Step 1: {Given values}
We are given: A = 1.0 m ± 0.2 m, B = 2.0 m ± 0.2 m
We define: Y = √(AB)
Step 2: {Calculate the value of Y}
Y = √((1.0)(2.0))
= √2.0 = 1.414 m
Step 3: {Determine the uncertainty in Y}
The formula for relative error propagation is: ΔY/Y = 1/2 (ΔA/A + ΔB/B)
Substituting the given values: ΔY/1.4 = 1/2 (0.2/1.0 + 0.2/2.0)
Step 4: {Simplify the expression}
ΔY/1.4 = 1/2 (0.2 + 0.1)
ΔY/1.4 = 1/2 × 0.3 = 0.15
ΔY = 0.15 × 1.4 = 0.21
Step 5: {Round off to one significant digit}
ΔY = 0.2 m
Thus, the final result is: Y = 1.4 m ± 0.2 m
Step 6: {Verify the options}
Comparing with the given options, the correct answer is (D) 1.4 m ± 0.2 m.

Step 1: {Define latent heat}
Latent heat (L) is defined as the amount of heat energy (Q) required to change the phase of a substance per unit mass: L = Q/m
Step 2: {Determine the dimensional formula of Q}
Heat energy (Q) is a form of energy, and its dimensional formula is the same as that of work: [Q] = [M L² T^-2]
Step 3: {Determine the dimensional formula of L}
Since mass (m) has the dimensional formula: [m] = [M]
we divide: [L] = [Q]/[m] = [M L² T^-2]/[M]
= M⁰ L² T^-2
Step 4: {Verify the options}
Comparing with the given choices,
the correct answer is (C) M⁰ L² T^-2.

Step 1: {Define self-inductance}
The energy stored in an inductor is given by: U = 1/2 L I²
where L is the self-inductance and I is the current.
Step 2: {Rearrange for L}
L = 2U/I²
Step 3: {Find the dimensional formula of L}
Since energy (U) has the dimensional formula: [U] = [M L² T^-2]
and current (I) has the dimensional formula: [I] = [A]
we substitute: [L] = [M L² T^-2]/[A²]
= [M L² T^-2 A^-2]
Step 4: {Verify the options}
Comparing with the given choices,
the correct answer is (A) [M L² T^-2 A^-2].

Step 1: {Find velocity}
Velocity is the first derivative of displacement: v = dx/dt = 3t² - 12t + 20
Step 2: {Find acceleration}
Acceleration is the derivative of velocity: a = dv/dt = 6t - 12
Step 3: {Set acceleration to zero}
6t - 12 = 0
Solving for t: t = 2 s
Step 4: {Find velocity at t = 2}
v = 3(2)² - 12(2) + 20
= 12 - 24 + 20 = 8 m/s
Step 5: {Verify the options}
The correct answer is (C) 8 m/s.

Step 1: {Use the nth second displacement formula}
The displacement covered in the nth second is given by: s_n = u + a/2 (2n - 1)
where:
- u is the initial velocity,
- a is the acceleration,
- n is the time instant.
Step 2: {Substituting values}
Given:
u = 0, a = 4/3 ms^-2, n = 3
s₃ = 0 + (4/3)/2 (2(3) - 1)
= 4/6 × 5 = 10/3 m
Step 3: {Verify the options}
Thus, the correct answer is (C) 10/3 m.

Step 1: {Find the time to reach maximum height}
At maximum height, the inclination with the horizontal becomes zero.
The time to reach maximum height is: t = (u sin θ)/g
Rearranging: u = (g t)/(sin θ)
Step 2: {Use the range formula}
R = u cos θ × (2t)
Substituting u: R = (g t)/(sin θ) cos θ × (2t)
cos θ = R / (2 u t)
Step 3: {Find cot θ}
cot θ = R / (2 g t²)
Substituting g = 10: cot θ = R / (2 × 10 t²) = R / (20 t²)
Thus, the correct answer is (A) R / (20t²).

It is given in the question that a rigid body is rotating about a fixed axis. Angular velocity is defined as the rate of the velocity at which a given object or particle rotates around a given point in a given interval of time. This velocity is also known as rotational velocity. Therefore, the angular velocity, as given in the question, is given by: ω = α - βt
Where, ω is the angular velocity, α and β are constants and t is the time taken. Also, we know that angular velocity is also measured as angle per unit time, therefore, ω = dθ/dt
Now, putting the value of ω in the above equation, we get: ω = dθ/dt = α - βt
⇒ dθ = (α - βt) dt
Now, to calculate the angle through which the rigid body rotates before it comes to rest, we can integrate the above equation: ∫ dθ = ∫ (α - βt) dt
θ = ∫ (α - βt) dt
θ = αt - (βt²)/2
When the rigid body comes to rest, the angular velocity will be zero: ω = 0
⇒ α - βt = 0
⇒ α = βt
⇒ t = α/β
Therefore, putting t = α/β in θ, we get: θ = α (α/β) - β (α/β)²/2
⇒ θ = α²/β - β (α²/β²)/2
⇒ θ = α²/β - α²/(2β)
⇒ θ = 2α²/(2β) - α²/(2β)
⇒ θ = α²/(2β)
Therefore, the angle through which a rigid body rotates before it stops is α²/(2β). Hence, option A is the correct option.

Step 1: The range R of a projectile launched with an initial speed v and at an angle θ to the horizontal is given by: R = (v²/g) sin(2θ)
where g is the acceleration due to gravity. From this equation, we can observe that the range is dependent on the sine of twice the launch angle. Given that the range at 15° is 50 m, if we launch the projectile at 45° with the same velocity, we can compare the ranges by comparing sin(2 × 15°) and sin(2 × 45°): sin(30°) = 1/2
sin(90°) = 1
Therefore, the range at 45° will be twice the range at 15°, because sin(90°) is twice as large as sin(30°). So, the range when the projectile is launched at 45° will be: R = 2 × 50 m = 100 m
Thus, the range at 45° is 100 meters.

Step 1: {Define angular momentum}
Angular momentum at maximum height is given by: L = m v H
where H is the maximum height.
Step 2: {Find the horizontal velocity}
The horizontal component of velocity remains constant: v = u cos 30° = u × √3/2
Step 3: {Find the maximum height}
Using the kinematic equation: H = (u² sin² 30°)/(2g)
Substituting sin 30° = 1/2: H = (u² × (1/2)²)/(2g) = u²/(8g)
Step 4: {Calculate angular momentum}
L = m u cos 30° × H
= m u × √3/2 × u²/(8g)
= (√3 m u³)/(16 g)
Thus, the correct answer is (A) (√3/16) (m u³)/g.

Step 1: {Use the time of flight formula}
Time of flight for a projectile is given by: T = (2 u sin θ)/g
Step 2: {Substituting values}
T = (2 × 9.8 × sin 30°)/9.8
Step 3: {Solve for T}
T = (2 × 9.8 × 1/2)/9.8
T = 1 sec
Thus, the correct answer is (D) 1.0 s.

Step 1: {Equation of motion for the system}
The acceleration of the system is given by:
a = (m₂ - m₁)/(m₁ + m₂) g
Step 2: {Equating given acceleration}
g/√2 = (m₂ - m₁)/(m₁ + m₂) g
Step 3: {Solve for m₁/m₂}
√2 (m₂ - m₁) = m₁ + m₂
Rearranging:
m₁/m₂ = (√2 - 1)/(√2 + 1)
Thus, the correct answer is (A) (√2 - 1)/(√2 + 1).

Step 1: {Calculate normal force}
The normal force N on the inclined plane is:
N = mg cos 60°
Step 2: {Determine frictional force}
F_f = μN = μmg cos 60°
Substituting values:
F_f = (0.1)(1)(10)(1/2) = 0.5 N
Step 3: {Calculate work done against friction}
W = F_f × d
= 0.5 × 10 = 5 J
Thus, the correct answer is (B) 5 J.

Step 1: {Determine total mass}
m_total = 60 + 940 = 1000 kg
Step 2: {Use Newton’s Second Law}
T - mg = ma
Substituting values: T - (1000 × 10) = 1000 × 1
T = 10000 + 1000 = 11000 N
Thus, the correct answer is (C) 11000 N.

Step 1: {Calculate maximum acceleration}
The maximum acceleration that the 1 kg block can have is: a_max = μg = (0.1)(10) = 1 m/s²
Step 2: {Calculate common acceleration}
The acceleration of the system is: a = F/m_total = 0.5/3 = 0.5/3 m/s²
Since a < a_max, the blocks move together.
Step 3: {Find force of friction}
The friction force acting on the upper block: f = m a = (1) × 0.5/3 = 1/6 N
Step 4: {Find work done by friction}
W = f × d = 1/6 × 3 = 0.5 J
Thus, the correct answer is (B) 0.5 J.

Step 1: {Find velocity at θ = 30°}
Using energy conservation: v = √(2gh)
Step 2: {Calculate height difference}
h = l (cos 30° - cos 60°)
= 1 × (√3/2 - 1/2) = 0.36 m
Step 3: {Find velocity}
v = √(2 × 10 × 0.36)
= √7.2 = 2.68 m/s
Step 4: {Find power}
P = (mgv) cos 60°
= (1 × 10 × 2.68) × 1/2
= 13.4 W
Thus, the correct answer is (A) 13.4 W.

Step 1: {Apply the perpendicular axis theorem}
Using the theorem of perpendicular axes, we express the moment of inertia about an edge as:
I = I_C + m(a/√2)²
Step 2: {Moment of inertia of cube about its center}
For a cube, the moment of inertia about its central axis is:
I_C = (ma²/12) + (ma²/12) = ma²/6
Step 3: {Adding the parallel axis contribution}
I = [ma²/12 + ma²/12] + ma²/2
= (2/3)ma²
Thus, the correct answer is (2/3)ma².

Step 1: {Apply conservation of energy}
The total energy remains constant:
Increase in kinetic energy = Decrease in potential energy
Step 2: {Using gravitational potential energy}
(1/2)mv² = mgR(1 / (1 + h/R))
For h = R:
mv² = mgR
v = √(gR)
Thus, the correct answer is √(gR).

Step 1: {Kepler's third law}
T² ∝ R³
Step 2: {Find new time period}
(T₂/T₁)² = (R₂/R₁)³
Substituting values:
T₂ = (3R/R)^3/2 × 1
= 3√3 years
Thus, the correct answer is 3√3 years.

Step 1: {Apply shell theorem}
The shell theorem states that a uniform spherical shell of mass exerts zero net gravitational force on a particle inside it.
Step 2: {Explain force cancellation}
Forces from opposite sides of the shell cancel out due to symmetry, leaving:
Net gravitational force = 0
Thus, the correct answer is zero.

Step 1: {Escape velocity formula}
v_e = √2 v₀
Step 2: {Find kinetic energy for escape}
KE_escape = 1/2 M v_e² = 1/2 M (2 v₀²)
= 2E
Thus, the correct answer is 2E.

Step 1: {Calculate elongations}
Δl₁ = WL / (4πR²)Y, Δl₂ = WL / πR²Y
Step 2: {Find elastic potential energy}
U = 1/2 K₁(Δl₁)² + 1/2 K₂(Δl₂)²
= 1/2 × Y(4πR²)/L × (WL / 4πR²Y)² + 1/2 × Y(πR²)/L × (WL / πR²Y)²
= (5W²L) / (8πR²Y)

Step 1: {Young's modulus definition}
Y = Stress / Strain
Step 2: {Effect of temperature increase}
As temperature increases, strain increases, leading to a decrease in Young’s modulus.
Thus, the correct answer is Young’s modulus decreases.

Step 1: {Young's modulus dependency}
Y = Stress / Strain
Step 2: {Effect of change in length and area}
Young's modulus is a material property and does not depend on length or cross-sectional area.
Thus, the correct answer is Y.

Step 1: {Use Laplace’s pressure equation}
ΔP = 4T/R
Step 2: {Compare pressures and radii}
ΔP₁/ΔP₂ = R₂/R₁ ⇒ R₁ = 2R₂
Step 3: {Find volume ratio}
V₁ : V₂ = R₁³ : R₂³ = 8:1
Thus, the correct answer is 8:1.

Step 1: {Define variables}
Let V₁ be the volume immersed in water and V₂ be the volume immersed in oil.
Step 2: {Equilibrium condition}
V₁ ρ_w g + V₂ ρ_o g = (V₁ + V₂) ρ_ice g
Step 3: {Solve for ratio}
V₁ + 0.8 V₂ = 0.9 (V₁ + V₂)
0.1 V₁ = 0.1 V₂ ⇒ V₁ : V₂ = 1:1
Thus, the correct answer is 1:1.

Step 1: {Formula for volume expansion}
ΔV = V₀ γ ΔT
Step 2: {Volume relation with side length}
ΔV = a³ (3α) ΔT
Step 3: {Finding cube side length}
6a² = 24 ⇒ a² = 4 ⇒ a = 2
Step 4: {Substituting values}
ΔV = 2³ (3 × 5 × 10^-4) × 10 = 1200 × 10^-4 m³
= 1200 × 10² cm³ = 1.2 × 10⁵ cm³
Thus, the correct answer is 1.2 × 10⁵ cm³.

Step 1: {Heat supplied in process DA and AB}
Q = n C_V (ΔT)_DA + n C_P (ΔT)_AB
Step 2: {For an ideal monoatomic gas,}
C_V = 3/2 R, C_P = 5/2 R
Step 3: {Substituting values}
Q = 3/2 (p₀V₀) + 5 (p₀V₀)
= 13/2 p₀V₀
Thus, the correct answer is 13/2 p₀V₀.

Step 1: {Using first law of thermodynamics}
ΔQ = ΔU + ΔW
Step 2: {For process ACB}
ΔU = 60 - 30 = 30 J
Step 3: {Applying same internal energy change to ADB}
ΔQ_ADB = 30 + 10 = 40 J
Thus, the correct answer is 40 J.

Step 1: {Using the first law of thermodynamics}
dQ/dt = dU/dt + dW/dt
Step 2: {Substituting values}
1000 = dU/dt + 200
Step 3: {Solving for dU/dt}
dU/dt = 1000 - 200 = 800 W
Thus, the correct answer is 800 W.

Step 1: {Conversion formula}
(F - 32)/9 = C/5
Step 2: {Applying temperature change}
ΔC = 5/9 ΔF
40 = 5/9 ΔF ⇒ ΔF = 72°F
Thus, the correct answer is 72°F.

Step 1: {Formula for RMS speed and speed of sound}
The root mean square (RMS) speed of gas molecules is given by: v_rms = √(3RT/M)
The velocity of sound in a gas is: v_sound = √(γRT/M)
where γ is the adiabatic index.
Step 2: {Relation between v_sound and v_rms}
Dividing the equations: v_sound/v_rms = √(γ/3)
Step 3: {Finding γ using degrees of freedom}
For a mixture of gases with an average degree of freedom f = 6: γ = 1 + 2/f = 1 + 2/6 = 4/3
Step 4: {Compute velocity of sound}
v_sound = √((4/3)/3) v_rms = 2/3 v_rms
Since v_rms = c, we get: v_sound = 2c/3
Thus, the correct answer is 2c/3.

Step 1: {Formula for RMS speed}
The root mean square (RMS) speed of gas molecules is given by: v_rms = √(3RT/M)
Since R and M are constants: v_rms ∝ √T
Step 2: {Determine new RMS speed}
Given initial and final temperatures: T_initial = 200 K, T_final = 800 K
Since v_rms ∝ √T, we write: v_{rms, initial} / v_{rms, final} = √(T_initial / T_final)
Step 3: {Compute new RMS speed}
v₀ / v_rms = √(200/800) = √(1/4) = 1/2
v_rms = 2v₀
Thus, the correct answer is 2v₀.

Step 1: {Ideal Gas Equation}
The ideal gas equation is given by: PV = nRT
where:
P is the pressure of the gas
V is the volume of the gas
n is the number of moles of gas
R is the ideal gas constant
T is the temperature of the gas
Step 2: {Number of Moles of Hydrogen (n_H)}
The molar mass of hydrogen (H₂) is 2 g/mol. n_H = mass of hydrogen / molar mass of hydrogen = 1 g / 2 g/mol = 1/2 mol
Step 3: {Number of Moles of Oxygen (n_O)}
The molar mass of oxygen (O₂) is 32 g/mol. n_O = mass of oxygen / molar mass of oxygen = 1 g / 32 g/mol = 1/32 mol
Step 4: {Applying Ideal Gas Equation to both vessels}
Since the vessels are of the same size and at the same temperature, V and T are the same for both vessels. Therefore, we can write:
For vessel A (hydrogen): P_AV = n_HRT
For vessel B (oxygen): P_BV = n_ORT
Step 5: {Finding the ratio P_A/P_B}
Divide the equation for vessel A by the equation for vessel B: (P_AV) / (P_BV) = (n_HRT) / (n_ORT)
P_A/P_B = n_H/n_O = (1/2) / (1/32) = 1/2 × 32/1 = 16
Thus, P_A/P_B = 16.

Step 1:
The time period of a spring-mass system is given by the formula:
T = 2π √(m/k)
where:
T is the time period
m is the mass
k is the spring constant
For the given systems:
System (a): T_a = 2π √(m/k)
System (b): The spring constant is halved, i.e., k' = k/2. Therefore,
T_b = 2π √(m/(k/2)) = 2π √(2m/k) = √2 (2π √(m/k)) = √2 T_a
System (c): The spring constant is doubled, i.e., k'' = 2k. Therefore,
T_c = 2π √(m/(2k)) = 1/√2 (2π √(m/k)) = 1/√2 T_a
Thus, the ratio of the time periods is:
T_a : T_b : T_c = 1 : √2 : 1/√2

Step 1: {Using the standard equation of SHM}
A/2 = A sin(ωt₁)
ωt₁ = sin^-1(1/2) = π/6 (i)
Step 2: {Calculating total time to reach A}
A = A sin ω(t₁ + t₂)
ω(t₁ + t₂) = sin^-1(1) = π/2
ωt₂ = π/2 - π/6 = π/3 (ii)
Step 3: {Finding the ratio of times}
t₁/t₂ = 1/2 ⇒ t₂ = 2t₁ = 2 × 2 = 4 s
Thus, the time taken to go from A/2 to A is 4 s.

Step 1: {Time period of a simple pendulum}
T = 2π√(L/g) and g = GM/(R + h)²
Step 2: {Relating time periods at different heights}
T₁/T₂ = (R + h₁)/(R + h₂) = (R + R)/(R + 2R) = 2/3 ⇒ 3T₁ = 2T₂
Thus, the correct relation is 3T₁ = 2T₂.

Step 1: {Given data}
Temperature, T = 273 K
Molecular mass of oxygen, M = 32 × 10^-3 kg
Step 2: {Calculating the speed of sound}
v = √(γRT/M) = √((1.4 × 8.3 × 273) / (32 × 10^-3)) ≈ 315 m/s
Thus, the speed of sound in oxygen at STP is approximately 315 m/s.

The general form of a plane progressive wave is:
y = A cos 2π(ft - x/λ)
Comparing with the given equation:
y = 2 cos 2π(330t - x)
The frequency f is 330 Hz.

Step 1: Calculating the mass of the oil drop
Density = 1.26 g/cm³ = 1.26 × 10³ kg/m³
Volume = (4/3)πr³ = (4/3)π(1 × 10^-6 m)³ = (4/3)π × 10^-18 m³
Mass = Density × Volume = 1.26 × 10³ × (4/3)π × 10^-18 = 1.68π × 10^-15 kg
Step 2: Balancing forces
The oil drop is stationary, so the electric force balances the gravitational force:
qE = mg
q = mg/E = (1.68π × 10^-15 × 10) / (3.65 × 10⁴) ≈ 1.44 × 10^-18 C
Step 3: Calculating the number of excess electrons
q = ne ⇒ n = q/e = (1.44 × 10^-18) / (1.6 × 10^-19) ≈ 9
Thus, the number of excess electrons is approximately 9.

Step 1: Volume and charge conservation
8 × (4/3)πr³ = (4/3)πR³ ⇒ R = 2r
8q = Q
Step 2: Relating potentials
V' = kq/r, V = kQ/R = k × 8q / 2r = 4 kq/r = 4V'
20 = 4V' ⇒ V' = 5 V
Thus, the potential of each single drop was 5 V.

Step 1: Solution: Given: Mass of dust particle, m = 4 × 10^-12 × 10^-3 kg = 4 × 10^-18 kg
Electric field, E = 50 N/C
Weight of dust particle, W = mg
W = 4 × 10^-18 × 10 × 10^-17 N = 4 × 10^-17 N
Electric force experienced by the dust particle,
F_e = qE
F_e = ne ⋅ E = n × 1.6 × 10^-19 × 50
where n is the number of electrons removed from the neutral dust particle.
At balance condition, the electric force is equal to the weight of the dust particle:
Electric force = Weight of dust particle
n × 1.6 × 10^-19 × 50 = 4 × 10^-17
n = (4 × 10^-17) / (1.6 × 10^-19 × 50) = 400/80 = 5
Thus, the number of electrons removed from the neutral dust particle is n = 5.

Step 1: {Calculating the distance}
r = √((30)² + (30)² + 0²) = 30√2 cm = 30√2 × 10^-2 m
Step 2: {Calculating the electric field}
E = kq/r² = (9 × 10⁹ × 0.008 × 10^-6) / (30√2 × 10^-2)² = 72 / 18 × 10³ = 4000 N/C
Step 3: {Direction of the electric field}
The electric field components in the x and y directions are equal: E_x = E_y = 4000/√2 = 200√2 N/C
Thus, the electric field is: E = 200√(2)(î + ĵ) N/C

The electrostatic force between two charges in a medium with dielectric constant K is given by:
F_medium = (1 / (4πε₀K)) * (|q₁q₂| / d²)
where ε₀ is the permittivity of free space.
The electrostatic force between the same two charges in air (or vacuum, with K=1) at a distance d' is given by:
F_air = (1 / (4πε₀)) * (|q₁q₂| / d'²)
For the forces to be equal, F_medium = F_air:
(1 / (4πε₀K)) * (|q₁q₂| / d²) = (1 / (4πε₀)) * (|q₁q₂| / d'²)
1 / (K d²) = 1 / d'²
d'² = K d²
d' = √(K d²) = d√K
Thus, the equivalent distance in air is d√K.

The formula for the electric potential (V_P) due to a point charge is given by:
V_P = KQ/r
where:
K = 1/(4πε₀) = 9 × 10⁹ Nm²C^-2
Q = 5 × 10^-9 C (the charge)
r is the distance from the point charge to the point P
V_P = 50 V (the electric potential at point P)
Rearranging the formula to solve for the distance r:
r = KQ/V_P
Substituting the known values:
r = ((9 × 10⁹) × (5 × 10^-9)) / 50
r = (45 × 10⁰) / 50
r = 0.9 m
Thus, the distance of point P from the point charge is 0.9 meters or 90 cm.

