Step 1: Recall the dot product formula for unit vectors.

The dot product of two unit vectors ^ˆa and ^ˆb is given by:

ˆa · ˆb = |ˆa||ˆb| cos θ,

ˆa · ˆb = cos θ.

Step 2: Use the given sine value and Pythagorean identity.

We are given sin θ = ³/₅. Using the Pythagorean identity:

sin²θ + cos²θ = 1,

(3/5)² + cos²θ = 1,
9/25 + cos²θ = 1,
cos²θ = 1 − 9/25 = 16/25.

Step 3: Solve for cos θ.

cos θ = ±√(16/25) = ±4/5.

Step 4: Conclusion.

The value of ^ˆa · ^ˆb is:

ˆa · ˆb = ±4/5.

Hence, the correct answer is (C) ±⁴/₅.

Step 1: Rewrite the given equation in standard linear form.

The given equation is:

x (dy/dx) − y = x4 − 3x.

Divide through by x to express the equation in standard linear form:

dy/dx − (1/x)y = x3 − 3.

Step 2: Identify the coefficient of y.

In standard form, the coefficient of y is −1/x. The integrating factor (IF) is given by:

IF = e∫P(x) dx,

Step 3: Calculate the integrating factor.

Integrate −1/x:

∫(−1/x) dx = −ln|x|.

Thus, the integrating factor is:

IF = e−ln|x|.

Simplify using the property e^ln(a) = a:

IF = |x|−1.

For positive x, this simplifies to:

IF = x−1.

Conclusion: The integrating factor is (C) x⁻¹.

Step 1: Recall the property of direction cosines.

For direction cosines l, m, n, the following property holds:

l² + m² + n² = 1.

Here, l = √3k, m = √3k, and n = √3k.

Step 2: Substitute the values of l, m, and n into the equation.

(√3k)² + (√3k)² + (√3k)² = 1.

Simplify:

3k² + 3k² + 3k² = 1
9k² = 1.

Step 3: Solve for k².

k² = 1/9.

Taking the square root:

k = ±1/3.

Conclusion: The value of k is (D) ±1/3.

Explanation:

Linear programming is a method for optimizing a linear objective function subject to linear constraints. The objective function and constraints take the general form:

Optimize: Z = c₁x₁ + c₂x₂ + ... + cₙxₙ
Subject to: a₁₁x₁ + a₁₂x₂ + ... + a₁ₙxₙ ≤ b₁
            a₂₁x₁ + a₂₂x₂ + ... + a₂ₙxₙ ≤ b₂, ...

Conclusion: Linear programming involves optimizing a linear function. The correct answer is (B) linear function.

Step 1: Write the definition of conditional probability.

The conditional probabilities are:

P(A | B) = P(A ∩ B) / P(B),
P(A′ | B) = P(A′ ∩ B) / P(B).

Since P(A | B) = P(A′ | B), we can equate:

P(A ∩ B) / P(B) = P(A′ ∩ B) / P(B).

Cancel P(B) (as P(B) > 0):

P(A ∩ B) = P(A′ ∩ B).

Step 2: Use the property of probabilities.

We know:

P(A ∩ B) + P(A′ ∩ B) = P(B).

Substitute P(A ∩ B) = P(A′ ∩ B):

P(A ∩ B) + P(A ∩ B) = P(B),
2P(A ∩ B) = P(B).

Divide by 2:

P(A ∩ B) = 1/2 P(B).

Conclusion: The correct answer is (C) P(A ∩ B) = 1/2 P(B).

Step 1: Recall a property of determinants.

For a matrix, the sum of the product of elements from one row and the cofactors from another row is always zero. That is:

∑ aijAkj = 0 (for i ≠ k).

Step 2: Apply the property to the given matrix.

We are calculating:

a11A21 + a12A22 + a13A23.

Here, the first row elements (a₁₁, a₁₂, a₁₃) are multiplied by the second row cofactors (A₂₁, A₂₂, A₂₃). By the property of determinants:

a11A21 + a12A22 + a13A23 = 0.

Conclusion: The correct answer is (A) 0.

Step 1: Write the function and differentiate it.

The given function is:

 f(x) = sin(x2). 

Using the chain rule, the derivative is:

 f'(x) = cos(x2) × d/dx(x2). 

Since d/dx(x²) = 2x, the derivative becomes:

 f'(x) = cos(x2) × 2x. 

Step 2: Substitute x = √π.

At x = √π, the expression becomes:

 f'(√π) = cos((√π)2) × 2√π = cos(π) × 2√π. 

Step 3: Simplify the expression.

Since cos(π) = −1:

 f'(√π) = −1 × 2√π = −2√π. 

Conclusion: The derivative of sin(x²) at x = √π is −2√π.

Step 1: Define order and degree.

• The order of a differential equation is the highest derivative in the equation.
• The degree is the power of the highest-order derivative, provided the equation is free from radicals and fractional powers of derivatives.

Step 2: Analyze the given equation.

 [1 + (dy/dx)2]3 = d2y/dx2. 

• The highest-order derivative is d²y/dx², so the order is 2.
• The power of d²y/dx² is 1, so the degree is 1.

Conclusion: The order is 2, and the degree is 1.

Step 1: Formula for a vector between two points.

