The dimension of Speed of Light (\(c\)) is the dimension of velocity.


Velocity is calculated as Distance/Time.


The dimensional formula for velocity is \([L][T^{-1}] = L T^{-1}\).


Expressed in terms of M, L, T, this is \(M^0LT^{-1}\).


Option (C) gives the dimension as \(M^0LT^{-2}\), which is the dimension of acceleration.


Therefore, option (C) lists the incorrect dimensions.
Quick Tip: Always verify the dimensions of fundamental physical quantities. Velocity (speed) is \(LT^{-1}\), Acceleration is \(LT^{-2}\), and Force is \(MLT^{-2}\). Kinetic Energy is \([M][LT^{-1}]^2 = ML^2T^{-2}\).

The length of the side of the cube is \(L = 1.1 \times 10^{-2}\) m.

\(L\) has 2 significant figures (SF).


The volume of the cube is \(V = L^3\).

\(V = (1.1)^3 \times (10^{-2})^3 \, m^3\).

\(V = 1.331 \times 10^{-6} \, m^3\).


In multiplication (or exponentiation), the result must have the same number of SF as the least precise input, which is 2 SF.


Rounding \(1.331\) to 2 significant figures gives \(1.3\).

\(V = 1.3 \times 10^{-6} \, m^3\).
Quick Tip: When performing multiplication or division, the final answer must be limited to the least number of significant figures found in the original numbers.

Let the distance between P and Q be \(D\).


Total distance traveled \(= D + D = 2D\).


Time taken from P to Q: \(t_1 = \frac{D}{u}\).


Time taken from Q to P: \(t_2 = \frac{D}{v}\).


Total time taken \(T = t_1 + t_2 = \frac{D}{u} + \frac{D}{v} = D \left( \frac{v+u}{uv} \right)\).


Average speed (\(\bar{v}\)) is \(\frac{Total Distance}{Total Time}\).

\(\bar{v} = \frac{2D}{D \left( \frac{u+v}{uv} \right)}\).

\(\bar{v} = \frac{2uv}{u+v}\).
Quick Tip: The average speed for two equal distances covered at speeds \(u\) and \(v\) is given by the harmonic mean \(\frac{2uv}{u+v}\). Note that this is different from the arithmetic mean \(\frac{u+v}{2}\) which applies to equal time intervals.

For \(\vec{a}\) to be a unit vector, its magnitude \(|\vec{a}|\) must be equal to 1.

\(|\vec{a}|^2 = (0.4)^2 + (0.3)^2 + b^2\).

\(1^2 = 0.16 + 0.09 + b^2\).

\(1 = 0.25 + b^2\).

\(b^2 = 1 - 0.25 = 0.75\).

\(b = \sqrt{0.75}\).

\(b = \sqrt{\frac{75}{100}} = \sqrt{\frac{3}{4}}\).

\(b = \frac{\sqrt{3}}{2}\).
Quick Tip: A vector \(\vec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k}\) is a unit vector if \(A_x^2 + A_y^2 + A_z^2 = 1\).

The acceleration (\(a\)) of a body is equal to the slope of its velocity-time graph.


The slope (\(m\)) of a line making an angle \(\theta\) with the horizontal axis (time axis) is \(m = \tan \theta\).


Given angle \(\theta = 60^\circ\).


Acceleration \(a = \tan(60^\circ)\).

\(a = \sqrt{3} \, m s^{-2}\).
Quick Tip: The slope of the \(v-t\) graph yields acceleration. If the line is a straight line, the acceleration is constant. If the angle is obtuse, the acceleration is negative (retardation).

The ceiling must support the total downward weight hanging from it.


The total weight supported by the ceiling is the sum of the weight of the body (\(W\)) and the weight of the chain (\(w\)).


Total downward force exerted by the chain/body system on the ceiling \(= W + w\).


By Newton's third law, the force exerted by the ceiling on the chain must be equal and opposite to the total downward force applied.


Force applied by the ceiling = \(W + w\).
Quick Tip: In a system in static equilibrium, the supporting force (tension in the chain at the ceiling) must balance the total load suspended beneath it.

For safe turning on a horizontal road, the necessary centripetal force must be provided by the force of static friction (\(F_s\)).

\(F_{centripetal} = F_{s, \max}\).

\(\frac{mv^2}{R} = \mu_s N\).


For a horizontal road, the normal force \(N = mg\).

\(\frac{mv^2}{R} = \mu_s mg\).

\(v^2 = \mu_s R g\).


Given \(\mu_s = 0.1\), \(R = 2\) m, \(g = 10 \, m s^{-2}\).

\(v^2 = (0.1) \times (2) \times (10) = 2\).

\(v = \sqrt{2} \, m s^{-1}\).
Quick Tip: The maximum safe speed for negotiating a curve on a horizontal road is independent of the mass of the vehicle/cyclist: \(v_{\max} = \sqrt{\mu_s R g}\).

Weight loss means the apparent weight (\(W'\)) is less than the actual weight (\(mg\)).

\(W' = m(g - a_{eff})\).


If the elevator moves with uniform velocity (up or down), \(a=0\), so \(W' = mg\) (no change).


If the elevator accelerates upwards, \(W' = m(g+a)\), resulting in weight gain.


If the elevator accelerates downwards (with constant acceleration \(a\)), \(W' = m(g-a)\).


Since \(a > 0\), \(W' < mg\), indicating weight loss.
Quick Tip: Apparent weight is calculated by \(W' = m(g - a)\), where \(a\) is the acceleration of the elevator (positive upwards). Downward acceleration means \(a\) is negative, leading to \(W' = m(g - (-|a|)) = m(g-|a|)\).

For any system undergoing Simple Harmonic Motion (SHM), mechanical energy (\(E\)) is conserved.


The total energy \(E\) is given by the maximum kinetic energy (\(KE_{\max}\)), which occurs at the equilibrium position.

\(E = KE_{\max}\).


The total energy \(E\) is also given by the maximum potential energy (\(PE_{\max}\)), which occurs at the extreme positions.

\(E = PE_{\max}\).


Therefore, \(KE_{\max} = PE_{\max}\).


The ratio \(\frac{KE_{\max}}{PE_{\max}} = \frac{1}{1}\).
Quick Tip: In SHM, the energy oscillates between kinetic and potential forms. The total energy is constant and equals the peak value of either the kinetic or the potential energy.

Force \(F = 6\) N.


Displacement \(d = 3\) m.


Time \(t = 2\) s.


Work done \(W = F \times d\) (assuming force is along displacement).

\(W = 6 \, N \times 3 \, m = 18 \, J\).


Average Power (\(P_{avg}\)) is the rate of work done: \(P_{avg} = \frac{W}{t}\).

\(P_{avg} = \frac{18 \, J}{2 \, s}\).

\(P_{avg} = 9 \, W\).
Quick Tip: Power is work per unit time. When calculating average power from total displacement, ensure both work (or energy change) and time interval are correctly calculated.

Mass \(m = 3\) kg.


Amplitude \(A = 0.1\) m.


Time period \(T = 3.14\) s. Use \(\pi \approx 3.14\).


Angular frequency \(\omega = \frac{2\pi}{T}\).

\(\omega = \frac{2\pi}{3.14} \approx \frac{2\pi}{\pi} = 2 \, rad s^{-1}\).


The maximum force (\(F_{\max}\)) occurs at the maximum displacement (amplitude \(A\)) and is given by \(F_{\max} = m a_{\max}\).


Maximum acceleration \(a_{\max} = \omega^2 A\).

\(F_{\max} = m \omega^2 A\).

\(F_{\max} = (3) \times (2)^2 \times (0.1)\).

\(F_{\max} = 3 \times 4 \times 0.1 = 1.2\) N.
Quick Tip: The maximum acceleration in SHM is \(a_{\max} = \omega^2 A\). The maximum restoring force is \(F_{\max} = m a_{\max}\). Always look for approximations (like \(3.14 \approx \pi\)) to simplify calculations.

A circus acrobat (or a diver) changes their rotational speed by changing their body posture (pulling arms/legs in or stretching them out).


When they pull their limbs in, their moment of inertia (\(I\)) decreases.


Since there is negligible external torque, the angular momentum (\(L = I\omega\)) must remain conserved.


As \(I\) decreases, the angular velocity (\(\omega\)) increases, allowing them to complete more rotations in the air.


This demonstrates the conservation of angular momentum.
Quick Tip: The concept of changing rotational speed by altering body configuration (moment of inertia) is a classic illustration of the conservation of angular momentum (\(L=constant\) if \(\tau_{ext}=0\)).

Angular velocity \(\omega\) is given by \(\omega = \frac{2\pi}{T}\), where \(T\) is the time period.


Time period of minute hand (\(T_M\)): \(T_M = 1\) hour.


Time period of hour hand (\(T_H\)): \(T_H = 12\) hours.


The ratio of angular velocities is \(\frac{\omega_M}{\omega_H} = \frac{2\pi/T_M}{2\pi/T_H} = \frac{T_H}{T_M}\).

\(\frac{\omega_M}{\omega_H} = \frac{12 \, hours}{1 \, hour} = 12\).


The ratio is \(12\).
Quick Tip: The ratio of angular velocities of rotating bodies with constant periods is equal to the inverse ratio of their time periods (\(\omega_1/\omega_2 = T_2/T_1\)).

The acceleration due to gravity at height \(h\) is given by \(g_h = g \left( \frac{R}{R+h} \right)^2\).


We require \(g_h = \frac{g}{2}\).

\(\frac{g}{2} = g \left( \frac{R}{R+h} \right)^2\).

\(\frac{1}{2} = \left( \frac{R}{R+h} \right)^2\).


Taking the square root: \(\frac{1}{\sqrt{2}} = \frac{R}{R+h}\).

\(R+h = R\sqrt{2}\).

\(h = R\sqrt{2} - R = R(\sqrt{2} - 1)\).

\(h \approx 0.414 R\).


Since \(0.414R\) is not among the options (A, B, C, D, E), the question was officially cancelled.
Quick Tip: For distances comparable to the Earth's radius, use the inverse square law for gravity \(g_h = g (R/(R+h))^2\). Do not use the linear approximation \(g_h \approx g (1 - 2h/R)\) unless \(h\) is very small.

Mass \(m = 100 \, g = 0.1\) kg.


Initial Kinetic Energy \(KE = 20\) J.


Acceleration due to gravity \(g = 10 \, m s^{-2}\).


By the principle of conservation of energy, the initial kinetic energy is entirely converted into gravitational potential energy (\(PE\)) at the maximum height \(H\).

\(KE = PE\).

\(KE = mgH\).

\(20 \, J = (0.1 \, kg) \times (10 \, m s^{-2}) \times H\).

\(20 = 1 \times H\).

\(H = 20\) m.
Quick Tip: Ensure units are consistent (S.I. units) before applying conservation laws. The maximum height occurs when all initial kinetic energy is converted to gravitational potential energy.

The figure shows that the streamlines are more crowded (denser) in region 2 (above the ball) than in region 1 (below the ball).


According to the continuity equation, higher density of streamlines implies higher fluid speed.


Thus, \(v_2 > v_1\).


According to Bernoulli's principle, for horizontal flow, \(P + \frac{1}{2}\rho v^2 = constant\).


Higher velocity corresponds to lower pressure.


Since \(v_2 > v_1\), it follows that \(P_2 < P_1\).


The correct relations are \(v_1 < v_2\) and \(P_1 > P_2\).
Quick Tip: Bernoulli's principle dictates an inverse relationship between dynamic pressure (\(\frac{1}{2}\rho v^2\)) and static pressure (\(P\)). Faster fluid flow means lower pressure.

The excess pressure (\(\Delta P\)) inside a soap bubble is given by \(\Delta P = \frac{4T}{R}\), where \(T\) is surface tension and \(R\) is the radius.


Thus, \(\Delta P\) is inversely proportional to the radius (\(R\)).

\(\Delta P \propto \frac{1}{R}\).


Given radii \(R_1 = 2\) cm and \(R_2 = 3\) cm.


The ratio of excess pressures is \(\frac{\Delta P_1}{\Delta P_2} = \frac{R_2}{R_1}\).

\(\frac{\Delta P_1}{\Delta P_2} = \frac{3 \, cm}{2 \, cm} = \frac{3}{2}\).


The ratio is \(3:2\).
Quick Tip: Always distinguish between a soap bubble (\(\Delta P = 4T/R\)) and a liquid drop (\(\Delta P = 2T/R\)). For the same liquid and temperature, the smaller radius always corresponds to higher excess pressure.

Elastic energy stored per unit volume (\(u\)) is defined as: \(u = \frac{1}{2} \times Stress \times Strain\).

\(u = \frac{1}{2} S \epsilon\), where \(\epsilon\) is the strain.


We use Hooke's law, which states \(Y = \frac{Stress}{Strain} = \frac{S}{\epsilon}\).


Rearranging for strain: \(\epsilon = \frac{S}{Y}\).


Substitute \(\epsilon\) into the energy density formula:

\(u = \frac{1}{2} S \left( \frac{S}{Y} \right)\).

\(u = \frac{1}{2} \frac{S^2}{Y}\).
Quick Tip: The three common forms for elastic energy density are: \(\frac{1}{2} S\epsilon\), \(\frac{1}{2} Y\epsilon^2\), and \(\frac{1}{2} S^2/Y\). Choose the form involving the variables specified in the question (S and Y).

The Zeroth Law of Thermodynamics establishes the concept of thermal equilibrium.


It states that if system A is in thermal equilibrium with system C, and system B is also in thermal equilibrium with system C, then A and B are in thermal equilibrium with each other.


This transitive property defines temperature as the state variable that determines whether systems are in thermal equilibrium.
Quick Tip: The four laws of thermodynamics are conceptually linked to specific properties: Zeroth Law \(\rightarrow\) Temperature; First Law \(\rightarrow\) Internal Energy/Energy Conservation; Second Law \(\rightarrow\) Entropy/Directionality; Third Law \(\rightarrow\) Absolute Zero.

The slope of a curve on a \(P-V\) diagram is \(m = \frac{dP}{dV}\).


1. Isothermal Process: \(PV = constant\). Differentiation yields \(P dV + V dP = 0\).

\(m_i = \left( \frac{dP}{dV} \right)_{iso} = -\frac{P}{V}\).


2. Adiabatic Process: \(PV^\gamma = constant\). Differentiation yields \(P (\gamma V^{\gamma-1} dV) + V^\gamma dP = 0\).


Dividing by \(V^{\gamma-1}\): \(P\gamma dV + V dP = 0\).

\(m_a = \left( \frac{dP}{dV} \right)_{adia} = -\gamma \frac{P}{V}\).


Comparing \(m_a\) and \(m_i\):

\(m_a = \gamma \left( -\frac{P}{V} \right) = \gamma m_i\).
Quick Tip: The adiabatic process curve is always steeper than the isothermal curve passing through the same point, with the ratio of slopes being the adiabatic index \(\gamma = C_p/C_v\).

The work done (\(W\)) during a thermodynamic process is defined as \(W = P \Delta V\).


In an isochoric process, the volume of the system is held constant.


Therefore, the change in volume is zero (\(\Delta V = 0\)).

\(W = P \times 0 = 0\).


The work done by the gas on the system in an isochoric process is zero.
Quick Tip: Work done in a pressure-volume process is zero if and only if the volume remains constant (isochoric process). This is visually represented by a vertical line on a P-V diagram.

The ratio of specific heats is \(\gamma = \frac{C_p}{C_v}\).


According to the Law of Equipartition of Energy, the molar heat capacity at constant volume is \(C_v = \frac{f}{2}R\), where \(f\) is the degrees of freedom.


By Mayer's relation, \(C_p - C_v = R\).

\(C_p = C_v + R = \frac{f}{2}R + R = \frac{f+2}{2}R\).


Thus, the ratio \(\gamma = \frac{C_p}{C_v} = \frac{(f+2)R/2}{fR/2} = \frac{f+2}{f} = 1 + \frac{2}{f}\).


For a polyatomic gas, \(f \ge 6\). For \(f=6\), \(\gamma = 1 + 2/6 = 4/3\).


Since the expected formula \(\frac{f+2}{f}\) does not match any given option, and given the complexity of degrees of freedom for real polyatomic gases, this question was likely cancelled due to ambiguity or incorrect options.
Quick Tip: For an ideal gas, the ratio of specific heats is strictly related to the degrees of freedom (\(f\)) by \(\gamma = 1 + 2/f\). Check if the question specifies monoatomic (\(f=3\)) or diatomic (\(f=5\) or \(7\)).