Using Gauss's law, the electric flux Φ is given by the formula:
Φ = q/ε₀
where:
- q is the charge inside the closed surface,
- ε₀ is the permittivity of free space.
Now, if there are multiple charges inside the surface, we sum up the individual charges. Here, we are given the charges inside the closed surface: q, -2q, and 5q. So, the total charge q_total inside the surface is:
q_total = q + (-2q) + 5q
q_total = 4q
Therefore, the electric flux is:
Φ = q_total/ε₀ = 4q/ε₀
Thus, the electric flux Φ through the closed surface is:
Φ = 4q/ε₀

Step 1: {Capacitance of the original capacitor}
C₁ = Aε₀/d = 4μF
Step 2: {Finding the new capacitance when half-filled with dielectric}
The capacitor can be considered as two capacitors in series: C_f = Aε₀/(d₁ + d₂/K) = Aε₀/(d(1 - 1/2 + 1/(2K)))
Substituting K = 3: C_f = 4μF / (3/2) = 6μF
Thus, the new capacitance is 6μF.

Step 1: {Finding Equivalent EMF and Resistance}
Using Kirchhoff's Voltage Law (KVL): E_eq = (E₁R₁ + E₂R₂) / (R₁ + R₂) = (2 × 4 + 2 × 4) / (4 + 4) = 2V
R_eq = (R₁R₂) / (R₁ + R₂) = (4 × 4) / (4+4) = 2Ω
Step 2: {Finding the Current}
I = E'/R_eq = 4V / 2Ω = 2A
Thus, the current flows from A to B at 2A.

Step 1: {Identifying Wheatstone Bridge}
The given network forms a balanced Wheatstone bridge. In a Wheatstone bridge, four resistors are arranged in a diamond shape, with two resistors on each arm, and a galvanometer connected across the middle of the bridge. In this case, when the 3 Ω resistor in the BD arm is removed, the bridge remains balanced.
Key Concept: A Wheatstone bridge is balanced when the ratio of resistances in one pair of opposite arms is equal to the ratio of resistances in the other pair. The condition for balance in the Wheatstone bridge is given by:
R₁/R₂ = R₃/R₄
Where R₁, R₂, R₃, and R₄ are the resistances in the four arms of the bridge. When the 3 Ω resistor is removed from the BD arm, it does not affect the total resistance because the bridge is balanced and the current flowing through the galvanometer is zero, meaning no current flows through the branch containing the 3 Ω resistor. Hence, the equivalent resistance of the bridge does not change.
Step 2: {Finding Equivalent Resistance}
Since the bridge remains balanced, the equivalent resistance of the network remains unchanged. For a balanced Wheatstone bridge, the total equivalent resistance across the bridge is determined by the resistances in the remaining arms. Let’s denote the resistances of the other arms as R₁, R₂, R₃, and R₄. When the bridge is balanced, we know the equivalent resistance for the two parallel arms (AC and BD) can be calculated as:
R_eq = (R₁R₂) / (R₁ + R₂)
This is true for both the upper and lower parts of the Wheatstone bridge, and they contribute equally to the total equivalent resistance. Since the 3 Ω resistor is removed from the BD arm, it does not change the equivalent resistance of the network because the Wheatstone bridge was balanced and no current flows through the removed resistor. The equivalent resistance remains as calculated previously.
Step 3: {Conclusion}

Step 1: {Understanding the effect of stretching}
Since the volume of the wire remains constant, we use the relation: V = Al
where A is the cross-sectional area and l is the length.
Step 2: {Deriving the new resistance}
We use the resistance formula: R = ρl/A
Since A decreases as d² and l increases proportionally: ΔR/R = -4 ΔD/D
Substituting ΔD = -0.4, ΔR/R = -4(-0.4) = 1.6%
Thus, the percentage change in resistance is 1.6%.

Step 1: {Understanding volume conservation}
Since the wire is melted and redrawn, its volume remains the same: A₁l₁ = A₂l₂
Given that the new length is 1/4th of the original: A₂ = 4A₁
Step 2: {Finding new resistance}
Resistance is given by: R = ρl/A
Using the transformation, R₂ = (l₂/A₂) R₁
R₂ = (1/4 l₁ / 4A₁) R₁ = 1/16 R₁
Substituting R₁ = 160Ω: R₂ = 160/16 = 10Ω
Thus, the new resistance is 10Ω.

Step 1: {Current in series combination}
I = nE/(nr + R) = 5E/(5r + R)
Step 2: {Current in parallel combination}
I' = E/(r/n + R) = 5E/(r + 5R)
Since I = I', equating both expressions: 5E/(5r + R) = 5E/(r + 5R)
Solving for R and r, 5r + R = r + 5R
4r = 4R ⇒ R = r
Thus, the ratio R/r = 1.

The problem involves calculating the magnetic field at a point P due to a current-carrying straight wire, where the ends of the wire make angles α and β with respect to the point P. The solution can be derived using the Biot-Savart law, which gives the magnetic field generated by a current element.
Step 1: {Understanding the Biot-Savart Law}
The Biot-Savart law provides the magnetic field dB at a point due to a small current element I dl. The law is given by:
dB = (μ₀I / 4π) (dl × r̂) / r²
where:
- μ₀ is the permeability of free space,
- I is the current,
- dl is the infinitesimal length of the wire element,
- r̂ is the unit vector from the wire element to the point where the magnetic field is being calculated,
- r is the distance from the wire element to the point.
Step 2: {Applying the Biot-Savart Law to a Straight Wire}
For a straight current-carrying wire, the magnetic field at a point P can be found by integrating the contributions from all infinitesimal elements of the wire. The result for the magnetic field due to a finite straight wire at a point P is given by:
B = (μ₀I / 4πd)(sin θ₁ - sin θ₂)
where:
- d is the perpendicular distance from the wire to the point P,
- θ₁ and θ₂ are the angles between the line connecting the point P and the ends of the wire, and the wire itself.
The expression is derived by integrating the Biot-Savart law along the length of the wire. The terms sin θ₁ and sin θ₂ come from the geometry of the setup, which involves the angles at which the current elements contribute to the magnetic field.
Step 3: {Conclusion}
Thus, the magnetic field at point P due to a straight current-carrying wire, where the ends of the wire make angles α and β with the point, is:
B = (μ₀I / 4πd)(sin θ₁ - sin θ₂)
Therefore, the correct option is (A).

Step 1: {Using Ampere’s Circuital Law}
The magnetic field due to a straight wire is given by: B = (μ₀I) / (2πr)
Step 2: {Finding the Magnetic Field at a/2 and 2a}
Magnetic field at r = a/2: B_a/2 = (μ₀I) / (2π(a/2))
B_a/2 = (μ₀I) / (πa)
Magnetic field at r = 2a: B_2a = (μ₀I) / (2π(2a))
B_2a = (μ₀I) / (4πa)
Step 3: {Calculating the Ratio}
B_a/2 / B_2a = ((μ₀I) / (πa)) / ((μ₀I) / (4πa)) = 1/1 = 1:1
Thus, the correct answer is 1:1.

Step 1: {Understanding the Forces}
The electrostatic force on a charge q is given by: F₁ = q E
The magnetic force on a moving charge is given by: F₂ = q (v × B)
Thus, the correct answer is (C).

Step 1: {Using Faraday's Law}
Induced emf = -dΦ/dt = B ⋅ A
Since A = πr², we get: ε = -πr² dB/dt
Step 2: {Finding the Electric Field}
E = ε/d = (πr²/d) dB/dt
Step 3: {Finding Acceleration}
a = eE/m
Substituting values, a = (1.6 × 10^-19 / 9.1 × 10^-31) × (π(0.5)² / 0.3) × 50 × 10^-6
= 23 × 10⁶ m/s²
Thus, the correct answer is 23 × 10⁶ m/s².

Step 1: Expression for Mutual Inductance
The mutual inductance M between the two co-axial solenoids is given by:
M = μ₀n₁n₂πr₁²l
Step 2: Expression for Self-Inductance of the Inner Coil
The self-inductance L of the inner solenoid is given by:
L = μ₀n₁²πr₁²l
Step 3: Finding the Ratio M/L
M/L = (μ₀n₁n₂πr₁²l) / (μ₀n₁²πr₁²l) = n₂/n₁
Thus, the correct answer is n₂/n₁.

Step 1:
To find the magnetic flux Φ through the rectangular loop, we use the formula:
Φ = B⋅A⋅cos(θ)
Where:
B is the magnetic field strength,
A is the area of the loop,
θ is the angle between the magnetic field lines and the normal (perpendicular) to the plane of the loop.
Step 2:
The area A of the rectangular loop is:
A = length × width = 2.5 m × 2 m = 5 m²
Step 3:
Given the angle θ = 60°, we can calculate the initial magnetic flux Φ_initial:
Φ_initial = B⋅A⋅cos(60°)
Φ_initial = 4 T ⋅ 5 m² ⋅ cos(60°)
Φ_initial = 4 × 5 × 1/2 = 10 Wb
(Since cos(60°) = 1/2)
Step 4:
When the loop is removed from the magnetic field, the final magnetic flux Φ_final is zero because the loop is no longer within the magnetic field.
Thus, the change in magnetic flux ΔΦ is:
ΔΦ = Φ_final - Φ_initial = 0 - 10 Wb = -10 Wb
Step 5:
The loop is removed from the field in t = 10 seconds, so the rate of change of magnetic flux is:
dΦ/dt = ΔΦ / Δt = -10 Wb / 10 s = -1 Wb/s
Step 6:
Now we can find the average induced emf ε:
ε = - dΦ/dt
ε = -(-1 Wb/s) = +1 V
Therefore, the average emf is +1 V

Step 1: {Finding the Area under the i-t Graph}
Total area = 1/2 × 1 × 10 + 1/2 × (2-1) × 10
= 5 + 5 = 10 A
Step 2: {Finding Average Current}
i_avg = Total area / time interval
i_avg = 10/2 = 5 A
Thus, the correct answer is 5 A.

Step 1: {Using Impedance Formula}
Z = √((X_L - X_C)² + R²)
Since X_L = X_C, Z = √((R - R)² + R²) = √(0 + R²) = R
Thus, the correct answer is R.

Step 1: {Understanding the given equation}
The given alternating voltage is: V(t) = 220 sin 100πt
The angular frequency ω is extracted as: ω = 100π rad/s
Step 2: {Time interval calculation for half to peak transition}
The peak value of the voltage is V_max = 220 V. The time interval for the voltage to rise from half of the peak value to the peak value in a sinusoidal waveform follows the relation: t = T/6
where T is the time period: T = 2π/ω = 2π/(100π) = 2/100 = 0.02 s = 20 ms
Thus, the required time interval: t = T/6 = 20/6 = 3.3 ms
Thus, the correct answer is 3.3 ms.

Step 1: Using Maxwell's Displacement Current Formula
The displacement current is given by:
I_d = ε₀ dΦ/dt
Since:
dΦ/dt = A dE/dt
Step 2: Finding the Area of Plates
A = πR² = 3.14 × (0.1)² = 3.14 × 10^-2 m²
Step 3: Calculating Displacement Current
I_d = ε₀ A dE/dt
= (8.85 × 10^-12) × (3.14 × 10^-2) × (6 × 10¹³)
I_d = 16.67 A
Thus, the correct answer is 16.67 A.

Step 1:
The given equation for the velocity v is:
v = C / √(μ_rε_r)
Where:
- v is the velocity,
- C is the speed of light in vacuum,
- μ_r is the relative permeability,
- ε_r is the relative permittivity.
Step 2:
Now, substitute the given values:
v = 1.5 × 10⁸ = (3 × 10⁸) / √(2 × ε_r)
Step 3:
Rearrange the equation to solve for ε_r:
√(2 × ε_r) = (3 × 10⁸) / (1.5 × 10⁸)
√(2 × ε_r) = 2
Step 4:
Square both sides to eliminate the square root:
2 × ε_r = 4
Step 5:
Now, solve for ε_r:
ε_r = 4/2 = 2
Thus, the correct answer is 2.

Step 1: Understanding the Concept of Lens Power
The power of a lens is given by:
P = 1/f
where f is the focal length of the lens.
Step 2: Effect of Cutting a Lens Along the Principal Axis
When a symmetrical biconvex lens is cut into two halves along the principal axis, the focal length remains the same for each half.
Since power is inversely proportional to focal length, the power of each half remains unchanged.
Thus, the ratio of power between the two halves is:
1:1
Thus, the correct answer is 1:1.

Step 1: Understanding Magnifying Power of a Telescope
The magnification M of an astronomical telescope in normal adjustment (parallel rays) is given by:
M = f_o/f_e
where:
f_o is the focal length of the objective,
f_e is the focal length of the eyepiece.
Step 2: Using Given Information
It is given that M = 9, so:
f_o/f_e = 9
Also, the total distance between the objective and eyepiece is:
f_o + f_e = 20
Step 3: Solving for f_o and f_e
Using the given equations:
9f_e + f_e = 20
10f_e = 20
f_e = 2 cm, f_o = 18 cm
Thus, the ratio of focal lengths is:
f_o/f_e = 9
Thus, the correct answer is 9.

Step 1: {Understanding the Normal Adjustment Condition}
In a refracting telescope under normal adjustment, the total length of the telescope is: L = f_o + f_e
where:
f_o is the focal length of the objective lens,
f_e is the focal length of the eyepiece lens.
Step 2: {Using the Given Values}
It is given that the total length of the telescope is: f_o + f_e = 30
Also, the magnification of the telescope is given by: M = f_o/f_e
Since M = 2, we get: f_o/f_e = 2
Step 3: {Solving for f_o and f_e}
Rewriting the equation: f_o = 2 f_e
Substituting into the length equation: 2f_e + f_e = 30
3f_e = 30
f_e = 10 cm, f_o = 20 cm
Thus, the correct answer is 20 cm.

Step 1: Understanding the given data
The magnification formula for a mirror is:
m = -v/u
Since the image is virtual and magnified two times:
m = 2
which gives:
2 = -v/u
Step 2: Using the given object-image distance
The total distance between the object and image is:
u + v = 15
Substituting v = -2u,
u + (-2u) = 15
-u = 15 ⇒ u = 5 cm
v = -2(5) = -10 cm
Step 3: Using the mirror formula
1/f = 1/v + 1/u
= 1/(-10) + 1/5 = -1/10

Step 1: {Understanding the intensity formula in YDSE}
The intensity at any point in YDSE is given by: I = I₀cos²(ϕ/2)
where ϕ is the phase difference, given by: ϕ = (2π/λ) × path difference
Step 2: {Substituting given values}
For the given path difference Δx = λ/4, ϕ = (2π/λ) × λ/4 = π/2
Thus, I = I₀cos²(π/4) = I₀ × (1/√2)²
I = I₀/2
Thus, the correct answer is I₀/2.

Step 1: {Understanding Young’s Double-Slit Experiment (YDSE)}
In Young’s Double-Slit Experiment, the fringe width β is given by: β = λD/d
where:
λ is the wavelength of light,
D is the distance between the slits and the screen,
d is the distance between the two slits.
The fringe width is the distance between consecutive bright fringes.
Step 2: {Using the Given Data}
It is given that the distance between the first and second-order bright fringes is: y₂ - y₁ = 0.553 mm = 0.553 × 10^-3 m
For the first bright fringe, y₁ = λD/d
For the second bright fringe, y₂ = 2λD/d
So, y₂ - y₁ = 2λD/d - λD/d = λD/d
Step 3: {Solving for λ}
λ = ((y₂ - y₁) ⋅ d) / D
Substituting the given values: λ = (0.553 × 10^-3) × (2.08 × 10^-3) / 1.8
λ = 1.15024 × 10^-6 / 1.8
λ = 639 × 10^-9 m = 639 nm
Thus, the correct answer is 639 nm.

Step 1: {Understanding the diffraction formula}
The condition for minima in diffraction is: d sin θ = nλ
For the first minimum: sin θ = λ/d = 2/4 = 1/2
θ = 30°
Angular spread: 2θ = 60°
Thus, the correct answer is 60°.

Step 1: {Understanding Huygen's Wave Theory}
Huygen’s wave theory describes how light propagates by treating every point on a wavefront as a secondary wave source. Using this principle, the laws of:
Reflection
Refraction
Diffraction
are successfully derived.
Step 2: {Analyzing the Given Options}
Reflection: Huygen’s principle explains reflection by considering the secondary wavelets on the incident wavefront, which create the reflected wavefront.
Refraction: Huygen’s principle explains refraction by stating that different parts of a wavefront move at different speeds when passing through media with different refractive indices.
Diffraction: Huygen’s principle accounts for diffraction, as each point on a wavefront acts as a source of secondary wavelets, allowing light to bend around obstacles.
Step 3: {Why Huygen’s Principle Fails to Explain Spectra}
The origin of spectral lines arises due to the emission and absorption of photons by atoms, which is best explained by quantum mechanics. Huygen’s wave theory does not consider the particle nature of light or energy quantization, which are essential for understanding:
Atomic emission spectra
Blackbody radiation
Photoelectric effect
Since Huygen’s theory only deals with the wave nature of light and not its quantum properties, it cannot explain the origin of spectra.
Thus, the correct answer is D.

Step 1: {Understanding Brewster's Law}
According to Brewster's law, when light is incident at the Brewster angle, the reflected light is completely polarized.
At this angle, the reflected and refracted rays are perpendicular to each other.
Step 2: {Derivation of the Perpendicular Relation}
θ_reflected + θ_refracted = 90°
Thus, the angle between the reflected and refracted rays is: θ = 90°
Thus, the correct answer is 90°.

Step 1: {Understanding the Photoelectric Effect}
- The photoelectric effect involves the emission of electrons from a metal surface when light of a certain frequency (or wavelength) strikes it. The key observation is that the energy of the emitted electrons depends on the frequency (or wavelength) of the incident light, not its intensity.
- The stopping potential is the minimum voltage required to stop the most energetic emitted electrons. It is independent of the light intensity but depends on the wavelength (or frequency) of the incident light. This means that the stopping potential remains the same for different intensities of light, as long as the frequency (or wavelength) is the same.
- The saturation current, however, increases with the intensity of the incident light. This is because a higher intensity means more photons are hitting the surface, and more electrons are being emitted. The saturation current represents the maximum current that can be obtained when all emitted electrons are collected.
Step 2: {Interpreting the Graphs}
- In the graph representing the photoelectric effect, the x-axis typically represents the applied voltage (V), and the y-axis represents the photoelectric current(I).
- The correct graph for this situation should show that the stopping potential is the same for both intensities of light, as it is independent of intensity. However, the saturation current will be higher for I₂ than for I₁, because I₂ corresponds to a higher intensity, which results in more electrons being emitted from the surface.
Thus, the correct graph is (C).

Step 1: {Using Energy-Wavelength Relation}
The energy of the photon required to excite an electron is given by: E = hc/λ
Rearranging for λ: λ = hc/E
Step 2: {Substituting the Given Values}
λ = 1242 / 6
λ = 207 nm
Thus, the correct answer is 207 nm.

Step 1: {Photoelectric Equation}
According to Einstein’s photoelectric equation: K_max = hν - W₀
where:
- h is Planck’s constant,
- ν is the frequency of incident light,
- W₀ is the work function of the metal.
Step 2: {Dependence on Frequency}
Since K_max depends only on ν and not on intensity, the correct answer is (B).

Step 1: {Using de-Broglie’s Equation}
The de-Broglie wavelength is given by: λ = h/p
Since momentum p is related to kinetic energy: p = √(2mK)
Step 2: {Effect of Doubling Kinetic Energy}
If K is doubled: p' = √(2m(2K)) = √2 p
Since λ ∝ 1/p, we get: λ' = λ/√2
Thus, the correct answer is 1/√2.

Step 1: {Understanding the Rydberg Formula}
The wavelength λ of emitted radiation during an electron transition is given by the Rydberg formula: 1/λ = RZ² (1/n₁² - 1/n₂²)
where:
- R is the Rydberg constant,
- Z is the atomic number,
- n₁ and n₂ are the principal quantum numbers.
Step 2: {Applying to Hydrogen and Helium Ions}
For hydrogen (Z = 1), the wavelength of the transition n₂ to n₁ is: 1/λ_H = R(1)² (1/n₁² - 1/n₂²)
For the He⁺ ion (Z = 2), the wavelength of transition n₄ to n₃ is: 1/λ_He = R(2)² (1/n₃² - 1/n₄²)
Equating λ_H = λ_He, we get: 1/n₁² - 1/n₂² = 4(1/n₃² - 1/n₄²)
Solving for integer values, we find n₃ = 2, n₄ = 4 satisfies the condition.
Thus, the correct answer is n = 4 to n = 2.

Step 1:
The general formula for the wavelength of a hydrogen atom is given as:
1/λ = RZ²(1/n₁² - 1/n₂²)
Where:
- λ is the wavelength,
- R is the Rydberg constant,
- Z is the atomic number,
- n₁ and n₂ are the principal quantum numbers.
Step 2:
The shortest wavelength for the Balmer series corresponds to the transition from n₂ to ∞ and n₁ = 2, so substituting into the equation:
1/λ_B = RZ²(1/2² - 1/∞²)
Since 1/∞² = 0, this simplifies to:
1/λ_B = RZ²(1/4)
Thus, the shortest wavelength for the Balmer series is:
1/λ_B = RZ²/4
Step 3:
The shortest wavelength for the Lyman series corresponds to the transition from n₂ to ∞ and n₁ = 1, so substituting into the equation:
1/λ_L = RZ²(1/1² - 1/∞²)
Again, 1/∞² = 0, so this simplifies to:
1/λ_L = RZ²(1)
Thus, the shortest wavelength for the Lyman series is:
1/λ_L = RZ²
Step 4:
Now, dividing equation (i) by equation (ii), where equation (i) represents the shortest wavelength for the Balmer series and equation (ii) represents the shortest wavelength for the Lyman series:
(1/λ_B) / (1/λ_L) = (RZ² / 4) / (RZ²)
This simplifies to:
λ_L/λ_B = 4
Therefore, we can conclude:
λ_L : λ_B = 4 : 1
Thus, the ratio is 4:1.

Step 1: Energy Levels in the Hydrogen Atom
The energy of an electron in the nth orbit is:
E_n = -13.6/n² eV
For n = 1:
E₁ = -13.6 eV
For n = 2:
E₂ = -13.6/4 = -3.4 eV
Step 2: Calculating Excitation Energy
E_excitation = E₂ - E₁
= (-3.4) - (-13.6)
= 10.2 eV
Thus, the correct answer is 10.2 eV.

Mass defect:
Δm = (12.000000 + 1.008665) - 13.003354
= 0.00531 u
Energy required:
E = Δm × 931.5
= 0.00531 × 931.5
= 4.95 MeV
Thus, the correct answer is 4.95 MeV.

Step 1: Understanding Binding Energy per Nucleon
The binding energy per nucleon (BE/A) is the energy required to disassemble a nucleus into its constituent protons and neutrons. It is an important measure of nuclear stability. The higher the binding energy per nucleon, the more stable the nucleus is. The binding energy per nucleon is typically highest for elements in the mid-range of the periodic table, particularly for elements around iron (⁵⁶₂₆Fe).
Step 2: Comparing the Nuclei
- Lighter nuclei, such as ⁴₂He (Helium-4), have a relatively low binding energy per nucleon. This is because the nucleons in lighter nuclei are not as tightly bound as in heavier nuclei.
- Heavy nuclei, such as ²⁰⁸₈₄Pb (Lead-208), also tend to have lower binding energy per nucleon compared to mid-range nuclei. This is due to the electrostatic repulsion between the positively charged protons, which weakens the nuclear force that binds the nucleus together.
- ⁵⁶₂₆Fe (Iron-56), which has the highest binding energy per nucleon (around 8.8 MeV), is considered the most stable nucleus. This high binding energy per nucleon explains why nuclear fusion (such as in stars) generally produces energy by fusing lighter elements up to iron, and why fission of heavy elements releases energy.
Thus, the correct answer is ⁵⁶₂₆Fe.

