Let \(M_E\) be the mass of the Earth and \(M_m\) be the mass of the Moon, with \(M_E = 81 M_m\).


Let \(r\) be the distance from the centre of the Earth where the net gravitational force on a test mass \(m\) is zero.


The distance from the Moon's centre is \((D-r)\).


For the force to be zero, \(F_E = F_m\): \(\frac{G M_E m}{r^2} = \frac{G M_m m}{(D-r)^2}\)


Substituting \(M_E = 81 M_m\): \(\frac{81 M_m}{r^2} = \frac{M_m}{(D-r)^2}\)

\(\frac{81}{r^2} = \frac{1}{(D-r)^2}\)


Taking the square root (since \(r > 0\) and \(D-r > 0\)): \(\frac{9}{r} = \frac{1}{D-r}\)

\(9(D-r) = r\)

\(9D - 9r = r\)

\(10r = 9D\)

\(r = \frac{9D}{10}\)
Quick Tip: For two masses \(M_1\) and \(M_2\) separated by distance \(D\), the null point distance \(r_1\) from \(M_1\) is \(r_1 = \frac{D \sqrt{M_1}}{\sqrt{M_1} + \sqrt{M_2}}\).
Here, \(r = \frac{D \sqrt{81 M_m}}{\sqrt{81 M_m} + \sqrt{M_m}} = \frac{9 D \sqrt{M_m}}{10 \sqrt{M_m}} = \frac{9D}{10}\).

The increase in length (\(\Delta L\)) under force \(F\) is given by \(\Delta L = \frac{F L}{A Y}\).


Since \(F\) and \(Y\) (same material) are constant, \(\Delta L \propto \frac{L}{A}\).


The area of cross-section \(A \propto d^2\), where \(d\) is the diameter. Thus, \(\Delta L \propto \frac{L}{d^2}\).


Given ratios: \(\frac{L_A}{L_B} = \frac{1}{2}\) and \(\frac{d_A}{d_B} = \frac{2}{1}\).


The ratio of elongation is: \(\frac{\Delta L_A}{\Delta L_B} = \frac{L_A}{L_B} \cdot \left(\frac{d_B}{d_A}\right)^2\)

\(\frac{\Delta L_A}{\Delta L_B} = \left(\frac{1}{2}\right) \cdot \left(\frac{1}{2}\right)^2\)

\(\frac{\Delta L_A}{\Delta L_B} = \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8}\)


The ratio is \(1:8\).
Quick Tip: If length is halved and diameter is doubled, \(\Delta L \propto \frac{L}{d^2}\) changes by a factor of \(\frac{(1/2)}{(2)^2} = \frac{1}{8}\).

By the Principle of Homogeneity, the unit of every term must match the unit of the LHS (\(x\)).


Unit of \(x\) is \(km\). Unit of \(t\) is \(s\).


We must have \(Unit of (bt^2) = Unit of x\).

\((Unit of b) \cdot (Unit of t)^2 = km\)

\((Unit of b) \cdot (s)^2 = km\)

\(Unit of b = \frac{km}{s^2}\) or \(km/s^2\).
Quick Tip: In physics, any term multiplied by \(t^2\) must have the units of acceleration (Length/Time\(^2\)) if the resulting dimension is Length.

The excess pressure inside a bubble of radius \(r\) is \(\Delta P = \frac{4T}{r}\).


Since \(r_1 < r_2\), the pressure in bubble 1 (\(\Delta P_1\)) is greater than the pressure in bubble 2 (\(\Delta P_2\)).


The common separating film (radius \(R\)) experiences a net pressure difference \(\Delta P_{net} = \Delta P_1 - \Delta P_2\).


This net pressure difference is balanced by the excess pressure of the common film: \(\Delta P_{net} = \frac{4T}{R}\).

\(\frac{4T}{R} = \frac{4T}{r_1} - \frac{4T}{r_2}\)

\(\frac{1}{R} = \frac{1}{r_1} - \frac{1}{r_2}\)

\(\frac{1}{R} = \frac{r_2 - r_1}{r_1 r_2}\)

\(R = \frac{r_1 r_2}{r_2 - r_1}\)
Quick Tip: The formula \(\frac{1}{R} = \frac{1}{r_{small}} - \frac{1}{r_{large}}\) always holds for the radius of the common interface separating two soap bubbles.

The magnetic Lorentz force (\(\vec{F}\)) on a charge \(q\) moving with velocity \(\vec{v}\) in a magnetic field \(\vec{B}\) is \(\vec{F} = q (\vec{v} \times \vec{B})\).