The vector from B to A is:

 A⃗ − B⃗ = (x2 − x1)î + (y2 − y1)ĵ + (z2 − z1)k̂. 

Step 2: Substitute the coordinates.

 A⃗ − B⃗ = (2 − 3)î + (−3 − (−4))ĵ + (5 − 7)k̂. 

Step 3: Simplify.

 A⃗ − B⃗ = −î + ĵ − 2k̂. 

Conclusion: The vector is −î + ĵ − 2k̂.

Step 1: Define the distance formula.

The distance of a point from the y-axis is the perpendicular distance to the plane x = 0, z = 0.

This is given by:

 Distance = √(a2 + c2). 

Step 2: Apply the formula.

For P(a, b, c):

 Distance from y-axis = √(a2 + c2). 

Conclusion: The distance is √(a² + c²).

Step 1: Analyze the constraints.

• x ≥ 0: The region lies to the right of the y-axis.
• y ≥ 0: The region lies above the x-axis.
• x + y ≥ 4: The region is above or on the line x + y = 4.

Step 2: Find the intersection points of the constraints.

• The line x + y = 4 intersects the x-axis at x = 4, y = 0.
• The line x + y = 4 intersects the y-axis at x = 0, y = 4.

Step 3: Identify the feasible region.

The feasible region is the part of the plane satisfying all constraints. It lies in the first quadrant and includes the line x + y ≥ 4.

Step 4: Count the corner points.

• The feasible region has two corner points: (4, 0) and (0, 4).

Conclusion: The number of corner points is 2.

Step 1: Expand the equation.

We are given:

(A + B)² = A² + B²

Expand the left-hand side using matrix multiplication:

(A + B)² = A² + AB + BA + B²

Step 2: Substitute into the given equation.

A² + AB + BA + B² = A² + B²

Step 3: Simplify.

Cancel A² and B² from both sides:

AB + BA = 0

Step 4: Rearrange the terms.

Rearrange to find the relationship between A and B:

AB = -BA

Conclusion: The matrices A and B are anti-commutative, satisfying AB = -BA.

The relation R = {(x, y) : x = y} is the equality relation, meaning x is equivalent to y only if x = y. The given set is:

A = {0, 1, 2, ..., 10}.

Step 1: Verify Equivalence Relation

• Reflexive: For all x ∈ A, (x, x) ∈ R. True, since x = x.
• Symmetric: If (x, y) ∈ R, then (y, x) ∈ R. True, since x = y implies y = x.
• Transitive: If (x, y) ∈ R and (y, z) ∈ R, then (x, z) ∈ R. True, since x = y and y = z imply x = z.

Therefore, R is an equivalence relation.

Step 2: Find Equivalence Classes

Each element of A forms its own equivalence class because x = y only holds for a single y. Hence, the equivalence classes are:

{0}, {1}, {2}, ..., {10}.

The total number of equivalence classes is equal to the number of elements in A, which is:

11.

Conclusion: The number of equivalence classes is 11.

The total number of elements in a matrix is given by the product of its number of rows and columns:

m × n = 36,

where m is the number of rows, and n is the number of columns.

Step 1: Find All Factor Pairs

To determine the possible orders of the matrix, find all pairs of positive integers (m, n) such that their product equals 36. These pairs are:

(1, 36), (2, 18), (3, 12), (4, 9), (6, 6), (9, 4), (12, 3), (18, 2), (36, 1).

There are 9 such pairs.

Conclusion: The number of possible orders is 9.

The function f(x) = ⌊x⌋ (greatest integer function) assigns the greatest integer less than or equal to x for any real number x.

Step 1: Behavior of ⌊x⌋

• The function ⌊x⌋ is constant in each open interval between consecutive integers, i.e., it takes the same value in n ≤ x < n + 1 for any integer n.
• However, at integer points x = n, the function has a discontinuity in its derivative because of the jump in its value.

Step 2: Points in the Interval 0 < x < 3

• The interval 0 < x < 3 contains the integers 1 and 2.
• At x = 1 and x = 2, the function f(x) = ⌊x⌋ is not differentiable due to the discontinuity in its derivative.

Conclusion: The function f(x) is not differentiable at x = 1 and x = 2. Therefore, the total number of non-differentiable points is 2.

Step 1: Strictly Increasing Function

A function f(x) is strictly increasing on an interval (a, b) if:

f'(x) > 0, ∀ x ∈ (a, b).

This means that the derivative of f(x) must be positive throughout the interval.

Step 2: Analysis of Options

• (A): f'(x) < 0, ∀ x ∈ (a, b) implies f(x) is strictly decreasing, not increasing. Hence, incorrect.
• (B): f'(x) > 0, ∀ x ∈ (a, b) correctly implies that f(x) is strictly increasing. Hence, correct.
• (C): f'(x) = 0, ∀ x ∈ (a, b) implies that f(x) is constant, not strictly increasing. Hence, incorrect.
• (D): f(x) > 0, ∀ x ∈ (a, b) does not guarantee that f(x) is strictly increasing because f(x) > 0 does not describe the behavior of the derivative. Hence, incorrect.

Conclusion: For f(x) to be strictly increasing on (a, b), the derivative f'(x) must satisfy:

f'(x) > 0, ∀ x ∈ (a, b).