The mean free path (\(\lambda\)) is inversely proportional to the number of molecules per unit volume (\(n\), number density).


The relation is \(\lambda = \frac{1}{\sqrt{2}\pi d^2 n}\).


Therefore, \(\lambda_1 n_1 = \lambda_2 n_2\).


We are given: \(\lambda_1 = 2.2 \times 10^{-5}\) m, \(n_1 = 0.8 \times 10^{24} \, m^{-3}\), and \(n_2 = 1.0 \times 10^{24} \, m^{-3}\).


We need to find \(\lambda_2\): \(\lambda_2 = \lambda_1 \left( \frac{n_1}{n_2} \right)\).

\(\lambda_2 = (2.2 \times 10^{-5}) \times \left( \frac{0.8 \times 10^{24}}{1.0 \times 10^{24}} \right)\).

\(\lambda_2 = 2.2 \times 10^{-5} \times 0.8\).

\(\lambda_2 = 1.76 \times 10^{-5}\) m.
Quick Tip: The mean free path is inversely proportional to the number density (\(n\)). If the number density increases, the molecules are closer together, leading to a shorter path length between collisions.

The equation for displacement in SHM is \(x = a \sin(\omega t)\).


The displacement amplitude is \(A_x = a\).


The magnitude of acceleration is \(A = \omega^2 x\).


The maximum acceleration (acceleration amplitude) occurs at maximum displacement \(x=a\):

\(A_{max} = \omega^2 a\).


The ratio requested is \(\frac{Acceleration Amplitude}{Displacement Amplitude}\).


Ratio \(= \frac{A_{max}}{a} = \frac{\omega^2 a}{a} = \omega^2\).
Quick Tip: In SHM, the maximum velocity is \(v_{\max} = a\omega\), and the maximum acceleration is \(a_{\max} = a\omega^2\). These relations highlight the dependence on amplitude (\(a\)) and angular frequency (\(\omega\)).

The speed (\(v\)) of a transverse wave traveling on a stretched string is determined by the restoring force (Tension, \(T\)) and the inertia of the medium (linear mass density, \(\mu\)).


The standard formula derived from wave mechanics is:

\(v = \sqrt{\frac{Tension}{Linear Mass Density}}\).

\(v = \sqrt{\frac{T}{\mu}}\).
Quick Tip: Wave speed generally follows the form \(v = \sqrt{Elastic Property/Inertial Property}\). For a string, \(T\) is the elastic property (restoring force) and \(\mu\) is the inertial property.

An open pipe is open at both ends, resulting in antinodes at both ends.


The fundamental mode (lowest frequency, \(f_1\)) corresponds to the standing wave pattern where the length of the pipe \(L\) equals half a wavelength (\(\lambda/2\)).

\(L = \frac{\lambda_1}{2} \implies \lambda_1 = 2L\).


Since frequency \(f = \frac{v}{\lambda}\), the lowest frequency is:

\(f_1 = \frac{v}{2L}\).
Quick Tip: For open pipes, resonant frequencies are \(f_n = n \frac{v}{2L}\) (\(n=1, 2, 3, \dots\)). For closed pipes, they are \(f_n = n \frac{v}{4L}\) (\(n=1, 3, 5, \dots\)).

The energy density (\(u\)) or energy stored per unit volume in a region where the electric field is \(E\) (in vacuum or free space) is given by the formula:

\(u = \frac{Energy Stored}{{Volume}} = \frac{1}{2} \varepsilon_0 E^2\).
Quick Tip: The energy stored in a capacitor (\(U = 1/2 C V^2\)) is distributed throughout the volume occupied by the electric field \(E\). Energy density is a critical concept relating energy storage directly to the field strength.

The force between two charges \(q_1\) and \(q_2\) separated by distance \(r\) in a medium with dielectric constant \(K\) is related to the force in air (\(F_{air}\)) by:

\(F_{medium} = \frac{F_{air}}{K}\).


Given \(F_{air} = F\) and \(F_{medium} = 0.5 F\).

\(0.5 F = \frac{F}{K}\).

\(K = \frac{F}{0.5 F} = \frac{1}{0.5}\).

\(K = 2\).
Quick Tip: The dielectric constant \(K\) (or relative permittivity \(\varepsilon_r\)) quantifies how much the medium reduces the electrostatic force compared to vacuum (or air). Since \(K\) is always \(\ge 1\), the force always decreases or stays the same.

The diagram shows a capacitor \(C\) connected in series with a battery providing potential difference \(V\) when the key K is closed.


After closing the key and allowing the circuit to reach steady state, the capacitor \(C\) charges fully.


The potential difference across the capacitor will be equal to the battery voltage \(V\).


The energy (\(U\)) stored in a capacitor \(C\) charged to a potential difference \(V\) is given by:

\(U = \frac{1}{2} C V^2\).
Quick Tip: Remember the three equivalent formulas for stored capacitor energy: \(U = \frac{1}{2} Q V = \frac{1}{2} C V^2 = \frac{Q^2}{2C}\).

Resistance \(R\) is given by \(R = \rho \frac{L}{A}\), where \(\rho\) is resistivity (constant).


The mass \(m\) is \(m = Volume \times Density = A L \rho_d\). (\(\rho_d\) is constant material density).


We express area \(A\) in terms of \(m\) and \(L\): \(A = \frac{m}{L \rho_d}\).


Substitute \(A\) into the resistance formula: \(R = \rho \frac{L}{(m / (L \rho_d))} = \frac{\rho \rho_d L^2}{m}\).


Since \(\rho\) and \(\rho_d\) are constant for copper wires, \(R \propto \frac{L^2}{m}\).


Given ratios: \(L_1: L_2: L_3 = 5:3:1\) and \(m_1: m_2: m_3 = 1:3:5\).

\(R_1: R_2: R_3 = \frac{L_1^2}{m_1} : \frac{L_2^2}{m_2} : \frac{L_3^2}{m_3}\).

\(R_1: R_2: R_3 = \frac{5^2}{1} : \frac{3^2}{3} : \frac{1^2}{5}\).

\(R_1: R_2: R_3 = 25 : 3 : \frac{1}{5}\).


Multiplying by 5 to clear the fraction: \(R_1: R_2: R_3 = 125 : 15 : 1\).
Quick Tip: When resistance problems involve mass or density, use the proportionality \(R \propto L^2/m\) (if length is given) or \(R \propto m/A^2\) (if area is given) to solve quickly, as resistivity and density are constant for the same material.

Mobility (\(\mu\)) is defined as the magnitude of the drift velocity (\(v_d\)) acquired per unit electric field (\(E\)).

\(\mu = \frac{v_d}{E}\).


The drift velocity of an electron is related to the relaxation time (\(\tau\)) by \(v_d = \frac{e E \tau}{m}\).


Substituting the expression for \(v_d\) into the mobility equation:

\(\mu = \frac{(e E \tau / m)}{E}\).

\(\mu = \frac{e\tau}{m}\).
Quick Tip: Mobility (\(\mu\)) connects microscopic parameters (\(e, \tau, m\)) to macroscopic behavior (conductivity), making it crucial for understanding charge transport in materials.

\(P\) is the power delivered at voltage \(V\). The current (\(I\)) flowing through the cable is \(I = \frac{P}{V}\).


The transmission cable has resistance \(R_c\).


The power dissipated (\(P_{dissipated}\)) in the cable is the Joule heating loss, given by \(P_{loss} = I^2 R_c\).


Substitute the expression for current \(I\):

\(P_{loss} = \left( \frac{P}{V} \right)^2 R_c\).

\(P_{loss} = \frac{P^2 R_c}{V^2}\).
Quick Tip: Power dissipation (loss) is proportional to \(I^2 R\). High voltage transmission (\(V\)) minimizes the current (\(I=P/V\)) for a given power (\(P\)), thereby significantly reducing heat loss.

When a charged particle moves perpendicular to a uniform magnetic field, the radius (\(r\)) of its circular path is given by \(r = \frac{mv}{qB}\).


Since the speed (\(v\)) and magnetic field (\(B\)) are the same for both particles, the radius is proportional to the mass-to-charge ratio: \(r \propto \frac{m}{q}\).


Proton (p): Mass \(m_p\), Charge \(q_p = e\).


Deuteron (d): Mass \(m_d \approx 2m_p\), Charge \(q_d = e\).


Ratio of radii: \(\frac{r_p}{r_d} = \frac{m_p/q_p}{m_d/q_d}\).

\(\frac{r_p}{r_d} = \frac{m_p/e}{2m_p/e} = \frac{m_p}{2m_p} = \frac{1}{2}\).


The ratio is \(1:2\).
Quick Tip: The motion of charged particles in a perpendicular magnetic field depends on the specific charge (\(q/m\)). Remember that a deuteron is essentially a heavy proton, having approximately twice the mass but the same charge.

Biot-Savart's law calculates the magnetic field produced by a current element (\(d\vec{B} = \frac{\mu_0}{4\pi} \frac{I d\vec{l} \times \hat{r}}{r^2}\)).


Ampere's circuital law (\(\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enclosed}\)) is another fundamental law in magnetostatics, which often provides an easier way to calculate magnetic fields for highly symmetric current distributions.


Ampere's circuital law is derived from the Biot-Savart law and is considered an alternative mathematical formulation for magnetostatics, similar to how Gauss's law is an alternative to Coulomb's law in electrostatics.
Quick Tip: In electromagnetism, integral laws (Ampere's, Gauss's) simplify calculations for symmetric setups, while differential laws (Biot-Savart, Coulomb's) describe the fundamental interaction elements.

The resonant angular frequency (\(\omega_0\)) for a series LCR circuit is given by the formula:

\(\omega_0 = \frac{1}{\sqrt{LC}}\).


Convert given values to S.I. units:

\(L = 1 μH= 1 \times 10^{-6}\) H.

\(C = 1 μF  = 1 \times 10^{-6}\) F.


(Note: Resistance \(R\) does not affect the resonant frequency, only the sharpness of resonance).

\(\omega_0 = \frac{1}{\sqrt{(1 \times 10^{-6}) \times (1 \times 10^{-6})}}\).

\(\omega_0 = \frac{1}{\sqrt{10^{-12}}} = \frac{1}{10^{-6}}\).

\(\omega_0 = 10^6 \, rad s^{-1}\).
Quick Tip: Resonance occurs when inductive reactance (\(X_L\)) equals capacitive reactance (\(X_C\)), leading to the condition \(\omega_0 L = 1/(\omega_0 C)\), which directly yields \(\omega_0 = 1/\sqrt{LC}\).

The visible spectrum ranges approximately from \(400\) nm (Violet) to \(700\) nm (Red).


Using the relation \(f = c/\lambda\) (where \(c = 3 \times 10^8 \, m/s\)):


Frequency range for visible light is roughly \(4.3 \times 10^{14}\) Hz (Red) to \(7.5 \times 10^{14}\) Hz (Violet).


The given frequency \(f = 5 \times 10^{14}\) Hz falls within this range. (This frequency corresponds roughly to green light).
Quick Tip: It is helpful to memorize the approximate frequency ranges for visible light (\(\approx 4 \times 10^{14}\) to \(7.5 \times 10^{14}\) Hz). Frequencies lower than visible are Infrared, Microwave, Radio; frequencies higher are UV, X-ray, Gamma ray.

When light travels from a rarer medium (lower refractive index \(n_1\)) to a denser medium (higher refractive index \(n_2\)).


1. Speed: The speed of light \(v\) decreases, as \(v_2 = v_1 / n_{rel}\).


2. Frequency: The frequency (\(f\)) of light is determined by the source and remains constant across mediums.


3. Wavelength: Since \(v = f\lambda\), and \(f\) is constant, if the speed \(v\) decreases, the wavelength \(\lambda\) must also decrease proportionally.


The wavelength decreases (\(\lambda_2 < \lambda_1\)).
Quick Tip: Frequency is invariant across media changes because it is determined by the source oscillation. Wavelength and velocity change inversely with the refractive index.

The fringe width (\(\beta\)) in Young's double-slit experiment is given by \(\beta = \frac{\lambda D}{d}\).


Since the setup parameters \(D\) (screen distance) and \(d\) (slit separation) are the same, \(\beta\) is directly proportional to the wavelength (\(\lambda\)).


The wavelengths for the given colors are ordered as: \(\lambda_{Red} > \lambda_{Green} > \lambda_{Blue}\).


Therefore, the fringe widths must follow the same order: \(\beta_R > \beta_G > \beta_B\).


In increasing order, this is \(\beta_B < \beta_G < \beta_R\).
Quick Tip: Fringe width is directly proportional to the wavelength. Red light, having the longest visible wavelength, produces the widest fringes, while blue light produces the narrowest fringes.

According to Bohr's model, supplemented by de Broglie's hypothesis, stationary orbits are formed only when the circumference of the orbit (\(2\pi r_n\)) is an integral multiple (\(n\)) of the electron's de Broglie wavelength (\(\lambda\)).


Bohr's quantization condition is \(2\pi r_n = n\lambda\).


Here, \(n\) represents the principal quantum number of the orbit and also the number of standing waves (de Broglie waves) that fit into the circumference.


For the third orbit, \(n=3\).


Therefore, the number of de Broglie waves associated with the electron is 3.
Quick Tip: The key takeaway from de Broglie's explanation of Bohr's orbits is that the number of standing electron waves around the nucleus is equal to the principal quantum number \(n\).

Nuclear reactions must conserve both mass number (\(A\)) and atomic number (\(Z\)).


Reactant (X): \(A = 198\), \(Z = 80\).


Product (Y): \(A' = 197\), \(Z' = 79\).


Let the emitted particle be \({}_{z}^{a}P\).


Conservation of mass number (\(A\)): \(198 = 197 + a \implies a = 1\).


Conservation of atomic number (\(Z\)): \(80 = 79 + z \implies z = 1\).


The emitted particle is \({}_{1}^{1}P\).


A particle with mass number 1 and atomic number 1 is a proton (\(p\)).
Quick Tip: Protons (\(^1_1p\)), neutrons (\(^1_0n\)), \(\alpha\)-particles (\(^4_2He\)), and \(\beta\)-particles (\(^0_{-1}e\) or \(^0_1e\)) are the standard components of nuclear reactions. Always ensure mass number (\(A\)) and charge (\(Z\)) balance.

Nuclear fusion is a process where two lighter nuclei (\(M_A\) and \(M_B\)) combine to form a single heavier product nucleus (\(M_C\)) and release a tremendous amount of energy (\(Q\)).


This released energy is due to a mass defect (\(\Delta M\)) according to Einstein's mass-energy equivalence \(Q = \Delta M c^2\).


The mass defect is \(\Delta M = (Mass of Reactants) - (Mass of Products)\).


Assuming no other particles are emitted: \(\Delta M = (M_A + M_B) - M_C\).


Since energy (\(Q\)) is released (exothermic), \(\Delta M\) must be positive (\(\Delta M > 0\)).


Therefore, \(M_A + M_B > M_C\), or \(M_C < M_A + M_B\).
Quick Tip: Both nuclear fusion and fission are exothermic (release energy). In any exothermic nuclear reaction, the total mass of the products is always less than the total mass of the reactants, satisfying the relation \(M_{products} < M_{reactants}\).

This relationship is known as the Law of Mass Action in semiconductors.


It states that under thermal equilibrium, the product of the concentration of electrons (\(n_e\)) and the concentration of holes (\(n_h\)) is constant for a given material and temperature, irrespective of whether the semiconductor is intrinsic or extrinsic.


This product is equal to the square of the intrinsic carrier concentration (\(n_i\)):

\(n_e n_h = n_i^2\).
Quick Tip: In intrinsic semiconductors, \(n_e = n_h = n_i\). In extrinsic semiconductors, \(n_e \neq n_h\), but their product remains \(n_i^2\). For example, in n-type, \(n_e > n_i\) and \(n_h = n_i^2/n_e\).

Half-life \(T_{1/2} = 2\) days.


The fraction remaining undecayed (\(\frac{N}{N_0}\)) after \(n\) half-lives is given by \(\frac{N}{N_0} = \left(\frac{1}{2}\right)^n\).


Given \(\frac{N}{N_0} = \frac{1}{32}\).


Since \(32 = 2^5\), we have \(\frac{1}{32} = \left(\frac{1}{2}\right)^5\).


Therefore, the number of half-lives passed is \(n=5\).


The total time elapsed \(t\) is \(t = n \times T_{1/2}\).

\(t = 5 \times 2 \, days = 10\) days.
Quick Tip: When the remaining fraction is given as \(1/2^n\), the number of half-lives (\(n\)) is simply the power of 2 in the denominator. Total time is then \(n \times T_{1/2}\).

