We are given the equation:

√3 sec(x) = -2

First, solve for sec(x):

sec(x) = -2/√3 = -2√3/3

Next, recall that sec(x) = 1/cos(x), which gives:

cos(x) = -√3/2

Now, determine the principal solutions for cos(x) = -√3/2, which occurs at:

x = 5π/6, 7π/6

From the available options, 5π/6 is the correct solution.

First, observe the structure of the integral and apply substitution. Let:

u = e^(3x) + 1 → du = 3e^(3x) dx

Thus, the integral becomes:

(e^(3x) / (e^(3x) + 1)) dx = (1/3) * (du / u)

Now, the integral simplifies to:

∫ (1/3) * (du / u) = (1/3) ln|u| + C

Substitute back u = e^(3x) + 1:

(1/3) ln|e^(3x) + 1| + C

Thus, the final answer is:

(1/3) ln(e^(3x) + 1) + C

We are given the differential equation:

(x dy/dx - y) sin(y/x) = x³ e^x.

To simplify, let v = y/x, so y = vx. Differentiate y = vx to get:

dy/dx = v + x dv/dx.

Substitute y = vx and dy/dx = v + x dv/dx into the equation:

x(v + x dv/dx) - vx = x³ e^x sin(v).

Simplify the terms:

x² dv/dx = x³ e^x sin(v).

Separate variables:

dv/sin(v) = x e^x dx.

Integrate both sides:

∫ dv/sin(v) = ∫ x e^x dx.

The integral of the left-hand side is ln|tan(v/2)|, and the right-hand side is computed as:

∫ x e^x dx = e^x (x - 1) + c.

Substitute back v = y/x to get the solution:

e^x (x - 1) + cos(y/x) + c = 0.

The equation of the parabola is:

y² = 4ax,

which represents a standard parabola opening to the right. Its vertex is at (0, 0), and the latus rectum is the vertical line x = a.

The area under the parabola from x = 0 to x = a is given by:

Area = 2 ∫₀ᵃ √(4ax) dx,

where the factor of 2 accounts for the symmetry of the parabola.

Substitute y = √(4ax):

Area = 2 ∫₀ᵃ √(4a) √(x) dx = 2 √(4a) ∫₀ᵃ √(x) dx.

Evaluate the integral:

∫₀ᵃ √(x) dx = ∫₀ᵃ x^(1/2) dx = (2/3)x^(3/2) |₀ᵃ = (2/3)a^(3/2).

Substitute back:

Area = 2 √(4a) × (2/3)a^(3/2) = (8/3)a².

We are given the following logical conditions:

1. p ∧ q = False: At least one of p or q must be False.

2. p → q = False: This occurs only when p is True and q is False.

From condition 2, we conclude:

p = True and q = False.

Thus, the truth values of p and q are True and False, respectively.

To find the inverse of the given matrix:

            [
                [1, 0, 0],
                [3, 3, 0],
                [5, 2, -1]
            ]
            

First, calculate the determinant:

det(A) = -3.

Next, compute the adjoint matrix:

            adj(A) = [
                [-3, 0, 0],
                [3, -1, 0],
                [-9, -2, 3]
            ]
            

Finally, divide the adjoint by the determinant:

            A⁻¹ = (-1/3) * adj(A)
            

Passing the output of an AND gate through a NOT gate inverts the AND operation:

            Output = NOT(A AND B) = NAND(A, B)
            

Thus, the circuit behaves as a NAND gate.

The fringe spacing (β) in Young's double-slit experiment is given by:

            β = (λ * L) / d
            

where:

• λ = wavelength of light
• L = distance to the screen
• d = distance between the slits

If d is doubled, β becomes:

            β' = (λ * L) / (2d) = β / 2
            

Thus, the fringe spacing is halved.

At resonance in a series LCR circuit, the total impedance (Z) is given by:

Z = sqrt(R² + (X_L - X_C)²)

where:

• R is the resistance,
• X_L = ωL is the inductive reactance,
• X_C = 1/(ωC) is the capacitive reactance.

At resonance, the inductive reactance (X_L) and the capacitive reactance (X_C) are equal, so their effects cancel each other out:

Z = sqrt(R² + (X_L - X_C)²) = sqrt(R²) = R

This results in the minimum impedance, which allows the maximum current to flow according to Ohm's law:

I = V / Z

Thus, the current is maximum because the inductive and capacitive reactances cancel each other out.

Butter is an example of a colloidal system in which liquid (typically water or milk) is dispersed within a solid matrix (fat). This type of colloid is referred to as "liquid in solid." In butter, the fat acts as the continuous phase, with water droplets dispersed within it.

Thus, butter is a colloid where the liquid phase is dispersed in a solid phase.

Transmission Electron Microscopy (TEM) is a highly effective technique for analyzing the size, shape, and morphology of nanoparticles at the nanometer scale. Unlike UV-Vis Spectroscopy, FTIR, and AAS, which provide information about optical properties, molecular vibrations, and elemental composition, respectively, TEM directly images nanoparticles, providing precise measurements of their size and structure.

Thus, TEM is the most suitable technique for determining the size and morphology of nanoparticles.

Hydrochloric acid (HCl) is a strong acid that dissociates completely in water:

HCl → H⁺ + Cl⁻

Since HCl dissociates fully, the concentration of hydrogen ions [H⁺] is equal to the concentration of HCl, which is 0.01 M.

To calculate the pH, we use the formula:

pH = -log [H⁺]

Substituting the values:

pH = -log (0.01) = 2

Thus, the pH of the solution is 2.