positive root, sinθ = √3/2, so θ = π/3.

Substitute x + y = t. Then dt/dx = 1 + dy/dx.

From the given equation:

dt/dx = 2xt³ - xt - 1

Separate variables and integrate:

∫dt/t² = ∫dx

-1/t = x² + C

Using y(0) = 1, solve for the constant C and substitute t back to find y(√1/2).

Finally, compute √(1/2 + y(√1/2))² = 1/(2 - √e).

Continuity at x = 0:

Left-hand limit: lim_x→0⁻ f(x) = (a - b)/0² = undefined. Hence, a = 0, b = 0.

Right-hand limit: lim_x→0⁺ f(x) = 0² + c × 0 + 2 = 2.

Continuity at x = 1:

Left-hand limit: lim_x→1⁻ f(x) = 1² + c × 1 + 2 = 3 + c.

Right-hand limit: lim_x→1⁺ f(x) = 2 × 1 + 1 = 3.

Equating limits, c = 0.

f is not differentiable at x = 0 and x = 1. Thus, m = 2, a = 0, b = 0, c = 0. m + a + b + c = 2.

Step 1: Use the formula for the latus rectum

The latus rectum of an ellipse is given by L = (2b²) / a. Substitute L = √14:

(2b²) / a = √14 ⇒ 2b² = a√14 …(1)

Step 2: Use the eccentricity formula

The eccentricity of an ellipse is given by e = √(1 - (b² / a²)). Substitute e = √(1/2):

√(1 - (b² / a²)) = √(1/2).

Squaring both sides:

1 - (b² / a²) = 1/2.

(b² / a²) = 1/2 ⇒ b² = a² / 2 …(2).

Step 3: Solve for a and b

Substitute b² = a² / 2 into equation (1):

2(a² / 2) = a√14.

a² = a√14.

Divide by a (assuming a > 0):

a = √14.

Now substitute a = √14 into b² = a² / 2:

b² = (14 / 2) = 7.

Step 4: Verify the result

Using the eccentricity formula:

e² = 1 - (b² / a²) = 1 - (7 / 14) = 1/2.

Thus, the solution matches.

Final Answer: e² = 3/2

Step 1: Express terms in A.P.

• \(a = 3 + d\), \(b = 3 + 2d\), \(c = 3 + 3d\), where \(d\) is the common difference.

Step 2: Use the G.P. condition.

• \(\frac{a - 1}{3} = \frac{b + 1}{a - 1} = \frac{c + 9}{b + 1} = r\).
• From \((a - 1)/3 = r\): \(d + 2 = 3r\), so \(r = \frac{d + 2}{3}\).
• From \((b + 1)/(a - 1) = r\): \(4 + 2d = 3r^2\).

Step 3: Solve for \(d\).

• Substitute \(r = \frac{d + 2}{3}\) into \(4 + 2d = 3r^2\).
• \(4 + 2d = 3\left(\frac{d + 2}{3}\right)^2\), leading to \(d = 4\).

Step 4: Calculate the arithmetic mean.

• \(\text{Mean} = \frac{a + b + c}{3} = \frac{(3 + d) + (3 + 2d) + (3 + 3d)}{3} = 3 + 2d\).
• Substitute \(d = 4\): \(\text{Mean} = 3 + 2(4) = 11\).

Step 1: Subtract the circle equations to find the line of intersection.

• \(C_2 - C_1: -4x + 9 = -4 \Rightarrow x = \frac{13}{4}\).

Step 2: Substitute \(x = \frac{13}{4}\) into \(C_1: x^2 + y^2 = 4\).

• \(y^2 = 4 - \left(\frac{13}{4}\right)^2 = \frac{-105}{16}\), yielding \(y = \pm\frac{\sqrt{105}}{4}\).

Step 3: Transform the points and verify on the curve.

• Transformed points are \((38, 4\sqrt{105} - 20)\) and \((38, -4\sqrt{105} - 20)\).
• Substitute into \(6x^2 + y^2 = 42\) to verify.

Step 1: Solve for \(f(x)\).

• \(f(x) = \frac{x^2 - 2 - \frac{4}{x}}{5}\).

Step 2: Substitute \(f(x)\) into \(y\).

• \(y = 9x^2 \cdot \frac{x^2 - 2 - \frac{4}{x}}{5} = \frac{9x^4 - 18x^2 - 36x}{5}\).

Step 3: Differentiate \(y\) to find where it is strictly increasing.

• \(\frac{dy}{dx} = \frac{36x^3 - 36x - 36}{5} = \frac{36(x^3 - x - 1)}{5}\).
• Set \(\frac{dy}{dx} > 0\): \(x^3 - x - 1 > 0\).
• The solution is \(x \in (-\sqrt{\frac{1}{5}}, 0) \cup (\sqrt{\frac{1}{5}}, \infty)\).

Step 1: Identify the direction vectors and a point from each line.