At absolute zero temperature (\(T=0\) K), there is no thermal energy available to break the covalent bonds in an intrinsic semiconductor.


Consequently, all valence electrons remain trapped in the valence band, leaving the conduction band completely empty.


Since there are no free charge carriers (electrons or holes) available for conduction, the semiconductor behaves as a perfect insulator.
Quick Tip: The conductivity of a semiconductor relies on thermal excitation. At \(T=0\) K, this excitation stops, and the material acts like a material with a large band gap, i.e., an insulator.

In reverse biasing, the positive terminal of the external battery is connected to the \(n\)-side and the negative terminal to the \(p\)-side.


This external voltage adds to the built-in potential barrier (\(\phi_B\)) of the \(p-n\) junction.


This results in an increase in the width of the depletion layer and an increase in the potential barrier height.


A higher barrier height effectively prevents the flow of majority carriers.
Quick Tip: Reverse bias increases the depletion width and barrier height, severely limiting current flow (only a small minority carrier current flows). Forward bias decreases the depletion width and barrier height, allowing majority carriers to flow easily.

Mass of solute (alcohol) \(= 10\) g.


Mass of solvent (water) \(= 90\) g.


Total mass of solution \(= Mass of solute + Mass of solvent = 10 \, g + 90 \, g = 100\) g.


Mass percentage of alcohol \(= \frac{Mass of alcohol}{Total mass of solution} \times 100\).


Percentage \(= \frac{10 \, g}{100 \, g} \times 100 = 10%\).
Quick Tip: Mass percentage concentration is calculated as the mass of the component divided by the total mass of the solution, multiplied by 100.

The allowed ranges for quantum numbers are: \(n \ge 1\); \(l = 0\) to \(n-1\); \(m_l = -l\) to \(+l\); \(m_s = \pm 1/2\).


(A) \(n=3, l=2\). \(m_l\) must be between \(-2\) and \(+2\). \(m_l=-4\) is impossible.


(B) \(n=2\). \(l\) must be 0 or 1. \(l=2\) is impossible.


(C) \(n=2\). \(l=2\) is impossible. Also, \(m_s=1\) is impossible.


(D) \(n=3\). \(l=2\) (possible since \(l \le n-1 = 2\)). \(m_l=-2\) (possible since \(|-2| \le 2\)). \(m_s=1/2\) (possible). This set is possible.


(E) \(n=3\). \(l\) must be \(\le 2\). \(l=3\) is impossible.
Quick Tip: Always check the dependencies: \(l\) depends on \(n\) (\(l_{max} = n-1\)), and \(m_l\) depends on \(l\) (\(|m_l| \le l\)). \(m_s\) is fixed at \(\pm 1/2\).

Palladium (Pd) has atomic number \(Z=46\). The noble gas core is Krypton (\(Z=36\)).


Expected configuration (following Aufbau): \([Kr] 4d^8 5s^2\).


However, Pd is a crucial exception in the transition metals. It achieves maximum stability by completely filling the \(d\)-subshell.


Two electrons from the \(5s\) orbital shift to the \(4d\) orbital to obtain the \(4d^{10}\) configuration.


The actual stable ground-state configuration is \([Kr] 4d^{10} 5s^0\).
Quick Tip: Chromium (Cr), Copper (Cu), and Palladium (Pd) are important exceptions to the Aufbau principle due to the enhanced stability associated with half-filled (\(d^5, s^1\)) or completely filled (\(d^{10}, s^1\) or \(d^{10}, s^0\)) \(d\)-orbitals.

We use VSEPR theory to determine geometry based on steric number (SN = bond pairs + lone pairs).


(A) \(NH_4^+\): SN = 4 (4 BP, 0 LP). Hybridization \(sp^3\). Geometry: Tetrahedral.


(B) \(XeF_4\): Xenon has 8 valence electrons. 4 bond pairs (to F), 2 lone pairs. SN = 6 (4 BP, 2 LP). Hybridization \(sp^3d^2\).


The geometry corresponding to SN=6 and 2 LP is Square Planar (lone pairs occupy axial positions).


(C), (D), (E) \(CCl_4, SiCl_4, CH_4\): SN = 4 (4 BP, 0 LP). Hybridization \(sp^3\). Geometry: Tetrahedral.
Quick Tip: Square planar geometry is typically associated with \(d^8\) metal complexes or molecules/ions with \(sp^3d^2\) hybridization having 4 bond pairs and 2 lone pairs, like \(XeF_4\).

A molecule is paramagnetic if it possesses unpaired electrons in its molecular orbitals.


We check the total number of electrons and apply Molecular Orbital Theory (MOT).


(A) \(O_2\): 16 electrons. MOT configuration: \((\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^1 (\pi^* 2p_y)^1\).


The last two electrons occupy the degenerate antibonding \(\pi^* 2p\) orbitals singly (Hund's rule). Since there are two unpaired electrons, \(O_2\) is paramagnetic.


(B) \(C_2\): 12 electrons (diamagnetic).


(C) \(N_2\): 14 electrons (diamagnetic).


(D) \(F_2\): 18 electrons (diamagnetic).


(E) \(H_2\): 2 electrons (diamagnetic).
Quick Tip: Molecular orbital diagrams for \(Z=8\) (like \(O_2\)) show that the last two electrons enter the \(\pi^*\) antibonding orbitals separately, leading to paramagnetism. Molecules with an odd number of total electrons are always paramagnetic.

Given: Pure solvent vapour pressure \(P^0 = 95\) mm Hg.


Solute (Sucrose): Mass \(W_2 = 176\) g, Molar mass \(M_2 = 342 \, g mol^{-1}\).


Solvent (Water): Mass \(W_1 = 900\) g, Molar mass \(M_1 = 18 \, g mol^{-1}\).


Calculate moles:

\(n_2 (sucrose) = \frac{176}{342} \approx 0.5146\) mol.

\(n_1 (water) = \frac{900}{18} = 50.0\) mol.


Total moles \(= n_1 + n_2 = 50.5146\) mol.


According to Raoult's Law, the vapour pressure of the solution is \(P_s = P^0 X_1\), where \(X_1\) is the mole fraction of the solvent.


Mole fraction of solvent \(X_1 = \frac{n_1}{n_1 + n_2} = \frac{50.0}{50.5146} \approx 0.989814\).

\(P_s = 95 \, mm \times 0.989814\).

\(P_s \approx 94.0323\) mm Hg.

\(94.03\) mm Hg is closest to \(94.06\) mm Hg.
Quick Tip: Raoult's Law states that the vapour pressure of a solution (\(P_s\)) is \(P^0 X_{solvent}\). Use the exact formula \(X_{solvent} = n_{solvent} / (n_{solvent} + n_{solute})\) unless the solution is extremely dilute.

Entropy (\(S\)) is a thermodynamic state function related to the degree of disorder or randomness in a system and the number of possible microscopic arrangements (microstates).


Option (A): Correct. Higher disorder corresponds to higher entropy.


Option (B): Correct. Crystalline solids are highly ordered, hence possess minimum entropy (Third Law of Thermodynamics).


Option (D): Correct. Gaseous state has maximum molecular freedom and randomness, therefore highest entropy.


Option (E): Correct. For a reversible process, \[ \Delta S = \frac{q_{rev}}{T} \]
which is the fundamental definition of entropy change.


Option (C): Incorrect.
Entropy \emph{is related to the randomness or chaotic nature of particle distribution in a system.
While temperature measures average kinetic energy, entropy represents the extent of disorder and randomness of molecular motion.
Hence, stating that entropy is \emph{not a measure of average chaotic motion is incorrect.


Therefore, the incorrect statement is (C). Quick Tip: Temperature is a measure of average particle kinetic energy (chaotic motion), whereas Entropy is a measure of the system's disorder or the number of accessible microstates. While correlated, they are distinct thermodynamic variables.

The equilibrium constant \(K_c\) expression is given by the ratio of product concentrations to reactant concentration, raised to their stoichiometric coefficients:

\(K_c = \frac{[PCl_3] [Cl_2]}{[PCl_5]}\)


Substitute the given equilibrium concentrations: \([PCl_3] = 4.0 \, M\), \([Cl_2] = 4.0 \, M\), and \([PCl_5] = 2.0 \, M\).

\(K_c = \frac{(4.0) \times (4.0)}{(2.0)}\)

\(K_c = \frac{16.0}{2.0}\).

\(K_c = 8.0 \, M\) (or \(8 \, mol dm^{-3}\)).
Quick Tip: When calculating \(K_c\), ensure that all concentrations are given in Molarity (\(mol L^{-1}\) or \(mol dm^{-3}\)) and that coefficients in the balanced equation are correctly used as exponents.

The Daniell cell reaction is: \(Zn (s) + Cu^{2+} (aq) \rightarrow Zn^{2+} (aq) + Cu (s)\).


Standard electrode potentials are: \(E^0(Zn^{2+}/Zn) = -0.76\) V and \(E^0(Cu^{2+}/Cu) = +0.34\) V.


The standard cell potential is \(E^0_{cell} = E^0_{cathode} - E^0_{anode}\).

\(E^0_{cell} = 0.34 \, V - (-0.76 \, V) = 1.10 \, V\).


Statement (C) is the correct standard potential for the Daniell cell.
Quick Tip: In a galvanic cell, \(E^0_{cell}\) is calculated by subtracting the standard potential of the anode (oxidation) from the standard potential of the cathode (reduction). For the Daniell cell, this value is a standard constant, 1.1 V.

Molar conductivity (\(\Lambda_m\)) is calculated using the formula \(\Lambda_m = \frac{\kappa}{c}\), where \(\kappa\) is conductivity and \(c\) is concentration in \(mol m^{-3}\).


Given conductivity \(\kappa = 0.248 \, S m^{-1}\).


Given concentration \(c = 0.02 \, mol L^{-1}\).


Convert concentration to \(mol m^{-3}\): \(c = 0.02 \times 1000 = 20 \, mol m^{-3}\).

\(\Lambda_m = \frac{0.248 \, S m^{-1}}{20 \, mol m^{-3}}\).

\(\Lambda_m = 0.0124 \, S m^2 mol^{-1}\).

\(\Lambda_m = 1.24 \times 10^{-2} \, S m^2 mol^{-1}\).
Quick Tip: Crucial unit conversion in electrochemistry: Concentration \(C\) in \(mol L^{-1}\) must be multiplied by 1000 to get \(C\) in \(mol m^{-3}\) when \(\kappa\) is in \(S m^{-1}\) to yield \(\Lambda_m\) in \(S m^2 mol^{-1}\).

Boiling point depends on the strength of intermolecular forces (IMFs).


Water and Ethanol exhibit strong Hydrogen Bonding (highest BP).


Benzene (non-polar, \(M \approx 78\)) and Chloroform (polar, \(M \approx 119\), dipole-dipole) have intermediate BPs.


Carbon Disulphide (\(CS_2\)) is linear and non-polar, relying only on relatively weak London Dispersion Forces (LDF). \(M(CS_2) = 76\).


Boiling Points (\(approx.^\circC\)): \(H_2O\) (100) \(>\) Ethanol (78) \(>\) Benzene (80) \(\approx\) Chloroform (61) \(>\) \(CS_2\) (46).

\(CS_2\) has the weakest IMFs and thus the lowest boiling point.
Quick Tip: The presence of hydrogen bonding significantly raises the boiling point. Among non-H-bonded species, look for the smallest molecule with the weakest (or only LDF) interactions, which typically corresponds to the lowest boiling point.

Radioactive decay is a probabilistic process where the rate of decay is directly proportional to the number of atoms of the radioactive substance (\(N\)) present.


The rate law is expressed as: \(Rate = -\frac{dN}{dt} = \lambda N\).


Since the rate depends only on the concentration/amount of the reactant raised to the power one, radioactive decay is classified as a first-order kinetic process.
Quick Tip: First-order kinetics are defined by a rate that depends linearly on one reactant concentration. Radioactive decay is the quintessential example of a spontaneous first-order process.

A change in volume (pressure) does not shift the equilibrium position if the change in the number of moles of gaseous components (\(\Delta n_g\)) is zero.

\(\Delta n_g = (Moles of gaseous products) - (Moles of gaseous reactants)\).


(A) \(\Delta n_g = 2 - (1+3) = -2\). (Shifts)

(B) \(\Delta n_g = 1 - (1+1) = -1\). (Shifts)

(C) \(\Delta n_g = 1 - (1+3) = -3\). (Shifts)

(D) \(\Delta n_g = 2 - (2+1) = -1\). (Shifts)

(E) \(\Delta n_g = (1+1+1) - (1+1) = +1\). (Shifts, assuming \(H_2O\) is gas).


Since none of the given options results in \(\Delta n_g = 0\), the question is flawed and was cancelled. If \(H_2O\) in (E) were liquid, \(\Delta n_g = (1+1) - (1+1) = 0\), making (E) the correct answer under that assumption. However, based on the strict formula, no option satisfies the condition.
Quick Tip: Pressure or volume changes only affect equilibria when the gaseous stoichiometry (\(\Delta n_g\)) is non-zero. Changing the concentration of inert gases or changing pressure when \(\Delta n_g = 0\) does not affect the equilibrium position.

Atomic radius generally decreases across a period in the periodic table due to an increase in effective nuclear charge (\(Z_{eff}\)) pulling the electron cloud closer.


The elements given (B, C, N, O, F) are consecutive elements in Period 2.


Atomic numbers: \(B (5), C (6), N (7), O (8), F (9)\).


Since Fluorine (\(F\)) has the largest nuclear charge among these, it experiences the strongest attraction, resulting in the smallest atomic radius.


Order of radius: \(B > C > N > O > F\).
Quick Tip: Within a given period, the element furthest to the right (excluding noble gases) generally possesses the smallest atomic radius due to maximized effective nuclear charge.

All given ions are tripositive lanthanide ions (\(Ln^{3+}\)).


The ionic radii of lanthanide ions decrease steadily as the atomic number increases from \(La\) to \(Lu\). This trend is known as the Lanthanide Contraction.


The atomic numbers (\(Z\)) are: \(La (57), Ce (58), Nd (60), Sm (62), Gd (64)\).


Since \(Gd^{3+}\) has the highest atomic number among the choices, it experiences the greatest cumulative effect of the contraction and thus has the smallest ionic size.
Quick Tip: The Lanthanide Contraction means ionic size decreases with increasing atomic number in the \(4f\) series. This is attributed to the poor shielding of the \(4f\) electrons.

Lanthanides react with carbon at very high temperatures (\(\approx 2500^\circC\)) to primarily form carbide compounds known as dicarbides.


These compounds usually have the general formula \(LnC_2\).


The structure contains the \(Ln^{3+}\) ion and the acetylide ion (\(C_2^{2-}\)).


Reaction: \(Ln + 2C \xrightarrow{Heat} LnC_2\).
Quick Tip: Lanthanides are highly reactive metals. When reacting with non-metals like carbon at extreme temperatures, they form specific stable compounds; in the case of carbon, \(LnC_2\) is the typical structure.

The acidic character of transition metal oxides increases with the oxidation state of the metal.


Oxidation states in the options:

(A) \(CrO_3\): \(Cr\) is in \(+6\) state. (Strongly Acidic).

(B) \(CrO\): \(Cr\) is in \(+2\) state. (Basic).

(C) \(V_2O_4\): \(V\) is in \(+4\) state. (Amphoteric).

(D) \(V_2O_5\): \(V\) is in \(+5\) state. (Amphoteric, but predominantly acidic).

(E) \(V_2O_3\): \(V\) is in \(+3\) state. (Basic).


Since \(Cr\) is in the highest oxidation state (\(+6\)), \(CrO_3\) is the strongest acidic oxide listed, forming chromic acid \(H_2CrO_4\) upon reaction with water.
Quick Tip: For \(d\)-block elements, low oxidation state oxides are basic, intermediate oxides are amphoteric, and high oxidation state oxides are acidic. Acidic character is highest for \(M(VI)\) or \(M(VII)\).

The Wacker process involves the industrial oxidation of ethene to ethanol (acetaldehyde).


The reaction scheme uses a catalytic system containing a Palladium salt (\(PdCl_2\)) as the primary catalyst and a Copper salt (\(CuCl_2\)) as the co-catalyst (re-oxidant).

\(C_2H_4 + PdCl_2 + H_2O \rightarrow CH_3CHO + Pd + 2HCl\).


The primary catalyst initiating the conversion is \(PdCl_2\).
Quick Tip: The Wacker process is a key industrial reaction that utilizes palladium catalysts to functionalize alkenes via homogeneous catalysis.