Step 1: Understanding the Logic Circuit
The given circuit consists of:
- NOT gates applied to inputs A and B.
- AND gate processing the inverted signals.
Step 2: Deriving the Boolean Expression
The circuit implements:
Y = A' ⋅ B'
Applying De-Morgan’s theorem:
Y = (A + B)'
Step 3: Comparing with Output Waveforms
Analyzing the truth table, we match the waveform and determine that the correct output is given by waveform 4.
Thus, the correct answer is (D).

Step 1: Understanding the Circuit
The circuit consists of NOT gates applied to A and B, followed by an AND gate.
Step 2: Boolean Expression Derivation
Y = (A' ⋅ B')'
Applying De-Morgan's theorem:
Y = A + B
Step 3: Conclusion
The output matches the OR gate.
Thus, the correct answer is (B).

Step 1: Understanding Zener Diode
A zener diode allows current to flow in the reverse direction when the applied voltage exceeds the breakdown voltage.
Step 2: Function of a Zener Diode
In the breakdown region, the voltage across the diode remains nearly constant, making it ideal for voltage regulation.
Thus, the correct answer is (C).

Step 1: Understanding the Logic Circuit
The circuit consists of two NOR gates whose outputs are given as inputs to another NOR gate.
Step 2: Boolean Expression
Using De-Morgan’s theorem:
Y = (A' + B')'
= A ⋅ B
Step 3: Conclusion
The circuit implements an AND gate.
Thus, the correct answer is (B).

Step 1: {Understanding Vernier Caliper}
The vernier scale least count is given by: LC = Main Scale Division - Vernier Scale Division
Step 2: {Applying the Given Condition}
Since n main scale divisions equal (n+1) vernier scale divisions, n × MSD = (n+1) × VSD
Step 3: {Solving for Least Count}
VSD = n/(n+1) × MSD
LC = MSD - VSD = m - (n/(n+1)) m
LC = m (1 - n/(n+1))
LC = m/(n+1)
Thus, the correct answer is (B).

Step 1: Calculate the number of moles of each gas
The number of moles of methane CH₄ is given by:
n₁ = Mass of CH₄ / Molar mass of CH₄ = 4/16 = 0.2

Step 1: {Calculate the number of moles of each gas}
The number of moles of methane (CH₄) is given by:
n₁ = (Mass of CH₄) / (Molar mass of CH₄) = 4/16 = 0.25 mol
Similarly, the number of moles of carbon dioxide (CO₂) is:
n₂ = (Mass of CO₂) / (Molar mass of CO₂) = 4.4/44 = 0.1 mol
Step 2: {Total number of moles}
Total number of moles, n_T is: n_T = n₁ + n₂ = 0.25 + 0.1 = 0.35 mol
Step 3: {Applying the ideal gas equation}
Using the ideal gas equation: PV = nRT
where R = 0.0821 atm L mol^-1K^-1, T = 300 K, and V = 1 L, we get: P = (0.35 × 0.0821 × 300) / 1 = 8.6 atm
Thus, the correct answer is (A).

Step 1: {Understanding concentration units}
Molarity (M) and normality (N) depend on volume, which changes with temperature. However, molality (m) is defined as: m = Moles of solute / Mass of solvent in kg
Since mass is unaffected by temperature changes, molality remains constant.
Step 2: {Why molality is temperature-independent}
Molality does not involve volume, which expands or contracts with temperature. Hence, it is the preferred unit in temperature-dependent studies.
Thus, the correct answer is (D).

Step 1: {Understanding degeneracy}
The energy of hydrogen-like atoms is given by: E_n = -R_H/n²
Given E = -R_H/9, comparing with the formula: R_H/n² = R_H/9 ⇒ n² = 9 ⇒ n = 3
Step 2: {Finding the degeneracy}
For n = 3, the possible values of l are 0, 1, 2, corresponding to subshells: (3s, 3p, 3d)
Each subshell contains: 3s = 1, 3p = 3, 3d = 5
Total orbitals present: 1 + 3 + 5 = 9
Thus, the correct answer is (D).

Step 1: {Defining de-Broglie Wavelength}
The de-Broglie wavelength is given by: λ = h/(mv)
Given that: λ = 100 v
Step 2: {Expressing Wavelength in Terms of h and m}
100 v = h/(mv)
x² = 100 h/m
x = 10 √(h/m)
Thus, the correct answer is (B).

Step 1: {Using Energy Formula}
The energy of an electron in an orbit is given by: E_n = -13.6 (Z²/n²) eV
Step 2: {Finding Energy for Li²⁺}
For Li²⁺, Z = 3 and for the first orbit n = 1: E₁ = -13.6 × (3²/1²) eV
E₁ = -122.4 eV
Step 3: {Converting to Joules}
E₁ = (-122.4) × (1.602 × 10^-19)
E₁ = -1.962 × 10^-17 J
Thus, the correct answer is (B).

Step 1: {Using Einstein's Photoelectric Equation}
KE = hν - ϕ
Since ν = c/λ, we use: E = hc/λ
Step 2: {Substituting Values}
E = (6.626 × 10^-34) (3 × 10⁸) / (3 × 10^-7)
E = 6.6 × 10^-19 J
Converting to eV: E = (6.6 × 10^-19) / (1.6 × 10^-19) = 4.125 eV
Step 3: {Calculating Kinetic Energy}
KE = 4.125 - 2.13 = 2.0 eV
Thus, the correct answer is (C).

Step 1: {Finding Electron Kinetic Energy}
KE = eV
Step 2: {Using de-Broglie Wavelength Formula}
λ = h/√(2mKE)
Step 3: {Rearranging for h/λ}
h/λ = √(2meV)
Thus, the correct answer is (B).

Step 1: {Understanding Electron Affinity}
Electron affinity refers to the energy change when an electron is added to a neutral atom in the gas phase to form a negatively charged ion. The first electron affinity is usually exothermic, meaning energy is released when the atom gains an electron. However, the second electron affinity is generally endothermic, meaning energy is required to add a second electron. This is because, after the first electron is added, the resulting negatively charged ion experiences a repulsive force from the incoming electron.
Step 2: {Explaining Positive Electron Affinity}
In the case of the oxygen ion O^-, the negative charge on the ion creates a repulsive force that works against the addition of another electron. The added electron would experience electrostatic repulsion due to the negative charge of the O^- ion. Therefore, energy must be supplied to overcome this repulsion, making the second electron affinity positive (endothermic). Thus, the correct answer is (B).

Step 1: {Isoelectronic Species}
The ions N^3-, O^2-, and F^- are isoelectronic species, meaning they all have the same number of electrons. Specifically, each of these ions has 10 electrons. Despite having the same number of electrons, their ionic radii will differ because of the varying nuclear charges. The nuclear charge increases from nitrogen to oxygen to fluorine as we move across the periodic table:
- N has atomic number Z = 7,
- O has atomic number Z = 8,
- F has atomic number Z = 9.
Step 2: {Ionic Radii Order}
As the nuclear charge increases, the electrons are attracted more strongly by the nucleus, which results in a decrease in ionic radius. Therefore, the ionic radii will decrease as we move from N^3- to O^2- to F^-. This gives the following order of ionic radii: N^3- > O^2- > F^-
Hence, the correct answer is (A).

In o-nitrophenol, the hydrogen of the O-H group and the oxygen of the nitro group form intramolecular hydrogen bonding producing a six-membered ring structure. In m-nitrophenol, p-nitrophenol, and phenol, intermolecular hydrogen bonding is present.
Hence, the correct answer is (A).

I₃^- is linear with the central iodine atom undergoing sp³d-hybridisation and three lone pairs occupying the equatorial position. Thus, this structure includes a d-orbital contribution to the hybridisation scheme.
Hence, the correct answer is (A).

Among the given species, Cr²⁺ and Mn³⁺ have the same number of unpaired electrons.
Unpaired electrons, n = 4
Magnetic moment, μ = √(n(n+2)) = √(4(4+2)) = √24 BM
Thus, the correct answer is (C).

Step 1: Electronic Configuration of Fe
The electronic configuration of Fe is Fe = [Ar] 3d⁶ 4s².
For Fe³⁺, it loses three electrons, resulting in the configuration:
Fe³⁺ = [Ar] 3d⁵
Step 2: Number of Unpaired Electrons
For Fe³⁺, the number of unpaired electrons, n, is 5 (since it has 5 electrons in the 3d orbitals).
Step 3: Spin Only Magnetic Moment Formula
The formula for calculating the spin-only magnetic moment is:
μ_s = √(n(n+2)) BM
where n = 5. Therefore:
μ_s = √(5(5+2)) = √(5 × 7) = √35 BM ≈ 6 BM
Thus, the correct answer is (C).

Step 1: Understanding Lone Pair-Lone Pair Repulsions
Lone pairs of electrons on the central atom experience repulsion from each other. The maximum repulsion occurs when the lone pairs are placed in positions that minimize their mutual interaction. The lone pair-lone pair repulsion increases with the number of lone pairs and the spatial arrangement.
Step 2: Analysis of Compounds
In ClF₃, there are 3 lone pairs on the central chlorine atom and 2 bonding pairs, with a T-shaped geometry.
In IF₅, there are 2 lone pairs on iodine and 5 bonding pairs, with a square pyramidal geometry.
In SF₄, there is 1 lone pair on the sulfur atom and 4 bonding pairs, with a seesaw geometry.
In XeF₂, there are 3 lone pairs on the central xenon atom and 2 bonding pairs, with a linear geometry.
Since XeF₂ has the most lone pairs (3), the lone pair-lone pair repulsions are the highest, and hence, XeF₂ experiences the maximum repulsion.
Thus, the correct answer is (D).

Step 1: Analysis of Canonical Forms
Canonical forms are the different resonance structures that a molecule can have, resulting in delocalization of electrons. The species with the maximum number of canonical forms has the most stable resonance structure.
Step 2: Analysis of Compounds
SO₃ has three resonance structures but lacks a π-bond between sulfur and oxygen in each structure.
O₂ has two resonance structures with a single π-bond between the two oxygen atoms, but it doesn’t have the maximum number of canonical forms.
SO₂ has two resonance structures, but again, it doesn’t maximize the number of canonical forms.
CO₃^2- has three resonance structures, each with a π-bond and delocalized electrons between the carbon and the oxygen atoms.
The carbonate ion (CO₃^2-) has the maximum number of canonical forms.
Thus, the correct answer is (D).

Step 1: {Hybridisation of the Central Atom}
In sp³d² hybridisation, the central atom undergoes hybridisation by combining one s, three p, and two d orbitals, resulting in six hybrid orbitals. This hybridisation occurs in species where the central atom is bonded to six other atoms.
Step 2: {Analysis of Compounds}
In BrF₅, bromine undergoes sp³d² hybridisation, as it is bonded to five fluorine atoms.
In SF₆, sulfur undergoes sp³d² hybridisation, as it is bonded to six fluorine atoms.
In [CrF₆]^3-, chromium undergoes sp³d² hybridisation due to the bonding with six fluoride ions.
In PF₅, phosphorus undergoes sp³d hybridisation, as it is bonded to five fluorine atoms, not six.
Thus, the correct answer is (D).

Step 1: The first equation representing the change in internal energy is: ΔU = Δq + ΔW
Step 2: As ΔU = 0, we have: Δq = - ΔW
Step 3: From the graph, work done ΔW is the area under the curve. The area under the curve is the area of a circle: ΔW = πr²
Step 4: Given the radius as (25 - 5)/2, we calculate the work done as: ΔW = π ((25 - 5)/2)² = π × 10²
Thus, the heat absorbed is: Δq = 100 π J

Step 1: {Using Given Data}
Let the bond dissociation energy of X₂ be a kJ/mol. Thus, the bond dissociation energy of Y₂ is 0.5a kJ/mol and the bond dissociation energy of XY is a kJ/mol. The formation reaction is: 1/2 X₂ + 1/2 Y₂ → XY, ΔH = -200 kJ/mol
Step 2: {Bond Energy Formula}
The bond dissociation energy can be calculated using the formula for enthalpy change: ΔH = BE(Reactants) - BE(Products)
ΔH = [1/2 BE(X₂) + 1/2 BE(Y₂)] - BE(XY)
ΔH = [a/2 + 0.5a/2] - a
-200 = (a + 0.5a)/2 - a
-200 = 0.75a - a
-200 = -0.25a
a = 800 kJ/mol
Thus, the correct answer is (D).

Step 1: {Understanding the Relations}
Relation (A) is the incorrect one.
The correct expression for enthalpy change is ΔH = ΔU + PΔV.
The expression provided in option (A) is incorrect because it subtracts PΔV instead of adding it.
Step 2: {Correct Relations}
Relation (B) is correct because the change in internal energy ΔU is the sum of heat added to the system q and the work done on the system W, i.e., ΔU = q + W.
Relation (C) is correct according to the second law of thermodynamics, which states that the total entropy change (system plus surroundings) for a spontaneous process is greater than or equal to zero.
Relation (D) is the correct form of the Gibbs free energy equation.
Thus, the correct answer is (A).

Step 1: Using the Gibbs Free Energy Relation
The standard Gibbs free energy change is related to the equilibrium constant K_c by the equation:
ΔG° = -RT ln K_c
where:
- R = 8.314 J/K⋅mol is the universal gas constant.
- T = 298 K is the standard temperature.
- K_c = 10¹².
Step 2: Substitute Values and Calculate
ΔG° = - (8.314 J/K⋅mol) × (298 K) × ln(10¹²)
ΔG° = - 8.314 × 298 × 2.303 × log 10¹²
ΔG° = - 68.47 kJ/mol
Thus, the correct answer is (C).

Step 1: {Understanding the Problem}
The given reaction is the combustion of benzene. We are given the heat of combustion at constant volume, and we need to find the heat of combustion at constant pressure. For the combustion of benzene: C₆H₆(l) + 15/2 O₂(g) → 6CO₂(g) + 3H₂O(l)
Step 2: {Using the Relation between Heat at Constant Volume and Pressure}
The change in enthalpy is related to the change in internal energy by the equation: ΔH = ΔU + Δn_gRT
where Δn_g is the change in the number of moles of gas between products and reactants. In the given reaction, we calculate the change in the number of moles of gas: Δn_g = (6 mol CO₂) - (15/2 mol O₂) = 6 - 7.5 = -1.5
Thus, we have: ΔH = ΔU + (-1.5) × (8.314 × 10^-3 × 298)
ΔH = -3263.9 + (-1.5) × 2.478 ≈ -3267.6 kJ/mol
Thus, the correct answer is (D).

Step 1: {Understanding Free Expansion of an Ideal Gas}
In free expansion, the gas expands without doing work and without heat exchange with the surroundings. Since the process is adiabatic, q = 0, meaning no heat is exchanged.
Step 2: {Work Done in Free Expansion}
Since the expansion is against a vacuum, no work is done on or by the system. Hence, w = 0.
Step 3: {Temperature Change in Free Expansion}
For an ideal gas undergoing free expansion, there is no change in temperature, so ΔT = 0.
Thus, the correct answer is (D).

Le-Chatelier's principle applies to systems where changes in temperature, pressure, or concentration cause a shift in equilibrium. However, it does not apply to pure solids and liquids, because their concentrations do not change significantly during a reaction.
Step 1: {Identifying the Pure Solids and Liquids}
In reaction (B), Fe and S are pure solids, and thus, their concentrations do not significantly change during the reaction. Therefore, Le-Chatelier's principle is not applicable here.
Thus, the correct answer is (B).

Step 1: {Relation Between K_p and K_c}
The equilibrium constant K_p and K_c are related by the following equation: K_p = K_c(RT)^Δn
where Δn is the change in the number of moles of gas between products and reactants.
Step 2: {Calculate Δn}
For the given reaction: Δn = moles of products - moles of reactants = 1 - (1 + 1/2) = -1/2
Step 3: {Substitute Δn}
Substitute Δn = -1/2 into the relation: K_p = K_c(RT)^-1/2
Thus, the correct answer is (D).

Step 1: {Understanding the Strength of Acids}
Stronger the acid, lower is the pH, and weaker the acid, higher the pH.
Step 2: {Order of Acid Strength}
The strength of the given acids is in the order: HCl > HCOOH > CH₃COOH
Step 3: {Order of pH}
Hence, their pH is in the order: CH₃COOH > HCOOH > HCl
Thus, the correct answer is (C).

Step 1: {Henderson-Hasselbalch Equation}
For an acidic buffer, the pH is given by: pH = pK_a + log([Salt]/[Acid])
Step 2: {Concentrations}
Let the concentration of potassium acetate solution be x M. 20 mL of 0.1 M acetic acid = 20 × 0.1 millimol = 2 millimol
50 mL of x M potassium acetate = x × 50 millimol = 50x millimol
Step 3: {Substitute Values}
Given, pH = 4.8 4.8 = pK_a + log(50x/2)
where pK_a = log 1.8 × 10^-5 = 4.74
Step 4: {Solve for x}
Substitute pK_a: 4.8 = 4.74 + log(50x/2)
0.06 = log(25x)
10^0.06 = 25x
x = 10^0.06/25 = 0.04
Thus, the correct answer is (B) 0.04 M.

Step 1: Oxidation Half Reaction
Write the oxidation half reaction:
SO₂(g) + 2H₂O(l) → HSO₄^-(aq) + 3H⁺(aq) + 2e^-
Step 2: Reduction Half Reaction
Write the reduction half reaction:
MnO₄^-(aq) + 8H⁺(aq) + 5e^- → Mn²⁺(aq) + 4H₂O(l)
Step 3: Balance the Electrons
To balance the electrons, multiply the oxidation half reaction by 5 and the reduction half reaction by 2:
5SO₂(g) + 10H₂O(l) → 5HSO₄^-(aq) + 15H⁺(aq) + 10e^-
2MnO₄^-(aq) + 16H⁺(aq) + 10e^- → 2Mn²⁺(aq) + 8H₂O(l)
Step 4: Combine and Simplify
Combine the two half reactions:
2MnO₄^-(aq) + 5SO₂(g) + 16H⁺(aq) → 2Mn²⁺(aq) + 5HSO₄^-(aq) + 8H₂O(l)
Thus, the stoichiometric coefficient of SO₂ is 5.

Step 1: Balance the Redox Reaction
The balanced ionic equation for the reaction is:
MnO₄^- + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
Step 2: Stoichiometric Relationship
From the equation, it is clear that 1 mole of KMnO₄ reacts with 5 moles of FeSO₄.
Step 3: Apply Molarity Equation
Use the molarity equation to balance the reaction:
(M₁V₁)/n₁ (KMnO₄) = (M₂V₂)/n₂ (FeSO₄)
(1 × V₁)/(8 × 1) = (25 × 5)/(4 × 5)
Step 4: Solve for V₁
V₁ = (1 × 25 × 8) / (4 × 5)
V₁ = 10 cm³ or V₁ = 10.0 mL
Thus, the correct answer is (D).

Step 1: Understanding Disproportionation and Redox Reactions
A disproportionation reaction is a special type of redox reaction in which a single element undergoes both oxidation and reduction. In contrast, a normal redox reaction involves oxidation and reduction of different elements.
Step 2: Oxidation States Analysis
Let's analyze the oxidation states of phosphorus (P) and sulfur (S) in the reaction:
P₄ + 8SOCl₂ → 4PCl₃ + 2S₂Cl₂ + 4SO₂
In elemental phosphorus (P₄), P has an oxidation state of 0.
In phosphorus trichloride (PCl₃), P has an oxidation state of +3.
In sulfur oxychloride (SOCl₂), S has an oxidation state of +4.
In disulfur dichloride (S₂Cl₂), S has an oxidation state of +2.
In sulfur dioxide (SO₂), S has an oxidation state of +4.
Step 3: Identify the Nature of the Reaction
Phosphorus (P) is oxidized from 0 to +3.
Sulfur (S) is reduced from +4 to +2.
Since phosphorus only undergoes oxidation and sulfur only undergoes reduction, this is a simple redox reaction, NOT a disproportionation reaction.
Step 4: Verify Other Options
Option (A) involves phosphorus undergoing both oxidation and reduction, making it a disproportionation reaction.
Option (B) involves oxygen undergoing disproportionation from -1 to both -2 and 0.
Option (C) also shows disproportionation of phosphorus.
Thus, the correct answer is (D).

Step 1: Understanding Saline Hydrides
Saline hydrides (ionic hydrides) are formed by alkali and alkaline earth metals with hydrogen.
Examples: LiH, BeH₂, and MgH₂. These hydrides have a significant ionic character, though lighter ones like BeH₂ and MgH₂ have some covalent character.
Thus, Statement I is correct.
Step 2: Are Saline Hydrides Volatile?
Volatility depends on weak intermolecular forces. Saline hydrides form strong ionic bonds, making them non-volatile.
Thus, Statement II is incorrect.
Step 3: Understanding Electron-Precise Hydrides
Electron-precise hydrides like CH₄ and SiH₄ have enough valence electrons to form covalent bonds. These are not Lewis bases because they do not have lone pairs to donate.
Thus, Statement III is incorrect.
Step 4: Chromium Hydride Formula
Chromium forms a hydride with the formula CrH.
Thus, Statement IV is correct.
Step 5: Correct Answer
The correct statements are (I) and (IV).
Thus, the correct answer is (C).

Step 1: Understanding the Role of H₂O₂ as an Oxidizing Agent
Hydrogen peroxide (H₂O₂) can act as both an oxidizing and a reducing agent, depending on the reaction conditions.
In acidic and neutral conditions, it tends to act as an oxidizing agent.
In alkaline medium, it can act as either an oxidizing or reducing agent.
Step 2: Completing the Given Reactions
(i) 2Fe²⁺ + H₂O₂ → 2Fe³⁺ + 2OH^-
Here, H₂O₂ acts as an oxidizing agent by converting Fe²⁺ to Fe³⁺.
(ii) 2MnO₄^- + 6H⁺ + 5H₂O₂ → 2Mn²⁺ + 5O₂ + 8H₂O
Here, H₂O₂ acts as a reducing agent because it is oxidized to O₂.
(iii) I₂ + H₂O₂ + 2OH^- → 2I^- + O₂ + 2H₂O
Here, H₂O₂ acts as a reducing agent by reducing iodine (I₂) to iodide (I^-).
(iv) Mn²⁺ + H₂O₂ → MnO₂ + 2H⁺
Here, H₂O₂ oxidizes Mn²⁺ to MnO₂, acting as an oxidizing agent.
Step 3: Identifying the Correct Answer
H₂O₂ acts as an oxidizing agent in reactions (i) and (iv).
Thus, the correct answer is (B).

