Step 1: Understand the constraints. The constraints define the feasible region on a graph:

1. x + y ≤ 10
2. 3y - 2x ≤ 15
3. x ≤ 6
4. x, y ≥ 0

Step 2: Identify the vertices of the feasible region by graphing the constraints. The vertices are:

1. (0, 0), (0, 5), (6, 4), (6, 0)

Step 3: Evaluate z = x + y at each vertex:

1. At (0, 0), z = 0
2. At (0, 5), z = 5
3. At (6, 4), z = 10
4. At (6, 0), z = 6

Step 4: The maximum value of z is 10, which occurs at (6, 4).

Step 1: Identify the discriminant condition for real roots. For a quadratic equation ax² + bx + c = 0, the discriminant b² - 4ac must be non-negative.

Step 2: For the given equation:

1. a = cos(p) - 1, b = cos(p), c = sin(p)

The discriminant condition becomes:

(cos(p))² - 4(cos(p) - 1)(sin(p)) ≥ 0

Step 3: Analyze the inequality. For p in (0, π), cos(p) > 0 and sin(p) > 0, satisfying the condition. Other intervals do not satisfy this inequality.

Step 4: Hence, p is in the interval (0, π).

Step 1: Expand the matrix equation into a system of equations:

1. Equation 1: x - y + z = 4
2. Equation 2: 2x + y - 3z = 0
3. Equation 3: x + y + z = 2

Step 2: Solve for y in terms of x and z using Equation 3:

y = 2 - x - z

Step 3: Substitute y into Equation 1:

x - (2 - x - z) + z = 4

Simplify to find x + z = 3.

Step 4: Substitute y into Equation 2:

2x + (2 - x - z) - 3z = 0

Simplify to find x = 4z - 2.

Step 5: Substitute x = 4z - 2 into x + z = 3:

4z - 2 + z = 3

Simplify to find z = 1.

Step 6: Solve for x and y:

1. x = 2
2. y = -1

Step 7: Substitute into 2x + y - z:

2 + (-1) - 1 = 2.

Therefore, 2x + y - z = 2.

Step 1: Understand the problem. The plane passes through (1, 1, 1) and is perpendicular to the planes 2x + y - 2z = 5 and 3x - 6y - 2z = 7.

Step 2: The normal vectors of the given planes are:

1. [2, 1, -2]
2. [3, -6, -2]

Step 3: Find the cross product of the normal vectors:

[2, 1, -2] × [3, -6, -2] = [14, 2, 15]

Step 4: Equation of the plane:

14(x - 1) + 2(y - 1) + 15(z - 1) = 0

Simplify to get:

14x + 2y + 15z = 31.

To negate (p ∧ q) ∧ (q ∨ ¬r), apply De Morgan's laws:

1. Negate the entire expression: ¬((p ∧ q) ∧ (q ∨ ¬r))
2. Using De Morgan's law, this becomes: ¬(p ∧ q) ∨ ¬(q ∨ ¬r)
3. Further apply De Morgan's law:
   1. ¬(p ∧ q) = ¬p ∨ ¬q
   2. ¬(q ∨ ¬r) = ¬q ∧ r
4. Combine results: (¬p ∨ ¬q) ∨ (¬q ∧ r)

Finally, simplify to get: ¬q ∧ (¬p ∨ r).

Step 1: Calculate the sum of the five given marks.

The given marks are 45, 54, 41, 57, and 43. Their sum is:

45 + 54 + 41 + 57 + 43 = 240.

Step 2: Find the sixth test score using the mean.

The mean score for the six tests is 48. The sum of all six marks is:

Sum of six marks = 48 × 6 = 288.

Therefore, the sixth test score is:

x₆ = 288 - 240 = 48.

Step 3: Calculate the variance.

The marks are 45, 54, 41, 57, 43, and 48, with the mean μ = 48. The variance is calculated as:

Variance = (1/6) Σ(xᵢ - μ)²

Now calculate (xᵢ - 48)² for each mark:

• (45 - 48)² = 9
• (54 - 48)² = 36
• (41 - 48)² = 49
• (57 - 48)² = 81
• (43 - 48)² = 25
• (48 - 48)² = 0

Variance = (9 + 36 + 49 + 81 + 25 + 0) / 6 = 200 / 6 = 100/3.