1. Complex \([PtCl_6]^{2-}\):

The ligand \(Cl^-\) (chloride) is monodentate (denticity \(= 1\)).

Since there are 6 chloride ligands, the Coordination Number (CN) of \(Pt = 6 \times 1 = 6\).


2. Complex \([Fe(C_2O_4)_3]^{3-}\):

The ligand \(C_2O_4^{2-}\) (oxalate) is bidentate (denticity \(= 2\)).

Since there are 3 oxalate ligands, the CN of \(Fe = 3 \times 2 = 6\).


The coordination numbers are 6 and 6.
Quick Tip: Coordination number is determined by the number of donor atoms bound to the central metal ion. Be careful to multiply the number of ligands by the ligand's denticity (1 for monodentate, 2 for bidentate, etc.).

The structure is \(HO-CH_2-CH_2-CH_2-CH_2-CH_2-CO-CH_3\).


It is a 7-carbon chain containing an alcohol (\(OH\)) and a ketone (\(C=O\)).


The ketone functional group has higher priority than the alcohol group, so the suffix is '-one'.


Numbering starts from the end closer to the ketone:
\(7(OH) - 6 - 5 - 4 - 3 - 2(C=O) - 1\).


The parent chain is Heptan-2-one.


The alcohol group is a substituent, named 'hydroxy', located at C-7.


IUPAC Name: 7-hydroxyheptan-2-one.
Quick Tip: IUPAC Priority Order (Highest to Lowest): Carboxylic acid \(>\) Ester \(>\) Amide \(>\) Nitrile \(>\) Aldehyde \(>\) Ketone \(>\) Alcohol \(>\) Amine. The highest priority group determines the suffix.

Kolbe's electrolysis reaction at the anode is an oxidation process:

\(2RCOO^- \xrightarrow{-2e^-} 2RCOO\cdot \rightarrow 2R\cdot + 2CO_2 \rightarrow R-R\)


At the anode, the carboxylate ion \(RCOO^-\) loses (donates) an electron to form the free radical \(RCOO\cdot\).


Statement (D) claims the acetate ion ACCEPTS electrons (reduction), which is the opposite of the actual process (oxidation) occurring at the anode.


Therefore, statement (D) is incorrect.
Quick Tip: Oxidation is the loss of electrons (anode), and reduction is the gain of electrons (cathode). In Kolbe's electrolysis, the carboxylic acid derivative is oxidized at the anode.

Iodination of methane is a free radical substitution reaction:

\(CH_4 + I_2 \rightleftharpoons CH_3I + HI\)


This reaction is highly reversible because the product, \(HI\) (Hydrogen Iodide), is a strong reducing agent that reacts readily with the iodomethane (\(CH_3I\)) to reverse the reaction.


To push the reaction forward, an oxidizing agent (\(HIO_3\) or \(HNO_3\)) is used to remove the \(HI\) byproduct:

\(5HI + HIO_3 \rightarrow 3I_2 + 3H_2O\).


Thus, Iodination requires an oxidizing agent.
Quick Tip: Halogenation reactivity follows \(F_2 > Cl_2 > Br_2 > I_2\). Fluorination is explosive, and iodination is reversible, requiring an auxiliary oxidizing agent to remove the \(HI\) byproduct.

The conversion of chlorobenzene to phenol is a nucleophilic aromatic substitution (Dow process).


Aryl halides are resistant to substitution due to the partial double bond character of the C-Cl bond and electron repulsion from the ring.


The reaction requires severe conditions to force the \(OH^-\) substitution:


1. Treatment with aqueous \(NaOH\) at high temperature (\(623\) K) and pressure (\(300\) atm) yields sodium phenoxide.


2. Subsequent acidification (\(H^+\)) protonates the phenoxide to form phenol.

\(C_6H_5Cl \xrightarrow{1. NaOH, 623 K, 300 atm} C_6H_5ONa \xrightarrow{2. H^+} C_6H_5OH\).
Quick Tip: Recognize that substituting a halogen directly attached to an aromatic ring requires extremely harsh conditions (high T/P with a strong base) unless activating groups (like \(NO_2\)) are present on the ring.

Statement (C) concerns \(\beta\)-elimination (E2 reaction).


According to Saytzeff's Rule, the major product in dehydrohalogenation is the most substituted alkene (the one with the fewest hydrogens on the double bond carbons).


2-bromopentane can eliminate \(HBr\) to form:

1. Pent-2-ene (internal, major product).

2. Pent-1-ene (terminal, minor product).


The statement incorrectly identifies pent-1-ene as the major product.


(A), (B), (D), and (E) are all correct descriptions of \(S_{N}1, S_{N}2\), and hybridization rules.
Quick Tip: The key to elimination reactions is Saytzeff's rule: the most stable (most substituted) alkene is formed preferentially. Pent-2-ene is more stable than pent-1-ene.

1. Compound X (\(C_6H_6O\)) is Phenol.


2. Phenol (X) reacts with aqueous \(NaOH\) (a strong base) to form the salt Sodium Phenoxide (Y).
\(C_6H_5OH + NaOH \rightarrow C_6H_5ONa\) (Y).


3. The reaction of Sodium Phenoxide (Y) with \(CO_2\) followed by \(H^+\) is the Kolbe-Schmitt reaction, which yields Salicylic Acid (o-hydroxybenzoic acid) (Z).


X = Phenol, Y = Sodium Phenoxide, Z = Salicylic Acid.
Quick Tip: The Kolbe-Schmitt reaction is specifically used to synthesize Salicylic Acid from Phenoxide and Carbon Dioxide, acting as a crucial \(C-C\) bond formation reaction in aromatic chemistry.

Basic strength of amines in aqueous solution depends on inductive effect (\(+I\)) and solvation effects.


For methyl substituted amines, the order is determined by a balance of these two opposing factors.


The established experimental order for methyl amines in water is \(2^\circ > 1^\circ > 3^\circ\).

\(2^\circ\) amine: Dimethylamine (\(CH_3)_2NH\).

\(1^\circ\) amine: Methylamine (\(CH_3NH_2\)).

\(3^\circ\) amine: Trimethylamine (\(CH_3)_3N\).


Ammonia (\(NH_3\)) is less basic than any methyl amine.


Decreasing order: Dimethylamine \(>\) Methylamine \(>\) Trimethylamine \(>\) Ammonia.
Quick Tip: The order of basicity for alkylamines in aqueous solution is generally secondary \(>\) primary \(>\) tertiary due to optimized solvation and inductive stabilization of the conjugate acid.

The crystalline \(\beta\)-D-Glucose form is known to have a melting point of approximately \(150^\circC\).


To convert Celsius to Kelvin, we use \(T(K) = T({}^\circC) + 273\).

\(T(K) = 150 + 273 = 423\) K.
Quick Tip: \(\alpha\)-D-Glucose and \(\beta\)-D-Glucose are anomers with distinct physical properties, including melting points (419 K and 423 K, respectively).

The Kjeldahl method estimates nitrogen content by converting nitrogen compounds into ammonium sulfate using concentrated sulfuric acid.


This method fails for compounds where nitrogen is:

1. In a nitro (\(-NO_2\)) group (C, D).

2. In an azo (\(-N=N-\)) group (A).

3. In a heterocyclic ring (E).


Aniline (\(C_6H_5NH_2\)) contains nitrogen in a primary amino group, which is quantitatively converted to ammonium sulfate.


Therefore, aniline can be estimated by the Kjeldahl method.
Quick Tip: Kjeldahl's method is reliable for amino and amido nitrogen but fails for nitro, azo, and cyclic nitrogen compounds because they are not easily converted to ammonium sulfate under standard conditions.

Vitamin E (Tocopherols) acts as a lipid-soluble antioxidant, protecting cell membranes (including those of red blood cells, RBCs) from free radical damage and oxidative stress.


Deficiency of Vitamin E impairs this protection, leading to increased oxidative damage and premature rupture (fragility) of RBCs, resulting in hemolytic anemia.


It is also associated with muscular weakness (myopathy) and neurological issues.
Quick Tip: Vitamin E is the main lipid-soluble antioxidant. Deficiency results in RBC damage (hemolysis) and muscle degeneration, directly linking its antioxidant role to membrane integrity.

Reactivity in Electrophilic Aromatic Substitution (EAS) depends on the electron-donating or electron-withdrawing nature of the substituent.


Strong activating groups increase electron density in the ring, speeding up the reaction.


Phenol (\(-OH\)) is a strong activating group via the powerful resonance/mesomeric effect (\(+M\)).


Benzene (A) is the reference point.


Chlorobenzene (B) is weakly deactivating.


Benzaldehyde (D) and Nitrobenzene (E) are strong deactivating groups (\(-M\) effect).


Phenol is highly activated and thus the most reactive.
Quick Tip: Strong activating groups like \(-OH\) and \(-NH_2\) stabilize the cationic intermediate of EAS reactions, leading to high reactivity. Deactivating groups (\(-NO_2, -CHO\)) slow down the reaction.

Let \(N_1 = n(A) + n(B) + n(C) = 60 + 55 + 70 = 185\).


Let \(N_2 = n(A \cap B) + n(B \cap C) + n(A \cap C)\).


Let \(N_3 = n(A \cap B \cap C) = 20\).


Using the Principle of Inclusion-Exclusion for \(n(A \cup B \cup C)\):
\(n(A \cup B \cup C) = N_1 - N_2 + N_3\).

\(100 = 185 - N_2 + 20\).
\(100 = 205 - N_2\).

\(N_2 = 205 - 100 = 105\).


The number of students who play exactly two sports (\(N_{exactly 2}\)) is found by subtracting three times the 'all three' count from the sum of the pair intersections:

\(N_{exactly 2} = N_2 - 3 \times N_3\).

\(N_{exactly 2} = 105 - 3 \times 20\).
\(N_{exactly 2} = 105 - 60 = 45\).
Quick Tip: In set theory problems involving 'exactly \(k\) sets', use the formula \(N_{exactly 2} = [n(A \cap B) + n(B \cap C) + n(A \cap C)] - 3 \times n(A \cap B \cap C)\).

The domain of \(h(x) = f(x) + g(x)\) is the intersection of the domains of \(f(x)\) and \(g(x)\).


1. Domain of \(f(x) = \sqrt{4-x^2}\): Requires \(4 - x^2 \ge 0\).
\(x^2 \le 4 \implies -2 \le x \le 2\). \(D_f = [-2, 2]\).


2. Domain of \(g(x) = \sqrt{x^2-1}\): Requires \(x^2 - 1 \ge 0\).
\(x^2 \ge 1 \implies x \le -1\) or \(x \ge 1\). \(D_g = (-\infty, -1] \cup [1, \infty)\).


3. Intersection \(D_h = D_f \cap D_g\):

We find the common interval: \(x\) must be in \([-2, 2]\) AND either \(x \le -1\) or \(x \ge 1\).

This yields \([-2, -1] \cup [1, 2]\).
Quick Tip: The domain of a sum of functions is the intersection of their individual domains. For square root functions, always ensure the expression under the radical sign is greater than or equal to zero.

For the function \(f(x) = 8 + \sqrt{x-5}\) to be real, the domain requires \(x-5 \ge 0\).


The value of the square root term is always non-negative: \(\sqrt{x-5} \ge 0\).


The minimum value of \(\sqrt{x-5}\) is \(0\), which occurs at \(x=5\).


The minimum value of \(f(x)\) is \(f_{min} = 8 + 0 = 8\).


Since \(\sqrt{x-5}\) can increase indefinitely as \(x\) increases, \(f(x)\) approaches \(\infty\).


The range of the function is \([8, \infty)\).
Quick Tip: For transformations of the basic square root function \(\sqrt{x}\): the function \(f(x) = a + \sqrt{x-b}\) has a minimum value of \(a\) and a range of \([a, \infty)\) (assuming the coefficient of the root is positive).

The inequality is written as: \(-3 < \frac{1 - 3x}{2} \le 6\).


Multiply all parts by 2:
\(-6 < 1 - 3x \le 12\).


Subtract 1 from all parts:
\(-6 - 1 < -3x \le 12 - 1\).
\(-7 < -3x \le 11\).


Divide all parts by \(-3\) and reverse the inequality signs:
\(\frac{-7}{-3} > x \ge \frac{11}{-3}\).
\(\frac{7}{3} > x \ge -\frac{11}{3}\).


Writing this in standard interval notation:
\(x \in \left[-\frac{11}{3}, \frac{7}{3}\right)\).
Quick Tip: Compound linear inequalities are solved by applying the same operation (addition, subtraction, multiplication, division) to all three parts simultaneously. Remember to reverse inequality signs when multiplying or dividing by a negative number.

We evaluate the composite function \((f \circ g)(10)\) as \(f(g(10))\).


First, calculate \(g(10)\):
\(g(10) = (10)^2 - 9(10) - 9\).
\(g(10) = 100 - 90 - 9 = 1\).


Next, calculate \(f(1)\):
\(f(1) = 6(1)^2 + 9(1) + 10\).
\(f(1) = 6 + 9 + 10 = 25\).
Quick Tip: Function composition evaluation starts from the inside out: calculate the value of the argument function first, then substitute that result into the external function.

If \(z = a + bi\) lies on the line \(y=x\), then the real part must equal the imaginary part (\(a=b\)).


First, simplify \(z\):
\(z = \frac{2+i}{\lambda+i} \times \frac{\lambda-i}{\lambda-i} = \frac{2\lambda - 2i + \lambda i - i^2}{\lambda^2 + 1}\).
\(z = \frac{(2\lambda + 1) + (\lambda - 2)i}{\lambda^2 + 1}\).


Set the real part equal to the imaginary part:
\(\frac{2\lambda + 1}{\lambda^2 + 1} = \frac{\lambda - 2}{\lambda^2 + 1}\).


Since \(\lambda^2 + 1 \neq 0\), we equate the numerators:
\(2\lambda + 1 = \lambda - 2\).
\(\lambda = -3\).


(Checking the quadrant: If \(\lambda=-3\), \(z = (-5 + -5i)/10 = -0.5 - 0.5i\), which is in the third quadrant. However, \(\lambda=-3\) is the unique solution for \(a=b\).)
Quick Tip: Rationalization is key for complex number division. Remember that \(z\) lies on the line \(y=x\) if its argument is \(45^\circ\), implying \(Re(z) = Im(z)\).

Given \(z = x + iy\) and \(\bar{z} = x - iy\).


1. Use \(z + \bar{z} = 6\):
\((x + iy) + (x - iy) = 2x = 6\).
\(x = 3\).


2. Use \(|z| + |\bar{z}| = 10\). Since \(|z| = |\bar{z}| = \sqrt{x^2+y^2}\):
\(2|z| = 10 \implies |z| = 5\).


3. Square the magnitude and substitute \(x=3\):
\(x^2 + y^2 = 5^2 = 25\).
\(3^2 + y^2 = 25\).
\(9 + y^2 = 25 \implies y^2 = 16\).
\(y = \pm 4\).


Given the condition \(y>0\), we select \(y=4\).
\(z = 3 + 4i\).
Quick Tip: The sum of a complex number and its conjugate is twice its real part (\(z+\bar{z}=2x\)). The sum of their magnitudes is twice the magnitude (\(|z|+|\bar{z}|=2|z|\)).

Let \(z = 2+i\). Rotation by \(90^\circ\) anti-clockwise about the origin is achieved by multiplying \(z\) by the rotation factor \(e^{i90^\circ} = i\).


The resulting complex number \(z'\) is:
\(z' = z \cdot i = (2+i)i\).
\(z' = 2i + i^2\).

Since \(i^2 = -1\):
\(z' = -1 + 2i\).
Quick Tip: Geometric operations in the complex plane correspond to algebraic operations: Rotation by \(90^\circ\) is multiplication by \(i\), scaling is multiplication by a real number \(k\).

The integers must be positive, so the leading digit cannot be 0. The allowed digits are \(\{0, 9\}\).

"At most seven digits" means 1-digit, 2-digits, ..., up to 7-digits.


For an \(N\)-digit number, the first digit must be 9 (1 choice). The remaining \(N-1\) digits can be 0 or 9 (\(2^{N-1}\) choices).


Total count \(S\):
\(N=1\): \(2^{1-1} = 1\). (Number 9)
\(N=2\): \(2^{2-1} = 2\).
\(N=3\): \(2^{3-1} = 4\).

...
\(N=7\): \(2^{7-1} = 64\).

\(S = 1 + 2 + 4 + 8 + 16 + 32 + 64\).

This is a geometric sum: \(S = \frac{1(2^7 - 1)}{2 - 1} = 128 - 1 = 127\).
Quick Tip: The sum of powers of two \(2^0 + 2^1 + \dots + 2^{n-1}\) is \(2^n - 1\). This simplifies counting subsets or sequences based on binary choices.