Step 1: Understanding Reducing Agent Strength
A reducing agent donates electrons and gets oxidized.
- The strength of a reducing agent increases as we move down the group in the periodic table.
- This happens because the atomic size increases, and the bond strength between the central atom and hydrogen weakens, making electron donation easier.
Step 2: Trend in Group 15 Hydrides
- Group 15 hydrides: NH₃, PH₃, AsH₃, SbH₃, BiH₃.
- The reducing character increases as: NH₃ < PH₃ < AsH₃ < SbH₃ < BiH₃
- Thus, BiH₃ is the strongest reducing agent among the given choices.
Step 3: Correct Answer
Thus, the correct answer is (C).

Step 1: {Understanding Melting Point Trends in Ionic Compounds}
The melting point of an ionic compound depends on:
1. Lattice Energy: Higher lattice energy means a higher melting point.
2. Size of the Anion: Smaller anions lead to stronger lattice energy.
Step 2: {Analyzing the Given Salts}
Fluoride (F^-) is the smallest anion, followed by chloride (Cl^-) and bromide (Br^-).
The order of lattice energy is: LiF > LiCl > LiBr
Since lattice energy determines melting point, we get: LiF > LiCl > LiBr
Step 3: {Correct Answer}
Thus, the correct order is: LiF (II) > LiCl (I) > LiBr (III)
Thus, the correct answer is (B).

Step 1: {Understanding the Role of Detergents}
Detergents are surfactants that help in emulsifying grease and oils, making them soluble in water.
Common detergents include anionic detergents such as sodium lauryl sulphate (CH₃(CH₂)₁₀CH₂ - OSO₃^-Na⁺).
Step 2: {Why is Sodium Lauryl Sulphate Used?}
Sodium stearate is used in soaps, not in detergents.
Calcium stearate is insoluble and not used in detergents.
Sodium lauryl sulphate is widely used in detergents due to its excellent cleansing properties.
Thus, the correct answer is (D).

Step 1: {Thermal Decomposition of Lithium Nitrate}
The decomposition reaction for lithium nitrate is: 4LiNO₃ → 2Li₂O + 4NO₂ + O₂
Step 2: {Explanation of Products}
Lithium nitrate decomposes to form lithium oxide (Li₂O), nitrogen dioxide (NO₂), and oxygen (O₂).
The presence of NO₂ and O₂ confirms the decomposition follows a distinct pathway compared to other nitrates.
Thus, the correct answer is (D).

Step 1: Identifying the Double Bonds
The given compound has two double bonds.
The key factor in determining geometrical isomerism is whether each double bond has two different substituents.
Step 2: Checking for Geometrical Isomerism
The double bond at position C₂ - C₃ has two different groups, allowing cis-trans isomerism.
The double bond at C₄ - C₅ has identical hydrogen atoms, preventing cis-trans isomerism.
Step 3: Number of Isomers
Only one double bond exhibits cis-trans isomerism.
Therefore, only two geometrical isomers exist.
Thus, the correct answer is (A).

Step 1: Evaluating Stability Factors
Carbanion stability depends on resonance, hybridization, and inductive effects.
Aromatic stabilization makes carbanions more stable.
Step 2: Stability Order
Compound (D) is aromatic, making it the most stable.
Compound (A) is anti-aromatic and the least stable.
Compounds (B) and (C) are less stable due to lack of delocalization.
Thus, the correct answer is (D).

Step 1:

Step 1: {Understanding IUPAC Nomenclature Rules}
The longest chain in the benzene ring is chosen as the parent chain. The substituents are assigned numbers based on priority groups to ensure the lowest possible sum.
Step 2: {Numbering the Benzene Ring}
Methyl (-CH₃) is given the highest priority and assigned position 1. Chlorine (-Cl) is then assigned position 2, and nitro (-NO₂) is assigned position 4.
Step 3: {Naming the Compound}
The correct name following IUPAC rules is 2-Chloro-1-methyl-4-nitrobenzene.
Thus, the correct answer is (C).

Step 1: {Understanding Free Radical Stability Factors}
The stability of free radicals is influenced by:
1. Resonance stabilization
2. Hyperconjugation
3. Inductive effects
Step 2: {Evaluating Stability of Each Compound}
Free radicals (II) and (I) have stabilizing effects from resonance and hyperconjugation.
Free radical (III) has a strong -I effect from -NO₂, which destabilizes it.
Free radical (IV) has a weak electron-withdrawing effect from -CN, making it more stable than (III).
Step 3: {Ranking Stability Order}
The stability order follows: II > I > IV > III
Thus, the correct answer is (B).

Step 1: {Understanding Bronsted Bases}
A Bronsted base is a substance that can accept protons. The strength of a Bronsted base depends on:
1. The availability of a lone pair for protonation.
2. The electron-donating or withdrawing effects of the surrounding groups.
Step 2: {Evaluating Each Structure}
Structure 1 has electron-withdrawing effects, reducing basicity.
Structure 2 has delocalization of the lone pair, reducing its basicity.
Structure 3 is in a constrained ring, affecting electron availability.
Structure 4 has a localized lone pair, making it the most basic.
Thus, the correct answer is (D).

Step 1: {Understanding Reactivity of Benzyl Halides}
Benzyl halides are more reactive than aryl halides because the benzylic position stabilizes carbocation intermediates.
Step 2: {Analyzing the Reaction Mechanism}
The amine group (NH₃) undergoes nucleophilic substitution at the benzylic position.
The bromine remains intact since the benzyl chloride reacts first due to higher reactivity.
Step 3: {Identifying the Correct Product}
The correct product is the one where amination occurs at the benzylic carbon, leading to o-bromobenzylamine.
Thus, the correct answer is (C).

Step 1: {Understanding the Reaction Mechanism}
The given reaction involves sodium ethoxide (CH₃CH₂ONa) and tert-butyl chloride [(CH₃)₃CCl] in ethanol.
Tertiary alkyl halides undergo elimination (E2) rather than substitution due to steric hindrance.
Step 2: {Identifying the Major Product}
Since sodium ethoxide is a strong base, elimination occurs via the E2 mechanism, leading to the formation of an alkene. The product formed is isobutene (CH₂ = C(CH₃)₂).
Thus, the correct answer is (B).

Step 1: {Understanding Nitrogen Oxide Formation}
Dinitrogen (N₂) reacts with oxygen (O₂) at high altitudes to form nitrogen oxides. The reaction occurs as follows: N₂ (g) + O₂ (g) → 2NO (g)
Step 2: {Conversion to NO₂}
Nitric oxide (NO) further reacts with oxygen: 2NO (g) + O₂ (g) → 2NO₂ (g)
Nitrogen dioxide (NO₂) is a toxic gas that damages plant leaves and hinders photosynthesis.
Thus, the correct answer is (C).

Step 1: {Degree of Dissociation (α) and Van't Hoff Factor (i)}
The reaction for dissociation of K₄[Fe(CN)₆] is: K₄[Fe(CN)₆] ⇌ 4K⁺ + [Fe(CN)₆]^4-
Since the dissociation is 50% (α = 0.5), the total number of particles: i = 1 + 4α = 1 + 4(0.5) = 3
Step 2: {Applying the Osmotic Pressure Formula}
π = iCRT
Substituting values: π = (3)(0.1)(8.314)(300)
π = 7.48 atm
Thus, the correct answer is (A).

Step 1: {Calculating Moles of Solute}
Moles of NaCl = 58.5/58.5 = 1 mol
Moles of Glucose = 180/180 = 1 mol
Step 2: {Dissociation of NaCl}
NaCl dissociates into two ions (Na⁺ and Cl^-), increasing the effective particle concentration. Glucose does not dissociate, so the number of particles remains the same.
Thus, NaCl solution will have a higher elevation in boiling point.
Thus, the correct answer is (A).

Definition of Molarity
Molarity (M) is defined as: M = moles of solute / liters of solution
Thus, 1 M solution means 1 mole of solute is present in 1 L of solution.
Thus, the correct answer is (B).

Step 1: Understanding Colligative Properties
- Colligative properties depend on the number of particles in the solution rather than their nature.
- The van't Hoff factor (i) represents the number of ions each molecule dissociates into.
Step 2: Calculating the Number of Ions
BaCl₂ dissociates into 3 ions: Ba²⁺ and 2Cl^-.
AgNO₃ dissociates into 2 ions: Ag⁺ and NO₃^-.
Urea does not dissociate, so i = 1.
(NH₄)₃PO₄ dissociates into 4 ions: 3 NH₄⁺ and 1 PO₄^3-.
Step 3: Identifying the Substance with the Highest Colligative Property
The greater the number of ions, the higher the colligative properties.
Since (NH₄)₃PO₄ produces 4 ions, it exhibits the highest colligative effect.
Thus, the correct answer is (D).

Step 1: Understanding the Electrolysis of NaCl Solution
Electrolysis of NaCl solution produces NaOH.
The reaction at the cathode generates OH^- ions, increasing pH.
Step 2: Calculating the Number of Moles of NaOH Formed
Total charge passed:
Q = It = 5.0 × 965 = 4825 C
Moles of NaOH produced:
4825/96500 = 0.05 mol
Step 3: Calculating pH
Molarity of NaOH:
0. 05/0.5 = 0.1 M
pOH:
pOH = -log(0.1) = 1
pH:
pH = 14 - 1 = 13
Thus, the correct answer is (D).

Step 1: Understanding the H₂ - O₂ Fuel Cell
The fuel cell reaction is:
2H₂ + O₂ → 2H₂O
Step 2: Calculating the Charge Passed
Charge passed:
Q = It = 1 × 595.1 × 60 × 60 = 2142360 C
Step 3: Finding Water Produced
Water produced:
(36 / (4 × 96500)) × 2142360 ≈ 200 mL
Step 4: Calculating Molarity of NaOH
Moles of NaOH:
5/40 = 0.125 mol
Molarity:
0. 125/0.2 = 0.625 M
Thus, the correct answer is (A).

Step 1: Understanding Faraday’s Laws of Electrolysis
According to Faraday’s first law of electrolysis, the mass of metal deposited on the cathode is directly proportional to the charge passed:
m = ZQ/F
where:
m is the mass of the deposited metal,
Z is the electrochemical equivalent of the metal,
Q is the charge passed (which is the same for all solutions),
F is Faraday’s constant 96500 C/mol.
Step 2: Electrochemical Equivalent and Equivalent Weight
The electrochemical equivalent (Z) is given by:
Z = Atomic weight / (Valency × F)
The more the electrochemical equivalent (Z), the greater the mass of metal deposited.
The equivalent weight is given by:
Equivalent weight = Atomic weight / Valency
Step 3: Calculating Equivalent Weights for Given Options
Zn from ZnSO₄
Atomic weight of Zn = 65
Valency = 2
Equivalent weight = 65/2 = 32.5
Fe from FeCl₃
Atomic weight of Fe = 56
Valency = 3
Equivalent weight = 56/3 ≈ 18.67
Ag from AgNO₃
Atomic weight of Ag = 108
Valency = 1
Equivalent weight = 108/1 = 108
Ni from NiCl₂
Atomic weight of Ni = 58.7
Valency = 2
Equivalent weight = 58.7/2 = 29.35
Step 4: Identifying the Metal Deposited in Maximum Amount
Since the mass deposited is directly proportional to equivalent weight, the metal with the highest equivalent weight will be deposited in the maximum amount.
Silver (Ag) has the highest equivalent weight (108 g/mol), so Ag will be deposited in the greatest amount.
Thus, the correct answer is (C) AgNO₃.

Step 1: Understanding the Reaction Stoichiometry
The given reaction is:
2SO₂ + O₂ ⇌ 2SO₃
Step 2: Relating the Rate of Disappearance and Appearance
The rate of disappearance of O₂ is related to the rate of appearance of SO₃ using stoichiometry:
-d[O₂]/dt = 1/2 d[SO₃]/dt
Step 3: Substituting Values and Solving
Given:
-d[O₂]/dt = 2 × 10^-4 mol L^-1 s^-1
Solving for d[SO₃]/dt:
d[SO₃]/dt = 2 × (2 × 10^-4)
= 4 × 10^-4 mol L^-1 s^-1
Thus, the correct answer is (B).

Step 1: Understanding the Arrhenius Equation
The Arrhenius equation is:
log k = log A - E_a/(2.303RT)
Step 2: Finding the Rate Constant k
For a first-order reaction, the half-life is given by:
k = 0.693/t_1/2
Substituting t_1/2 = 10 minutes = 600 s:
k = 0.693/600 = 1.1 × 10^-3 s^-1
Step 3: Substituting into the Arrhenius Equation
log (1.1 × 10^-3) = log (4 × 10¹³) - (98.6 × 10³) / (2.303 × 8.314 × T)
Step 4: Solving for T
T = 311.15 K
Thus, the correct answer is (D).

Step 1: Understanding the Rate Law
The rate of a reaction is given by:
r = k[A]ⁿ
where k is the rate constant, [A] is the concentration of reactant, and n is the order of the reaction.
Step 2: Applying Given Conditions
If the concentration of A is increased four times, the rate doubles. Mathematically,
2r₁ = k[4A]ⁿ
Using the original rate equation:
r₁ = k[A]ⁿ
Step 3: Dividing the Equations
2r₁/r₁ = k(4[A])ⁿ / k[A]ⁿ
2 = 4ⁿ
Step 4: Solving for n
Taking logarithm on both sides,
log 2 = n log 4
log 2 = n (2 log 2)
n = (log 2) / (2 log 2) = 1/2 = 0.5
Final Answer: The order of reaction is 0.5.

Step 1:
The equation given for the rate constant k is:
log(k₂/k₁) = E_a / (2.303 R) [1/T₁ - 1/T₂]
Where:
- k₂ and k₁ are the rate constants at temperatures T₂ and T₁,
- E_a is the activation energy,
- R is the universal gas constant.
Step 2:
Now, substitute the given values:
log(2k/k) = E_a / (2.303 × 8.3) × [(600 - 300) / (300 × 600)]
Step 3:
Simplifying the expression:
log(2) = E_a / (2.303 × 8.3) × 300/180000
Step 4:
After calculating the logarithmic value of 2, and simplifying further:
log(2) = 0.3010
Now, solving for E_a:
0.3010 = E_a / (2.303 × 8.3) × 1/600
Step 5:
Solving for E_a:
E_a ≈ 3.45 kJ/mol
Final Answer: The activation energy is 3.45 kJ/mol.

Step 1: Understanding the Rate Law
The general rate law is:
r = k[A]ⁿ
where n is the order of the reaction.
Step 2: Given Condition
When the concentration of A is doubled, the rate remains unchanged.
Mathematically,
r₂ = k[2A]ⁿ
Since r₁ = r₂, we equate:
k[A]ⁿ = k[2A]ⁿ
Step 3: Solving for n
Dividing both sides:
1 = 2ⁿ
2ⁿ = 1
n = 0
Final Answer: The reaction follows zero order kinetics.

Step 1: {Understanding Half-Life in First-Order Reactions}
For a first-order reaction, the half-life (t_1/2) is constant and given by: t_1/2 = 0.693/k
From the given data, the concentration falls from 0.8 M to 0.4 M in 15 minutes.
Since t_1/2 is constant, another half-life will reduce it from 0.4 M to 0.2 M in another 15 minutes.
Another half-life will further reduce it from 0.1 M to 0.025 M, which takes another 15 minutes.
Step 2: {Total Time Required}
Two half-lives are needed to reduce 0.1 M to 0.025 M.
2 × 15 = 30 minutes.
Final Answer: 30 minutes.

Step 1: {Understanding Colloidal Systems}
Colloidal particles are heterogeneous mixtures with dispersed-phase particles ranging from 1 nm to 1000 nm.
They exhibit electrical charge, which helps maintain stability by preventing coagulation.
Step 2: {Mechanism of Charge Development}
The charge on colloidal particles arises primarily due to the adsorption of ions from the solution.
When a colloidal particle is in contact with a solution, selective adsorption of a particular ion occurs.
Step 3: {Examples of Charge Formation}
- If AgI sol is prepared in an excess of KI, the colloidal particles adsorb I^- ions and acquire a negative charge: AgI + I^- → Negatively charged AgI colloid
- If AgI sol is prepared in an excess of AgNO₃, the colloidal particles adsorb Ag⁺ ions and acquire a positive charge: AgI + Ag⁺ → Positively charged AgI colloid
Step 4: {Importance of Charge in Colloidal Stability}
- The mutual repulsion between similarly charged colloidal particles prevents coagulation and keeps them dispersed.
- Oppositely charged colloids can undergo coagulation (precipitation) when mixed.
Final Answer: The charge on colloidal particles is due to the adsorption of ions from the solution.

Step 1: {Understanding Smelting in Copper Extraction}
- Copper is extracted from copper pyrites (CuFeS₂) in a blast furnace.
- The ore contains iron as an impurity, which must be removed as slag.
Step 2: {Chemical Reactions During Smelting}
- Iron present as FeS gets oxidized to iron oxide (FeO): 2FeS + 3O₂ → 2FeO + 2SO₂
- FeO then reacts with silica (SiO₂) to form ferrous silicate (FeSiO₃), which is the slag: FeO + SiO₂ → FeSiO₃
Step 3: {Role of Slag}
- The formation of FeSiO₃ helps in removing iron impurities from the molten copper.
- The lighter FeSiO₃ floats on top of the molten copper and is separated easily.
Final Answer: The slag formed is ferrous silicate (FeSiO₃).

Step 1: {Understanding Mineral Composition}
Different ores and minerals have characteristic chemical compositions:
1. Calamine (ZnCO₃):
- It is the main ore of zinc and contains zinc carbonate.
- When heated, it decomposes to form zinc oxide: ZnCO₃ → ZnO + CO₂
2. Malachite (CuCO₃·Cu(OH)₂):
- It is a basic copper carbonate mineral.
- It gives green color to many copper-rich rocks.
3. Magnetite (Fe₃O₄):
- It is one of the most common iron ores.
- It contains mixed oxidation states of iron.
4. Cryolite (Na₃AlF₆):
- It is a fluoride mineral used in the Hall-Héroult process for aluminum extraction.
- It acts as a flux to lower the melting point of alumina.
Step 2: {Identifying the Correct Answer}
- The correct composition is ZnCO₃, CuCO₃·Cu(OH)₂, Fe₃O₄, Na₃AlF₆.
Final Answer: The correct answer is (A).

Step 1: Understanding the Molecular Geometry
PCl₅ has a trigonal bipyramidal structure with two types of bond positions: axial and equatorial.
The three chlorine atoms in the equatorial plane are bonded at 120°, while the two axial chlorine atoms are at 90° with respect to the equatorial bonds.
Step 2: Bond Length Difference
Due to repulsion effects, axial bonds are longer than equatorial bonds.
The axial bonds experience greater repulsion from equatorial bonds, causing them to be longer.
Final Answer: PCl₅ has unequal bond lengths due to its trigonal bipyramidal shape.

Step 1: Identify the Reaction Type
The reaction involves arsenic(III) oxide reacting with hydrogen sulfide to form arsenic sulfide (As₂S₃).
Step 2: Balancing the Reaction
As₂O₃ + 3H₂S → As₂S₃ + 3H₂O
Here, arsenic oxide is reduced to arsenic sulfide while hydrogen sulfide is oxidized.
Final Answer: The sol formed in this reaction is arsenic sulfide (As₂S₃).

Step 1: Understanding Reactivity Series
The tendency of a metal to liberate hydrogen from acids depends on its position in the electrochemical series.
More electropositive metals react with acids to release H₂ gas.
Step 2: Analyzing the Given Metals
Zn and Mn are highly reactive and readily react with acids.
Ni has moderate reactivity.
Cu lies below hydrogen in the reactivity series, meaning it does not react with acids to liberate H₂.
Final Answer: Copper (Cu) has the least tendency to liberate H₂ from acids.

Step 1: Understanding Oxidation States of Transition Metals
Transition metals exhibit variable oxidation states due to the availability of d-orbitals for bonding.
Step 2: Oxidation States of Given Metals
Mn (Manganese) shows oxidation states ranging from +2 to +7, making it the metal with the highest oxidation states.
Fe (Iron) shows +2, +3, +6.
Co (Cobalt) shows +2, +3, +4.
Ti (Titanium) shows +2, +3, +4.
Final Answer: Mn exhibits the highest number of oxidation states.

Step 1: {Identify the Metal Ion}
Nickel in [Ni(CN)₄]^2- has an oxidation state of +2.
Its electronic configuration: 3d⁸ 4s⁰.
Step 2: {Effect of Ligand}
CN^- is a strong field ligand, which causes pairing of electrons in the d-orbitals.
This leads to dsp² hybridization, forming a square planar geometry.
Final Answer: [Ni(CN)₄]^2- undergoes dsp² hybridization, leading to a square planar shape.

Step 1: Oxidation Number Calculation Ni(CO)₄:
Carbonyl (CO) is a neutral ligand. Since Ni has no charge, oxidation state is 0.
[Fe(H₂O)₅NO]²⁺: NO ligand contributes +1 oxidation state. Fe must be +1 to balance the charge.
[Co(CO)₅]^2-: Since CO is neutral, oxidation state of Co is -2.
[Cr₂(CO)₁₀]^2-: CO is neutral, so the total oxidation state of both Cr atoms is -2. Each Cr has -1 oxidation state.
Final Answer: A-II, B-I, C-IV, D-III

Step 1: Understanding Ligand Field Strength
Ligand field strength follows the spectrochemical series.
The stronger the ligand field, the greater the splitting of d-orbitals in transition metal complexes.
Step 2: Spectrochemical Series Order
S^2- < C₂O₄^2- < NH₃ < en < CO
Explanation:
Sulfide (S^2-) is a weak field ligand.
Oxalate (C₂O₄^2-) is stronger than S^2- but still weak.
Ammonia (NH₃) is a moderate field ligand.
Ethylenediamine (en) is stronger than NH₃ due to chelation effect.
CO is the strongest field ligand, leading to the greatest crystal field splitting.

Step 1: Structure of Ferrocene
Ferrocene is Fe(η⁵ - C₅H₅)₂.
It consists of two η⁵-cyclopentadienyl (Cp^-) anions bonded to an Fe²⁺ ion.
It exhibits a "sandwich" structure with iron between two parallel Cp rings.

Step 1: Understanding the Complex
- The chemical formula of nitropentammine chromium (III) chloride is:
[Cr(NH₃)₅(NO₂)]Cl₂
This complex contains the nitro ligand (NO₂), which can bind through either the nitrogen or oxygen.
Step 2: Types of Isomerism
Optical Isomerism: Seen in chiral molecules with non-superimposable mirror images. Not applicable here.
Ionization Isomerism: When exchange of anions leads to different compounds in solution. Not applicable.
Polymerization Isomerism: Occurs when two complexes have the same empirical formula but different molecular formulas. Not applicable.
Linkage Isomerism: Occurs when a ligand can bind through two different atoms. (Correct Answer)
Step 3: Linkage Isomerism in This Complex
The NO₂ group can bind via:
Nitro form (NO₂-N)
Nitrito form (ONO)
Final Answer: The correct isomerism is linkage isomerism due to NO₂ binding through either N or O.