Step 1: Equate Corresponding Elements

• From the first row, first column: x + y = 6.
• From the second row, second column: xy = 8.

Step 2: Expression to Evaluate

We need to find:

(24 / x) + (24 / y).

Simplify using the identity:

(a / x) + (a / y) = a · (x + y) / (xy).

Substitute known values: x + y = 6, xy = 8.

(24 / x) + (24 / y) = 24 · (6 / 8) = 24 · (3 / 4) = 18.

Conclusion: The value of (24 / x) + (24 / y) is 18.

Step 1: Standard Integral Form

The integral of 1 / √(a² + x²) is:

∫ 1 / √(a² + x²) dx = (1 / a) · tan⁻¹(x / a) + C.

Here, a² = 4, so a = 2. The integral becomes:

∫ 1 / √(4 + x²) dx = (1 / 2) · tan⁻¹(x / 2).

Step 2: Evaluate the Definite Integral

Substitute the limits 0 to a:

∫₀^a 1 / √(4 + x²) dx = (1 / 2) · [tan⁻¹(a / 2) - tan⁻¹(0)].

Since tan⁻¹(0) = 0, we have:

(1 / 2) · tan⁻¹(a / 2) = π / 6.

Step 3: Solve for a

Multiply through by 2:

tan⁻¹(a / 2) = π / 3.

Take the tangent on both sides:

a / 2 = tan(π / 3).

The value of tan(π / 3) is √3, so:

a / 2 = √3.

Multiply through by 2:

a = 2√3.

Conclusion: The value of a is 2√3.

Step 1: Analyze the Assertion

A line in three-dimensional space cannot be perpendicular to all three axes simultaneously. If a line is perpendicular to all three axes, the direction cosines cosα, cosβ, cosγ would all be zero, which would violate the fundamental relation of direction cosines:

cos²α + cos²β + cos²γ = 1.

Step 2: Analyze the Reason

The given equation cos²α + cos²β + cos²γ = 1 ensures that at least one of the direction cosines is non-zero, indicating that the line cannot be simultaneously perpendicular to x, y, and z axes.

Conclusion: Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Step 1: Compute the Determinant of A

Using cofactor expansion along the first row:

|A| = 1 × |[1   cosθ], [-cosθ   1]| - cosθ × |[-cosθ   cosθ], [-1   1]| + 1 × |[-cosθ   1], [-1   -cosθ]|.

• First minor: |[1   cosθ], [-cosθ   1]| = (1)(1) - (-cosθ)(cosθ) = 1 + cos²θ.
• Second minor: |[-cosθ   cosθ], [-1   1]| = (-cosθ)(1) - (cosθ)(-1) = 0.
• Third minor: |[-cosθ   1], [-1   -cosθ]| = (-cosθ)(-cosθ) - (1)(-1) = cos²θ + 1.

Substitute back:

|A| = 1 × (1 + cos²θ) - cosθ × 0 + 1 × (1 + cos²θ).

|A| = (1 + cos²θ) + (1 + cos²θ) = 2 + 2cos²θ.

Step 2: Determine Range of |A|

Since cosθ ∈ [-1, 1], cos²θ ∈ [0, 1].

Therefore, |A| = 2 + 2cos²θ ∈ [2, 4].

Step 3: Verify the Assertion and Reason

• Assertion (A): The determinant |A| lies in the interval [2, 4], so the assertion is true.
• Reason (R): The cosine function satisfies cosθ ∈ [-1, 1] for all θ ∈ [0, 2π], so the reason is true.

Conclusion: Both Assertion (A) and Reason (R) are true, and the Reason (R) correctly explains the Assertion (A).

Step 1: Substitution

Let u = 1 + 2x. Then, du = 2 dx and x = (u - 1) / 2.

Substitute into the integral:

I = ∫ [(u - 1) / 2] √u · (1 / 2) du = (1 / 4) ∫ (u - 1)u^1/2 du.

Step 2: Simplify the Integral

I = (1 / 4) ∫ (u^3/2 - u^1/2) du.

Split the integral:

I = (1 / 4) [∫ u^3/2 du - ∫ u^1/2 du].

Step 3: Integrate

∫ u^3/2 du = (2 / 5) u^5/2, and ∫ u^1/2 du = (2 / 3) u^3/2.

Substitute back:

I = (1 / 4) [(2 / 5) u^5/2 - (2 / 3) u^3/2].

Substitute u = 1 + 2x:

I = (1 / 10) (1 + 2x)^5/2 - (1 / 6) (1 + 2x)^3/2 + C.

Final Answer:

∫ x √(1 + 2x) dx = (1 / 10) (1 + 2x)^5/2 - (1 / 6) (1 + 2x)^3/2 + C.

Step 1: Substitution

Let t = √x. Then, x = t², dx = 2t dt, and √x = t.

Substitute into the integral:

I = ∫_π/4^π/2 (sin t / t) · 2t dt.

Simplify:

I = 2 ∫_π/4^π/2 sin t dt.

Step 2: Integrate

I = 2 [-cos t]_π/4^π/2.

Evaluate:

I = 2 [(-cos π/2) + cos(π/4)].

Since cos π/2 = 0 and cos(π/4) = 1, we have:

I = 2 [0 + 1] = 2.