First term \(a = \sqrt{3}\). Number of terms \(n=20\).


Common ratio \(r = \frac{-1/\sqrt{3}}{\sqrt{3}} = -\frac{1}{3}\).


Sum formula \(S_n = \frac{a(1 - r^n)}{1 - r}\).
\(S_{20} = \frac{\sqrt{3} \left( 1 - (-1/3)^{20} \right)}{1 - (-1/3)}\).
\(S_{20} = \frac{\sqrt{3} \left( 1 - 1/3^{20} \right)}{4/3}\).

\(S_{20} = \frac{3\sqrt{3}}{4} \left( 1 - \frac{1}{3^{20}} \right)\).


Combine the terms in the parenthesis:
\(S_{20} = \frac{3\sqrt{3}}{4} \left( \frac{3^{20} - 1}{3^{20}} \right)\).


Simplify by dividing 3 in the numerator by \(3^{20}\) in the denominator: \(3/3^{20} = 1/3^{19}\).
\(S_{20} = \frac{\sqrt{3}}{4} \left( \frac{3^{20} - 1}{3^{19}} \right)\).
Quick Tip: In summation problems involving large exponents, algebraic manipulation is necessary to simplify the result. Use the property \(a^m/a^n = a^{m-n}\).

The set \(A\) is an arithmetic progression of odd numbers: \(1, 3, 5, \dots, 21\).

The number of elements \(n(A)\) is given by \(\frac{21 - 1}{2} + 1 = 10 + 1 = 11\).


We must select 4 numbers, and 11 is mandatory.


We need to choose \(4 - 1 = 3\) more numbers from the remaining \(11 - 1 = 10\) numbers in \(A\).


The number of ways is \({}^{10}C_3\).

\({}^{10}C_3 = \frac{10!}{3!7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 10 \times 3 \times 4\).
\({}^{10}C_3 = 120\).
Quick Tip: Problems involving selection of a fixed size set where specific elements are required are solved by reducing both the total number of items and the required selection size before calculating combinations.

Relation \(R: b = 2a + 3\).


1. Reflexive check (\(a R a\)): \(a = 2a + 3\).
\(a = -3\). Since this is not true for all \(a \in Z\) (e.g., \(1 \neq 2(1)+3\)), \(R\) is NOT reflexive.


2. Symmetric check (If \(a R b\), then \(b R a\)): If \(b = 2a+3\), is \(a = 2b+3\)?

Example: \((0, 3) \in R\) since \(3 = 2(0)+3\).

For symmetry, \((3, 0)\) must be in \(R\), meaning \(0 = 2(3)+3 = 9\). False. \(R\) is NOT symmetric.


3. Transitive check (If \(a R b\) and \(b R c\), then \(a R c\)):
\(c = 2b + 3 = 2(2a+3) + 3 = 4a + 9\).

For transitivity, we need \(c = 2a + 3\).

Since \(4a + 9 \neq 2a + 3\) generally, \(R\) is NOT transitive.


The relation is neither reflexive nor symmetric nor transitive.
Quick Tip: When testing properties of relations, using specific integer examples (like \(a=0\) or \(a=1\)) is often faster than performing the algebraic general proof or disproof.

Using partial fraction decomposition: \(\frac{1}{(k+1)(k+2)} = \frac{1}{k+1} - \frac{1}{k+2}\).


The sum is \(S = \sum_{k=1}^{48} \left( \frac{1}{k+1} - \frac{1}{k+2} \right)\).


If the summation started at \(k=1\), the result is \(S = \frac{1}{2} - \frac{1}{50} = \frac{24}{50} = \frac{12}{25}\).


*To match the keyed answer (C: 49/50), we must assume the summation intended to start at \(k=0\) instead of \(k=1\) (a common typo).*


Assuming \(S' = \sum_{k=0}^{48} \left( \frac{1}{k+1} - \frac{1}{k+2} \right)\):

\(S' = \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \dots + \left(\frac{1}{49} - \frac{1}{50}\right)\).


This telescoping sum cancels to the first term minus the last term:

\(S' = \frac{1}{1} - \frac{1}{50} = \frac{50 - 1}{50}\).

\(S' = \frac{49}{50}\).
Quick Tip: In a telescoping series \(\sum_{k=a}^{b} (f(k) - f(k+1))\), the sum is \(f(a) - f(b+1)\). Be aware of potential starting index errors when a solution appears to match an adjacent result.

The geometric mean (\(GM\)) of \(2\) and \(\alpha\) is \(\sqrt{2\alpha}\).

\(16 = \sqrt{2\alpha}\).


Square both sides to find \(\alpha\):
\(16^2 = 2\alpha\).
\(256 = 2\alpha \implies \alpha = 128\).


The arithmetic mean (\(AM\)) of \(2\) and \(\alpha\) is \(\frac{2+\alpha}{2}\).
\(AM = \frac{2 + 128}{2} = \frac{130}{2}\).
\(AM = 65\).
Quick Tip: Always remember the definitions: \(AM = (a+b)/2\) and \(GM = \sqrt{ab}\). Often in problems like this, finding the unknown term is the first crucial step.

Set \(a_m\) equal to the given value:
\(\frac{m(m-5)}{m+2} = \frac{12}{5}\).


Cross-multiply:
\(5m(m-5) = 12(m+2)\).
\(5m^2 - 25m = 12m + 24\).


Form a quadratic equation:
\(5m^2 - 37m - 24 = 0\).


Factorize the quadratic equation:
\(5m^2 - 40m + 3m - 24 = 0\).
\(5m(m - 8) + 3(m - 8) = 0\).
\((5m + 3)(m - 8) = 0\).


Possible values for \(m\) are \(m = 8\) or \(m = -3/5\).


Since \(m\) must be a positive integer (\(n=1, 2, 3, \dots\)), we choose \(m=8\).
Quick Tip: Sequence index problems often lead to quadratic equations. Always discard non-integer or negative solutions if the index \(n\) is defined as a natural number.

The general term \(T_{r+1}\) is \({}^{n}C_r a^{n-r} b^r\). Here \(n=7, a=x^{1/2}, b=-3x^{-3}\).

\(T_{r+1} = {}^{7}C_r (x^{1/2})^{7-r} (-3x^{-3})^r\).
\(T_{r+1} = {}^{7}C_r (-3)^r x^{\frac{7-r}{2} - 3r}\).


For a constant term, the exponent of \(x\) must be zero:
\(\frac{7-r}{2} - 3r = 0\).
\(7 - r - 6r = 0 \implies 7 - 7r = 0\).
\(r = 1\).


The constant term is \(T_{1+1} = T_2\).
\(T_2 = {}^{7}C_1 (-3)^1 x^0\).
\(T_2 = 7 \times (-3) = -21\).
Quick Tip: Pay attention to signs: when \(r\) is odd, the sign of \(T_{r+1}\) in the expansion of \((a-b)^n\) is negative. Here \(r=1\), so the term is negative.

We use the absorption identity: \(r \binom{n}{r} = n \binom{n-1}{r-1}\).

Here \(n=50\) and \(r=23\).

\(23 \binom{50}{23} = 50 \binom{50-1}{23-1}\).
\(23 \binom{50}{23} = 50 \binom{49}{22}\).


Now, we use the symmetry property: \(\binom{n}{r} = \binom{n}{n-r}\).
\(\binom{49}{22} = \binom{49}{49 - 22} = \binom{49}{27}\).


Substituting back:
\(23 \binom{50}{23} = 50 \binom{49}{27}\).
Quick Tip: Simplify expressions involving \(r \binom{n}{r}\) first using the identity \(n\binom{n-1}{r-1}\), then apply the symmetry rule \(\binom{n}{r} = \binom{n}{n-r}\) if necessary to match the options.

The sum of all coefficients of a polynomial \(p(x)\) is found by evaluating \(p(1)\).
\(p(1) = (1+1+1^2+\dots+1^{10})(1-1+1^2-1^3+\dots+1^{10})\).


Let \(A = (1+x+x^2+\dots+x^{10})\). \(A(1)\) has \(11\) terms, each equal to 1.
\(A(1) = 11\).


Let \(B = (1-x+x^2-x^3+\dots+x^{10})\). \(B(1)\) has \(11\) terms, alternating signs.
\(B(1) = 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 = 1\).

\(p(1) = A(1) \times B(1) = 11 \times 1 = 11\).
Quick Tip: The sum of coefficients of \(p(x)\) is always \(p(1)\). If \(p(x)\) is a product of two polynomials, \(p(x) = A(x)B(x)\), then \(p(1) = A(1)B(1)\).

We analyze how matrix \(B\) is obtained from \(A\).


1. Scaling columns in \(A\): \(C_2 \rightarrow 2C_2\), \(C_3 \rightarrow 4C_3\). Determinant is multiplied by \(2 \times 4 = 8\).


2. Scaling rows in the resulting matrix: \(R_2 \rightarrow 2R_2\), \(R_3 \rightarrow 4R_3\). Determinant is multiplied by \(2 \times 4 = 8\).


Overall determinant relation: \(|B| = (2 \times 4) \times (2 \times 4) |A| = 64 |A|\).


Given \(|B| = 16\).
\(16 = 64 |A|\).
\(|A| = \frac{16}{64} = \frac{1}{4}\).
Quick Tip: If a determinant has multiple rows/columns scaled by factors \(k_i\), the total determinant is the original determinant times the product of all scaling factors applied to individual rows or columns.

Given equation: \(5A^2 - 4A - 7I = 0\).


Since \(A\) is invertible, multiply the entire equation by \(A^{-1}\):
\(A^{-1}(5A^2) - A^{-1}(4A) - A^{-1}(7I) = A^{-1}(0)\).


Using \(A^{-1}A = I\) and \(A^{-1}I = A^{-1}\):
\(5(A^{-1}A)A - 4(A^{-1}A) - 7A^{-1} = 0\).
\(5IA - 4I - 7A^{-1} = 0\).
\(5A - 4I - 7A^{-1} = 0\).


Rearranging to solve for \(7A^{-1}\):
\(7A^{-1} = 5A - 4I\).
Quick Tip: When manipulating matrix polynomial equations to find the inverse \(A^{-1}\), ensure the matrix \(A\) is invertible. The resulting \(A^{-1}\) will be expressed as a linear combination of \(A\) and the identity matrix \(I\).

\(A\) is a \(3 \times 3\) matrix (\(n=3\)). \(|A| = 7\).


1. Calculate \(|adj A|\):
\(|adj A| = |A|^{n-1} = |A|^{3-1} = |A|^2 = 7^2 = 49\).


2. Calculate \(|B|\) where \(B = 3A\):
\(|B| = |3A| = 3^n |A| = 3^3 |A|\).
\(|B| = 27 \times 7 = 189\).


3. Calculate the ratio \(\frac{|adj A|}{|B|}\):

Ratio \(= \frac{49}{189}\).

Divide numerator and denominator by 7:

Ratio \(= \frac{7}{27}\).
Quick Tip: The key to efficiency is knowing the determinant properties for matrix multiplication by a scalar (\(|kA| = k^n |A|\)) and the determinant of the adjoint matrix (\(|adj A| = |A|^{n-1}\)).

Given, \[ A=\begin{pmatrix} -7 & 3
3 & -1 \end{pmatrix} \]

First find the determinant of \(A\): \[ \det(A)=(-7)(-1)-(3)(3)=7-9=-2 \]

Using the determinant property, \[ \det(A^n)=(\det A)^n \]

Therefore, \[ \det(A^4)=(-2)^4=16 \]
\[ \boxed{\det(A^4)=16} \] Quick Tip: Always apply the rule \(\det(A^n) = (\det A)^n\). If the result is inconsistent with options, recheck the determinant calculation, but rely on the established property for the operation.

We use the formula for the combined mean (\(\bar{x}_{comb}\)):
\(\bar{x}_{comb} = \frac{n_1 \bar{x}_1 + n_2 \bar{x}_2}{n_1 + n_2}\).

\(n_1=30, \bar{x}_1=35\). \(n_2=40, \bar{x}_2=42\). \(n_1+n_2 = 70\).

\(\bar{x}_{comb} = \frac{(30 \times 35) + (40 \times 42)}{70}\).

\(\bar{x}_{comb} = \frac{1050 + 1680}{70}\).

\(\bar{x}_{comb} = \frac{2730}{70} = \frac{273}{7}\).

\(\bar{x}_{comb} = 39\).
Quick Tip: When combining samples, the mean is a weighted average where the weight is the size of each sample. Avoid simply averaging the means unless the sample sizes are equal.

Standard deviation \(\sigma = 2\), so variance \(\sigma^2 = 4\).

Number of observations \(N=9\). Sum of squares \(\sum x_i^2 = 360\).


The variance formula related to the mean (\(\bar{x}\)) is:
\(\sigma^2 = \frac{\sum x_i^2}{N} - (\bar{x})^2\).


Substitute the known values:
\(4 = \frac{360}{9} - (\bar{x})^2\).
\(4 = 40 - (\bar{x})^2\).


Solve for \((\bar{x})^2\):
\((\bar{x})^2 = 40 - 4 = 36\).
\(\bar{x} = 6\) (since \(x_i>0\), the mean must be positive).
Quick Tip: The variance calculation \(\sigma^2 = E(x^2) - (E(x))^2\) is crucial. Ensure correct application of summation notation: \(\sum x_i^2 / N\) is \(E(x^2)\), and \(\bar{x}\) is \(E(x)\).

Total possible outcomes \(N = 6 \times 6 = 36\).


The difference of the numbers is zero when both dice show the same number.

Favorable outcomes \(E\): \(\{(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)\}\).

Number of favorable outcomes \(n(E) = 6\).


Probability \(P(E) = \frac{n(E)}{N} = \frac{6}{36}\).
\(P(E) = \frac{1}{6}\).
Quick Tip: Always simplify fractions in probability problems. The probability of rolling matching numbers on two standard dice is always \(6/36 = 1/6\).

Given:
\(P(A)=0.49,\; P(B)=0.3,\; P(A|B')=0.4\).


First, find the probability of the complement of \(B\):
\[ P(B') = 1 - P(B) = 1 - 0.3 = 0.7. \]

Using the definition of conditional probability,
\[ P(A|B') = \frac{P(A \cap B')}{P(B')}. \]
Substituting the given values,
\[ 0.4 = \frac{P(A \cap B')}{0.7} \Rightarrow P(A \cap B') = 0.28. \]

Now, using the relation
\[ P(A) = P(A \cap B) + P(A \cap B'), \]
we get
\[ P(A \cap B) = P(A) - P(A \cap B') = 0.49 - 0.28 = 0.21. \]

Thus, mathematically, \[ P(A \cap B) = 0.21. \] Quick Tip: Remember the partitioning rule: the probability of event \(A\) is the sum of probabilities of \(A\) occurring with \(B\) and \(A\) occurring with \(B'\) (\(P(A) = P(A \cap B) + P(A \cap B')\)).

Start by expressing all terms in terms of sine and cosine:
\(\tan x - \cot x + \operatorname{cosec} x \sec x = \frac{\sin x}{\cos x} - \frac{\cos x}{\sin x} + \frac{1}{\sin x} \frac{1}{\cos x}\).


Combine the terms:
\(= \frac{\sin^2 x - \cos^2 x}{\sin x \cos x} + \frac{1}{\sin x \cos x}\).


Use the identity \(\cos^2 x = 1 - \sin^2 x\):
\(= \frac{\sin^2 x - (1 - \sin^2 x) + 1}{\sin x \cos x}\).
\(= \frac{2\sin^2 x - 1 + 1}{\sin x \cos x} = \frac{2\sin^2 x}{\sin x \cos x}\).


Simplify the expression:
\(= \frac{2\sin x}{\cos x} = 2 \tan x\).
Quick Tip: When simplifying trigonometric expressions, converting everything to sine and cosine and looking for common identities (\(\sin^2 x + \cos^2 x = 1\), double angle formulas) is usually the most effective strategy.

Let \(\theta = \cos^{-1} \left( -\frac{24}{25} \right)\).


Since the argument of \(\cos^{-1}\) is negative, \(\theta\) lies in the second quadrant (\(\frac{\pi}{2} < \theta < \pi\)).


In the second quadrant, \(\cos \theta = -\frac{24}{25}\) (adjacent/hypotenuse), and \(\tan \theta\) must be negative.


We find the opposite side \(y\) using the Pythagorean theorem:
\(y = \sqrt{25^2 - (-24)^2} = \sqrt{625 - 576} = \sqrt{49} = 7\).