The magnitude is \(F = q v B \sin\theta\), where \(\theta\) is the angle between \(\vec{v}\) and \(\vec{B}\).


Since the charge is moving parallel to the magnetic field, \(\theta = 0^\circ\).

\(F = q v B \sin(0^\circ)\)


Since \(\sin(0^\circ) = 0\), the force \(F = 0\).
Quick Tip: The magnetic force acts only if the charge has a velocity component perpendicular to the magnetic field. If motion is parallel to the field, the force is zero.

Mass of A (\(m_A\)) = \(m\). Radius of A (\(r_A\)) = \(2R\). Center of A at \(x_A = 0\).


Mass of B (\(m_B\)) = \(2m\). Radius of B (\(r_B\)) = \(R\). Center of B at \(x_B = r_A + r_B = 3R\).


The position of the Center of Mass (\(x_{com}\)) is:

\(x_{com} = \frac{m_A x_A + m_B x_B}{m_A + m_B}\)

\(x_{com} = \frac{m(0) + 2m(3R)}{m + 2m}\)

\(x_{com} = \frac{6mR}{3m} = 2R\)


The COM is located at a distance \(2R\) from the center of A.


Since \(r_A = 2R\), the center of mass is exactly on the circumference of sphere A, which is the point of contact with sphere B.
Quick Tip: The center of mass for two masses \(m_A\) and \(m_B\) separated by distance \(D\) is located at a distance \(D \cdot \frac{m_B}{m_A+m_B}\) from \(m_A\).

According to Amontons' laws of friction, frictional forces are independent of the apparent area of contact. Thus (A) and (B) are incorrect.


The coefficient of static friction (\(\mu_s\)) is typically greater than or equal to the coefficient of kinetic friction (\(\mu_k\)).


It is generally true that \(\mu_k < \mu_s\).


Thus, the coefficient of kinetic friction is less than the coefficient of static friction. (D) is correct.
Quick Tip: Static friction (\(\mu_s N\)) must overcome inter-atomic forces to initiate motion, making it usually greater than kinetic friction (\(\mu_k N\)), which sustains motion.

Initial velocity \(u = 0\). Acceleration \(a = \frac{4}{3} m/s^2\).


The distance traveled in the \(n\)-th second (\(S_n\)) is given by: \(S_n = u + \frac{a}{2}(2n - 1)\).


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

\(S_3 = 0 + \frac{4/3}{2}(2(3) - 1)\)

\(S_3 = \frac{2}{3}(6 - 1)\)

\(S_3 = \frac{2}{3}(5)\)

\(S_3 = \frac{10}{3} m\)
Quick Tip: The distance in the \(n\)-th second is not the displacement after \(n\) seconds (\(S_n\)); it is the distance between \(t=n-1\) and \(t=n\).

Work function \(W = 1 eV\). Wavelength \(\lambda = 3000 \AA\).


Energy of incident photon \(E = \frac{12400 eV \cdot \AA}{3000 \AA} \approx 4.13 eV\).


Maximum kinetic energy \(K_{\max} = E - W = 4.13 eV - 1 eV = 3.13 eV\).

\(K_{\max} = 3.13 \times (1.6 \times 10^{-19}) J \approx 5.0 \times 10^{-19} J\).


Using \(K_{\max} = \frac{1}{2} m_e v_{\max}^2\) and \(m_e \approx 9.1 \times 10^{-31} kg\):

\(v_{\max}^2 = \frac{2 K_{\max}}{m_e} = \frac{2 \times 5.0 \times 10^{-19}}{9.1 \times 10^{-31}} \approx 1.1 \times 10^{12} m^2/s^2\)

\(v_{\max} = \sqrt{1.1 \times 10^{12}} \approx 1.05 \times 10^6 m/s\)


The maximum velocity is of the order of \(10^6 metres/sec\).
Quick Tip: The quick formula \(E(eV) = 1240 / \lambda(nm)\) helps calculate photon energy quickly. \(3000 \AA = 300 nm\). \(E = 1240/300 \approx 4.13 eV\).

The relation between true coefficient of volume expansion of mercury (\(\gamma_m\)) and apparent expansion in a container (\(\gamma_a\)) is \(\gamma_a = \gamma_m - \gamma_c\).


1. Calculate \(\gamma_{steel}\): \(\gamma_s = 3 \alpha_s = 3 \times (12 \times 10^{-6}) = 36 \times 10^{-6}/^\circC\).


2. Find \(\gamma_m\) using steel data:
\(\gamma_{a, s} = 144 \times 10^{-6}/^\circC\).
\(144 \times 10^{-6} = \gamma_m - 36 \times 10^{-6}\)
\(\gamma_m = (144 + 36) \times 10^{-6} = 180 \times 10^{-6}/^\circC\).