Final Answer:

∫_π/4^π/2 (sin √x) / √x dx = 2.

Step 1: Analyze the given conditions

• (a→+ b→) ⊥a→ implies that the dot product (a→ + b→) ·a→ = 0.
• (2⃗a + ⃗b) ⊥ ⃗b implies that the dot product (2⃗a + ⃗b) · ⃗b = 0.

Step 2: Expand and simplify the dot products

From (a→ +b→) ·a→ = 0:

⃗a · ⃗a + ⃗b · ⃗a = 0.

Using |a→|² = a→ ·a→:

|a→|² +b→ · a→ = 0.

Thus:

⃗b · ⃗a = −|⃗a|². (1)

From (2⃗a + ⃗b) · ⃗b = 0:

(2⃗a) · ⃗b + ⃗b · ⃗b = 0.

Using |b→|² =b→ ·b→:

2(a→ ·b→) + |b→|² = 0.

Substitute a→ ·b→ = −|a→|² from equation (1):

2(−|a→|²) + |b→|² = 0.

Simplify:

−2|a→|² + |b→|² = 0.

Thus:

|b→|² = 2|a→|². (2)

Step 3: Solve for |b→|

Take the square root on both sides of equation (2):

|b→| = √2|a→|.

Conclusion:

It is proven that |b→| = √2|a→|.

Step 1: Calculate AD→

Using the relationship AD→= AB→+ DB→:

AD→ = (2î − 4ĵ + 5k̂) + (3î − 6ĵ + 2k̂).

Simplify:

AD→ = (2 + 3)î + (−4 − 6)ĵ + (5 + 2)k̂ = 5î − 10ĵ + 7k̂.

Step 2: Find the Area of the Parallelogram

The area is given by:

Area = |AB→ × AD→|.

Calculate the cross product:

AB⃗ × AD⃗ = | î   ĵ   k̂ |
             |  2   -4   5 |
             |  5  -10   7 |.
        

Expand the determinant:

AB→ × AD→ = î((-4)(7) - (5)(-10)) - ĵ((2)(7) - (5)(5)) + k̂((2)(-10) - (-4)(5)).

Simplify:

• î = (−28 + 50) = 22,
• ĵ = (14 − 25) = −11,
• k̂ = (−20 + 20) = 0.

Thus, AB⃗ × AD⃗ = 22î − 11ĵ + 0k̂.

Step 3: Magnitude of the Cross Product

The magnitude is:

|AB⃗ × AD⃗| = √(22² + (−11)² + 0²) = √(484 + 121) = √605.

Final Answer: The area of parallelogram ABCD is √605.

Step 1: Square Both Sides

Start with:

y = √(cos x + y).

Square both sides:

y² = cos x + y.

Step 2: Differentiate Both Sides

Differentiating with respect to x:

2y (dy/dx) = −sin x + dy/dx.

Step 3: Rearrange

Collect dy/dx terms:

2y (dy/dx) − dy/dx = −sin x.

Factorize:

dy/dx (2y − 1) = −sin x.

Step 4: Solve for dy/dx

dy/dx = −sin x / (2y − 1).

Since y = √(cos x + y), y is always positive. This simplifies to:

dy/dx = sin x / (1 − 2y).

Final Answer: dy/dx = sin x / (1 − 2y).

Step 1: Rewrite the Function

The function can be written as:

f(x) = {x³, if x ≥ 0; −x³, if x < 0}.

Step 2: Continuity

At x = 0, check the left-hand and right-hand limits:

• lim_x→0⁻ f(x) = (−x)³ = 0,
• lim_x→0⁺ f(x) = x³ = 0.

Since both limits and f(0) = 0, the function is continuous at x = 0.

Step 3: Differentiability

Differentiate f(x) for x > 0 and x < 0:

f'(x) = {3x², if x > 0; 3x², if x < 0}.

At x = 0, compute the left-hand and right-hand derivatives:

• f'(0⁻) = lim_x→0⁻ f(x) / x = lim_x→0⁻ (−x³ / x) = 0,
• f'(0⁺) = lim_x→0⁺ f(x) / x = lim_x→0⁺ (x³ / x) = 0.

Since f'(0⁻) = f'(0⁺) = 0, the derivative exists at x = 0.

Final Answer: The function f(x) = |x|³ is differentiable at all points of its domain.

Step 1: Partial Fraction Decomposition

Let:

x² / [(x² + 4)(x² + 9)] = A / (x² + 4) + B / (x² + 9).

Multiply through by (x² + 4)(x² + 9) to get:

x² = A(x² + 9) + B(x² + 4).

Simplify:

x² = (A + B)x² + (9A + 4B).

Equating coefficients:

• A + B = 1
• 9A + 4B = 0

From the first equation, B = 1 − A. Substitute into the second equation:

9A + 4(1 − A) = 0 → 9A + 4 − 4A = 0 → 5A = −4 → A = −4/5.

Substitute A = −4/5 into B = 1 − A:

B = 1 + 4/5 = 9/5.

Thus:

x² / [(x² + 4)(x² + 9)] = −(4/5) / (x² + 4) + (9/5) / (x² + 9).