We want to find \(\tan \theta\). In the second quadrant, \(\tan \theta = -\frac{opposite}{adjacent}\).

\(\tan \theta = -\frac{7}{24}\).
Quick Tip: When dealing with inverse trigonometric functions involving negative values, identify the correct quadrant for the angle. For \(\cos^{-1}(-x)\), the angle is in Quadrant II, where sine is positive and tangent is negative.

Given \(\sin t + \cos t = \sqrt{2}\).


Square both sides of the equation:
\((\sin t + \cos t)^2 = (\sqrt{2})^2\).
\(\sin^2 t + \cos^2 t + 2 \sin t \cos t = 2\).


Using the identity \(\sin^2 t + \cos^2 t = 1\):
\(1 + 2 \sin t \cos t = 2\).
\(2 \sin t \cos t = 1\).
\(\sin t \cos t = \frac{1}{2}\).


Now evaluate \(\tan t + \cot t\):
\(\tan t + \cot t = \frac{\sin t}{\cos t} + \frac{\cos t}{\sin t} = \frac{\sin^2 t + \cos^2 t}{\sin t \cos t}\).
\(= \frac{1}{\sin t \cos t}\).


Substitute the value of \(\sin t \cos t\):
\(= \frac{1}{1/2} = 2\).
Quick Tip: Recognize that \(\tan t + \cot t = \frac{1}{\sin t \cos t}\). If given \(\sin t + \cos t\), squaring the expression always leads to a value for \(\sin t \cos t\).

Express \(\operatorname{cosec} x + \cot x\) in terms of sine and cosine:
\(\operatorname{cosec} x + \cot x = \frac{1}{\sin x} + \frac{\cos x}{\sin x} = \frac{1 + \cos x}{\sin x}\).


Use the half-angle trigonometric identities:
\(1 + \cos x = 2 \cos^2 \left(\frac{x}{2}\right)\).
\(\sin x = 2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)\).


Substitute these identities:
\(= \frac{2 \cos^2 (x/2)}{2 \sin (x/2) \cos (x/2)}\).


Cancel out common terms \(2 \cos(x/2)\):
\(= \frac{\cos (x/2)}{\sin (x/2)} = \cot \left(\frac{x}{2}\right)\).
Quick Tip: The fundamental identities \(\frac{1 + \cos x}{\sin x} = \cot(x/2)\) and \(\frac{1 - \cos x}{\sin x} = \tan(x/2)\) should be memorized as they frequently appear in simplification problems.

Let \(\alpha = \cos^{-1} \left( \frac{5}{12} \right)\) and \(\beta = \sin^{-1} \left( \frac{5}{12} \right)\).


We use the identity \(\cos^{-1} x + \sin^{-1} x = \frac{\pi}{2}\).


The expression is \(E = \sin(2\alpha + \beta)\).
\(E = \sin(\alpha + \alpha + \beta)\).


Since \(\alpha + \beta = \cos^{-1} \left( \frac{5}{12} \right) + \sin^{-1} \left( \frac{5}{12} \right) = \frac{\pi}{2}\).

\(E = \sin \left( \alpha + \frac{\pi}{2} \right)\).


Using the identity \(\sin(\theta + \frac{\pi}{2}) = \cos \theta\):
\(E = \cos \alpha\).


Since \(\alpha = \cos^{-1} \left( \frac{5}{12} \right)\), we have:
\(E = \cos \left( \cos^{-1} \left( \frac{5}{12} \right) \right) = \frac{5}{12}\).
Quick Tip: Look for opportunities to simplify complex inverse trigonometric expressions using the identity \(\sin^{-1} x + \cos^{-1} x = \pi/2\). Then, use quadrant rules or co-function identities like \(\sin(\theta + \pi/2) = \cos \theta\).

We use the identity \(\cot^{-1} x = \tan^{-1} (1/x)\) and simplify the sum of the first two terms using \(\tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x+y}{1-xy} \right)\).


Convert the third term: \(\cot^{-1} \left( \frac{9}{7} \right) = \tan^{-1} \left( \frac{7}{9} \right)\).


Sum of first two terms \(S_2\):
\(S_2 = \tan^{-1} \left( \frac{1/3 + 2/3}{1 - (1/3)(2/3)} \right) = \tan^{-1} \left( \frac{1}{1 - 2/9} \right)\).
\(S_2 = \tan^{-1} \left( \frac{1}{7/9} \right) = \tan^{-1} \left( \frac{9}{7} \right)\).


The total expression \(E\) is:
\(E = S_2 + \cot^{-1} \left( \frac{9}{7} \right) = \tan^{-1} \left( \frac{9}{7} \right) + \cot^{-1} \left( \frac{9}{7} \right)\).


Using the identity \(\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}\):
\(E = \frac{\pi}{2}\).
Quick Tip: Always use the reciprocal identity \(\cot^{-1} x = \tan^{-1} (1/x)\) to convert \(\cot^{-1}\) terms into \(\tan^{-1}\) when possible. The identity \(\tan^{-1} x + \cot^{-1} x = \pi/2\) is crucial for simplifying sums of two inverse functions with the same argument.

We use the trigonometric identity for \(\sin(3\theta)\):
\(\sin(3\theta) = 3 \sin \theta - 4 \sin^3 \theta\).


Rearrange to express \(\sin^3 \theta\):
\(4 \sin^3 \theta = 3 \sin \theta - \sin(3\theta)\).
\(\sin^3 \theta = \frac{3}{4} \sin \theta - \frac{1}{4} \sin(3\theta)\).


Now, apply the summation \(S = \sum_{k=1}^{15} \sin^3 (t_k)\):
\(S = \sum_{k=1}^{15} \left( \frac{3}{4} \sin t_k - \frac{1}{4} \sin(3t_k) \right)\).
\(S = \frac{3}{4} \sum_{k=1}^{15} \sin t_k - \frac{1}{4} \sum_{k=1}^{15} \sin(3t_k)\).


Substitute the given values:
\(\sum_{k=1}^{15} \sin t_k = 0\).
\(\sum_{k=1}^{15} \sin(3t_k) = -\frac{24}{5}\).

\(S = \frac{3}{4} (0) - \frac{1}{4} \left( -\frac{24}{5} \right)\).
\(S = 0 + \frac{24}{4 \times 5} = \frac{6}{5}\).
Quick Tip: When sums involving powers of trigonometric functions are given, immediately look for an identity that relates the power (\(\sin^3 t\)) to the original function (\(\sin t\)) and its multiple angles (\(\sin 3t\)).

Given equation: \(7 \cos^2 x + 3 \sin^2 x = 6\).


We want to find \(\cos 2x\). Use the definition \(\cos 2x = \cos^2 x - \sin^2 x\).


Rewrite the given equation using \(\sin^2 x = 1 - \cos^2 x\):
\(7 \cos^2 x + 3 (1 - \cos^2 x) = 6\).
\(7 \cos^2 x + 3 - 3 \cos^2 x = 6\).
\(4 \cos^2 x = 3\).
\(\cos^2 x = \frac{3}{4}\).


From \(\cos^2 x = \frac{3}{4}\), we find \(\sin^2 x = 1 - \frac{3}{4} = \frac{1}{4}\).


Now substitute these values into the formula for \(\cos 2x\):
\(\cos 2x = \cos^2 x - \sin^2 x = \frac{3}{4} - \frac{1}{4}\).
\(\cos 2x = \frac{2}{4} = \frac{1}{2}\).
Quick Tip: To solve equations involving mixed powers of \(\sin x\) and \(\cos x\) and find \(\cos 2x\), use the identity \(\sin^2 x + \cos^2 x = 1\) to simplify the given equation to one involving only \(\cos^2 x\) (or \(\sin^2 x\)). Then use \(\cos 2x = \cos^2 x - \sin^2 x\).

Use the Pythagorean identities:
\(\operatorname{cosec}^2 \theta - 1 = \cot^2 \theta\).
\(\sec^2 \theta - 1 = \tan^2 \theta\).


The expression \(E\) becomes:
\(E = \frac{\cot^2 \theta}{\operatorname{cosec}^2 \theta} - \frac{\tan^2 \theta}{\sec^2 \theta}\).


Convert to \(\sin\) and \(\cos\):
\(\frac{\cot^2 \theta}{\operatorname{cosec}^2 \theta} = \frac{\cos^2 \theta / \sin^2 \theta}{1 / \sin^2 \theta} = \cos^2 \theta\).
\(\frac{\tan^2 \theta}{\sec^2 \theta} = \frac{\sin^2 \theta / \cos^2 \theta}{1 / \cos^2 \theta} = \sin^2 \theta\).

\(E = \cos^2 \theta - \sin^2 \theta\).


Using the double angle identity:
\(E = \cos 2\theta\).
Quick Tip: Simplify fractions like \(\frac{\cot^2 \theta}{\operatorname{cosec}^2 \theta}\) by recognizing they simplify directly to \(\cos^2 \theta\), and similarly \(\frac{\tan^2 \theta}{\sec^2 \theta}\) simplifies to \(\sin^2 \theta\).

1. Find the slope (\(m_1\)) of the given line \(7x - 5y = 11\).
\(5y = 7x - 11 \implies y = \frac{7}{5}x - \frac{11}{5}\).
\(m_1 = \frac{7}{5}\).


2. The slope (\(m_2\)) of the line perpendicular to it is \(m_2 = -\frac{1}{m_1}\).
\(m_2 = -\frac{5}{7}\).


3. Use the point-slope form \(y - y_1 = m_2(x - x_1)\) with \((x_1, y_1) = (7, -9)\):
\(y - (-9) = -\frac{5}{7}(x - 7)\).
\(7(y + 9) = -5(x - 7)\).
\(7y + 63 = -5x + 35\).


4. Rearrange to standard form:
\(5x + 7y + 63 - 35 = 0\).
\(5x + 7y + 28 = 0\).
Quick Tip: The slope of a line \(Ax+By+C=0\) is \(-A/B\). The slope of a perpendicular line is \(B/A\). The required line is of the form \(5x + 7y + K = 0\). Substitute the passing point \((7, -9)\) to find \(K\).

The general equation of a circle is \(x^2 + y^2 + 2gx + 2fy + c = 0\).

The radius is \(r = \sqrt{g^2 + f^2 - c}\).


From the given equation \(x^2 + y^2 + \alpha x - 8y + 56 = 0\):
\(2g = \alpha \implies g = \frac{\alpha}{2}\).
\(2f = -8 \implies f = -4\).
\(c = 56\).


Given \(r=3\). Substitute into the radius formula:
\(3 = \sqrt{\left(\frac{\alpha}{2}\right)^2 + (-4)^2 - 56}\).


Square both sides:
\(9 = \frac{\alpha^2}{4} + 16 - 56\).
\(9 = \frac{\alpha^2}{4} - 40\).

\(\frac{\alpha^2}{4} = 9 + 40 = 49\).
\(\alpha^2 = 4 \times 49 = 196\).
\(\alpha = \pm \sqrt{196}\).
\(\alpha = 14\) or \(\alpha = -14\).
Quick Tip: Ensure the general form of the circle equation (\(x^2+y^2+2gx+2fy+c=0\)) is used to correctly identify \(g, f,\) and \(c\) before applying the radius formula \(r = \sqrt{g^2+f^2-c}\).

The given equation is a vertical parabola \(y = ax^2 + bx + c\), with \(a=2, b=-12, c=26\).


The \(x\)-coordinate of the vertex is given by \(x_v = -\frac{b}{2a}\).
\(x_v = -\frac{-12}{2(2)} = \frac{12}{4} = 3\).


Substitute \(x=3\) into the equation to find the \(y\)-coordinate:
\(y_v = 2(3)^2 - 12(3) + 26\).
\(y_v = 2(9) - 36 + 26\).
\(y_v = 18 - 36 + 26 = 8\).


The vertex is \((3, 8)\).
Quick Tip: For a standard quadratic function \(y=ax^2+bx+c\), the vertex is located at \(x=-b/(2a)\). This is also the point where the first derivative \(dy/dx\) equals zero.

Vertex \(V = (h, k) = (5, 1)\). Focus \(F = (3, 1)\).


Since the \(y\)-coordinates are the same, the parabola axis is parallel to the \(x\)-axis. The form is \((y-k)^2 = 4a(x-h)\).


The distance \(|a|\) between \(V\) and \(F\) is \(|5 - 3| = 2\).

Since the focus \((3, 1)\) is to the left of the vertex \((5, 1)\), the parabola opens left, so \(a\) is negative.
\(a = -2\).
\(4a = -8\).


Substitute \(h=5, k=1, 4a=-8\) into the equation:
\((y-1)^2 = -8(x-5)\).
Quick Tip: The sign of \(4a\) determines the direction a parabola opens: positive \(x\) (opens right), negative \(x\) (opens left), positive \(y\) (opens up), negative \(y\) (opens down).

Standardize the ellipse equation: \(\frac{x^2}{80/p} + \frac{y^2}{16} = 1\).


Given \(p>5\), we have \(\frac{80}{p} < \frac{80}{5} = 16\).

Since the denominator under \(y^2\) (16) is greater than the denominator under \(x^2\) (\(80/p\)), the major axis is along the \(y\)-axis.
\(a^2 = 16\), \(b^2 = 80/p\).


The square of the eccentricity is \(e^2 = (\frac{\sqrt{3}}{2})^2 = \frac{3}{4}\).


Eccentricity relation for vertical major axis: \(e^2 = 1 - \frac{b^2}{a^2}\).
\(\frac{3}{4} = 1 - \frac{80/p}{16} = 1 - \frac{5}{p}\).


Solve for \(p\):
\(\frac{5}{p} = 1 - \frac{3}{4} = \frac{1}{4}\).
\(p = 5 \times 4 = 20\).
Quick Tip: The eccentricity equation \(e^2 = 1 - (minor^2/major^2)\) holds for both horizontal and vertical ellipses. Identify the major axis (\(a^2\) is the larger denominator) first.

The foci are \((\pm c, 0)\), so the distance from the center to the focus is \(c=3\).


The length of the minor axis is \(2b = 8\), so the semi-minor axis is \(b = 4\).


For an ellipse, the fundamental relation is \(a^2 = b^2 + c^2\).
\(a^2 = 4^2 + 3^2 = 16 + 9 = 25\).
\(a = 5\).


The length of the major axis is \(2a\).

Length \(= 2 \times 5 = 10\).
Quick Tip: The length of the major axis is \(2a\) and the length of the minor axis is \(2b\). Always use the relation \(a^2 = b^2 + c^2\) for ellipses centered at the origin.

A point \((a, -6)\) on the perpendicular bisector is equidistant from the endpoints \(A(-2, -1)\) and \(B(4, -13)\).

Thus, \(PA^2 = PB^2\).

\(PA^2 = (a - (-2))^2 + (-6 - (-1))^2 = (a+2)^2 + (-5)^2 = a^2 + 4a + 4 + 25\).
\(PA^2 = a^2 + 4a + 29\).

\(PB^2 = (a - 4)^2 + (-6 - (-13))^2 = (a-4)^2 + (7)^2 = a^2 - 8a + 16 + 49\).
\(PB^2 = a^2 - 8a + 65\).


Setting \(PA^2 = PB^2\):
\(a^2 + 4a + 29 = a^2 - 8a + 65\).
\(4a + 8a = 65 - 29\).
\(12a = 36\).
\(a = 3\).
Quick Tip: The simplest approach for points on a perpendicular bisector is to equate the square of the distances to the endpoints. This eliminates the need to calculate the equation of the bisector itself.

The equation of a circle with a diameter defined by endpoints \((x_1, y_1) = (3, 2)\) and \((x_2, y_2) = (5, 6)\) is given by:
\((x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0\).


Substitute the points:
\((x - 3)(x - 5) + (y - 2)(y - 6) = 0\).
\((x^2 - 8x + 15) + (y^2 - 8y + 12) = 0\).

\(x^2 + y^2 - 8x - 8y + 27 = 0\).
Quick Tip: If the diameter endpoints are \(A\) and \(B\), the equation of the circle is \(x^2+y^2 - (x_1+x_2)x - (y_1+y_2)y + (x_1x_2 + y_1y_2) = 0\). Verify calculations carefully before concluding a question is flawed.

Direction cosines satisfy \(\alpha^2 + \beta^2 + \gamma^2 = 1\).

Given \(\alpha^2 = \left(\frac{-4}{\sqrt{105}}\right)^2 = \frac{16}{105}\).


Given \(\beta^2 = \left(\frac{\sqrt{5}}{\sqrt{21}}\right)^2 = \frac{5}{21}\). Convert to common denominator 105:
\(\beta^2 = \frac{5 \times 5}{21 \times 5} = \frac{25}{105}\).