Step 1: Identifying the Electronic Configuration
Nickel (Ni) has an atomic number 28.
The electronic configuration of Ni is:
[Ar] 3d⁸ 4s²
The oxidation state of Ni in the given complexes needs to be determined.
Step 2: Evaluating Each Complex
(A) [Ni(Cl)₄]^2-
Cl^- is a weak field ligand.
The tetrahedral geometry follows an sp³ hybridization.
Paramagnetic due to unpaired electrons. Not the correct answer.
(B) [Co(C₂O₄)₃]^3-
Co³⁺ has a strong field ligand (oxalate).
Octahedral structure, not tetrahedral. Incorrect choice.
(C) [Ni(CN)₄]^2-
CN^- is a strong field ligand.
Follows dsp² hybridization, giving square planar geometry.
Diamagnetic, but not tetrahedral. Incorrect choice.
(D) [Ni(CO)₄]
CO is a strong field ligand.
Causes electron pairing and sp³ hybridization, forming a tetrahedral structure.
No unpaired electrons = Diamagnetic. Correct answer.
Final Answer: The correct diamagnetic, tetrahedral complex is [Ni(CO)₄].

Step 1: Understanding Ferrocene
Ferrocene is an organometallic compound with an iron (Fe²⁺) center.
It consists of two cyclopentadienyl (C₅H₅^-) rings bonded in a sandwich-like structure.
Step 2: Type of Bonding
The hapticity (η⁵) indicates that all five carbon atoms of the ring interact with the metal center.
This forms a stable and symmetrical structure.
Final Answer: The correct formula of ferrocene is Fe(η⁵ - C₅H₅)₂.

Step 1: understanding what is catalog
Calgon is a water softener used to prevent the formation of scale in boilers.
It is chemically known as sodium hexametaphosphate (Na₆P₆O₁₈).
Step 2: Role in Water Softening
It sequesters Ca²⁺ and Mg²⁺ ions by forming soluble complexes, preventing their precipitation.
Final Answer: The chemical name of calgon is sodium hexametaphosphate.

Step 1: Factors Affecting Crystal Field Splitting
The magnitude of Δ₀ depends on the oxidation state, metal size, and ligand type.
Higher oxidation states lead to stronger field splitting.
Step 2: Comparing the Ions
Cr³⁺ has a smaller ionic radius than Mn³⁺, Fe³⁺, and Ti³⁺.
Smaller cations have a stronger ligand interaction, increasing Δ₀.
Final Answer: The complex [Cr(OH)₆]³⁺ has the highest crystal field splitting energy.

Step 1: Identify the Ligands
NH₃ (Diammine)
Cl^- (Chloro)
NH₂CH₃ (Methanamine)
Step 2: Determine the Oxidation State
Pt is in the +2 oxidation state.
Step 3: Naming the Complex
Ligands are named alphabetically.
The correct name is Diamminechloro (methanamine) platinum (II) chloride.

Step 1: Understanding Magnetic Properties
Paramagnetic complexes have unpaired electrons.
The more unpaired electrons, the stronger the attraction to a magnetic field.
Step 2: Compare Electron Configurations
Zn²⁺ (d¹⁰) → 0 unpaired electrons (diamagnetic).
Co²⁺ (d⁷) → 3 unpaired electrons (paramagnetic).
Co³⁺ (d⁶, low spin) → 0 unpaired electrons (diamagnetic).
Ni²⁺ (d⁸) → 2 unpaired electrons (paramagnetic, but weaker than Co²⁺).
Final Answer: [Co(H₂O)₆]²⁺ has the highest paramagnetism due to 3 unpaired electrons.

Step 1: Understanding the S_N2 Mechanism
The S_N2 reaction proceeds via a backside attack, where steric hindrance significantly affects the rate of reaction.
The lower the steric hindrance around the leaving group, the faster the reaction.
Step 2: Evaluating Steric Hindrance
Option (A) and (B): Both have sterically hindered secondary and tertiary carbons, making S_N2 difficult.
Option (D): A primary halide but still has a longer chain.
Option (C): A primary halide with the least steric hindrance, making it the fastest in S_N2.
Final Answer: CH₃CH₂Br has the highest relative rate.

Step 1: Understanding the Victor-Meyer Test
This test differentiates primary, secondary, and tertiary alcohols based on their reactions with HI, AgNO₂, and KOH.
Step 2: Color Development
Primary Alcohols: Convert to red nitroalkane.
Secondary Alcohols: Convert to blue pseudonitrole.
Tertiary Alcohols: Do not form a stable compound and remain colorless.
Final Answer: Red, blue, colorless.

Step 1: Understanding the Industrial Synthesis
The reaction represents the industrial production of methanol from CO and H₂.
It is performed over ZnO-Cr₂O₃ catalyst at high temperature and pressure.
Final Answer: ZnO-Cr₂O₃ is used as the catalyst under specified conditions.

Step 1: Understanding Lucas Test
Lucas test differentiates alcohols based on their reaction with ZnCl₂/HCl.
Step 2: Order of Reactivity
Tertiary Alcohols react fastest via the S_N1 mechanism.
Final Answer: Tertiary alcohol via S_N1.

Step 1: Understanding Aldol Condensation
Aldol condensation occurs in aldehydes/ketones that contain at least one α-hydrogen.
The α-hydrogen is removed by a base to form an enolate ion, which then attacks another carbonyl compound.
Step 2: Analyzing the Given Compounds
CH₃CH₂CHO (Propionaldehyde) contains an α-hydrogen, allowing aldol condensation.
CH₂ = CH - CHO (Acrolein) and CH ≡ C - CHO do not easily form enolates due to resonance stabilization.
C₆H₅CHO (Benzaldehyde) lacks α-hydrogen, preventing aldol condensation.
Final Answer: CH₃CH₂CHO undergoes aldol condensation.

Step 1: Understanding Ozonolysis
Ozonolysis of an alkene breaks the double bond and forms aldehydes and ketones.
Identifying the correct alkene requires reversing this process.
Step 2: Identifying the Structure of X
The products given are Propan-2-one (CH₃COCH₃) and methanal (HCHO).
The only alkene that produces these fragments is 2-Methylpropene (CH₃ − C(CH₃) = CH₂).
Final Answer: The correct alkene is 2-Methylpropene.

Step 1: Understanding Cheilosis
Cheilosis causes cracked lips, inflammation of the mouth, and digestive issues.
It results from riboflavin (Vitamin B₂) deficiency.
Step 2: Role of Riboflavin
Riboflavin plays a crucial role in metabolism and energy production.
It helps in enzyme functions essential for digestive health.
Final Answer: Riboflavin (Vitamin B₂) deficiency causes cheilosis and digestive disorders.

Step 1: Understanding Peptide Sequence Formation
A tetrapeptide consists of 4 amino acids.
The C-terminal amino acid is fixed as alanine.
The N-terminal must be chiral (valine or serine).
Step 2: Listing Possible Sequences
Since glycine is achiral, the N-terminal choices are valine or serine, leading to 4 possible arrangements:
1. Val-Gly-Ser-Ala
2. Val-Ser-Gly-Ala
3. Ser-Gly-Val-Ala
4. Ser-Val-Gly-Ala
Final Answer: The correct number of sequences is 4.

Step 1: Understanding Water-Soluble Vitamins
Water-soluble vitamins (B-complex and C) dissolve in water and are excreted in urine.
However, Vitamin B₁₂ is an exception.
Step 2: Storage of Vitamin B₁₂
Unlike other B vitamins, B₁₂ is stored in the liver.
It plays a vital role in red blood cell formation and neurological functions.
Final Answer: Vitamin B₁₂ is water-soluble but not excreted easily.

Step 1: Understanding Glycosidic Linkage
Glycosidic bonds are covalent bonds formed between sugar molecules through dehydration synthesis.
The type of glycosidic bond determines the digestibility and function of the disaccharide.
Step 2: Comparison of Linkages
Maltose consists of two glucose units linked by a C₁-C₄ α-glycosidic bond.
Sucrose has a glycosidic bond between C₁ of α-glucose and C₂ of β-fructose.
Lactose is made up of galactose and glucose with a C₁-C₄ glycosidic bond.
Amylose consists of glucose monomers linked by C₁-C₄ α-glycosidic bonds.
Final Answer: Since sucrose contains an α-glucose and a β-fructose linked at C₁-C₂, the correct answer is (B).

Step 1: Understanding Functional Groups in Amino Acids
Amino acids contain different functional groups that determine their chemical behavior.
Basic functional groups include amine (-NH₂) and imidazole rings.
Step 2: Classification of Given Options
Arginine, Lysine, and Histidine contain multiple basic groups due to their side chains.
Asparagine contains only one amine group, making it the correct answer.
Final Answer: Asparagine contains a single amine functional group, making it the correct choice.

Step 1: Understanding Semi-Synthetic Polymers
Semi-synthetic polymers are derived by chemically modifying natural polymers.
Examples include cellulose derivatives and modified rubber.
Step 2: Classification of Given Options
Cis-polyisoprene is natural rubber, not a semi-synthetic polymer.
Cellulose nitrate and Cellulose acetate are modified cellulose derivatives.
Vulcanized rubber is chemically treated natural rubber, making it semi-synthetic.
Final Answer: Since cis-polyisoprene is natural rubber, the correct answer is (A).

Step 1: Understanding the Polymerization Process
Zinc acetate and antimony trioxide catalyze the formation of polyesters.
These catalysts are used in the production of Terylene (Dacron), a polyester.
Step 2: Comparison of Polymers
High-density polyethylene, Teflon, and PVC do not require these catalysts.
Terylene is synthesized using zinc acetate-antimony trioxide as a catalyst.
Final Answer: Since Terylene is produced using this catalyst system, the correct answer is (C).

Step 1: Understanding Vasodilators
Vasodilators are substances that widen blood vessels, reducing blood pressure.
Histamine plays a key role in immune response and vasodilation.
Step 2: Classification of Given Options
Histamine causes vasodilation and regulates stomach acid secretion.
Serotonin affects mood and blood clotting but is not a vasodilator.
Codeine is an opioid pain reliever.
Cimetidine is a histamine receptor blocker.
Final Answer: Since histamine is the primary vasodilator, the correct answer is (A).

Step 1: {Let the complex number \( z = x + iy \)
The points corresponding to \( z, \bar{z}, -z, -\bar{z} \) will form a rectangle with vertices \( (x, y), (x, -y), (-x, -y), (-x, y) \).

Step 2: {Find the area of the rectangle
Area of rectangle = \( 2x \times 2y = 4xy \).

Step 3: {Given that the area is \( 2\sqrt{3} \)
Thus, we have:
\( 4xy = 2\sqrt{3} \Rightarrow 2xy = \sqrt{3} \).

Step 4: {Solve for \( x \) and \( y \)
We know that the rectangle's sides are formed by \( x \) and \( y \). Solving this equation will give us:
\( x = \frac{1}{2}, \quad y = \sqrt{3}. \)
Therefore, \( z = \frac{1}{2} + \sqrt{3}i \).

Step 5: {Verify the answer
Thus, the correct value for \( z \) is \( \frac{1}{2} + \sqrt{3}i \), which matches option (A).
Quick Tip: In problems involving rectangles formed by complex numbers, the vertices of the rectangle are represented by the complex number and its conjugate, as well as their negatives.

Step 1: {Given that \( |z_1| = |z_2| = \dots = |z_n| = 1 \)
Thus, we know that \( |z_1| = |z_2| = \dots = |z_n| = 1 \).

Step 2: {Write the sum of complex numbers
Now, we have:
\( z_1 + z_2 + \dots + z_n = z_1 + z_2 + \dots + z_n. \)
Step 3: {Conclusion
By calculating the sum and using the properties of the magnitudes, we find:
\( |z_1 + z_2 + \dots + z_n| = \frac{1}{|z_1|} + \frac{1}{|z_2|} + \dots + \frac{1}{|z_n|}. \)
Step 4: {Verify the result
Thus, the correct value is \( \frac{1}{|z_1|} + \frac{1}{|z_2|} + \dots + \frac{1}{|z_n|} \), which matches option (C).
Quick Tip: In problems involving the sum of complex numbers with equal magnitudes, symmetry and geometric interpretation help simplify the result.

We are given the equation \( |8z_2z_3 + 27z_1z_3 + 64z_1z_2| = |z_1| |z_2| |z_3| \). First, we break down the terms as follows: \[ \left| \frac{8}{z_1} + \frac{27}{z_2} + \frac{64}{z_3} \right| \] This can be rewritten as: \[= (2)(3)(4) \left| \frac{8z_1}{|z_1|^2} + \frac{27z_2}{|z_2|^2} + \frac{64z_3}{|z_3|^2} \right| \] Simplifying the equation: \[= 24 \left| 2\overline{z_1} + 3\overline{z_2} + 4\overline{z_3} \right| \] Finally, we calculate: \[= 24 \left| 2z_1 + 3z_2 + 4z_3 \right| \] This results in: \[= 24 \times 4 = 96 \] Thus, the final answer is 96. Quick Tip: For problems involving absolute values of complex expressions, use the properties of magnitudes and simplify each term before combining them.

Step 1: {Total number of people
The total number of people is \( 10 \), and they are divided into two groups: one group of 4 persons and another group of 6 persons.

Step 2: {Number of ways to form groups
The number of ways to form the two groups is \( \frac{10!}{4!6!} \).

Step 3: {Arrangements on the round tables
For the group of 4 persons, the number of arrangements on a round table is \( (4-1)! = 3! = 6 \). For the group of 6 persons, the number of arrangements on a round table is \( (6-1)! = 5! = 120 \).

Step 4: {Total ways to arrange the guests
The total number of ways to arrange the guests is: \[ \frac{10!}{4!6!} \times 6 \times 120 = \frac{10!}{24}. \] Thus, the total number of ways is \( \frac{10!}{24} \), which matches option (B).
Quick Tip: In problems involving round table arrangements, subtract 1 from the number of people to account for rotational symmetry.

Step 1: {Odd and even digits
The digits of the number are \( 2, 2, 3, 3, 5, 5, 8, 8, 8 \). We have 4 odd digits \( 3, 3, 5, 5 \) and 5 even digits \( 2, 2, 8, 8, 8 \).

Step 2: {Placing odd digits in even positions
Here we have 4 odd digits (3, 3, 5, 5) and 5 even digits (2, 2, 8, 8, 8).
The arrangement is as follows: \[ O \, E \, O \, E \, O \, E \, O \, E \, O \] where, \( E \) represents even places and \( O \) represents odd places.
\[ \Rightarrow \text{Number of ways odd digits will be placed on even places} \] This is given by: \[ \frac{4!}{2!2!} \] \[ \Rightarrow \text{Number of ways even digits will be placed on odd places} \] This is given by: \[ \frac{5!}{2!3!} \] Thus, the total number of ways is: \[ \frac{4!}{2!2!} \times \frac{5!}{2!3!} = 60 \] , which matches option (C).
Quick Tip: When rearranging digits with repetitions, use the formula for permutations of multiset: \( \frac{n!}{k_1!k_2!...k_m!} \), where \( k_1, k_2, \dots \) are the frequencies of the distinct elements.

We are given: \[ \frac{22 P_{r+1}}{20 P_{r+2}} = \frac{11}{52}. \] which can be expressed as: \[ \frac{{^{22}P_{r+1}}}{{^{20}P_{r+2}}} = \frac{11}{52} \] Using the formula for permutations, we know that: \[ ^{22}P_{r+1} = \frac{22!}{(22 - (r + 1))!} \quad \text{and} \quad ^{20}P_{r+2} = \frac{20!}{(20 - (r + 2))!} \] Thus, we have: \[ \frac{\frac{22!}{(22 - (r + 1))!}}{\frac{20!}{(20 - (r + 2))!}} = \frac{11}{52} \] This simplifies to: \[ \frac{{(22)!}}{(21 - r)!} \times \frac{(18 - r)!}{(20)!} = \frac{11}{52} \] \[ \frac{22 \times 21 \times (20 - r)!}{(21 - r)!} = \frac{11}{52} \] Simplifying further, we get: \[ \frac{(21 - r)(20 - r)}{20} = \frac{11}{52} \] \[ (21 - r)(20 - r) = 14 \times 13 \] \[ (21 - r)(20 - r) = (21 - 7)(20 - 7) \] This gives: \[ (21 - r)(20 - r) = 14 \times 13 \] Now, we solve the equation: \[ (21 - r)(20 - r) = (21 - 7)(20 - 7) \] Thus, we find: \[ r = 7 \] Thus, the value of \( r \) is 7, which matches option (C). Quick Tip: For problems involving permutations, simplify the expressions and solve step-by-step to isolate the variable.

We are given 12 contested candidates, and we need to select 4 candidates for the election. The total number of ways of selections where voter votes for at least one candidate is calculated by summing over the possible choices of 1, 2, 3, and 4 candidates: \[ \text{Total selections} = ^1^2C_1 + ^1^2C_2 + ^1^2C_3 + ^1^2C_4 \] Step 1: Calculate individual combinations \[ ^1^2C_1 = 12, \quad ^1^2C_2 = 66, \quad ^1^2C_3 = 220, \quad ^1^2C_4 = 495. \] Step 2: Add them up \[ 12 + 66 + 220 + 495 = 793. \] Thus, the total number of ways of selection is 793, which matches option (A). Quick Tip: For combinations, use the formula \( ^nC_r = \frac{n!}{r!(n-r)!} \) to calculate the number of ways to choose \( r \) objects from \( n \) objects.

We are given the number 12345, and we need to find how many arrangements of its digits will result in at least 3 digits not being in their original position. This is a problem of derangements where we calculate how many digits do not appear in their original position. Step 1: Calculate the total number of arrangements of 5 digits The total number of arrangements of 5 digits is: \[ 5! = 120. \] Step 2: Calculate derangements for different cases (3 digits, 4 digits, and 5 digits out of position) Using inclusion-exclusion, the number of ways that at least 3 digits do not appear in their original position is: \[ 5C3 \times 2 + 5C4 \times 9 + 5C5 \times 44 = 20 + 45 + 44 = 109. \] Thus, the number of arrangements is 109, which matches option (B). Quick Tip: Derangements are used to count the number of permutations where no element appears in its original position. Use inclusion-exclusion for such problems.

We are given \( a > 0, b > 0, c > 0 \) and the condition that \( a, b, c \) are distinct. The goal is to find which of the following expressions \( (a + b)(b + c)(c + a) \) is greater than. Step 1: Use AM-GM inequality We know that the Arithmetic Mean is greater than or equal to the Geometric Mean: \[ AM \geq GM. \] Applying this to the terms \( a + b, b + c, c + a \), we get: \[ (a + b)(b + c)(c + a) \geq 8abc. \] Thus, the correct answer is \( 8abc \), which matches option (D). Quick Tip: Use the AM-GM inequality to compare products of sums for positive distinct values. This is helpful in many problems involving inequalities.

Given, \[ \sum_{k=1}^{n} k(k+1)(k-1) = pn^4 + qn^3 + tn^2 + sn \] Therefore, \[ \sum_{k=1}^{n} (k^3 - k) = pn^4 + qn^3 + tn^2 + sn \] \[ \Rightarrow \left( \frac{n(n+1)}{2} \right)^2 - \frac{n(n+1)}{2} = pn^4 + qn^3 + tn^2 + sn \] On dividing both sides by \( n \), we get \[ \frac{(n(n+1))^2}{4n} - \frac{n(n+1)}{2n} = pn^3 + qn^2 + tn + s \] Put \( n = 0 \), we get \[ 0 - \frac{1}{2} = s \] \[ \Rightarrow s = - \frac{1}{2} \] Thus, the value of \( s \) is \( -1/2 \), which matches option (B). Quick Tip: Use the standard formulas for summations of powers of integers to simplify complex summation expressions.

Solution: Let 3 numbers in AP be \(a, (a + 6), (a + 12)\). Also, the first and last number out of 4 numbers are equal \( \therefore \) 4 numbers are \( (a + 12), a, (a + 6), (a + 12) \). Given that the first 3 numbers are in GP. \begin{align*} \Rightarrow a^2 &= (a + 12)(a + 6) \\ \Rightarrow a^2 &= a^2 + 18a + 72 \\ \Rightarrow a &= \frac{-72}{18} = -4 \end{align*} Thus, the numbers are \( -4, 2 \). Quick Tip: For sequences involving both GP and AP, always check consistency in conditions provided, especially when sequences must terminate at the same number.

We are given the following two infinite geometric series: For \( A \): \[ A = \frac{1}{1 - r^a} \] This implies that: \[ 1 - r^a = \frac{1}{A} \] Rearranging this: \[ r^a = 1 - \frac{1}{A} \] Thus, we can write: \[ r^a = \frac{A - 1}{A} \] For \( B \): \[ B = \frac{1}{1 - r^b} \] This implies that: \[ 1 - r^b = \frac{1}{B} \] Rearranging this: \[ r^b = 1 - \frac{1}{B} \] Thus, we can write: \[ r^b = \frac{B - 1}{B} \] Now, to solve for \( \frac{a}{b} \), we take the logarithm of both sides of each equation. \[ a \log r = \log \left( \frac{A - 1}{A} \right) \] \[ b \log r = \log \left( \frac{B - 1}{B} \right) \] Next, we divide the two equations to find \( \frac{a}{b} \): \[ \frac{a}{b} = \frac{\log \left( \frac{A - 1}{A} \right)}{\log \left( \frac{B - 1}{B} \right)} \] Thus, we have: \[ \frac{a}{b} = \log_r \left( \frac{A - 1}{A} \right) = \log_r \left( \frac{B - 1}{B} \right) \] Therefore, the answer matches option (C). Quick Tip: For problems involving geometric series and logs, transform the series into their sum formula first, then apply logarithmic identities.

Step 1: Let the series be defined as: \[ S = 1 + \frac{5}{6} + \frac{12}{6^2} + \frac{22}{6^3} + \frac{35}{6^4} + \cdots \quad \text{(i)} \] Now, define another series: \[ \frac{S}{6} = \frac{1}{6} + \frac{5}{6^2} + \frac{12}{6^3} + \frac{22}{6^4} + \cdots \quad \text{(ii)} \] Step 2: On subtracting equation (ii) from equation (i): \[ S - \frac{S}{6} = \left(1 + \frac{5}{6} + \frac{12}{6^2} + \frac{22}{6^3} + \frac{35}{6^4} + \cdots \right) - \left(\frac{1}{6} + \frac{5}{6^2} + \frac{12}{6^3} + \frac{22}{6^4} + \cdots \right) \] Step 3: This simplifies to: \[ \frac{5S}{6} = 1 + \frac{4}{6} + \frac{7}{6^2} + \frac{10}{6^3} + \frac{13}{6^4} + \cdots \] Now, multiply both sides by 6: \[ S - \frac{S}{6} = 1 + \frac{5}{6} + \frac{12}{6^2} + \frac{22}{6^3} + \cdots \] Step 4: Now, \[ \frac{5S}{6} - \frac{1}{6} \left(\frac{5S}{6}\right) = 1 + \frac{4}{6} + \frac{7}{6^2} + \frac{10}{6^3} + \frac{13}{6^4} + \dots - \left( \frac{4}{6^2} + \frac{7}{6^3} + \frac{10}{6^4} + \frac{13}{6^5} \dots \right) \] \[ \Rightarrow \frac{25S}{36} = 1 + \frac{3}{6} + \frac{3}{6^2} + \frac{3}{6^3} + \dots \] \[ = 1 + \frac{3}{6} + \frac{\frac{3}{6^2}}{1 - \frac{1}{6}} = 1 + \frac{1}{2} + \frac{\frac{1}{12}}{\frac{5}{6}} = 1 + \frac{1}{2} + \frac{1}{10} = \frac{8}{5} \] Thus, the sum of the series is: \[ S = \frac{288}{125} \] Thus, the value of \( S \) matches option (C). Quick Tip: Use series transformation and subtraction techniques to find closed forms for complex series.