Step 2: Integrate Each Term

The integral becomes:

∫ x² / [(x² + 4)(x² + 9)] dx = −(4/5) ∫ 1 / (x² + 4) dx + (9/5) ∫ 1 / (x² + 9) dx.

Using the standard formula ∫ 1 / (x² + a²) dx = (1/a) tan⁻¹(x/a):

• ∫ 1 / (x² + 4) dx = (1/2) tan⁻¹(x/2)
• ∫ 1 / (x² + 9) dx = (1/3) tan⁻¹(x/3)

Substitute back:

∫ x² / [(x² + 4)(x² + 9)] dx = −(4/5)(1/2) tan⁻¹(x/2) + (9/5)(1/3) tan⁻¹(x/3) + C.

Simplify:

∫ x² / [(x² + 4)(x² + 9)] dx = −(2/5) tan⁻¹(x/2) + (3/5) tan⁻¹(x/3) + C.

Final Answer:

∫ x² / [(x² + 4)(x² + 9)] dx = −(2/5) tan⁻¹(x/2) + (3/5) tan⁻¹(x/3) + C.

Step 1: Find the derivative of f(x)

The given function is:

f(x) = 12x^4/3 − 6x^1/3.

Differentiate with respect to x:

f'(x) = 12 · (4/3)x^1/3 − 6 · (1/3)x^−2/3.

Simplify:

f'(x) = 16x^1/3 − 2x^−2/3.

Step 2: Find critical points

Set f'(x) = 0:

16x^1/3 − 2x^−2/3 = 0.

Factorize:

2x^−2/3(8x − 1) = 0.

This gives:

• x^−2/3 = 0 (not valid as x ≠ 0).
• 8x − 1 = 0 → x = 1/8.

Step 3: Evaluate f(x) at critical points and endpoints

f(0) = 12(0)^4/3 − 6(0)^1/3 = 0.

f(1) = 12(1)^4/3 − 6(1)^1/3 = 12 − 6 = 6.

f(1/8) = 12(1/8)^4/3 − 6(1/8)^1/3.

Simplify:

(1/8)^1/3 = 1/2, (1/8)^4/3 = (1/2)⁴ = 1/16.

f(1/8) = 12(1/16) − 6(1/2) = 3/4 − 3 = −9/4.

• At x = 0:
• At x = 1:
• At x = 1/8:

Step 4: Determine the maximum and minimum values

• f(0) = 0
• f(1) = 6
• f(1/8) = −9/4

Conclusion:

• Absolute maximum value: 6 at x = 1.
• Absolute minimum value: −9/4 at x = 1/8.

Step 1: Analyze the Absolute Values

Break the interval [1, 3] at critical points x = 1, x = 2, and x = 3. The intervals are:

• [1, 2]
• [2, 3]

For each interval, rewrite the absolute values:

• For x ∈ [1, 2]: |x − 1| = x − 1, |x − 2| = 2 − x, |x − 3| = 3 − x.
• For x ∈ [2, 3]: |x − 1| = x − 1, |x − 2| = x − 2, |x − 3| = 3 − x.

Thus, the integrand becomes:

• For x ∈ [1, 2]: (x − 1) + (2 − x) + (3 − x) = 4 − x.
• For x ∈ [2, 3]: (x − 1) + (x − 2) + (3 − x) = x.

Step 2: Integrate Over Each Interval

∫₁² (4 − x) dx = [4x − (x² / 2)]₁².

Evaluate:

4(2) − (2² / 2) − [4(1) − (1² / 2)] = (8 − 2) − (4 − 0.5) = 6 − 3.5 = 2.5.

∫₂³ x dx = [(x² / 2)]₂³.

Evaluate:

(3² / 2) − (2² / 2) = (9 / 2) − (4 / 2) = 5 / 2.

• For x ∈ [1, 2]:
• For x ∈ [2, 3]:

Step 3: Add the Results

Total Integral:

∫₁³ (|x − 1| + |x − 2| + |x − 3|) dx = 2.5 + 2.5 = 5.

Final Answer:

∫₁³ (|x − 1| + |x − 2| + |x − 3|) dx = 5.

Step 1: Simplify the Equation

Rewrite by dividing numerator and denominator by x²:

dy/dx = (1 + (y/x)²) / (2(y/x)).

Let v = y/x, so y = vx and dy/dx = v + x(dv/dx).

Substitute into the equation:

v + x(dv/dx) = (1 + v²) / (2v).

Step 2: Separate Variables

Rearrange to isolate dv/dx:

x(dv/dx) = (1 + v²) / (2v) − v.

Simplify:

x(dv/dx) = (1 + v² − 2v²) / (2v) = (1 − v²) / (2v).

Separate variables:

2v / (1 − v²) dv = 1/x dx.

Step 3: Integrate Both Sides

∫ 2v / (1 − v²) dv = ∫ −du/u = −ln|u| = −ln|1 − v²|.

∫ 1/x dx = ln|x|.

• For the left-hand side, use substitution: u = 1 − v² → du = −2v dv.
• For the right-hand side:

Combine results:

−ln|1 − v²| = ln|x| + C.

Exponentiate both sides:

|1 − v²| = Kx, where K = e^−C is a constant.

Step 4: Substitute Back v = y/x

Replace v with y/x:

1 − (y/x)² = Kx.

Multiply through by x²:

x² − y² = Kx³.