Step 4: Calculate the standard deviation.

Standard Deviation = √(Variance) = √(100/3) = 10/√3.

Step 1: Calculate the number of ways to form the committee with at least 6 males.

• 6 males and 5 females: C(8, 6) × C(5, 5) = 28 × 1 = 28
• 7 males and 4 females: C(8, 7) × C(5, 4) = 8 × 5 = 40
• 8 males and 3 females: C(8, 8) × C(5, 3) = 1 × 10 = 10

Total m = 28 + 40 + 10 = 78.

Step 2: Calculate the number of ways to form the committee with at least 3 females.

• 8 males and 3 females: C(8, 8) × C(5, 3) = 1 × 10 = 10
• 7 males and 4 females: C(8, 7) × C(5, 4) = 8 × 5 = 40
• 6 males and 5 females: C(8, 6) × C(5, 5) = 28 × 1 = 28

Total n = 10 + 40 + 28 = 78.

Step 3: Conclusion.

Since both m and n are 78, the correct answer is m = n = 78.

Step 1: Substitution.

Let u = 1 + log(tan(x/2)).

Differentiating both sides: du/dx = (1 / tan(x/2)) * sec²(x/2) * (1/2).

Thus: du = (csc(x) / 2) dx → 2 du = (csc(x) / cos²(u)) dx.

Step 2: Substituting into the integral.

I = 2 ∫ sec²(u) du.

Step 3: Solving the integral.

The integral of sec²(u) is tan(u), so:

I = 2 tan(u) + C.

Step 4: Substitute back the value of u.

Substituting u = 1 + log(tan(x/2)) gives:

I = tan(1 + log(tan(x/2))) + C.

Step 1: Express the terms using basic trigonometric identities.

We know csc(θ) = 1 / sin(θ) and sec(θ) = 1 / cos(θ).

So: √3 csc(20°) - sec(20°) = √3 × (1 / sin(20°)) - (1 / cos(20°)).

Step 2: Approximate the values of sin(20°) and cos(20°).

Using known values: sin(20°) ≈ 0.3420, cos(20°) ≈ 0.9397.

Substitute these into the expression:

√3 × (1 / 0.3420) - (1 / 0.9397).

Step 3: Simplify the expression.

√3 ≈ 1.732.

(1.732 / 0.3420) ≈ 5.06, and (1 / 0.9397) ≈ 1.064.

So: 5.06 - 1.064 = 4.

Step 4: Conclusion.

The value of √3 csc(20°) - sec(20°) is approximately 4.

Step 1: Understand ionization energy.

Ionization energy is the energy required to remove one mole of electrons from one mole of atoms in the gaseous state. It increases across a period due to increasing nuclear charge and decreases down a group due to increasing atomic size and shielding effect.

Step 2: Compare the elements.

• Lithium (Li): Group 1, Period 2, low ionization energy.
• Sodium (Na): Group 1, Period 3, lower ionization energy than Li.
• Neon (Ne): Group 18, Period 2, noble gas with a stable configuration and the highest ionization energy in its period.
• Magnesium (Mg): Group 2, Period 3, lower ionization energy than Ne.

Step 3: Conclusion.

Neon (Ne) has the highest first ionization energy.

Step 1: Write the formula for a first-order reaction.

The fraction of reactant remaining is given by:

[A]_t / [A]₀ = e^-kt.

Here, [A]_t is the concentration at time t, [A]₀ is the initial concentration, k is the rate constant, and t is the time elapsed.

Step 2: Find the rate constant k.

The relationship between half-life and k for a first-order reaction is:

t_1/2 = 0.693 / k.

Given t_1/2 = 1 hour, we calculate:

k = 0.693 / 1 = 0.693 hr^-1.

Step 3: Calculate the fraction remaining after 3 hours.

[A]_t / [A]₀ = e^{-0.693 × 3} = e^-2.079.