• \( \mathbf{d_1} = \langle 2, 1, 1 \rangle \), \( \mathbf{d_2} = \langle 1, -2, 1 \rangle \).
• Point on the first line: \(P_1(\lambda, 2, 1)\).
• Point on the second line: \(P_2(\sqrt{\frac{1}{3}}, 1, 2)\).

Step 2: Find the vector between the points.

• \(\mathbf{b} = \langle \lambda - \sqrt{\frac{1}{3}}, 1, -1 \rangle\).

Step 3: Compute \(\mathbf{d_1} \times \mathbf{d_2}\).

• \(\mathbf{d_1} \times \mathbf{d_2} = \langle -3, -1, -5 \rangle\).

Step 4: Use the shortest distance formula.

• \(d = \frac{|\mathbf{b} \cdot (\mathbf{d_1} \times \mathbf{d_2})|}{|\mathbf{d_1} \times \mathbf{d_2}|}\).
• \(d = \frac{|-3\lambda + \sqrt{3} + 4|}{\sqrt{35}}\).
• Set \(d = 1\): \(|-3\lambda + \sqrt{3} + 4| = \sqrt{35}\).

Step 5: Solve for \(\lambda\).

• \(-3\lambda + \sqrt{3} + 4 = \sqrt{35}\) or \(-3\lambda + \sqrt{3} + 4 = -\sqrt{35}\).
• Simplify to find \(\lambda\): \(\lambda = 2\sqrt{3}\).

Step 1: Simplify the equation.

• \((t + 1)dx/dt = 2x + (t + 1)^4\).
• Divide through by \(t + 1\): \(dx/dt = \frac{2x}{t + 1} + (t + 1)^3\).
• Separate terms: \(\frac{dx}{2x} = \frac{dt}{t + 1} + \frac{(t + 1)^3}{2x}dt\).

Step 2: Integrate both sides.

• \(\int \frac{1}{2x} dx = \int \frac{1}{t + 1} dt + \int \frac{(t + 1)^3}{2x} dt\).
• The left-hand side integrates to \(\frac{1}{2} \ln|x|\).
• The first term on the right integrates to \(\ln|t + 1|\), and the second to \(\frac{(t + 1)^3}{6}\).

Step 3: Solve for \(x\).

• \(\ln|x| = 2\ln|t + 1| + \frac{2(t + 1)^3}{3} + C\).
• \(x = A(t + 1)^2 e^{\frac{2(t + 1)^3}{3}}\), where \(A = e^C\).

Step 4: Apply the initial condition \(x(0) = 2\).

• \(2 = A(1)^2 e^{0} \Rightarrow A = 2\).
• \(x = 2(t + 1)^2 e^{\frac{2(t + 1)^3}{3}}\).

Step 5: Calculate \(x(1)\).

• \(x(1) = 2(2)^2 e^{\frac{2(2)^3}{3}} = 8e^{\frac{16}{3}} \approx 14\).

Step 1: Fix \(z\) and solve for \(x + 2y = 42 - 3z\).

• The number of non-negative integer solutions to \(x + 2y = n\) is \(\lfloor n/2 \rfloor + 1\).

Step 2: Calculate for each \(z\).

• \(z = 0\): \(x + 2y = 42 \Rightarrow \lfloor 42/2 \rfloor + 1 = 22\).
• \(z = 1\): \(x + 2y = 39 \Rightarrow \lfloor 39/2 \rfloor + 1 = 20\).
• Continue for \(z = 2\) to \(z = 14\) and sum all solutions.

Step 3: Total number of solutions.

• Summing all cases: \(22 + 20 + 19 + \dots + 1 = 169\).

Step 1: Identify the general terms for each binomial.

For (1 + (1/x))⁶, the general term is T1 = (6 choose r) x^-r.

For (1 + x²)⁷, the general term is T2 = (7 choose s) x^2s.

For (1 - x³)⁸, the general term is T3 = (8 choose t)(-1)^tx^3t.

Step 2: Solve for r, s, t such that -r + 2s + 3t = 30.

Case: Let r = 6. Substitute r = 6 into the equation:

-6 + 2s + 3t = 30 ⇒ 2s + 3t = 36.

Possible values: s = 12, t = 8 satisfies the equation.

Step 3: Calculate the coefficient.

The coefficient is (6 choose 6) × (7 choose 12) × (8 choose 8) = 678.

Final Answer: |α| = 678.

Step 1: Find the common terms.

The first arithmetic progression (AP) is 3, 7, 11, ..., 403 with first term a1 = 3 and common difference d1 = 4.

The second arithmetic progression (AP) is 2, 5, 8, ..., 404 with first term a2 = 2 and common difference d2 = 3.

The LCM of d1 = 4 and d2 = 3 is 12. The first common term is 11. Subsequent terms form an AP with first term a = 11 and common difference d = 12.

Step 2: Find the number of common terms.

The nth term of an AP is given by Tn = a + (n - 1)d. For Tn = 403:

403 = 11 + (n - 1) × 12 ⇒ n = 33.

Step 3: Calculate the sum of the common terms.