Final Answer:

x² − y² = Kx³, where K is an arbitrary constant.

Step 1: Convert the inequalities into equations for graphing

• x + 2y = 4 → y = (4 − x) / 2
• 3x + y = 6 → y = 6 − 3x
• x + y = 4 → y = 4 − x

Step 2: Determine the feasible region

The feasible region is determined by the intersection of the half-planes defined by the inequalities:

• x + 2y ≥ 4: Region above the line y = (4 − x) / 2
• 3x + y ≤ 6: Region below the line y = 6 − 3x
• x + y ≤ 4: Region below the line y = 4 − x
• x, y ≥ 0: First quadrant

Graph these constraints and shade the common region satisfying all the inequalities.

Step 3: Identify the corner points of the feasible region

Find the intersection points of the boundary lines:

From x + 2y = 4 → y = (4 − x) / 2. Substitute into 3x + y = 6:

3x + (4 − x) / 2 = 6 → 6x + 4 − x = 12 → 5x + 4 = 12 → x = 8/5.

Substitute x = 8/5 into y = (4 − x) / 2:

y = (4 − 8/5) / 2 = (20/5 − 8/5) / 2 = 12/5 / 2 = 6/5.

Corner point: (8/5, 6/5).

From x + y = 4 → y = 4 − x. Substitute into x + 2y = 4:

x + 2(4 − x) = 4 → x + 8 − 2x = 4 → −x + 8 = 4 → x = 4.

Substitute x = 4 into y = 4 − x:

y = 4 − 4 = 0.

Corner point: (4, 0).

Subtract x + y = 4 from 3x + y = 6:

3x + y − (x + y) = 6 − 4 → 2x = 2 → x = 1.

Substitute x = 1 into x + y = 4:

y = 4 − 1 = 3.

Corner point: (1, 3).

• Intersection of x + 2y = 4 and 3x + y = 6:
• Intersection of x + 2y = 4 and x + y = 4:
• Intersection of 3x + y = 6 and x + y = 4:

Step 4: Evaluate z = 5x + 4y at the corner points

• At (8/5, 6/5): z = 5(8/5) + 4(6/5) = 8 + 24/5 = 40/5 + 24/5 = 64/5 = 12.8
• At (4, 0): z = 5(4) + 4(0) = 20
• At (1, 3): z = 5(1) + 4(3) = 5 + 12 = 17

Conclusion:

• Maximum value of z is 20 at (4, 0).
• Minimum value of z is 17 at (1, 3).

Step 1: Solve for the coefficients a and b

From the given conditions:

• f(1) = a(1) + b = 1 → a + b = 1 (1)
• f(2) = a(2) + b = 3 → 2a + b = 3 (2)

Solve equations (1) and (2) simultaneously:

From equation (1): b = 1 − a.

Substitute b = 1 − a into equation (2):

2a + (1 − a) = 3 → 2a + 1 − a = 3 → a = 2.

Substitute a = 2 into equation (1):

2 + b = 1 → b = −1.

Thus, the function is:

f(x) = 2x − 1.

Step 2: Check if f(x) is one-one

A function is one-one (injective) if distinct inputs lead to distinct outputs.

For f(x) = 2x − 1, the slope is 2 (non-zero), which implies it is a strictly increasing linear function.

Therefore, f(x) is one-one.

Step 3: Check if f(x) is onto

A function is onto (surjective) if for every y ∈ R, there exists x ∈ R such that f(x) = y.

For f(x) = 2x − 1, solve for x:

y = 2x − 1 → x = (y + 1) / 2.

Since x ∈ R for all y ∈ R, the function is onto.

Conclusion:

• The function is f(x) = 2x − 1.
• f(x) is both one-one and onto.

Step 1: Differentiate both sides

The given equation is:

√(1 − x²) + √(1 − y²) = a(x − y).

Differentiating both sides with respect to x:

(d/dx)√(1 − x²) + (d/dx)√(1 − y²) = (d/dx)[a(x − y)].

Using the chain rule:

−x / √(1 − x²) − y / √(1 − y²) · (dy/dx) = a(1 − dy/dx).

Step 2: Rearrange terms

Rearrange to isolate dy/dx:

−y / √(1 − y²) · (dy/dx) + a(dy/dx) = a − x / √(1 − x²).

Factorize dy/dx:

dy/dx [a − y / √(1 − y²)] = a − x / √(1 − x²).

Solve for dy/dx:

dy/dx = [a − x / √(1 − x²)] / [a − y / √(1 − y²)].

Step 3: Simplify for a = 1

If a = 1, the equation simplifies to:

dy/dx = √(1 − y²) / √(1 − x²).

Conclusion:

It is proved that dy/dx = √(1 − y²) / √(1 − x²).

Step 1: Take the natural logarithm

The given function is:

y = (tan x)^x.

Take the natural logarithm of both sides:

ln y = x ln(tan x).

Step 2: Differentiate both sides

Differentiating with respect to x:

(1 / y)(dy/dx) = ln(tan x) + x · (1 / tan x) · sec²x.

Step 3: Solve for dy/dx

Multiply through by y = (tan x)^x:

dy/dx = (tan x)^x[ln(tan x) + x(sec²x / tan x)].