Given the series: \[ \tan^{-1} \left( \frac{1}{1 + 1 \cdot 2} \right) + \tan^{-1} \left( \frac{1}{1 + 2 \cdot 3} \right) + \ldots + \tan^{-1} \left( \frac{1}{1 + n(n+1)} \right) = \tan^{-1}(x) \] We can simplify the terms as follows: \[ \tan^{-1} \left( \frac{(2-1)}{1 + 1 \cdot 2} \right) + \tan^{-1} \left( \frac{(3-2)}{1 + 2 \cdot 3} \right) + \ldots + \tan^{-1} \left( \frac{(n+1)-n}{1 + n(n+1)} \right) = \tan^{-1}(x) \] Using the difference identity for inverse tangents: \[ \tan^{-1}(A) - \tan^{-1}(B) = \tan^{-1}\left( \frac{A - B}{1 + AB} \right) \] Applying this identity for each pair, we simplify the entire expression to: \[ \tan^{-1}(2) - \tan^{-1}(1) + \tan^{-1}(3) - \tan^{-1}(2) + \ldots + \tan^{-1}(n+1) - \tan^{-1}(n) = \tan^{-1}(x) \] Thus, we are left with: \[ \tan^{-1} \left(\frac{n}{n+1}\right) = \tan^{-1}(x) \] Therefore, \[ x = \frac{n}{n+1} \] Thus, the value is \( \frac{n}{n+

By the given condition, \[ \frac{a + b}{2} = 2 \sqrt{ab} \] \[ \Rightarrow a + b = 4 \sqrt{ab} \] Now, \((a - b)^2 = (a + b)^2 - 4ab\) \[ = 16ab - 4ab \] \[ = 12ab \] \[ \therefore a - b = \sqrt{12ab} = 2\sqrt{3} \sqrt{ab} \] (Taking \(+ve\) sign only as \(a > b\)) \[ \frac{a + b}{a - b} = \frac{4\sqrt{ab}}{2\sqrt{3ab}} = \frac{2}{\sqrt{3}} \] By componendo and dividendo, \[ \frac{a + b + a - b}{a + b - a + b} = \frac{2 + \sqrt{3}}{2 - \sqrt{3}} \] \[ \frac{2a}{2b} = \frac{2 + \sqrt{3}}{2 - \sqrt{3}} \] Thus, the ratio is \((2 + \sqrt{3}) : (2 - \sqrt{3})\), matching option (A). Quick Tip: For problems involving means, always test edge cases and simplify radical expressions to find recognizable patterns.

Step 1: {Write the series terms
The general term of the given expression is: \[ y = \sum_{r=0}^{n-1} \tan^{-1} \left( \frac{1}{x^2 + (2r+1)x + (r^2 + 1)} \right) \] \[ = \sum_{r=0}^{n-1} \tan^{-1} \left( \frac{1}{1+(x + r)(x + r + 1)} \right) \] Then we can further write: \[ \tan^{-1} \left( \frac{(x+r+1) - (x+r)}{1 + (x+r)(x+r+1)} \right) \] Step 2: Simplify by using the property: \[ \tan^{-1} A - \tan^{-1} B = \tan^{-1}\left( \frac{A - B}{1 + AB} \right) \] \[ \sum_{r=0}^{n-1} \tan^{-1} \left( \frac{(x+r+1) - (x+r)}{1 + (x+r)(x+r+1)} \right) = \sum_{r=0}^{n-1} \left( \tan^{-1}(x + r + 1) - \tan^{-1}(x + r) \right) \] Step 3: Apply the series and simplify
\[ = \left[ \tan^{-1}(x+1) - \tan^{-1} x \right] + \left[ \tan^{-1}(x+2) - \tan^{-1}(x+1) \right] + \dots + \left[ \tan^{-1}(x+n) - \tan^{-1}(x+n-1) \right] \] \[ = \tan^{-1}(x+n) - \tan^{-1} x \] Step 4: Differentiate with respect to \(x\) \[ \frac{dy}{dx} = \frac{1}{1 + (x + n)^2} - \frac{1}{1 + x^2} \] Thus, the derivative is: \[ \frac{dy}{dx} = \frac{1}{(x + n)^2 + 1} - \frac{1}{x^2 + 1} \] This matches option (B). Quick Tip: For trigonometric series involving inverse functions and variables, look for telescoping patterns to simplify the expression before differentiating.

To find the coefficient of \(x^2\) in the expansion, identify the appropriate terms from the expansion that contribute to \(x^2\) when multiplied. Step 1: Identify relevant terms
The general term in the expansion can be written as: \[ T_{r+1} = {^{10}C_{r} \left(\frac{1}{3}x^{\frac{1}{2}}\right)^{10-r} \left(x^{-\frac{1}{4}}\right)^{r} \] \[ = {^{10}C_{r} \left(\frac{1}{3}\right)^{10-r} x^{\frac{10-r}{2} - \frac{r}{4}} \] We have to find the coefficient of \(x^2\). \[ \frac{10-r}{2} - \frac{r}{4} = 2 \Rightarrow 20 - 2r - r = 8 \Rightarrow 3r = 12 \] \[ \Rightarrow r = 4 \] \[ T_{4+1} = {}^{10}C_{4} \left(\frac{1}{3}\right)^{6} x^2 \] Step 2: Calculate the coefficient
\[ = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} \times \frac{1}{3^6} = \frac{210}{729} \] \[ \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} \times \frac{1}{3^6} = \frac{70}{243} \] Thus, the coefficient of \(x^2\) is \(\frac{70}{243}\), which matches option (A). Quick Tip: Always set the power of \(x\) in the general term equal to the desired power, and solve for \(r\) to find the specific term contributing to that power.

The given expression can be expanded using the Maclaurin series for \(e^x\): Given,
\[\frac{e^{7x} + e^x}{e^{3x}} \Rightarrow e^{4x} + e^{-2x}\] \[\text{Series of } e^a = 1 + \frac{a}{1!} + \frac{a^2}{2!} + \frac{a^3}{3!} + \dots\] \[\Rightarrow e^{4x} + e^{-2x} = \left(1 + \frac{4x}{1!} + \frac{(4x)^2}{2!} + \frac{(4x)^3}{3!} + \dots\right) + \left(1 + \frac{(-2x)}{1!} + \frac{(-2x)^2}{2!} + \dots\right)\] \[\text{Here, coefficient of } x^n = \frac{4^n}{n!} + \frac{(-2)^n}{n!}\] \[x^2 \equiv \frac{4^2}{2!} + \frac{(-2)^2}{2!} = \frac{16}{2!} + \frac{4}{2!}\]
\[x^3 \equiv \frac{4^3}{3!} + \frac{(-2)^3}{3!}\] \vdots \[x^n = \frac{4^n}{n!} + \frac{(-2)^n}{n!}\] This matches option (C), providing the correct coefficient for \(x^n\). Quick Tip: When combining terms from exponential expansions, ensure consistent base and power adjustments to match given options.

This problem involves simplifying two terms raised to the eighth power. The terms inside the parentheses are structured such that they effectively represent hyperbolic cosine functions. Step 1: Identify the simplification strategy \[ \text{Since } (x+\sqrt{x^2-1})^8 + (x-\sqrt{x^2-1})^8 \] \[= 2\{{}^8C_0 x^8 + {}^8C_2 x^6 (x^2-1) + {}^8C_4 x^4 (x^2-1)^2 + {}^8C_6 x^2 (x^2-1)^3 + {}^8C_8 (x^2-1)^4 }\] \[\text{So coefficient of highest power of } x \text{ is } \] \[= 2\{{}^8C_0 + {}^8C_2 + {}^8C_4 + {}^8C_6 + {}^8C_8 }\] \[= 2^8 = 256\] This corresponds to the highest power and matches option (C). Quick Tip: Always consider trigonometric and hyperbolic identities when dealing with complex binomial expressions to simplify calculation.

Given, 17^th and 18^th terms in the expansion \( (2 + a)^{50} \) are equal: \[ T_{17} = T_{18} \] \[ \Rightarrow {}^{50} C_{16} (2)^{34} (a)^{16} = {}^{50} C_{17} (2)^{33} (a)^{17} \] \[ \Rightarrow a = \frac{{}^{50} C_{16}}{{}^{50} C_{17}} \times 2 = \frac{17}{34} \times 2 = 1 \] Now, the coefficient of \( x^{35} \) in the expansion of \[ (1 + x)^{-2} \] The coefficient of \( x^r \) in the expansion of \( (1 - x)^{-2} \) is \( r + 1 \). Thus, the coefficient of \( x^{35} \) in \( (1 + x)^{-2} \) is \( -(35 + 1) = -36 \). Quick Tip: In problems involving equal terms of a binomial expansion, equating the terms helps solve for unknowns. For negative binomial expansions, use properties of the binomial theorem extended to negative exponents.

Using the angle sum property of a triangle: \[ A + B + C = 180^\circ \] \[ \Rightarrow C = 180^\circ - A - B \] \[ \Rightarrow \frac{C}{2} = 90^\circ - \frac{A+B}{2} \] \[ \Rightarrow \tan \frac{C}{2} = \tan(90^\circ - \frac{A+B}{2}) \] \[ \Rightarrow \tan \frac{C}{2} = \cot\left(\frac{A + B}{2}\right) \] \[ \Rightarrow \tan \frac{C}{2} = \frac{1 - \tan \frac{A}{2} \tan \frac{B}{2}}{\tan \frac{A}{2} + \tan \frac{B}{2}} \] Given that: \[ \tan \frac{A}{2} = \frac{1}{3} \quad \text{and} \quad \tan \frac{B}{2} = \frac{2}{3} \] Substituting the values: \[ \tan \frac{C}{2} = \frac{1 - \frac{1}{3} \times \frac{2}{3}}{\frac{1}{3} + \frac{2}{3}} \] \[ \tan \frac{C}{2} = \frac{1 - \frac{2}{9}}{1} \] \[ \tan \frac{C}{2} = \frac{7}{9} \] Quick Tip: Always verify angle sum properties and recalculations when dealing with trigonometric identities in geometry.

We are given the equation: \[ \sin x + \sin 2x + \sin 3x + \sin 4x = 0 \] We can rearrange and group terms: \[ (\sin x + \sin 3x) + (\sin 2x + \sin 4x) = 0 \] Using the sum-to-product identity: \[ \sin A + \sin B = 2 \sin \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right) \] Applying this identity to both pairs of sines: \[ 2 \sin \left( \frac{x + 3x}{2} \right) \cos \left( \frac{x - 3x}{2} \right) + 2 \sin \left( \frac{2x + 4x}{2} \right) \cos \left( \frac{2x - 4x}{2} \right) = 0 \] \[ 2 \sin 2x \cos(-x) + 2 \sin 3x \cos(-x) = 0 \] Since \( \cos(-x) = \cos x \): \[ 2 \cos x(\sin 2x + \sin 3x) = 0 \] Using the sum-to-product identity again: \[ 2 \cos x \left( 2 \sin \left( \frac{2x + 3x}{2} \right) \cos \left( \frac{2x - 3x}{2} \right) \right) = 0 \] \[ 4 \cos x \sin \left( \frac{5x}{2} \right) \cos \left( \frac{-x}{2} \right) = 0 \] Since \( \cos(-x) = \cos x \): \[ 4 \cos x \sin \left( \frac{5x}{2} \right) \cos \left( \frac{x}{2} \right) = 0 \] For the product to be zero, either: \[ \cos x = 0, \quad \sin \left( \frac{5x}{2} \right) = 0, \quad \text{or} \quad \cos \left( \frac{x}{2} \right) = 0 \] We consider the solutions within \( [0, 2\pi] \). If \( \cos x = 0 \), then \( x = \frac{\pi}{2}, \frac{3\pi}{2} \). If \( \sin \left( \frac{5x}{2} \right) = 0 \), then \( \frac{5x}{2} = n\pi \) where \( n \) is an integer. Thus, \( x = \frac{2n\pi}{5} \). For \( n = 0 \), \( x = 0 \). For \( n = 1 \), \( x = \frac{2\pi}{5} \). For \( n = 2 \), \( x = \frac{4\pi}{5} \). For \( n = 3 \), \( x = \frac{6\pi}{5} \). For \( n = 4 \), \( x = \frac{8\pi}{5} \). If \( \cos \left( \frac{x}{2} \right) = 0 \), then \( \frac{x}{2} = \frac{\pi}{2} + n\pi \) where \( n \) is an integer. Thus, \( x = \pi, 3\pi, \dots \). But \( x \in [0, 2\pi] \), so \( x = \pi \). The values of \( x \) are: \[ 0, \frac{\pi}{2}, \frac{2\pi}{5}, \frac{3\pi}{2}, \frac{4\pi}{5}, \pi, \frac{6\pi}{5}, \frac{8\pi}{5}, 2\pi \] Sum of these values: \[ 0 + \frac{\pi}{2} + \frac{2\pi}{5} + \frac{3\pi}{2} + \frac{4\pi}{5} + \pi + \frac{6\pi}{5} + \frac{8\pi}{5} + 2\pi \] \[ = \pi \left( \frac{1}{2} + \frac{2}{5} + \frac{3}{2} + \frac{4}{5} + 1 + \frac{6}{5} + \frac{8}{5} + 2 \right) \] \[ = \pi \left( 4 + \frac{20}{5} \right) = \pi(4 + 4) = 9\pi \] Quick Tip: Use the sum-to-product identity to transform sums of trigonometric functions, and find the solutions in the given interval to determine their sum.

We are given that: \[ \sin 9\theta = \sin \theta \] which can be rewritten as: \[ \sin 9\theta - \sin \theta = 0 \] We use the identity for the difference of sines: \[ \sin A - \sin B = 2 \cos \left( \frac{A + B}{2} \right) \sin \left( \frac{A - B}{2} \right) \] This leads to: \[ 2 \cos \left( \frac{9\theta + \theta}{2} \right) \sin \left( \frac{9\theta - \theta}{2} \right) = 0 \] \[ 2 \cos 5\theta \sin 4\theta = 0 \] Thus, we have two conditions: \[ \cos 5\theta = 0 \quad \text{or} \quad \sin 4\theta = 0 \] For \( \cos 5\theta = 0 \): \[ 5\theta = (2n + 1) \frac{\pi}{2} \] \[ \theta = \frac{(2n + 1) \pi}{10} \] For \( \sin 4\theta = 0 \): \[ 4\theta = n\pi \] \[ \theta = \frac{n\pi}{4} \] Now, substituting \( n = 0, 1, 2, \dots \), we get the solutions for \( \theta \) from \( \cos 5\theta = 0 \): \[ \theta = \frac{\pi}{10}, \frac{3\pi}{10}, \frac{5\pi}{10}, \frac{7\pi}{10}, \frac{9\pi}{10}, \frac{11\pi}{10}, \frac{13\pi}{10}, \frac{15\pi}{10}, \frac{17\pi}{10}, \frac{19\pi}{10} \] For the second set of solutions \( \sin 4\theta = 0 \): \[ \theta = 0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi, \frac{5\pi}{4}, \frac{3\pi}{2}, \frac{7\pi}{4}, 2\pi \] Thus, we have a total of 17 solutions, and the common solutions are \( \frac{\pi}{2} \) and \( \frac{3\pi}{2} \). Total number of solutions = 17 Quick Tip: Use graphical or numerical methods to verify the count of solutions for trigonometric equations.

We can write \( 8\sin\theta + 6\cos\theta \) as \( 10\left(\frac{4}{5}\sin\theta + \frac{3}{5}\cos\theta\right) \), which is in the form \( a \sin\theta + b \cos\theta \) with \( a = 8 \) and \( b = 6 \). We know that the maximum and minimum values of \( a \sin \theta + b \cos \theta \) are \( \pm\sqrt{a^2 + b^2} \), respectively, and the value of \[ -\sqrt{8^2 + 6^2} \leq 8\sin\theta + 6\cos\theta \leq \sqrt{8^2 + 6^2} \] \[ \Rightarrow -10 \leq 8\sin\theta + 6\cos\theta \leq 10 \] Now: \[ 0 \leq (8\sin\theta + 6\cos\theta)^2 \leq 100 \] Adding 2 to the entire inequality: \[ 2 \leq (8 \sin \theta + 6 \cos \theta)^2 + 2 \leq 102 \] The correct range is \([2, 102]\). Quick Tip: Always adjust range calculations by considering the transformations applied to the trigonometric functions (scaling and translation).

Given that: \[ x = a \left( \frac{1 - t^2}{1 + t^2} \right) \quad \text{and} \quad y = \frac{2at}{1 + t^2} \] Let \( t = \tan \theta \). Thus: \[ x = a \left( \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \right) \quad \text{and} \quad y = \frac{2a \tan \theta}{1 + \tan^2 \theta} \] From this, we have: \[ x = a \cos 2\theta \quad \text{and} \quad y = a \sin 2\theta \] We can write: \[ \cos 2\theta = \frac{x}{a} \quad \text{and} \quad \sin 2\theta = \frac{y}{a} \] Squaring both sides: \[ \cos^2 2\theta = \frac{x^2}{a^2} \quad \text{and} \quad \sin^2 2\theta = \frac{y^2}{a^2} \] Adding these equations: \[ \cos^2 2\theta + \sin^2 2\theta = \frac{x^2}{a^2} + \frac{y^2}{a^2} \] Using the Pythagorean identity \( \cos^2 \theta + \sin^2 \theta = 1 \), we get: \[ 1 = \frac{x^2 + y^2}{a^2} \] Thus, the equation becomes: \[ x^2 + y^2 = a^2 \] This represents the equation of a circle with center at the origin and radius \( a \). Therefore, the locus of the point is a circle having center at the origin and radius \( a \). Quick Tip: Understanding the geometrical interpretation of parametric equations simplifies the identification of loci.

Here, point \( A \) is the intersection of line \( AB \) and \( AC \). So, the equation of the line passing through \( A \) is given by: \begin{equation} (x + 2y - 1) + \lambda (2x + 3y - 1) = 0 \end{equation} This line passes through the orthocentre \( (0,0) \), hence substituting \( (0,0) \): \[ -1 + \lambda (-1) = 0 \] \[ -1 - \lambda = 0 \] \[ \lambda = -1 \] Substituting \( \lambda = -1 \) in Eq. (1), we get: \[ x + y = 0 \] Thus, the equation of \( AD \) is: \begin{equation} x + y = 0 \end{equation} Since \( AD \perp BC \), therefore: \[ a x + by = 0 \] Comparing the equation with this, we get \( a + b = 0 \). Similarly, by applying the condition that \( BE \) is perpendicular to \( CA \), we obtain: \( a + 2b = 8 \). From Eqs. (3) and (4), we obtain: \[ a = -8, \quad b = 8. \] Thus, the values are: \[ a = -8, \quad b = 8 \] Quick Tip: Understanding orthocentre properties helps simplify the problem-solving approach.

Using the formula for the image of a point $(x_1,y_1)$ with respect to the line $Ax + By + C = 0$: \[ x' = x_1 - \frac{2A(Ax_1 + By_1 + C)}{A^2 + B^2}, \quad y' = y_1 - \frac{2B(Ax_1 + By_1 + C)}{A^2 + B^2} \] Substituting values, we get the image as $(-6,-6)$. The required distance from the origin: \[ D = \sqrt{(-6 - 0)^2 + (-6 - 0)^2} = \sqrt{36 + 36} = 6\sqrt{2} \] Quick Tip: Using the reflection formula helps determine the image of a point easily.

Since \( AD \) is the bisector of \( \angle BAC \), we use the angle bisector theorem: \begin{equation \frac{BD{DC = \frac{AB{AC \quad \text{...(i) \end{equation Now, we calculate the lengths of \( AB \) and \( AC \). \subsection*{Calculating \( AB \) \[ AB = \sqrt{(5 - 3)^2 + (3 - 2)^2 + (2 - 0)^2} \] \[ = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \] \subsection*{Calculating \( AC \) \[ AC = \sqrt{(-9 - 3)^2 + (6 - 2)^2 + (-3 - 0)^2} \] \[ = \sqrt{144 + 16 + 9} \] \[ = \sqrt{169} = 13 \] \subsection*{Finding the Ratio of \( BD : DC \) From equation (i): \[ \frac{BD}{DC} = \frac{3}{13} \] Thus, \( D \) divides \( BC \) in the ratio \( 3:13 \). \subsection*{Finding the Coordinates of \( D \) Using the section formula: \[ D \left( \frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n}, \frac{m z_2 + n z_1}{m+n} \right) \] where \( B(-9,6,-3) \) and \( C(5,3,2) \), and the ratio \( m:n = 3:13 \), \[ x = \frac{3(-9) + 13(5)}{3+13}, \quad y = \frac{3(6) + 13(3)}{3+13}, \quad z = \frac{3(-3) + 13(2)}{3+13} \] \[ = \left( \frac{-27 + 65}{16}, \frac{18 + 39}{16}, \frac{-9 + 26}{16} \right) \] \[ = \left( \frac{19}{8}, \frac{57}{16}, \frac{17}{16} \right) \] Thus, the coordinates of \( D \) are: \[ D \left( \frac{19}{8}, \frac{57}{16}, \frac{17}{16} \right) \] Quick Tip: The angle bisector theorem is a powerful tool for solving triangle-related coordinate geometry problems.

Let the mid-point of the chord be $(h,k)$. The perpendicular from the origin to the chord satisfies: \[ OC = \sqrt{h^2 + k^2}, \] Using trigonometry and given conditions, we derive: \[ h^2 + k^2 = 2 \] Quick Tip: Understanding the perpendicularity condition simplifies locus problems.

The given equation of the curve is: \[ x^2 + y^2 - 10x - 14y - 51 = 0 \] We rewrite it in standard circle form by completing the square. \[ (x^2 - 10x) + (y^2 - 14y) = 51 \] Completing squares: \[ (x - 5)^2 - 25 + (y - 7)^2 - 49 = 51 \] \[ (x - 5)^2 + (y - 7)^2 = 5\sqrt{5}^2 \] So, the center \( C(5,7) \) and radius \( r = 5\sqrt{5} \). Now, the distance of point \( (-7,2) \) from the center \( C(5,7) \): \[ PC = \sqrt{(5 + 7)^2 + (7 - 2)^2} \] \[ PC = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \] Thus, the farthest and closest distances: \[ p = 13 + 5\sqrt{5}, \quad q = 13 - 5\sqrt{5} \] The geometric mean: \[ \sqrt{pq} = \sqrt{(13 - 5\sqrt{5}) (13 + 5\sqrt{5})} \] Using the identity \( (a - b)(a + b) = a^2 - b^2 \): \[ \sqrt{169 - 125} = \sqrt{44} = 2\sqrt{11} \] Quick Tip: Understanding the concept of distance from a point to a circle helps in solving such problems efficiently.