3. Find \(\gamma_{glass}\) using glass data:
\(\gamma_{a, g} = 153 \times 10^{-6}/^\circC\).
\(153 \times 10^{-6} = \gamma_m - \gamma_g\)
\(\gamma_g = 180 \times 10^{-6} - 153 \times 10^{-6}\)
\(\gamma_g = 27 \times 10^{-6}/^\circC\)
Quick Tip: Apparent expansion is always less than true expansion because the container itself expands upon heating. If \(\gamma_a\) is given for two different containers, you must first calculate the true expansion coefficient of the liquid.

Given: Ratio of primary to secondary windings \(\frac{N_p}{N_s} = \frac{1}{25}\).


Secondary current \(I_s = 2 A\).


For an ideal transformer, the power is conserved (\(P_p = P_s\)), leading to the inverse relationship between current and turns ratio:

\(\frac{I_p}{I_s} = \frac{N_s}{N_p}\)

\(\frac{I_p}{2 A} = \frac{25}{1}\)

\(I_p = 25 \times 2 A\)

\(I_p = 50 A\)
Quick Tip: A step-up transformer increases voltage (\(N_s > N_p\)) but decreases current (\(I_s < I_p\)). If the turns ratio is \(1:25\), the current ratio must be \(25:1\).

The horizontal range \(R\) of a projectile launched at velocity \(u\) and angle \(\theta\) is \(R = \frac{u^2 \sin(2\theta)}{g}\).


Since \(u\) and \(g\) are constant, \(R \propto \sin(2\theta)\).


For the first launch: \(\theta_1 = 30^\circ\). \(2\theta_1 = 60^\circ\). \(R_1 \propto \sin(60^\circ)\).


For the second launch: \(\theta_2 = 60^\circ\). \(2\theta_2 = 120^\circ\). \(R_2 \propto \sin(120^\circ)\).


Since \(\sin(120^\circ) = \sin(180^\circ - 60^\circ) = \sin(60^\circ)\),

\(R_1\) and \(R_2\) are proportional to the same value.


The ratio \(\frac{R_1}{R_2} = \frac{\sin(60^\circ)}{\sin(120^\circ)} = \frac{1}{1}\).


The ratio is \(1:1\).
Quick Tip: Maximum range is achieved at \(\theta=45^\circ\). Any two angles \(\theta\) and \(90^\circ - \theta\) (complementary angles) yield the same range, provided the initial velocity is the same.

The electrostatic force \(F\) is proportional to the product of charges: \(F \propto q_1 q_2\).


Initial charges: \(q_1 = +3\), \(q_2 = +8\). Product \(P_{initial} = 3 \times 8 = 24\).


Initial force \(F_{initial} = +40 N\) (Repulsive).


New charges \(q'_1 = 3 - 5 = -2\). \(q'_2 = 8 - 5 = +3\).


New product \(P_{final} = (-2) \times (+3) = -6\).


The ratio of forces is proportional to the ratio of charge products:

\(\frac{F_{final}}{F_{initial}} = \frac{P_{final}}{P_{initial}} = \frac{-6}{24} = -\frac{1}{4}\)

\(F_{final} = -\frac{1}{4} \times (40 N) = -10 N\).


The negative sign indicates attraction.
Quick Tip: Since the distance \(r\) is constant, you only need to compare the initial and final product of the charges. The force direction flips if the final product is negative (attraction).

The linear acceleration \(a\) for a body rolling without slipping down an incline is:

\(a = \frac{g \sin\theta}{1 + I / (M R^2)}\)


where \(I\) is the moment of inertia and \(M R^2\) is the product of mass and radius squared.


For a solid sphere, the moment of inertia is \(I = \frac{2}{5} M R^2\).


Substituting the value of \(I\):

\(a = \frac{g \sin\theta}{1 + (\frac{2}{5} M R^2) / (M R^2)}\)

\(a = \frac{g \sin\theta}{1 + 2/5}\)

\(a = \frac{g \sin\theta}{7/5}\)

\(a = \frac{5}{7} g \sin\theta\)
Quick Tip: The acceleration for a rolling body is always less than \(g \sin\theta\). The larger the fraction \(I/MR^2\) (hollow shapes), the smaller the acceleration.

The deviation of a ray passing through a prism depends on the angle of incidence (\(i\)), angle of emergence (\(e\)), and prism angle (\(A\)). \(\delta = i + e - A\).