Simplify further:

dy/dx = (tan x)^x[ln(tan x) + x csc x sec x].

Final Answer:

dy/dx = (tan x)^x[ln(tan x) + x csc x sec x].

Step 1: Simplify the integrand

Rewrite using trigonometric identities:

• 1 + cos x = 2cos²(x/2)
• 1 + sin x = 2cos²(π/4 − x/2)

Substitute these into the integrand:

(1 + sin x) / (1 + cos x) = cos²(π/4 − x/2) / cos²(x/2).

The integral becomes:

∫₀^π/2 e^x · [cos²(π/4 − x/2) / cos²(x/2)] dx.

Step 2: Solve the integral

This integral requires advanced techniques or numerical evaluation for simplification. Computational tools or approximation methods are suggested for practical evaluation.

Final Answer: Requires further computation using advanced methods or numerical tools.

Step 1: Simplify the integrand

Use the identity sin 2x = 2sin x cos x. Then:

√(sin 2x) = √(2sin x cos x).

Substitute into the integral:

∫_π/6^π/3 (sin x + cos x) / √(2sin x cos x) dx = (1/√2) ∫_π/6^π/3 (sin x + cos x) / √(sin x cos x) dx.

Step 2: Simplify further

Factorize sin x + cos x using the identity:

sin x + cos x = √2 sin(x + π/4).

Substitute into the integral:

I = (1/√2) ∫_π/6^π/3 √2 sin(x + π/4) / √(sin x cos x) dx = ∫_π/6^π/3 sin(x + π/4) / √(sin x cos x) dx.

Step 3: Solve the integral

Further substitutions or advanced techniques are required for exact evaluation. Approximation methods might be applied for practical computation.

Final Answer: Requires further computation using advanced methods or numerical tools.

Step 1: Solve for y

The equation of the ellipse is:

x²/16 + y²/4 = 1.

Rearrange to solve for y²:

y²/4 = 1 − x²/16 → y² = 4(1 − x²/16) = 4 − x²/4.

y = ±√(4 − x²/4).

Step 2: Use symmetry to simplify the calculation

The ellipse is symmetric about the x-axis. The area between x = −2 and x = 2 can be calculated as twice the area above the x-axis:

Area = 2 ∫₋₂² √(4 − x²/4) dx.

Since the integrand is even, further simplify:

Area = 4 ∫₀² √(4 − x²/4) dx.

Step 3: Substitution

Let u = 4 − x²/4 → du = −(x/2) dx and x dx = −2 du.

When x = 0, u = 4; when x = 2, u = 4 − 4/4 = 3.

The integral becomes:

∫₀² √(4 − x²/4) dx = ∫₄³ √u (−2 du).

Change limits and simplify:

∫₀² √(4 − x²/4) dx = 2 ∫₃⁴ √u du.

Step 4: Evaluate the integral

The integral of √u is:

∫ √u du = (2/3)u^3/2.

Evaluate from u = 3 to u = 4:

2 ∫₃⁴ √u du = 2[(2/3)(4^3/2) − (2/3)(3^3/2)].

Simplify:

• 4^3/2 = 8
• 3^3/2 = √27

Area = 2[(16/3) − (2/3)√27] = (32/3) − (4/3)√27.

Final Answer:

Area = (32/3) − (4/3)√27.

Step 1: Parametric equations of AB and CD

From AB: (x + 1)/1 = (y − 2)/−2 = (z − 1)/2, let t₁ be the parameter:

x = −1 + t₁, y = 2 − 2t₁, z = 1 + 2t₁.

From CD: (x − 4)/1 = (y + 7)/−2 = (z − 8)/2, let t₂ be the parameter:

x = 4 + t₂, y = −7 − 2t₂, z = 8 + 2t₂.

Step 2: Coordinates of points B and D

From BC: (x − 1)/3 = (y + 2)/−5 = (z − 5)/3, let t₃ be the parameter:

x = 1 + 3t₃, y = −2 − 5t₃, z = 5 + 3t₃.

At t₃ = 1, the coordinates of point B are:

B: (4, −7, 8).

From DA: (x − 2)/3 = (y + 3)/−5 = (z − 4)/3, let t₄ be the parameter:

x = 2 + 3t₄, y = −3 − 5t₄, z = 4 + 3t₄.

At t₄ = 1, the coordinates of point D are:

D: (5, −8, 7).

Step 3: Equation of diagonal BD

The equation of a line passing through points B(4, −7, 8) and D(5, −8, 7) is given by:

(x − 4)/(5 − 4) = (y + 7)/(−8 + 7) = (z − 8)/(7 − 8).

Simplify:

(x − 4)/1 = (y + 7)/−1 = (z − 8)/−1.

Parametrize:

x = 4 + t, y = −7 − t, z = 8 − t.

Final Answer: The equation of diagonal BD is:

(x − 4)/1 = (y + 7)/−1 = (z − 8)/−1.

Step 1: Represent the system in matrix form

The system can be written as:

A ·

        [ x ]
        [ y ]
        [ z ]

        [ 10 ]
        [  8 ]
        [  7 ]

where:

A =

        [  1 −2  0 ]
        [  2 −1 −1 ]
        [  0 −2  1 ]

Step 2: Find A⁻¹

(a) Compute det(A):

Using cofactor expansion, det(A) = 1.