The given equation of the circle is: \[ (x + 2)^2 + (y - 3)^2 = 4 \] Let the coordinates of \( Q(h,k) \). Since \( AQ = 2AB \), the midpoint \( B \) of segment \( AQ \) satisfies: \[ B = \left( \frac{0 + h}{2}, \frac{3 + k}{2} \right) \] Since point \( B \) lies on the given circle: \[ \left( \frac{h}{2} + 2 \right)^2 + \left( \frac{k}{2} - 3 \right)^2 = 4 \] Expanding: \[ \left( \frac{h + 4}{2} \right)^2 + \left( \frac{k - 3}{2} \right)^2 = 4 \] Multiplying both sides by 4: \[ (h + 4)^2 + (k - 3)^2 = 16 \] Thus, the required locus of \( Q(h,k) \) is: \[ (x + 4)^2 + (y - 3)^2 = 16 \] Quick Tip: Using the midpoint formula correctly ensures accurate derivation of the locus equation.

The standard form of the given parabola is: \[ (y - k)^2 = 4(x - h) \] The focus of the parabola is given by: \[ (h + 1, k) \] Since the focus must lie between the lines \(x + y = 1\) and \(x + y = 3\), we substitute the focus into the inequalities: \[ 1 < (h+1) + k < 3 \] \[ 0 < h + k < 2 \] Thus, the required range for \( h + k \) is: \[ 0 < h + k < 2 \] Quick Tip: Understanding how a parabola’s focus is derived from its equation is key to solving locus-related problems.

The given equations are: \[ x^2 + y^2 = 9 \] \[ y^2 = 8x \] Solving for the intersection points: \[ x^2 + 8x = 9 \] Rearrange: \[ x^2 + 8x - 9 = 0 \] Factoring: \[ (x + 9)(x - 1) = 0 \] So, \( x = -9, 1 \). For \( x = 1 \): \[ y^2 = 8(1) = 8 \] \[ y = \pm 2\sqrt{2} \] Thus, length of the common chord: \[ L_1 = \sqrt{(2\sqrt{2})^2 + (2\sqrt{2})^2} = 4\sqrt{2} \] Now, the length of the latus rectum of the parabola \( y^2 = 8x \) is: \[ L_2 = 4a = 4 \times 2 = 8 \] Since \( L_1 = 4\sqrt{2} \approx 5.66 \) and \( L_2 = 8 \), we get: \[ L_1 < L_2 \] Quick Tip: The length of the latus rectum of a parabola is always \(4a\), which helps in quick calculations.

Given equation: \[ 4x^2 - 9y^2 - 1 = 0 \] Rearrange: \[ \frac{x^2}{\frac{1}{4}} - \frac{y^2}{\frac{1}{9}} = 1 \] \[ \frac{x^2}{\left(\frac{1}{2}\right)^2} - \frac{y^2}{\left(\frac{1}{3}\right)^2} = 1 \] Comparing with the standard form: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \] we get: \[ a = \frac{1}{2}, \quad b = \frac{1}{3} \] The eccentricity of a hyperbola is: \[ e = \sqrt{1 + \frac{b^2}{a^2}} \] \[ = \sqrt{1 + \frac{\frac{1}{9}}{\frac{1}{4}}} = \sqrt{1 + \frac{4}{9}} = \sqrt{\frac{13}{9}} = \frac{\sqrt{13}}{3} \] Foci are given by: \[ (\pm ae, 0) = \left( \pm \frac{1}{2} \times \frac{\sqrt{13}}{3}, 0 \right) \] \[ = \left( \pm \frac{\sqrt{13}}{6}, 0 \right) \] Thus, the foci are: \[ \left( \pm \frac{\sqrt{13}}{6}, 0 \right) \] Quick Tip: For hyperbolas, the formula for foci is \( (\pm ae, 0) \), and eccentricity is found using \( e = \sqrt{1 + \frac{b^2}{a^2}} \).

- Right-hand limit (RHL): Approaching \( x \to 0^+ \), we substitute in the first case: \[ \lim\limits_{h \to 0^+} \frac{\tan^2 h}{h^2} = \lim\limits_{h \to 0^+} \frac{\tan^2 h}{h^2} = 1 \] - Left-hand limit (LHL): Approaching \( x \to 0^- \), we substitute in the third case: \[ \lim\limits_{h \to 0^-} \sqrt{(-h) \cot(-h)} \] Using \( \cot(-h) = -\cot h \), we get: \[ \lim\limits_{h \to 0^-} \sqrt{(1-h) \cot(1-h)} \] which is not equal to \( \lim\limits_{x \to 0^+} f(x) \), thus: \[ \lim\limits_{x \to 0} f(x) \text{ does not exist}. \] Quick Tip: For piecewise functions, always check left-hand and right-hand limits separately.

Given: \( f(x) = \sin x \), \( g(x) = \cos x \), and \( h(x) = x^2 \) \[ \lim\limits_{x \to 1} \frac{f(g(h(x))) - f(g(h(1)))}{x - 1} \] Substituting \( h(x) = x^2 \) and evaluating at \( x = 1 \): \[ \lim\limits_{x \to 1} \frac{\sin(\cos(x^2)) - \sin(\cos 1)}{x - 1} \] Using L'Hôpital's rule: \[ \lim\limits_{x \to 1} \frac{\cos(\cos(x^2)) \cdot (-\sin(x^2)) \cdot 2x}{1} \] Evaluating at \( x = 1 \): \[ = -2\sin 1 \cos(\cos 1) \] Quick Tip: Using chain rule and L'Hôpital's rule simplifies limit calculations for composite functions.

Applying De Morgan's Law: \[ \sim (p \vee q) = \sim p \wedge \sim q \] Thus: \[ (\sim p \wedge \sim q) \vee (\sim p \wedge q) \] Using distributive law: \[ \sim p \wedge (\sim q \vee q) \] Since \( \sim q \vee q = 1 \) (tautology): \[ \sim p \wedge 1 = \sim p \] Thus, the Boolean expression simplifies to: \[ \sim p \] Quick Tip: Using De Morgan's law and distribution simplifies complex Boolean expressions.

The given symbolic statement: \[ p \rightarrow (q \vee r) \] By definition of implication: \[ p \rightarrow (q \vee r) \equiv \sim p \vee (q \vee r) \] Since \( p \) represents "2 is an even number," the statement translates to: "If 2 is an even number, then 2 is a prime number or \( 2 + 2 = 2^2 \)." Quick Tip: Understanding logical implications helps in interpreting symbolic statements correctly.

The given statement is: \[ (A \wedge C) \rightarrow B \] The negation of an implication \( P \rightarrow Q \) is given by: \[ \sim (P \rightarrow Q) \equiv P \wedge \sim Q \] Thus: \[ \sim [(A \wedge C) \rightarrow B] \equiv (A \wedge C) \wedge \sim B \] which matches option (B). Quick Tip: Negation of an implication \( P \rightarrow Q \) is always \( P \wedge \sim Q \).

The given statement translates as: \[ \text{"If it does not rain or I do not go to school, then I shall meet my friend and go for a movie."} \] The phrase "does not rain or do not go to school" can be written as: \[ \sim (p \wedge q) \] The phrase "I shall meet my friend and go for a movie" translates to: \[ r \wedge s \] Thus, the logical expression becomes: \[ \sim (p \wedge q) \Rightarrow (r \wedge s) \] Quick Tip: Use logical operators systematically to convert English statements into symbolic form.

Expanding \( S_1 \): \[ (\sim p \vee q) \vee (\sim p \vee r) \equiv \sim p \vee (q \vee r) \] Expanding \( S_2 \): \[ p \rightarrow (q \vee r) \equiv \sim p \vee (q \vee r) \] Since both \( S_1 \) and \( S_2 \) are equivalent, if \( S_2 \) is false, then \( S_1 \) should also be false, contradicting option (C). Quick Tip: Equivalent logical statements will always have the same truth value.

We begin by constructing a truth table for the given expressions. The statement \( p \rightarrow ((\sim p) \vee q) \) is FALSE only when \( p = T \) and \( q = F \), which gives: \begin{tabular{|c|c|c|c|c|c|c|c|c|c| \hline $\mathbf{p$ & $\mathbf{q$ & $\mathbf{\sim p$ & $\mathbf{\sim q$ & $\mathbf{\sim p \lor q$ & $\mathbf{p \rightarrow (\sim p \lor q)$ & $\mathbf{p \rightarrow \sim q$ & $\mathbf{\sim (p \rightarrow \sim q)$ & $\mathbf{p \land \sim q$ & $\mathbf{p_2$
\hline T & T & F & F & T & T & F & T & F & F
\hline T & F & F & T & F & F & T & F & T & F
\hline F & T & T & F & T & T & T & F & F & F
\hline F & F & T & T & T & T & T & F & F & F
\hline \end{tabular \[ p \rightarrow ((\sim p) \vee q) = F \] This condition leads to \( P_1 \) and \( P_2 \) both being FALSE. Quick Tip: Constructing a truth table simplifies logical evaluation.

Given observations: \( 2, 3, 5, 8, 12 \). 1. Calculate Mean: \[ \text{Mean} = \frac{2 + 3 + 5 + 8 + 12}{5} = 6 \] 2. Calculate Variance: \[ \sigma^2 = 13.2 \] 3. Find Median: Since the number of observations is odd, the median is the middle value: \[ \text{Median} = 5 \] 4. Calculate Mean Deviation about Median: \[ M = \frac{|2 - 5| + |3 - 5| + |5 - 5| + |8 - 5| + |12 - 5|}{5} = 3 \] 5. Final Calculation: \[ \sigma^2 - M = 13.2 - 3 = 10.2 \] Quick Tip: Variance measures spread, while mean deviation measures absolute dispersion.

Let the items be \( a_1, a_2, ..., a_n \). \[ \bar{X} = \frac{a_1 + a_2 + ... + a_n}{n} \] Now, given the condition: \[ \bar{X}_{\text{new}} = \frac{(a_1+1) + (a_2+2) + ... + (a_n+n)}{n} \] Using the sum of the first \( n \) natural numbers: \[ \bar{X}_{\text{new}} = \bar{X} + \frac{n(n+1)}{2n} = \bar{X} + \frac{n+1}{2} \] Quick Tip: Use summation formulas for sequences to simplify mean calculations.

Given variance: \[ \sigma^2 = 5 \] If each observation is multiplied by a constant \( k = 2 \), the variance changes as: \[\text{Variance } (\sigma^2) = \frac{1}{n} \sum_{i=1}^{20} (x_i - \overline{x})^2\] \[\text{i.e. } 5 = \frac{1}{20} \sum_{i=1}^{20} (x_i - \overline{x})^2\] \[\text{or } \sum_{i=1}^{20} (x_i - \overline{x})^2 = 100 \quad .....(i)\] \[\text{If each observation is multiplied by 2 and the new resulting observations are } y_i, \text{ then }\] \[y_i = 2x_i \text{ i.e., } x_i = \frac{1}{2} y_i\] \[\text{Therefore, }\] \[\overline{y} = \frac{1}{n} \sum_{i=1}^{20} y_i = \frac{1}{20} \sum_{i=1}^{20} 2x_i = 2 \cdot \frac{1}{20} \sum_{i=1}^{20} x_i\] \[\text{i.e., } \overline{y} = 2\overline{x} \text{ or } \overline{x} = \frac{1}{2} \overline{y}\] \[\text{On substituting the values of } x_i \text{ and } \overline{x} \text{ in eq. (i), we get }\] \[\sum_{i=1}^{20} \left(\frac{1}{2} y_i - \frac{1}{2} \overline{y}\right)^2 = 100\] \[\text{i.e., } \sum_{i=1}^{20} (y_i - \overline{y})^2 = 400\] \[\text{Thus, the variance of new observations }\] \[= \frac{1}{20} \times 400 = 20 = 2^2 \times 5\] Quick Tip: If data is scaled by \( k \), variance scales by \( k^2 \).

Since \( f(x) \) is continuous on \([0,8]\), it must be continuous at \( x = 2 \) and \( x = 4 \). ### At \( x = 2 \), \[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) \] \[ \lim_{x \to 2^-} (x^2 + ax + b) = \lim_{x \to 2^+} (3x + 2) \] \[ 4 + 2a + b = 3(2) + 2 \] \[ 2a + b = 4 \quad \text{(Equation 1)} \] ### At \( x = 4 \), \[ \lim_{x \to 4^-} f(x) = \lim_{x \to 4^+} f(x) \] \[ \lim_{x \to 4^-} (3x + 2) = \lim_{x \to 4^+} (2ax + 5b) \] \[ 3(4) + 2 = 2a(4) + 5b \] \[ 8a + 5b = 14 \quad \text{(Equation 2)} \] Solving Equations 1 and 2, we get: \[ a = 3, \quad b = -2 \] Quick Tip: Continuity at a point requires matching left-hand and right-hand limits.

Let the height of the tower be \( h \). Using the tangent function in \( \triangle ABD \): \[ \tan 45^\circ = \frac{AB}{BD} \Rightarrow BD = 50 \text{ m} \] Now, in \( \triangle ACC' \): \[ \tan 30^\circ = \frac{AC'}{C'C} \] \[ \frac{1}{\sqrt{3}} = \frac{50 - h}{50} \] Solving for \( h \): \[ 50 = 50\sqrt{3} - h\sqrt{3} \] \[ h\sqrt{3} = 50(\sqrt{3} - 1) \] \[ h = 50 \left( 1 - \frac{\sqrt{3}}{3} \right) \text{ m} \] Quick Tip: Trigonometry is useful in solving real-world height and distance problems.

Let \( DE = h \) and \( CD = DB = x \). In \( \triangle EBD \): \[ \tan 60^\circ = \frac{h}{x} \Rightarrow x = \frac{h}{\sqrt{3}} \] Now, in \( \triangle ADE \): \[ \tan 45^\circ = \frac{ED}{DA} \Rightarrow DA = h \] Applying Pythagoras in \( \triangle ABD \): \[ \left( \frac{h}{\sqrt{3}} \right)^2 + h^2 = 100^2 \] \[ \frac{4h^2}{3} = 10000 \] \[ h = 50\sqrt{3} \] Quick Tip: Using Pythagoras’ theorem and trigonometry helps in height and distance calculations.

Using set theory: \[ n(R) = 794, \quad n(T) = 187, \quad n(R \cup T)' = 63 \] \[ n(Total) = n(R \cup T) + n(R \cup T)' \Rightarrow 1003 = n(R \cup T) + 63 \] \[ n(R \cup T) = 940 \] Using formula: \[ n(R \cup T) = n(R) + n(T) - n(R \cap T) \] \[ 940 = 794 + 187 - n(R \cap T) \] \[ n(R \cap T) = 41 \] Quick Tip: Use set operations to solve logical counting problems.

Step 1: {Check for Reflexivity
A relation is reflexive if every element is related to itself. In this case, every triangle is congruent to itself. So, \( \triangle A \cong \triangle A \). Thus, the relation R is reflexive. Step 2: {Check for Symmetry
A relation is symmetric if for every \( a \) related to \( b \), \( b \) is also related to \( a \). If \( \triangle A \cong \triangle B \), then \( \triangle B \cong \triangle A \). Thus, the relation R is symmetric. Step 3: {Check for Transitivity
A relation is transitive if whenever \( a \) is related to \( b \) and \( b \) is related to \( c \), then \( a \) is also related to \( c \). If \( \triangle A \cong \triangle B \) and \( \triangle B \cong \triangle C \), then \( \triangle A \cong \triangle C \). Thus, the relation R is transitive. Step 4: {Conclusion
Since the relation R is reflexive, symmetric, and transitive, it is an equivalence relation. Therefore, the correct answer is (D).
Quick Tip: A relation is an equivalence relation if it is reflexive, symmetric, and transitive.

Step 1: {Identify the distinct letters
The word 'MONOTONE' has the following distinct letters: M, O, N, T, E. Step 2: {Count the distinct letters
There are 5 distinct letters. Step 3: {Use the formula for the number of subsets
The number of subsets of a set with \( n \) elements is \( 2^n \). In this case, \( n = 5 \). Step 4: {Calculate the number of subsets
Number of subsets = \( 2^5 = 32 \).
Quick Tip: The number of subsets of a set with \( n \) elements is \( 2^n \).

Step 1: {Calculate the percentage of candidates who failed in either English or Mathematics or both
Percentage failed in English or Mathematics or both = Percentage failed in English + Percentage failed in Mathematics - Percentage failed in both = 62% + 42% - 20% = 84% Step 2: {Calculate the percentage of candidates who passed in both subjects
Percentage passed in both subjects = 100% - Percentage failed in English or Mathematics or both = 100% - 84% = 16% Step 3: {Assume the total number of candidates
Let the total number of candidates be \( x \). Step 4: {Calculate the number of candidates who passed in both subjects
Number of candidates who passed in both subjects = 16% of \( x \) = \( \frac{16}{100} \times x \) Step 5: {Relate the number to the options
Since the options are whole numbers, we can assume \( x = 100 \) for simplicity. Then, the number of candidates who passed in both subjects = \( \frac{16}{100} \times 100 = 16 \).
Quick Tip: Use the principle of inclusion-exclusion to find the percentage of candidates who failed in either English or Mathematics or both.

Step 1: {Recall the property of orthogonal matrices
A matrix \( A \) is orthogonal if its transpose \( A^T \) is equal to its inverse \( A^{-1} \), i.e., \( A^T = A^{-1} \). This implies that \( AA^T = I \), where \( I \) is the identity matrix. Step 2: {Find the transpose of matrix \( A \)
\[ A^T = \frac{1}{3} \begin{bmatrix} 1 & 2 & a
2 & 1 & 2
2 & -2 & b \end{bmatrix} \] Step 3: {Multiply \( A \) and \( A^T \)
\[ AA^T = \frac{1}{9} \begin{bmatrix} 1 & 2 & 2
2 & 1 & -2
a & 2 & b \end{bmatrix} \begin{bmatrix} 1 & 2 & a
2 & 1 & 2
2 & -2 & b \end{bmatrix} = \frac{1}{9} \begin{bmatrix} 9 & 0 & a+4+2b
0 & 9 & 2a+2-2b
a+4+2b & 2a+2-2b & a^2+4+b^2 \end{bmatrix} \] Step 4: {Set \( AA^T = I \)
For \( AA^T = I \), we must have: \[ \frac{1}{9} \begin{bmatrix} 9 & 0 & a+4+2b
0 & 9 & 2a+2-2b
a+4+2b & 2a+2-2b & a^2+4+b^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0
0 & 1 & 0
0 & 0 & 1 \end{bmatrix} \] Step 5: {Solve the resulting equations
From the above equation, we get the following equations: \( a + 4 + 2b = 0 \) \( 2a + 2 - 2b = 0 \) \( a^2 + 4 + b^2 = 9 \) Step 6: {Solve for \( a \) and \( b \)
Adding equations (1) and (2), we get \( 3a + 6 = 0 \), which gives \( a = -2 \). Substituting \( a = -2 \) into equation (1), we get \( -2 + 4 + 2b = 0 \), which gives \( 2b = -2 \), so \( b = -1 \). Step 7: {Verify the solution
Substituting \( a = -2 \) and \( b = -1 \) into equation (3), we get \( (-2)^2 + 4 + (-1)^2 = 4 + 4 + 1 = 9 \), which is true. Therefore, \( a = -2 \) and \( b = -1 \).
Quick Tip: Remember that for an orthogonal matrix \( A \), \( AA^T = I \).

Step 1: {Recall the relationship between \( A^{-1} \) and \( adj(A) \)
The inverse of a matrix \( A \) is given by \( A^{-1} = \frac{1}{det(A)} adj(A) \), where \( det(A) \) is the determinant of \( A \) and \( adj(A) \) is the adjugate of \( A \). Step 2: {Compare with the given equation
We are given that \( A^{-1} = \frac{1}{k} adj(A) \). Comparing this with the general formula, we see that \( k = det(A) \). Step 3: {Calculate the determinant of \( A \)
\[ \begin{vmatrix} 3 & -2 & 4
1 & 2 & -1
0 & 1 & 1 \end{vmatrix} = 3 \begin{vmatrix} 2 & -1
1 & 1 \end{vmatrix} - (-2) \begin{vmatrix} 1 & -1
0 & 1 \end{vmatrix} + 4 \begin{vmatrix} 1 & 2
0 & 1 \end{vmatrix} \] \[ = 3(2(1) - (-1)(1)) + 2(1(1) - (-1)(0)) + 4(1(1) - 2(0)) \] \[ = 3(2 + 1) + 2(1 - 0) + 4(1 - 0) \] \[ = 3(3) + 2(1) + 4(1) \] \[ = 9 + 2 + 4 = 15 \] Step 4: {Identify the value of \( k \)
Since \( k = det(A) \), we have \( k = 15 \).
Quick Tip: Remember the formula \( A^{-1} = \frac{1}{det(A)} adj(A) \).

Step 1: {Use the given equations
We are given \( AB + BA = X \) and \( AB - BA = Y \). Step 2: {Find \( X^T \) and \( Y^T \)
Since A and B are symmetric matrices, \( A^T = A \) and \( B^T = B \). \[ X^T = (AB + BA)^T = (AB)^T + (BA)^T = B^TA^T + A^TB^T = BA + AB = X \] \[ Y^T = (AB - BA)^T = (AB)^T - (BA)^T = B^TA^T - A^TB^T = BA - AB = -(AB - BA) = -Y \] So, \( X \) is symmetric and \( Y \) is skew-symmetric. Step 3: {Compute \( XY \)
\[ XY = (AB + BA)(AB - BA) = (AB)^2 - (BA)^2 \] Step 4: {Compute \( (XY)^T \)
\[ (XY)^T = ((AB)^2 - (BA)^2)^T = ((AB)^2)^T - ((BA)^2)^T = (B^TA^T)^2 - (A^TB^T)^2 = (BA)^2 - (AB)^2 = -(XY) \] Step 5: {Compute \( YX \)
\[ YX = (AB - BA)(AB + BA) = (AB)^2 - (BA)^2 = -(BA)^2 + (AB)^2 = -((BA)^2 - (AB)^2) = -XY \] Step 6: {Compare \( (XY)^T \) with \( YX \)
We found that \( (XY)^T = -YX \) .
Quick Tip: Remember that \( (AB)^T = B^TA^T \), and if A is symmetric, \( A^T = A \).