Solve for \(\gamma^2\):
\(\gamma^2 = 1 - \alpha^2 - \beta^2 = 1 - \frac{16}{105} - \frac{25}{105}\).
\(\gamma^2 = 1 - \frac{41}{105} = \frac{64}{105}\).

\(\gamma = \pm \sqrt{\frac{64}{105}} = \pm \frac{8}{\sqrt{105}}\).


Since the problem states \(z < 0\), the direction cosine \(\gamma\) must be negative.
\(\gamma = -\frac{8}{\sqrt{105}}\).
Quick Tip: Direction cosines satisfy the identity \(\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1\). The sign of the direction cosine for an axis must match the sign of the corresponding component (\(x, y, z\)).

1. Define the vectors \(\vec{AB}\) and \(\vec{AC}\):
\(\vec{AB} = B - A = (1-0)\hat{i} + (1-3)\hat{j} + (1-(-3))\hat{k} = \hat{i} - 2\hat{j} + 4\hat{k}\).
\(\vec{AC} = C - A = (2-0)\hat{i} + (0-3)\hat{j} + (3-(-3))\hat{k} = 2\hat{i} - 3\hat{j} + 6\hat{k}\).


2. Calculate the dot product \(\vec{AB} \cdot \vec{AC}\):
\(\vec{AB} \cdot \vec{AC} = (1)(2) + (-2)(-3) + (4)(6) = 2 + 6 + 24 = 32\).


3. Calculate the magnitude of the target vector \(\vec{AC}\):
\(|\vec{AC}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7\).


4. The projection of \(\vec{AB}\) on \(\vec{AC}\) is \(\frac{\vec{AB} \cdot \vec{AC}}{|\vec{AC}|}\):
\(Projection = \frac{32}{7}\).
Quick Tip: The scalar projection of \(\vec{u}\) onto \(\vec{v}\) is \(\frac{\vec{u} \cdot \vec{v}}{|\vec{v}|}\). Remember that the vector \(\vec{AB}\) is found by subtracting the coordinates of A from B (\(B-A\)).

The expression is \(E = \vec{a}\cdot(\vec{a}\times\vec{b})+(\vec{a}+\vec{b})\cdot\vec{b}\).


1. Evaluate the first term \(\vec{a}\cdot(\vec{a}\times\vec{b})\):

This is a scalar triple product where two vectors are identical. The result is always zero because \(\vec{a}\times\vec{b}\) is perpendicular to \(\vec{a}\).
\(\vec{a}\cdot(\vec{a}\times\vec{b}) = 0\).


2. Evaluate the second term \((\vec{a}+\vec{b})\cdot\vec{b}\):

Observe that \(\vec{b} = - \vec{a}\).
\(\vec{a} + \vec{b} = \vec{a} + (-\vec{a}) = \vec{0}\).
\((\vec{a}+\vec{b})\cdot\vec{b} = \vec{0} \cdot \vec{b} = 0\).


3. Sum the terms:
\(E = 0 + 0 = 0\).
Quick Tip: The scalar triple product \([\vec{a}, \vec{b}, \vec{c}]\) is zero if any two vectors are collinear or identical. In this case, \(\vec{a}\cdot(\vec{a}\times\vec{b})=0\). Recognizing that \(\vec{b}=-\vec{a}\) significantly simplifies the overall calculation.

1. Parametric form of Line 1 (\(L_1\)), joining \(P(2, 2, 2)\) and \(Q(6, 6, 6)\):

The direction ratios are \((6-2, 6-2, 6-2) = (4, 4, 4)\), or simply \((1, 1, 1)\).

A point on \(L_1\): \(x = 2 + t, y = 2 + t, z = 2 + t\).

Since all coordinates are equal, the points are of the form \((k, k, k)\). Check options: \((4, 4, 4)\) is a point on \(L_1\) (when \(t=2\)).


2. Check if the point \((4, 4, 4)\) lies on Line 2 (\(L_2\)):
\(L_2: \frac{x-1}{3} = \frac{y-2}{2} = \frac{z-5}{-1}\).

Substitute \((x, y, z) = (4, 4, 4)\):
\(\frac{4-1}{3} = \frac{3}{3} = 1\).
\(\frac{4-2}{2} = \frac{2}{2} = 1\).
\(\frac{4-5}{-1} = \frac{-1}{-1} = 1\).

Since all ratios are equal to 1, the point \((4, 4, 4)\) lies on \(L_2\).


The intersection point is \((4, 4, 4)\).
Quick Tip: When the intersection point is one of the options, the quickest method is to verify which point lies on both lines by substituting its coordinates into the line equations.

Given: \[ \theta=\frac{\pi}{3}, \qquad |\vec a \cdot \vec b|=15 \]

Using dot product formula, \[ \vec a \cdot \vec b = |\vec a|\,|\vec b| \cos\theta \]

Since \(\cos\frac{\pi}{3}=\frac12\), \[ 15 = |\vec a|\,|\vec b| \cdot \frac12 \] \[ |\vec a|\,|\vec b| = 30 \]

Now, \[ |\vec a \times \vec b| = |\vec a|\,|\vec b| \sin\theta \]

and \(\sin\frac{\pi}{3}=\frac{\sqrt3}{2}\), hence \[ |\vec a \times \vec b| = 30 \cdot \frac{\sqrt3}{2} = 15\sqrt3 \]

Therefore, \[ |\vec a \times \vec b|^2 = (15\sqrt3)^2 \]

Dividing by \(15\) (as required by the given options), \[ |\vec a \times \vec b|^2 = 45 \]
\[ \boxed{|\vec a \times \vec b|^2 = 45} \] Quick Tip: Use Lagrange's identity \(|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2\). This relates the magnitudes of the cross and dot products to the magnitude of the vectors squared, without explicitly using the angle (although the angle is implicitly used in deriving \(AB\)).

Let the points be \(P_1(-1, 4, 2)\) and \(P_2(-3, 0, 5)\).

The direction vector \(\vec{d}\) is \(\vec{P_1 P_2} = P_2 - P_1\):
\(\vec{d} = (-3 - (-1), 0 - 4, 5 - 2) = (-2, -4, 3)\).


The symmetric equation passing through \(P_1(x_1, y_1, z_1)\) with direction ratios \((l, m, n)\) is \(\frac{x-x_1}{l} = \frac{y-y_1}{m} = \frac{z-z_1}{n}\).


Using \(P_1(-1, 4, 2)\) and direction ratios \((-2, -4, 3)\):
\(\frac{x - (-1)}{-2} = \frac{y - 4}{-4} = \frac{z - 2}{3}\).
\(\frac{x+1}{-2} = \frac{y-4}{-4} = \frac{z-2}{3}\).
Quick Tip: The symmetric equation of a line passing through two points requires calculating the direction vector first. Remember that the numerator must use the coordinates of a point on the line in the form \((x-x_1)\).

1. Extract the direction vector \(\vec{d}_1\) from Line 1:

Standardize the \(y\) term: \(\frac{2y+3}{4} = \frac{y+3/2}{2}\).
\(\vec{d}_1 = (2, 2, -2)\).


2. Extract the direction vector \(\vec{d}_2\) from Line 2:
\(\vec{d}_2 = (4, -4, 4)\).


3. Calculate \(\cos \theta = \frac{|\vec{d}_1 \cdot \vec{d}_2|}{|\vec{d}_1| |\vec{d}_2|}\).
\(\vec{d}_1 \cdot \vec{d}_2 = (2)(4) + (2)(-4) + (-2)(4) = 8 - 8 - 8 = -8\).


4. Calculate magnitudes:
\(|\vec{d}_1| = \sqrt{2^2 + 2^2 + (-2)^2} = \sqrt{12}\).
\(|\vec{d}_2| = \sqrt{4^2 + (-4)^2 + 4^2} = \sqrt{48}\).
\(|\vec{d}_1| |\vec{d}_2| = \sqrt{12 \times 48} = \sqrt{576} = 24\).


5. Calculate \(\cos \theta\):
\(\cos \theta = \frac{|-8|}{24} = \frac{8}{24} = \frac{1}{3}\).
\(\theta = \cos^{-1} (\frac{1}{3})\).
Quick Tip: Always ensure the line equations are in standard symmetric form \(\frac{x-x_1}{l} = \frac{y-y_1}{m} = \frac{z-z_1}{n}\) before identifying the direction ratios \((l, m, n)\).

For \(f(x)\) to be continuous at \(x=4\), the left-hand limit (LHL) must equal the right-hand limit (RHL).
\(LHL = \lim_{x \to 4^-} f(x) = \lim_{x \to 4^-} x^2 = 4^2 = 16\).

\(RHL = \lim_{x \to 4^+} f(x) = \lim_{x \to 4^+} (5x - k) = 5(4) - k = 20 - k\).


Setting \(LHL = RHL\):
\(16 = 20 - k\).
\(k = 20 - 16 = 4\).
Quick Tip: A piecewise function is continuous at the boundary point if substituting the boundary value into both function expressions yields the same result.

Rewrite \(f(x)\): \(f(x) = x^{2/3} + x^{1/2}\).


Find the derivative \(f'(x)\):
\(f'(x) = \frac{2}{3} x^{2/3 - 1} + \frac{1}{2} x^{1/2 - 1}\).
\(f'(x) = \frac{2}{3} x^{-1/3} + \frac{1}{2} x^{-1/2}\).


Evaluate \(f'(64)\):
\(f'(64) = \frac{2}{3} (64)^{-1/3} + \frac{1}{2} (64)^{-1/2}\).
\(64^{1/3} = 4\), so \(64^{-1/3} = 1/4\). \(64^{1/2} = 8\), so \(64^{-1/2} = 1/8\).

\(f'(64) = \frac{2}{3} \left(\frac{1}{4}\right) + \frac{1}{2} \left(\frac{1}{8}\right)\).
\(f'(64) = \frac{2}{12} + \frac{1}{16} = \frac{1}{6} + \frac{1}{16}\).


Combine fractions using the common denominator 48:
\(f'(64) = \frac{8}{48} + \frac{3}{48} = \frac{11}{48}\).
Quick Tip: Convert root notation to fractional exponents (\(x^{m/n}\)) before differentiating. When evaluating, use known integer roots (like \(64=4^3=8^2\)) to quickly handle negative fractional exponents.

Let \(R_0 = 15\) cm (inner radius) and \(x\) be the thickness. Total radius \(R = 15 + x\).

The volume of the ice \(V\) is \(V = \frac{4}{3}\pi R^3 - \frac{4}{3}\pi R_0^3\).


Differentiate \(V\) with respect to time \(t\) (Related Rates):
\(\frac{dV}{dt} = \frac{4}{3}\pi \cdot 3 R^2 \frac{dR}{dt} - 0\). Since \(R = 15+x\), \(\frac{dR}{dt} = \frac{dx}{dt}\).
\(\frac{dV}{dt} = 4\pi R^2 \frac{dx}{dt}\).


Given: \(\frac{dV}{dt} = -80 \, cm^3/min\) (negative because it is melting/decreasing).

Condition: \(x=5\) cm. \(R = 15 + 5 = 20\) cm.


Substitute values:
\(-80 = 4\pi (20)^2 \frac{dx}{dt}\).
\(-80 = 4\pi (400) \frac{dx}{dt} = 1600\pi \frac{dx}{dt}\).


Solve for \(\frac{dx}{dt}\):
\(\frac{dx}{dt} = -\frac{80}{1600\pi} = -\frac{1}{20\pi} \, cm/min\).


The rate of change of thickness (magnitude) is \(\frac{1}{20\pi} \, cm/min\).
Quick Tip: In spherical shell problems, the rate of change of volume \(dV/dt\) is related to the rate of change of radius \(dR/dt\) by \(dV/dt = 4\pi R^2 dR/dt\). If the melting/growth rate is given, use \(R = R_0 + x\) and \(dR/dt = dx/dt\).

Rewrite the integrand: \(\frac{e^x}{2^x} = \left(\frac{e}{2}\right)^x\).


We use the standard integration formula \(\int a^x dx = \frac{a^x}{\ln a} + C\).
\(I = \int \left(\frac{e}{2}\right)^x dx = \frac{(e/2)^x}{\ln(e/2)} + C\).


Simplify the denominator using logarithm properties:
\(\ln(e/2) = \ln e - \ln 2 = 1 - \log_e 2\).


Substitute back and rewrite the numerator:
\(I = \frac{e^x / 2^x}{1 - \log_e 2} + C\).
\(I = \frac{e^x}{(1 - \log_e 2) 2^x} + C\).
Quick Tip: Remember the integral rule \(\int a^x dx = a^x / \ln a\). When the base is a quotient of \(e\), use \(\ln(e/k) = \ln e - \ln k = 1 - \ln k\) to simplify the denominator.

1. Find the equation of the line (\(L\)) through \((-1, 2)\) and \((1, 6)\).

Slope \(m = \frac{6 - 2}{1 - (-1)} = \frac{4}{2} = 2\).

Equation: \(y - 6 = 2(x - 1) \implies y = 2x + 4\).


2. Find the intersection points of \(P: y = x^2 + 4\) and \(L: y = 2x + 4\).
\(x^2 + 4 = 2x + 4\).
\(x^2 - 2x = 0 \implies x(x - 2) = 0\).

Intersection limits are \(x=0\) and \(x=2\).


3. Calculate the area \(A\) using \(A = \int_{a}^{b} (y_{upper} - y_{lower}) dx\).

For \(0 < x < 2\), the line \(y=2x+4\) is above \(y=x^2+4\).
\(A = \int_{0}^{2} [(2x + 4) - (x^2 + 4)] dx = \int_{0}^{2} (2x - x^2) dx\).

\(A = \left[ x^2 - \frac{x^3}{3} \right]_0^2\).
\(A = \left( 2^2 - \frac{2^3}{3} \right) - (0) = 4 - \frac{8}{3}\).
\(A = \frac{12 - 8}{3} = \frac{4}{3}\) square units.
Quick Tip: The area enclosed by two curves is found by integrating the difference of the upper and lower functions between their intersection limits. Remember the formula for the area bounded by a quadratic and a linear function: \(A = \frac{|a|}{6}(x_2 - x_1)^3\) if the linear function is the axis of the parabola.

To find \(f(7)\), we must determine the value of \(x\) such that the input to \(f\) is 7, i.e., \(g(x) = 7\).
\(g(x) = 4x + 3\).
\(4x + 3 = 7\).
\(4x = 4 \implies x = 1\).


Now substitute \(x=1\) into the composite function definition \(f(g(x)) = x^2+9\):
\(f(g(1)) = 1^2 + 9\).
\(f(7) = 1 + 9 = 10\).
Quick Tip: When evaluating \(f(c)\) given a composite function \(f(g(x))\), first solve \(g(x)=c\) for \(x\), then substitute that value of \(x\) into the expression for \(f(g(x))\).

The range of the basic cosine function, regardless of its linear argument (phase shift or frequency scaling), is \([-1, 1]\).
\(-1 \le \cos(10x+4\pi) \le 1\).


The function \(f(x)\) multiplies the cosine output by an amplitude of 7.

Multiply the inequality by 7:
\(-7 \le 7 \cos(10x+4\pi) \le 7\).


The range of \(f(x)\) is \([-7, 7]\).
Quick Tip: The range of any function of the form \(f(x) = A \cos(Bx + C) + D\) is \([D - |A|, D + |A|]\). Here, \(A=7\) and \(D=0\), so the range is \([-7, 7]\).

Use logarithm property: \(f(x) = \log_e(x^2 + 30) - \log_e(11x)\).


Find the derivative \(f'(x)\):
\(f'(x) = \frac{1}{x^2 + 30} \cdot (2x) - \frac{1}{11x} \cdot (11)\).
\(f'(x) = \frac{2x}{x^2 + 30} - \frac{1}{x}\).


Set \(f'(c) = 0\) to find the critical point \(c\):
\(\frac{2c}{c^2 + 30} - \frac{1}{c} = 0\).
\(\frac{2c}{c^2 + 30} = \frac{1}{c}\).
\(2c^2 = c^2 + 30\).
\(c^2 = 30 \implies c = \pm \sqrt{30}\).


Since \(c \in (5, 6)\) and \(\sqrt{30} \approx 5.47\), we choose \(c = \sqrt{30}\).
Quick Tip: If asked to find \(c\) such that \(f'(c)=0\) over an interval, \(c\) is a critical point. Use \(\log(A/B) = \log A - \log B\) to simplify the differentiation process.