(b) Compute adj(A):

adj(A) =

        [ −3  1  4 ]
        [  2  1  2 ]
        [  4 −2  5 ]

(c) Compute A⁻¹:

A⁻¹ = adj(A).

A⁻¹ =

        [ −3  1  4 ]
        [  2  1  2 ]
        [  4 −2  5 ]

Step 3: Solve for [x, y, z]

Compute:

        [x]
        [y]
        [z]
        = A⁻¹ · 
        [ 10 ]
        [  8 ]
        [  7 ]

The solution is:

[x, y, z] = [6, 42, 59].

Final Answer: x = 6, y = 42, z = 59.

Given Information:

• P(Severe) = P(Moderate) = P(Light) = 1/3.
• P(Late | Severe) = 0.55, P(Late | Moderate) = 0.37, P(Late | Light) = 0.17.

Part (i): Find the probability that an airplane reached its destination late.

Using the law of total probability:

P(Late) = P(Late | Severe)P(Severe) + P(Late | Moderate)P(Moderate) + P(Late | Light)P(Light).

Substitute the values:

P(Late) = (0.55 × 1/3) + (0.37 × 1/3) + (0.17 × 1/3).

Calculate:

P(Late) = (0.55 + 0.37 + 0.17) / 3 = 1.09 / 3 = 0.3633 (approximately).

Part (ii): Find the probability that the airplane was late due to moderate turbulence.

Using Bayes' theorem:

P(Moderate | Late) = P(Late | Moderate)P(Moderate) / P(Late).

Substitute the values:

P(Moderate | Late) = (0.37 × 1/3) / 0.3633.

Calculate:

P(Moderate | Late) = 0.37 / (3 × 0.3633) = 0.37 / 1.09 = 0.3394 (approximately).

Final Answers:

1. The probability that an airplane reached its destination late is P(Late) = 0.3633.
2. The probability that the airplane was late due to moderate turbulence is P(Moderate | Late) = 0.3394.

Part (i): Example of an interval other than the principal value branch

The principal value branch for sin(x) is [−π/2, π/2]. Another interval where sin(x) is one-one and onto [−1, 1] is:

A = [π/2, 3π/2].

Part (ii): Compute sin⁻¹(−1/2) − sin⁻¹(1)

• sin⁻¹(−1/2): The angle θ in [−π/2, π/2] such that sin(θ) = −1/2 is θ = −π/6.
• sin⁻¹(1): The angle θ in [−π/2, π/2] such that sin(θ) = 1 is θ = π/2.

sin⁻¹(−1/2) − sin⁻¹(1) = −π/6 − π/2 = −π/6 − 3π/6 = −4π/6 = −2π/3.

Part (iii): Graph of sin⁻¹(x) from [−1, 1] to its principal value branch

The graph of y = sin⁻¹(x) is obtained by reflecting the graph of y = sin(x) (restricted to [−π/2, π/2]) across the line y = x. The graph is continuous and increases from −π/2 to π/2.

Part (iv): Domain and range of f(x) = 2sin⁻¹(1 − x)

−1 ≤ 1 − x ≤ 1 → −1 − 1 ≤ −x ≤ 1 − 1 → −2 ≤ −x ≤ 0 → 0 ≤ x ≤ 2.

[2(−π/2), 2(π/2)] = [−π, π].

• Domain: For sin⁻¹ to exist, 1 − x must lie in [−1, 1].
• Range: The range of sin⁻¹(x) is [−π/2, π/2]. Thus, the range of 2sin⁻¹(1 − x) is:

Final Answers:

• Interval other than principal value branch: [π/2, 3π/2].
• sin⁻¹(−1/2) − sin⁻¹(1) = −2π/3.
• Graph: Reflect y = sin(x) across y = x (restricted to [−π/2, π/2]).
• Domain: [0, 2]; Range: [−π, π].

Part (i): Express θ in terms of the height and x

From the right triangle formed, tan(θ) = opposite/adjacent = 5/x.

θ = tan⁻¹(5/x).

Part (ii): Find dθ/dx

Differentiate θ = tan⁻¹(5/x):

dθ/dx = (1 / (1 + (5/x)²)) · (d/dx)(5/x).

dθ/dx = (1 / (1 + 25/x²)) · (−5/x²).

dθ/dx = −5 / (x² + 25).

Part (iii-a): Rate of change of angle of elevation at x = 50 m

The rate of change of angle with respect to time is given by:

dθ/dt = dθ/dx · dx/dt.

Substitute dθ/dx = −5 / (x² + 25) and dx/dt = 20 m/s:

dθ/dt = (−5 / (50² + 25)) · 20 = (−5 / 2525) · 20 = −20 / 505 = −4 / 101 rad/s.

Part (iii-b): Speed of another car with dθ/dt = 3/101 rad/s

Using dθ/dt = dθ/dx · dx/dt, solve for dx/dt:

3/101 = (−5 / 2525) · v.

v = (3/101) · (2525/5) = (3 × 505) / 101 = 15 m/s.

Final Answers:

• θ = tan⁻¹(5/x).
• dθ/dx = −5 / (x² + 25).
• (a) dθ/dt = −4/101 rad/s.
• (b) Speed of the car is 15 m/s.