Step 1: {Show that \( A \) is an orthogonal matrix
Since \( A A^T = I \), the matrix \( A \) is orthogonal.
Step 2: {Simplify \( A^T X^{50} A \)
\[ A^T X^{50} A = A^T X^{49} (A P A^T) A \] \[ = A^T X^{49} A P (A^T A) = A^T X^{49} A P \] \[ = A^T X^{48} (A P A^T) A P = A^T X^{48} A P^2 \dots \] \[ = A^T A P^{50} = I P^{50} = P^{50}. \] Step 3: {Compute \( P^{50} \)
\[ P = \begin{bmatrix} 1 & 1
0 & 1 \end{bmatrix} \] \[ P^2 = \begin{bmatrix} 1 & 2
0 & 1 \end{bmatrix} \] \[ P^3 = \begin{bmatrix} 1 & 3
0 & 1 \end{bmatrix} \] \[ \vdots \] \[ P^{50} = \begin{bmatrix} 1 & 50
0 & 1 \end{bmatrix} \] Step 4: {Conclusion
\[ A^T X^{50} A = P^{50} = \begin{bmatrix} 1 & 50
0 & 1 \end{bmatrix}. \] Quick Tip: For a matrix of the form \( P = \begin{bmatrix} 1 & k
0 & 1 \end{bmatrix} \), its \( n \)th power is given by: \[ P^n = \begin{bmatrix} 1 & nk
0 & 1 \end{bmatrix}. \]

Step 1: {Use the determinant property of adjugate matrices
For a square matrix \( A \) of order \( n \), the determinant of its adjugate is given by: \[ |\text{adj } A| = |A|^{n-1}. \] Since \( A \) is of order 3, we get: \[ |\text{adj } A^2| = |A^2|^{3-1} = |A^2|^2. \] Step 2: {Simplify further
\[ |A^2| = (|A|^2), \] \[ |\text{adj } A^2| = (|A|^2)^2 = |A|^4. \] Step 3: {Compute \( |\text{Adj}(\text{Adj } A^2)| \)
\[ |\text{Adj}(\text{Adj } A^2)| = (|A|^4)^{3-1} = (|A|^4)^2 = |A|^8. \] Step 4: {Conclusion
Thus, the correct answer is \( |A|^8 \).
Quick Tip: For any square matrix \( A \) of order \( n \), we have: \[ |\text{Adj } A| = |A|^{n-1}. \]

Step 1: {Construct the determinant of the coefficient matrix
\[ |A| = \begin{vmatrix} p+a & b & c
a & q+b & c
a & b & r+c \end{vmatrix}. \] Since the system has a non-trivial solution, the determinant must be zero: \[ |A| = 0. \] Step 2: {Expand determinant using row operations
\[ (p + a)[(q + b)(r + c) - bc] - b[a(r + c) - ca] + c[ab - a(q + b)] = 0. \] \[ (p + a)(qr + qc + br) - b(ar) + c[-aq] = 0. \] Step 3: {Divide by \( pqr \)
\[ \frac{pqr}{pqr} + \frac{pqc}{pqr} + \frac{prb}{pqr} + \frac{qra}{pqr} = 0. \] \[ 1 + \frac{c}{r} + \frac{b}{q} + \frac{a}{p} = 0. \] Step 4: {Conclusion
\[ \frac{a}{p} + \frac{b}{q} + \frac{c}{r} = -1. \] Quick Tip: For a homogeneous system \( AX = 0 \) to have a non-trivial solution, the determinant \( |A| \) must be zero.

Step 1: {Expand both determinants
\[ (1 - 3x^2 + 2x^3) + (3x^2 - x^3) = 0. \] Step 2: {Solve for \( x \)
\[ x^3 + 1 = 0. \] \[ x^3 = -1. \] \[ x = -\omega, -\omega^2, -1. \] Step 3: {Compute \( x^{2007} + x^{-2007} \)
Since \( x^3 = -1 \), \[ x^{2007} = (-1)^{669} = -1. \] \[ x^{-2007} = -1. \] Step 4: {Conclusion
\[ x^{2007} + x^{-2007} = -1 - 1 = -2. \] Quick Tip: If \( x \) is a cube root of unity, then it satisfies \( x^3 = -1 \), which helps simplify large exponents.

Step 1: {Compute the determinant of the coefficient matrix
\[ \Delta = \begin{vmatrix} 1 & -1 & 2
3 & 1 & 4
1 & 1 & 1 \end{vmatrix}. \] Expanding along the first row: \[ \Delta = 1(1 \times 1 - 4 \times 1) + (-1)(3 \times 1 - 4 \times 1) + 2(3 \times 1 - 1 \times 1). \] \[ = (1 - 4) + (-3 + 4) + 2(3 - 1). \] \[ = -3 + 1 + 4 = 0. \] Step 2: {Compute determinant of augmented matrix
\[ \Delta_1 = \begin{vmatrix} 4 & -1 & 2
6 & 1 & 4
1 & 1 & 1 \end{vmatrix}. \] Expanding along the first row: \[ \Delta_1 = 4(1 \times 1 - 4 \times 1) + (-1)(6 \times 1 - 4 \times 1) + 2(6 \times 1 - 1 \times 1). \] \[ = 4(1 - 4) + (-6 + 4) + 2(6 - 1). \] \[ = -12 + 2 + 10 = 0. \] Step 3: {Conclusion
Since \( \Delta = 0 \) and \( \Delta_1 = 0 \), the system has infinitely many solutions. Quick Tip: If \( \Delta = 0 \) and all augmented determinants are also zero, the system has infinitely many solutions.

Step 1: {Compute determinant of the coefficient matrix
\[ \Delta = \begin{vmatrix} 2 & 1 & -1
1 & -3 & 2
1 & 4 & \delta \end{vmatrix}. \] Since the system has infinitely many solutions, \( \Delta = 0 \). Expanding along the first row: \[ \Delta = 2(-3 \times \delta - 2 \times 4) - 1(1 \times \delta - 2 \times 1) + (-1)(1 \times 4 + 3 \times 1). \] Solving for \( \delta \): \[ -6\delta - 8 - \delta + 2 - 4 - 3 = 0. \] \[ -7\delta - 13 = 0. \] \[ \delta = -3. \] Step 2: {Compute augmented determinant
\[ \Delta_1 = \begin{vmatrix} 7 & 1 & -1
1 & -3 & 2
k & 4 & -3 \end{vmatrix}. \] Setting \( \Delta_1 = 0 \), solving for \( k \): \[ k = 6. \] Step 3: {Conclusion
\[ \delta + k = -3 + 6 = 3. \] Quick Tip: For infinitely many solutions, the determinant of the coefficient matrix and all augmented determinants must be zero.

Step 1: {Express cotangent and secant
\[ \cot(\cos^{-1} x) = \frac{x}{\sqrt{1 - x^2}}. \] \[ \sec \left( \tan^{-1} \left( \frac{a}{\sqrt{b^2 - a^2}} \right) \right) = \sqrt{1 + \left( \frac{a}{\sqrt{b^2 - a^2}} \right)^2}. \] Step 2: {Equating expressions
\[ \frac{x}{\sqrt{1 - x^2}} = \sqrt{1 + \frac{a^2}{b^2 - a^2}}. \] Simplify: \[ \frac{x}{\sqrt{1 - x^2}} = \sqrt{\frac{b^2 - a^2 + a^2}{b^2 - a^2}}. \] \[ \frac{x}{\sqrt{1 - x^2}} = \frac{b}{\sqrt{b^2 - a^2}}. \] Squaring both sides: \[ x^2(2b^2 - a^2) = b^2. \] Solving for \( x \): \[ x = \frac{b}{\sqrt{2b^2 - a^2}}. \] Quick Tip: For inverse trigonometric functions, rewrite in terms of known trigonometric identities to simplify expressions.

Step 1: {Express \( \cot^{-1} \) in terms of cosine
Let \[ \alpha = \cot^{-1} \left( \frac{1}{2} \right). \] Then, \[ \cot \alpha = \frac{1}{2} \Rightarrow \cos \alpha = \frac{1}{\sqrt{5}}. \] Step 2: {Use cotangent identity
\[ \cos (\cos^{-1} x) = \cot \left( \cos^{-1} x \right). \] Using the identity: \[ \cot (\cos^{-1} x) = \frac{x}{\sqrt{1 - x^2}}. \] Step 3: {Equating both sides
\[ \frac{1}{\sqrt{5}} = \frac{x}{\sqrt{1 - x^2}}. \] Squaring both sides: \[ 1 - x^2 = 5x^2. \] Step 4: {Solve for \( x \)
\[ 1 = 6x^2. \] \[ x = \pm \frac{1}{\sqrt{6}}. \] Step 5: {Select the correct sign
Ignoring the negative root: \[ x = \frac{1}{\sqrt{6}}. \] Quick Tip: To convert inverse trigonometric expressions, use the basic definitions of trigonometric functions in right-angled triangles.

Step 1: {Compute \( f(-8/5) \)
\[ f \left( \frac{-8}{5} \right) = \left\lfloor \frac{-8}{5} \right\rfloor = -2. \] Step 2: {Compute \( g(8/5) \) and \( g(-8/5) \)
\[ g \left( \frac{8}{5} \right) = \left| \frac{8}{5} \right| = \frac{8}{5}. \] \[ g \left( \frac{-8}{5} \right) = \left| \frac{-8}{5} \right| = \frac{8}{5}. \] Step 3: {Compute \( f(g(8/5)) \) and \( g(f(-8/5)) \)
\[ f \left( g \left( \frac{8}{5} \right) \right) = f \left( \frac{8}{5} \right) = \left\lfloor \frac{8}{5} \right\rfloor = 1. \] \[ g \left( f \left( \frac{-8}{5} \right) \right) = g(-2) = | -2 | = 2. \] Step 4: {Final computation
\[ f \left( g \left( \frac{8}{5} \right) \right) - g \left( f \left( \frac{-8}{5} \right) \right) = 1 - 2 = -1. \] Step 5: {Conclusion
Thus, the correct answer is \( -1 \). Quick Tip: The greatest integer function \( [x] \) returns the largest integer less than or equal to \( x \). The absolute function \( |x| \) removes the sign.

Step 1: {Substituting \( \log_2 |x| = t \)
Let \[ \log_2 |x| = t. \] Thus, the equation becomes: \[ \sqrt{5 - t} = 3 - t. \] Step 2: {Squaring both sides
\[ 5 - t = (3 - t)^2. \] Expanding: \[ 5 - t = 9 + t^2 - 6t. \] \[ t^2 - 5t + 4 = 0. \] Step 3: {Solving for \( t \)
\[ (t - 4)(t - 1) = 0. \] \[ t = 4 \quad \text{or} \quad t = 1. \] Rejecting \( t = 4 \) as it violates the equation. Step 4: {Finding \( x \)
\[ \log_2 |x| = 1. \] \[ |x| = 2. \] \[ x = \pm 2. \] Step 5: {Conclusion
Thus, there are \( 2 \) real solutions: \( x = 2, -2 \). Quick Tip: When solving logarithmic equations, check that all solutions satisfy the original equation.

Step 1: {Compute \( f(-x) \)
\[ f(-x) = \frac{\cos(-x)}{\left\lfloor \frac{2(-x)}{\pi} \right\rfloor + \frac{1}{2}}. \] Using \( \cos(-x) = \cos x \) and property of floor function: \[ \left\lfloor \frac{2(-x)}{\pi} \right\rfloor = - \left\lfloor \frac{2x}{\pi} \right\rfloor - 1. \] Step 2: {Compare \( f(-x) \) with \( -f(x) \)
\[ f(-x) = -f(x). \] Step 3: {Conclusion
Since \( f(-x) = -f(x) \), the function is odd. Quick Tip: A function is odd if \( f(-x) = -f(x) \) and even if \( f(-x) = f(x) \).

Step 1: {Check injectivity
For \( x_1, x_2 \) such that: \[ f(x_1) = f(x_2), \] \[ \frac{x_1}{\sqrt{1 + x_1^2}} = \frac{x_2}{\sqrt{1 + x_2^2}}. \] Squaring both sides: \[ x_1^2(1 + x_2^2) = x_2^2(1 + x_1^2). \] Solving, we get: \[ x_1 = x_2. \] Thus, \( f(x) \) is injective. Step 2: {Check surjectivity
Solving for \( y \): \[ y = \frac{x}{\sqrt{1 + x^2}}. \] This implies: \[ y^2(1 + x^2) = x^2. \] Solving, we find \( y \in (-1,1) \), meaning \( f(x) \) is not surjective. Step 3: {Conclusion
\( f(x) \) is injective but not surjective. Quick Tip: A function is injective if distinct inputs yield distinct outputs. It is surjective if its range covers the entire codomain.

Step 1: {Find \( f^{-1}(3) \)
\[ y = f(x) = 5x - 3. \] \[ x = \frac{y + 3}{5}. \] \[ f^{-1}(3) = \frac{6}{5}. \] Step 2: {Compute \( g(f^{-1}(3)) \)
\[ g(x) = x^2 + 3. \] \[ g \left( \frac{6}{5} \right) = \left( \frac{6}{5} \right)^2 + 3. \] \[ = \frac{36}{25} + 3 = \frac{111}{25}. \] Step 3: {Conclusion
Thus, \( g \circ f^{-1}(3) = \frac{111}{25} \). Quick Tip: To find \( g \circ f^{-1} \), first determine \( f^{-1}(x) \), then substitute into \( g(x) \).

Step 1: {Find the domain restrictions
For the function \( f(x) \) to be defined, the expression inside the square root must be non-negative: \[ \frac{2x^2 - 7x + 5}{3x^2 - 5x - 2} \geq 0. \] Also, the denominator must not be zero, i.e., \[ 3x^2 - 5x - 2 \neq 0. \] Step 2: {Find the zeros of the numerator
Solving: \[ 2x^2 - 7x + 5 = 0. \] Factoring: \[ (2x - 5)(x - 1) = 0. \] \[ x = \frac{5}{2}, 1. \] Step 3: {Find the zeros of the denominator
Solving: \[ 3x^2 - 5x - 2 = 0. \] Factoring: \[ (3x + 1)(x - 2) = 0. \] \[ x = -\frac{1}{3}, 2. \] Step 4: {Analyze sign changes using a number line
The critical points partition the number line into intervals: \[ (-\infty, -\frac{1}{3}), (-\frac{1}{3}, 1), (1,2), (2, \frac{5}{2}), (\frac{5}{2}, \infty). \] By testing values in each interval, the function is non-negative in: \[ (-\infty, -\frac{1}{3}) \cup [1,2) \cup [\frac{5}{2}, \infty). \] Step 5: {Conclusion
Thus, the domain is: \[ (-\infty, -\frac{1}{3}) \cup [1,2) \cup [\frac{5}{2}, \infty). \] Quick Tip: For rational functions inside a square root, ensure the numerator is non-negative while avoiding zeros of the denominator.

Step 1: Identify the value \( x \) cannot take
The function \( f(x) = \frac{x+3}{x-2} \) is undefined when the denominator is equal to zero. \[ x - 2 = 0 \] \[ x = 2 \] Therefore, \( x \) cannot take the value 2. So, \( l = 2 \). Step 2: Determine the values \( f(x) \) cannot take
Let \( y = f(x) \). We want to find the values that \( y \) cannot take. \[ y = \frac{x+3}{x-2} \] \[ y(x-2) = x+3 \] \[ xy - 2y = x+3 \] \[ xy - x = 2y + 3 \] \[ x(y-1) = 2y + 3 \] \[ x = \frac{2y+3}{y-1} \] From this expression for \( x \) in terms of \( y \), we can see that \( y \) cannot take the value 1, as the denominator would be zero. Therefore, \( m = 1 \). Step 3: Calculate \( 3l + 2m \)
We have \( l = 2 \) and \( m = 1 \). \[ 3l + 2m = 3(2) + 2(1) = 6 + 2 = 8 \] Therefore, \( 3l + 2m = 8 \).
Quick Tip: For bijections involving rational functions, ensure the function is not undefined at any point within its intended domain or that it avoids specific values in its codomain.

Step 1: Find \( g[f(x)] \)
We are given \( f(x) = \sin x + \cos x \) and \( g(x) = x^2 - 1 \). We need to find \( g[f(x)] \), which means \( g(f(x)) \). \[ g[f(x)] = g(\sin x + \cos x) \] Substitute \( \sin x + \cos x \) in place of \( x \) in the expression for \( g(x) \): \[ g[f(x)] = (\sin x + \cos x)^2 - 1 \] Step 2: Expand and simplify \( g[f(x)] \)
Expand the square: \[ g[f(x)] = (\sin^2 x + 2\sin x \cos x + \cos^2 x) - 1 \] Recall the trigonometric identities: \[ \sin^2 x + \cos^2 x = 1 \] \[ 2\sin x \cos x = \sin 2x \] Substitute these identities into the expression: \[ g[f(x)] = (1 + \sin 2x) - 1 \] \[ g[f(x)] = \sin 2x \] Step 3: Analyze the monotonicity of \( \sin 2x \)
The function \( \sin 2x \) is a sinusoidal function. We need to find an interval where it is strictly increasing or strictly decreasing (i.e., monotonic) to ensure invertibility. Step 4: Determine an interval for invertibility
The sine function is strictly increasing in the interval \( [-\frac{\pi}{2}, \frac{\pi}{2}] \). Since we have \( \sin 2x \), we need to consider the argument \( 2x \). For \( \sin 2x \) to be strictly increasing, we need \( 2x \) to be in the interval \( [-\frac{\pi}{2}, \frac{\pi}{2}] \). \[ -\frac{\pi}{2} \le 2x \le \frac{\pi}{2} \] Divide by 2: \[ -\frac{\pi}{4} \le x \le \frac{\pi}{4} \] In this interval \( [-\frac{\pi}{4}, \frac{\pi}{4}] \), \( \sin 2x \) is strictly increasing, and therefore invertible. Conclusion: \( g[f(x)] = \sin 2x \). The function \( g[f(x)] \) is invertible in the interval \( [-\frac{\pi}{4}, \frac{\pi}{4}] \).
Quick Tip: Invertibility requires the function to be monotonic (either strictly increasing or strictly decreasing) over the interval.

Step 1: {Analyze evenness or oddness
\[ g(-u) = 2 \tan^{-1}(e^{-u}) - \frac{\pi}{2} \] \[ = 2 \left( \frac{\pi}{2} - \tan^{-1}(e^u) \right) - \frac{\pi}{2} \] \[ = \pi - 2 \tan^{-1}(e^u) - \frac{\pi}{2} = -2 \tan^{-1}(e^u) + \frac{\pi}{2} \] \[ = -(2 \tan^{-1}(e^u) - \frac{\pi}{2}) = -g(u) \] Thus, \( g \) is an odd function. Step 2: {Verify increasing nature
The derivative \( g'(u) = 2\frac{1}{1+e^{2u}}e^u > 0 \) for all \( u \), indicating \( g \) is strictly increasing over \( (-\infty, \infty) \). Quick Tip: Remember, for a function to be odd, \( f(-x) = -f(x) \) must hold true, and the function's derivative should be positive for increasing nature.

Step 1: To determine the continuity of \( f(x) \) at \( x = 1 \), we need to check if the left-hand limit (LHL), the right-hand limit (RHL), and the function's value at \( x = 1 \) are all equal. Step 1: Calculate the Left-Hand Limit (LHL)
\[ \text{LHL} = \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} \frac{x^2 - 1}{x^2 + 2x - 3} \] Factor the numerator and denominator: \[ \lim_{x \to 1^-} \frac{(x - 1)(x + 1)}{(x + 3)(x - 1)} \] Cancel the common factor \( (x - 1) \): \[ \lim_{x \to 1^-} \frac{x + 1}{x + 3} \] Substitute \( x = 1 \): \[ \frac{1 + 1}{1 + 3} = \frac{2}{4} = \frac{1}{2} \] Step 2: Calculate the Right-Hand Limit (RHL)
\[ \text{RHL} = \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} \frac{x^2 - 1}{x^2 - 2x + 1} \] Factor the numerator and denominator: \[ \lim_{x \to 1^+} \frac{(x - 1)(x + 1)}{(x - 1)^2} \] Cancel the common factor \( (x - 1) \): \[ \lim_{x \to 1^+} \frac{x + 1}{x - 1} \] As \( x \) approaches 1 from the right (i.e., \( x > 1 \)), the numerator approaches 2, and the denominator approaches 0 from the positive side. Thus, the limit is \( +\infty \). \[ \lim_{x \to 1^+} \frac{x + 1}{x - 1} = +\infty \] Step 3: Evaluate the function at \( x = 1 \)
From the definition of \( f(x) \), we have \( f(1) = \frac{1}{2} \). Step 4: Compare LHL, RHL, and \( f(1) \)
We have: \[ \text{LHL} = \frac{1}{2} \] \[ \text{RHL} = +\infty \] \[ f(1) = \frac{1}{2} \] Conclusion:
Since LHL \( \ne \) RHL, the limit \( \lim_{x \to 1} f(x) \) does not exist. Therefore, the function \( f(x) \) is discontinuous at \( x = 1 \). Also, although LHL = \( f(1) \), the function is still discontinuous at \( x = 1 \) because the RHL is not equal to these values. For a function to be continuous at a point, the LHL, RHL, and the value of the function at that point must all be equal.
Quick Tip: For a function to be continuous at a point, the left-hand limit, right-hand limit, and the function's value at that point must all exist and be equal.

Step 1: {Check for continuity at \( x = 0 \)
We need to find the limit of \( f(x) \) as \( x \to 0 \): \[ \lim_{x \to 0} \frac{x^2 \log(\cos x)}{\log(1 + x)} = \lim_{x \to 0} x^2 \cdot \log(\cos x) = 0 \cdot \log(1) = 0 \] Thus, \( f(x) \) is continuous at \( x = 0 \). Step 2: {Check for differentiability at \( x = 0 \)
We now check if \( f(x) \) is differentiable at \( x = 0 \) by finding the derivative at this point: \[ f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{h^2 \log(\cos h)}{h \log(1 + h)} = 0 \] Thus, \( f(x) \) is differentiable at \( x = 0 \). Quick Tip: To check if a function is differentiable at a point, verify both its continuity and the existence of the derivative at that point.

Step 1: {Check the points of non-differentiability
We know that for a function to be differentiable at a point, both the left-hand derivative (LHD) and right-hand derivative (RHD) must be equal at that point. At \( x = -\sqrt{5} \), the left-hand derivative is 0, but the right-hand derivative is \( 2x = -2\sqrt{5} \). Therefore, \( f(x) \) is not differentiable at \( x = -\sqrt{5} \). Similarly, at \( x = \sqrt{5} \), the left-hand derivative is \( 2x = 2\sqrt{5} \), and the right-hand derivative is 0, meaning \( f(x) \) is not differentiable at \( x = \sqrt{5} \). Thus, \( k = 2 \). Step 2: {Calculate \( k - 2 \)
Since \( k = 2 \), we find: \[ k - 2 = 2 - 2 = 0 \] Quick Tip: For piecewise functions, check the left-hand and right-hand derivatives at the boundary points to determine non-differentiability.

Step 1: {Implicit Differentiation
Given the equation \( x\sqrt{1 + y} + y\sqrt{1 + x} = 0 \), differentiate both sides with respect to \( x \): \[ \frac{d}{dx} \left(x\sqrt{1 + y}\right) + \frac{d}{dx} \left(y\sqrt{1 + x}\right) = 0 \] Using the product rule: \[ \frac{d}{dx} \left(x\sqrt{1 + y}\right) = \sqrt{1 + y} + x \cdot \frac{1}{2\sqrt{1 + y}} \cdot \frac{dy}{dx} \] \[ \frac{d}{dx} \left(y\sqrt{1 + x}\right) = \sqrt{1 + x} \cdot \frac{dy}{dx} + y \cdot \frac{1}{2\sqrt{1 + x}} \] Now, substitute and solve for \( \frac{dy}{dx} \). Step 2: {Solve for \( \frac{dy}{dx} \)
After simplifying, we find: \[ \frac{dy}{dx} = -\frac{1}{(1 + x)^2} \] Quick Tip: When differentiating implicitly, apply the product and chain rule carefully, and isolate \( \frac{dy}{dx} \) to solve for it.