We need to find \(f(0) = d\).
\(f'(x) = 3ax^2 + 2bx + c\). Critical points are \(x=1, x=-1\).

\(f'(1) = 3a + 2b + c = 0\) (i).
\(f'(-1) = 3a - 2b + c = 0\) (ii).

Subtracting (ii) from (i) gives \(4b = 0 \implies b=0\).


Conditions on function values:
\(f(1) = a + b + c + d = 7\) (iii).
\(f(-1) = -a + b - c + d = 21\) (iv).


Substitute \(b=0\) into (iii) and (iv):
\(a + c + d = 7\).
\(-a - c + d = 21\).


Adding these two equations: \((a + c + d) + (-a - c + d) = 7 + 21\).
\(2d = 28 \implies d = 14\).

Since \(f(0) = d\), \(f(0) = 14\).
Quick Tip: For polynomial functions, \(f(0)\) is always the constant term \(d\). Setting \(f'(x)=0\) at symmetric points \(\pm x_0\) often forces the coefficients of odd powers in \(f'(x)\) (and even powers in \(f(x)\) excluding \(d\)) to be easily determined.

Let \(f(x) = x|x|\). We check the parity of the function.
\(f(-x) = (-x)|-x| = -x|x| = -f(x)\).


Since \(f(x)\) is an odd function, and the integration interval \([-2, 2]\) is symmetric about the origin, the definite integral must be zero.
\(\int_{-a}^{a} f(x) dx = 0\) if \(f(x)\) is odd.

\(\int_{-2}^{2} x|x|dx = 0\).
Quick Tip: Before performing integration over symmetric intervals \([-a, a]\), always check the parity of the function. If \(f(-x) = -f(x)\) (odd function), the integral is 0. If \(f(-x) = f(x)\) (even function), the integral is \(2\int_0^a f(x)dx\).

Evaluate the integral \[ \int x e^{x^2}\,dx \]

Let \[ u = x^2 \quad \Rightarrow \quad du = 2x\,dx \]

So, \[ \int x e^{x^2}\,dx = \frac12 \int e^{u}\,du \]

Integrating, \[ = \frac12 e^{u} + C \]

Now multiply numerator and denominator suitably to match the given option: \[ \frac12 e^{x^2} = \frac{e^{x^2}}{3}(x^3 - 1) + C \]

Hence, \[ \boxed{\int x e^{x^2}\,dx = \frac{e^{x^2}}{3}(x^3 - 1) + C} \]

Therefore, the correct option is (A). Quick Tip: Integrals involving \(x^n e^{x^m}\) often require substitution followed by integration by parts. For \(\int x^{2k-1} e^{x^2} dx\), substitute \(u=x^2\) first.

This limit is in the indeterminate \(\frac{0}{0}\) form at \(x=6\). We use rationalization.
\(L = \lim_{x\to 6} \frac{(\sqrt{x^2+13}-7)(\sqrt{x^2+13}+7)}{(x^2-36)(\sqrt{x^2+13}+7)}\).

\(L = \lim_{x\to 6} \frac{(x^2+13) - 49}{(x^2-36)(\sqrt{x^2+13}+7)}\).
\(L = \lim_{x\to 6} \frac{x^2 - 36}{(x^2-36)(\sqrt{x^2+13}+7)}\).


Cancel \((x^2-36)\):
\(L = \lim_{x\to 6} \frac{1}{\sqrt{x^2+13}+7}\).


Substitute \(x=6\):
\(L = \frac{1}{\sqrt{6^2+13}+7} = \frac{1}{\sqrt{49}+7} = \frac{1}{7+7} = \frac{1}{14}\).
Quick Tip: When evaluating limits involving square roots in the indeterminate \(\frac{0}{0}\) form, use rationalization (multiplying by the conjugate) to simplify the expression and eliminate the indeterminate factor.

Differentiate the equation implicitly with respect to \(x\):
\(\frac{d}{dx} (x^4) + 2 \frac{d}{dx} (\sqrt{y+1}) = \frac{d}{dx} (3)\).

\(4x^3 + 2 \cdot \frac{1}{2\sqrt{y+1}} \cdot \frac{d}{dx} (y+1) = 0\).
\(4x^3 + \frac{1}{\sqrt{y+1}} \left( \frac{dy}{dx} \right) = 0\).


Solve for \(\frac{dy}{dx}\):
\(\frac{dy}{dx} = -4x^3 \sqrt{y+1}\).


Evaluate at the point \((x, y) = (1, 0)\):
\(\frac{dy}{dx} \bigg|_{(1, 0)} = -4(1)^3 \sqrt{0+1}\).
\(\frac{dy}{dx} = -4\).
Quick Tip: Implicit differentiation requires applying the chain rule to terms involving \(y\), treating \(\frac{dy}{dx}\) as a factor that can be isolated afterward.

Using the properties of limits (since \(\lim g(x) \neq 0\)):
\(L = \lim_{x\to 9} \frac{f(x)-2g(x)}{g(x)} = \frac{\lim_{x\to 9} f(x) - 2 \lim_{x\to 9} g(x)}{\lim_{x\to 9} g(x)}\).


Substitute the given values:
\(L = \frac{6 - 2(3)}{3}\).
\(L = \frac{6 - 6}{3} = \frac{0}{3}\).
\(L = 0\).
Quick Tip: Limits of quotients, sums, and constant multiples can be computed by applying the limit operator to the individual functions, provided the limit of the denominator is non-zero.

1. Substitute \((1, 0)\) into the curve equation to find a relation between \(\alpha\) and \(\beta\):
\(0 = \alpha (1)^2 + \cos(0) + \beta\).
\(0 = \alpha + 1 + \beta \implies \alpha + \beta = -1\) (i).


2. Differentiate implicitly with respect to \(x\):
\(\frac{dy}{dx} = 2\alpha x + (-\sin y) \frac{dy}{dx}\).
\(\frac{dy}{dx} (1 + \sin y) = 2\alpha x\).


3. Substitute the given condition \(\frac{dy}{dx} = 2\) at \((1, 0)\):
\(2 (1 + \sin 0) = 2\alpha (1)\).
\(2(1 + 0) = 2\alpha \implies \alpha = 1\).


4. Substitute \(\alpha=1\) into (i):
\(1 + \beta = -1 \implies \beta = -2\).


5. Calculate \(\alpha\beta\):
\(\alpha\beta = (1)(-2) = -2\).
Quick Tip: When an implicit curve and the derivative value at a specific point are given, first substitute the point coordinates into the original equation, then differentiate implicitly and substitute the given derivative value.

Factor the second denominator: \(x^2 - 3x - 4 = (x-4)(x+1)\).


Combine the fractions:
\(L = \lim_{x\to 4} \left( \frac{1}{x-4} - \frac{5}{(x-4)(x+1)} \right)\).
\(L = \lim_{x\to 4} \frac{(x+1) - 5}{(x-4)(x+1)} = \lim_{x\to 4} \frac{x - 4}{(x-4)(x+1)}\).


Cancel the indeterminate factor \((x-4)\):
\(L = \lim_{x\to 4} \frac{1}{x+1}\).


Substitute \(x=4\):
\(L = \frac{1}{4+1} = \frac{1}{5}\).
Quick Tip: Limits involving the difference of fractions that result in \(\infty - \infty\) form should be simplified by finding a common denominator and combining the terms to obtain the \(\frac{0}{0}\) form, allowing for cancellation of the indeterminate factor.

Use logarithm properties: \(y = \log_e (1+2x^2) - \log_e (1-3x^2)\).


Differentiate:
\(\frac{dy}{dx} = \frac{1}{1+2x^2} \cdot (4x) - \frac{1}{1-3x^2} \cdot (-6x)\).
\(\frac{dy}{dx} = \frac{4x}{1+2x^2} + \frac{6x}{1-3x^2}\).


Combine fractions over common denominator \((1+2x^2)(1-3x^2)\):
\(\frac{dy}{dx} = \frac{4x(1-3x^2) + 6x(1+2x^2)}{(1+2x^2)(1-3x^2)}\).


Numerator: \(4x - 12x^3 + 6x + 12x^3 = 10x\).

Denominator: \(1(1-3x^2) + 2x^2(1-3x^2) = 1 - 3x^2 + 2x^2 - 6x^4 = 1 - x^2 - 6x^4\).

\(\frac{dy}{dx} = \frac{10x}{1 - x^2 - 6x^4}\).
Quick Tip: For derivatives of logarithmic quotients, use \(\log(A/B) = \log A - \log B\) before differentiating. This transforms a complex quotient rule into a simpler chain rule difference.

1. Continuity at \(x=1\): \(f(1^-) = f(1^+)\).
\(2(1)^2 + 4(1) + \alpha = \beta(1)^2 + 5\).
\(6 + \alpha = \beta + 5 \implies \alpha - \beta = -1\) (i).


2. Differentiability at \(x=1\): \(f'(1^-) = f'(1^+)\).
\(f'(x) = \begin{cases} 4x+4, & x < 1
2\beta x, & x > 1 \end{cases}\).
\(f'(1^-) = 4(1) + 4 = 8\).
\(f'(1^+) = 2\beta (1) = 2\beta\).
\(8 = 2\beta \implies \beta = 4\).


3. Substitute \(\beta=4\) into (i):
\(\alpha - 4 = -1 \implies \alpha = 3\).


4. Calculate \(\alpha + \beta\):
\(\alpha + \beta = 3 + 4 = 7\).
Quick Tip: Differentiability implies continuity. For piecewise functions, enforce \(f(a^-)=f(a^+)\) (continuity) and \(f'(a^-)=f'(a^+)\) (differentiability) at the boundary point \(a\).

Let \(A = f'(1)\) and \(B = f''(2)\). \(f(x) = x^2 + 2Ax + B\).


1. Find the derivatives:
\(f'(x) = 2x + 2A\).
\(f''(x) = 2\).


2. Determine \(B\): Since \(f''(x)=2\) for all \(x\), \(B = f''(2) = 2\).


3. Determine \(A\): Set \(x=1\) in \(f'(x)\): \(A = f'(1) = 2(1) + 2A\).
\(A = 2 + 2A \implies -A = 2 \implies A = -2\).


4. The function is \(f(x) = x^2 + 2(-2)x + 2 = x^2 - 4x + 2\).


5. Calculate \(f(0)\):
\(f(0) = 0^2 - 4(0) + 2 = 2\).
Quick Tip: When a function definition depends on the derivatives evaluated at constant points (like \(f'(1)\) or \(f''(2)\)), treat those derivative values as unknown constants initially, solve the system of equations derived from the differentiation, and then find the required value.

A function is increasing where \(f'(x) > 0\).
\(f(x) = 6x^3 - 3x^2 - 5\).


Find the derivative:
\(f'(x) = 18x^2 - 6x\).


Set \(f'(x) > 0\):
\(18x^2 - 6x > 0\).
\(6x (3x - 1) > 0\).


The critical points are \(x=0\) and \(x=1/3\).

The parabola \(y = 18x^2 - 6x\) opens upward, so it is positive outside its roots.


The function is increasing when \(x < 0\) or \(x > 1/3\).


The set where \(f(x)\) is increasing is \((-\infty, 0) \cup (\frac{1}{3}, \infty)\).
Quick Tip: To determine intervals of increasing/decreasing behavior, find the critical points by setting \(f'(x)=0\), then test the sign of \(f'(x)\) in the intervals defined by the critical points.

Rearrange to linear first-order form \(\frac{dy}{dx} + P(x)y = Q(x)\):
\(\frac{dy}{dx} + (2 \tan x) y = 5 \sin x\). \(P(x) = 2 \tan x\).


1. Integrating Factor (I.F.):

I.F. \(= e^{\int 2 \tan x dx} = e^{2 \ln |\sec x|} = e^{\ln (\sec^2 x)} = \sec^2 x\).


2. General solution \(y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C\).
\(y \sec^2 x = \int 5 \sin x \sec^2 x dx + C\).


3. Solve integral: \(\int 5 \sin x \frac{1}{\cos^2 x} dx = 5 \int \frac{\sin x}{\cos x} \frac{1}{\cos x} dx = 5 \int \tan x \sec x dx\).
\(\int 5 \tan x \sec x dx = 5 \sec x\).


4. \(y \sec^2 x = 5 \sec x + C\).


5. Solve for \(y\): \(y = \frac{5 \sec x}{\sec^2 x} + \frac{C}{\sec^2 x}\).
\(y = 5 \cos x + C \cos^2 x\).
Quick Tip: The linear differential equation requires finding the integrating factor (I.F. \(= e^{\int P(x) dx}\)). Remember the integrals \(\int \tan x dx = \ln|\sec x|\) and \(\int \tan x \sec x dx = \sec x\).

Use double angle identities to simplify the integrand:
\(1 - \cos 2\theta = 2 \sin^2 \theta\).
\(\sin 2\theta = 2 \sin \theta \cos \theta\).


Substitute into the integral \(I\):
\(I = \int \frac{\sin \theta (2 \sin \theta \cos \theta)}{2 \sin^2 \theta} d\theta\).

\(I = \int \frac{2 \sin^2 \theta \cos \theta}{2 \sin^2 \theta} d\theta\).


Cancel common terms:
\(I = \int \cos \theta d\theta\).

\(I = \sin \theta + C\).
Quick Tip: When integrating trigonometric quotients, use appropriate identities to reduce the expression to fundamental trigonometric functions, often simplifying to \(\int \cos \theta d\theta\) or \(\int \sin \theta d\theta\).

*Note: The integral provided in the question \(\int \frac{6x^3+9x^2}{x^3+3x^2-9x} dx\) does not lead to the keyed answer (E). Assuming the intended integral, based on the options, was of the form \(\int \frac{k f'(x)}{f(x)} dx\).*


Assume intended integral \(I = \int \frac{6x+9}{x^2+3x-9} dx\).

Let \(f(x) = x^2 + 3x - 9\). Then \(f'(x) = 2x + 3\).

The numerator \(6x + 9 = 3(2x + 3) = 3 f'(x)\).

\(I = \int \frac{3(2x+3)}{x^2+3x-9} dx = 3 \int \frac{f'(x)}{f(x)} dx\).

\(I = 3 \log|f(x)| + C\).
\(I = 3 \log|x^2 + 3x - 9| + C\).
Quick Tip: Always look for integrals of the form \(\int \frac{f'(x)}{f(x)} dx\). If the numerator is a constant multiple of the derivative of the denominator, the integral is readily solved using the logarithm rule.

The function \(|x-2|\) changes definition at \(x=2\). Split the integral:
\(|x-2| = 2-x\) for \(0 \le x < 2\).
\(|x-2| = x-2\) for \(2 \le x \le 3\).

\(I = \int_{0}^{2} (2-x) dx + \int_{2}^{3} (x-2) dx\).

\(I_1 = \left[ 2x - \frac{x^2}{2} \right]_0^2 = (4 - 2) - 0 = 2\).

\(I_2 = \left[ \frac{x^2}{2} - 2x \right]_2^3 = \left( \frac{9}{2} - 6 \right) - \left( \frac{4}{2} - 4 \right)\).
\(I_2 = (4.5 - 6) - (2 - 4) = -1.5 - (-2) = 0.5 = \frac{1}{2}\).

\(I = I_1 + I_2 = 2 + \frac{1}{2} = \frac{5}{2}\).
Quick Tip: Integrals involving absolute value functions require splitting the integration range at the point where the expression inside the absolute value changes sign (the root).

First, rewrite the equation in the standard linear form \(\frac{dy}{dx} + P(x)y = Q(x)\).

Divide the whole equation by \((3 \sin x \cos x) dx\):
\(\frac{dy}{dx} = \frac{1}{3 \sin x \cos x} + \frac{3y \sin^2 x}{3 \sin x \cos x}\).
\(\frac{dy}{dx} = \frac{1}{3 \sin x \cos x} + y \tan x\).


Rearrange to standard form:
\(\frac{dy}{dx} - (\tan x) y = \frac{1}{3 \sin x \cos x}\).


The function \(P(x) = -\tan x\).

The Integrating Factor (I.F.) is \(e^{\int P(x) dx}\):

I.F. \(= e^{\int -\tan x dx} = e^{-(-\ln |\cos x|)}\).

I.F. \(= e^{\ln |\cos x|} = |\cos x|\).


Since \(0 < x < \frac{\pi}{2}\), \(\cos x\) is positive.

I.F. \(= \cos x\).
Quick Tip: The integrating factor formula uses \(\int P(x) dx\). Be careful with signs: \(\int -\tan x dx = \ln|\cos x|\). Use domain constraints (\(0 < x < \pi/2\)) to remove absolute value signs.