Approximating e^-2.079, we find:

[A]_t / [A]₀ ≈ 1/8.

Step 4: Conclusion.

The fraction of reactant remaining after 3 hours is 1/8.

Step 1: Understand the Stephan Reaction.

The Stephan Reaction involves replacing a halogen (e.g., Cl) in an aromatic compound with a hydroxyl group (OH) using a strong base like NaOH.

The reaction is:

C₆H₅Cl + NaOH → C₆H₅OH + NaCl.

Step 2: Analyze the options.

Option (2) matches the definition of the Stephan Reaction, as it replaces Cl with OH in chlorobenzene to form phenol.

Step 1: Understand oxidation states.

The maximum oxidation state of an element corresponds to its group number for s- and p-block elements.

Step 2: Example of maximum oxidation state.

Manganese (Mn) in Group 7 has a maximum oxidation state of +7, as seen in MnO₄^- (permanganate ion).

Step 3: Conclusion.

The maximum oxidation state observed in the periodic table is +7.

Step 1: Relationship between radius and edge length.

In a BCC structure, the diagonal of the cube is equal to four times the radius of the atoms.

Mathematically: 4r = √3a, where a is the edge length of the unit cell.

Step 2: Solve for r.

r = (√3/4)a.

Step 3: Conclusion.

The radius of the atoms in a BCC structure is (√3/4)a.

Step 1: Recall the formula for freezing point depression.

The depression in freezing point is given by:

ΔT_f = K_f × m,

where:

• K_f is the cryoscopic constant (freezing point depression constant),
• m is the molality of the solution (moles of solute per kilogram of solvent).

Step 2: Conclusion.

Option (1) correctly represents the relationship.

Step 1: Analyze intermolecular forces.

Boiling point depends on the strength of intermolecular forces:

• Water (H₂O): Strong hydrogen bonding, highest boiling point.
• Ethanol (C₂H₅OH): Weaker hydrogen bonding than water.
• Acetone (CH₃COCH₃): Dipole-dipole interactions, weaker than hydrogen bonding.

Step 2: Conclusion.

The boiling points decrease in the order: Water > Ethanol > Acetone.

Step 1: Recall the formula for v_rms.

The root mean square velocity of gas molecules is given by:

v_rms = √(3kT/m),

where:

• k is the Boltzmann constant,
• T is the temperature in Kelvin,
• m is the mass of one molecule of the gas.

Step 2: Conclusion.

Option (3) is correct as it aligns with the formula from the kinetic theory of gases.

Step 1: Recall the conditions for ideal gas behavior.

The ideal gas equation PV = nRT assumes no intermolecular forces and negligible molecular volume. This is valid at high temperatures and low pressures.

Step 2: Analyze the options.

At high temperatures and low pressures, gas molecules are far apart, reducing intermolecular interactions and making the equation accurate.

Step 1: Recall the degrees of freedom for gases.

A monatomic gas has 3 translational degrees of freedom, giving C_p/C_v = 5/3.

A diatomic gas has translational and rotational degrees of freedom, giving C_p/C_v = 7/5.

Step 1: Recall the energy formula for SHM.

The total energy E in SHM is E = (1/2)kA², where k is the spring constant and A is the amplitude.

Step 2: Solve for A.

A = √(2E/k).

Step 1: Understand overtones and harmonics.

The fundamental frequency is the first harmonic. Overtones are higher-frequency modes of vibration:

• The first harmonic is the fundamental frequency.
• The second harmonic is the first overtone, with twice the frequency of the fundamental.
• The third harmonic is the second overtone, with three times the frequency of the fundamental.

Step 2: Conclusion.

The overtone refers to the second harmonic.

Step 1: Understand the function of different logic gates.

• AND Gate: Output is 1 only when both inputs are 1.
• OR Gate: Output is 1 if at least one input is 1.
• XOR Gate: Output is 1 if the inputs are different (one is 0, and the other is 1).
• NOT Gate: Inverts a single input.

Step 2: Conclusion.

The XOR Gate gives an output of 1 when the inputs are different.