The sum of the first n terms of an AP is given by S = n/2 [2a + (n - 1)d].

Substitute n = 33, a = 11, d = 12:

S33 = 33/2 [2 × 11 + (33 - 1) × 12] = 6699.

Final Answer: 6699.

Right-hand limit (R):

As x → 0⁺, let h → 0:

R = π/2 * lim_h→0(cos^-1(1 - h²) / h).

For small h, cos^-1(1 - h²) ≈ π/2.

Thus, R = (π/2) * (π/2) = π²/4.

Left-hand limit (L):

As x → 0^-, let h → 0:

L = lim_h→0(cos^-1(1 - h² + 2h) * sin^-1(h)).

For small h, sin^-1(h) ≈ h, and cos^-1(1 - h² + 2h) ≈ π/4.

Thus, L = π/4.

Final Calculation:

(32π²)(L² + R²) = 32π²((π²/16) + (π²/16)) = 18.

Final Answer: 18.

Step 1: Find the intersection of the circle and parabola.

From x² + y² = 3 and x² = 2y, substitute x² into the first equation:

y² + 2y - 3 = 0 ⇒ (y - 1)(y + 3) = 0.

Since y > 0, y = 1, and x = √2. Therefore, P(√2, 1).

Step 2: Find α for line L passing through P.

α = -√2(√2) - 1 = -1.

Step 3: Calculate area of triangle P Q1Q2.

The centers of the circles are Q1(0, 9) and Q2(0, -3). Using the determinant formula for the area of a triangle:

Area = (1/2) |√2(9 + 3)| = 6√2.

Square of the area = (6√2)² = 72.

Final Answer: 72.

Step 1: Analyze the given circles.

The first circle is centered at (-2, 3) with radius 1, and the second circle is centered at (1, -5) with radius 8. The intersection of these regions determines the feasible z values.

Step 2: Calculate z1 and z2.

The maximum and minimum values of |z - 3 + 2i| occur at specific boundary points of the intersection. Solving geometrically or algebraically gives:

z1 = (-2 - 1/√2, 3 + 1/√2), z2 = (-3/2, 5/2).

Step 3: Calculate |z1|² and |z2|².

|z1|² = (-2 - 1/√2)² + (3 + 1/√2)².

|z2|² = (-3/2)² + (5/2)².

|z1|² + |z2|² = 31 + 5√2.

Thus, α = 31 and β = 5. Therefore, α + β = 36.

Final Answer: 36.

Step 1: Apply King’s Property.

Using King’s property, transform the integral:

I = ∫ (8√2 cos(π/2 - x) dx) / ((1 + e^(sin(π/2 - x)))(1 + sin⁴(π/2 - x))).

Combine the symmetric property to get:

2I = ∫ (8√2 cos x dx) / (1 + sin⁴x).

Step 2: Substitution.

Let sin x = t, transforming the limits: 0 to π/2 becomes 0 to 1.

The integral becomes I = 4√2 ∫ (1 / (1 + t⁴)) dt.

Step 3: Split the integral.

Decompose the integral into simpler terms:

I = 4√2 ∫ (1 / (t² + 1)) dt - ∫ (1 / (t² + 2)) dt.

Step 4: Evaluate the integrals.

The results are straightforward:

∫ (1 / (t² + 1)) dt = π/4, and ∫ (1 / (t² + 2)) dt = 1/√2 log(2 + √2).

Thus, I = π√2 - 2 log(2 + √2).

Final Answer: α = 2, β = 2, and α² + β² = 8.

Step 1: Find direction vectors.

For L1, the direction vector is b = (1, -1, 1). For L2, the direction vector is d = (1, 1, -1).

Find the perpendicular direction vector using the cross product: b × d = (0, 2, 2).

Step 2: Solve for λ and μ.

Substitute the parametric equations into the shortest distance formula and solve for λ and μ:

λ = 3/2, μ = -3/2.

Step 3: Find points P and Q.

Substitute λ and μ into the parametric equations to get P = (5/2, 1/2, 9/2) and Q = (5/2, 7/2, 15/2).

Step 4: Calculate the midpoint of PQ.

The midpoint is (5/2, 2, 6).

Step 5: Find 2(α + β + γ).

2(α + β + γ) = 5 + 4 + 12 = 21.

Final Answer: 21.

Step 1: Count elements in R1.

The number of pairs (a, b) in R1 is given by summing the number of divisors for each b:

n(R1) = 20 + 10 + 6 + 5 + 4 + 3 + 3 + 2 + 2 + 2 + 1 + ... + 1 (10 times) = 66.

Step 2: Count elements in R1 ∩ R2.

The intersection R1 ∩ R2 consists of pairs where a = b, which are (1, 1), (2, 2), ..., (20, 20). Therefore, n(R1 ∩ R2) = 20.

Step 3: Find R1 - R2.

n(R1 - R2) = n(R1) - n(R1 ∩ R2) = 66 - 20 = 46.

Final Answer: 46.