For minimum deviation in prism P, the ray passes symmetrically, i.e., parallel to the base, and \(i=e\).


When prisms Q and R, which are identical to P and made of the same material, are added, the overall system maintains the same net refraction as the ray passes symmetrically through the entire structure.


The internal boundaries between P, Q, and R are parallel to the base of P. Since the ray is parallel to the base of P inside P (at minimum deviation), it passes undeviated across the interfaces P/Q and Q/R.


The ray emerges from R with the same overall deviation as if it had only passed through P.
Quick Tip: In a series of identical prisms combined such that the internal faces are parallel to the refracted ray path (which happens at minimum deviation), the overall deviation is simply that of the first or last prism alone.

The two \(4\Omega\) resistors are connected in parallel.


The equivalent resistance of the parallel \(4\Omega\) resistors (\(R_p\)):

\(R_p = \frac{4 \times 4}{4 + 4} = 2\Omega\)


This parallel combination (\(2\Omega\)) is in series with the \(1\Omega\) resistor and the \(6V\) source.


Total equivalent resistance of the circuit (\(R_{eq}\)):

\(R_{eq} = 1\Omega + R_p = 1\Omega + 2\Omega = 3\Omega\)


The total current (\(I_T\)) supplied by the source is:

\(I_T = \frac{V}{R_{eq}} = \frac{6 V}{3 \Omega} = 2 A\)


Since the \(1\Omega\) resistor is in the main line of the circuit, the total current \(I_T\) flows through it.


Current in the \(1\Omega\) resistor is \(2 A\).
Quick Tip: When calculating total current from a source, simplify the circuit step-by-step using series and parallel rules. The total current flows through all components that are in series with the battery.

The RMS velocity is given by \(v_{rms} = \sqrt{\frac{3 R T}{M}}\), where \(M\) is the molar mass.

\(v_{rms} \propto \sqrt{\frac{T}{M}}\).


Hydrogen (\(H_2\)): \(T_H = 300 K\), \(M_H = 2 g/mol\), \(v_H = 1930 m/s\).


Oxygen (\(O_2\)): \(T_O = 1200 K\), \(M_O = 32 g/mol\).


Ratio of velocities: \(\frac{v_O}{v_H} = \sqrt{\frac{T_O}{T_H} \cdot \frac{M_H}{M_O}}\)

\(\frac{v_O}{1930} = \sqrt{\frac{1200}{300} \cdot \frac{2}{32}}\)

\(\frac{v_O}{1930} = \sqrt{4 \cdot \frac{1}{16}} = \sqrt{\frac{1}{4}}\)

\(\frac{v_O}{1930} = \frac{1}{2}\)

\(v_O = \frac{1930}{2} = 965 m/s\)
Quick Tip: RMS velocity increases with temperature and decreases with molar mass. Here, the temperature ratio is \(4\) and the mass ratio is \(1/16\), resulting in a net velocity ratio of \(\sqrt{4/16} = 1/2\).

Faraday's Law of Induction calculates the magnitude of the induced EMF, \(\mathcal{E} = |\frac{d\Phi_B}{dt}|\).


Lenz's law is a consequence of the conservation of energy applied to electromagnetic induction.


It dictates that the induced current flows in a direction that creates a magnetic field opposing the change in flux that caused it.


Therefore, Lenz's law gives the direction of the induced current.
Quick Tip: Lenz's law is crucial for determining the polarity of the induced EMF or the flow direction of the induced current, fulfilling the 'minus sign' in Faraday's Law (\(\mathcal{E} = -d\Phi_B/dt\)).

Initial capacitance \(C_0 = 8 pF\). \(C_0 = \frac{\epsilon_0 A}{d}\).


New distance \(d' = d/2\). New dielectric constant \(\kappa = 6\).


The new capacitance \(C_{new}\) is given by: \(C_{new} = \frac{\kappa \epsilon_0 A}{d'}\)

\(C_{new} = \frac{6 \epsilon_0 A}{(d/2)}\)

\(C_{new} = 12 \frac{\epsilon_0 A}{d}\)


Since \(C_0 = \frac{\epsilon_0 A}{d}\):

\(C_{new} = 12 C_0\)

\(C_{new} = 12 \times 8 pF = 96 pF\)
Quick Tip: Capacitance \(C\) is inversely proportional to plate distance \(d\) and directly proportional to dielectric constant \(\kappa\). Changes multiply: \(C_{new} = C_0 \cdot \left(\frac{d}{d'}\right) \cdot \kappa\).

1. Potential Energy (P.E.) as a function of displacement (\(x\)):

\(P.E. = \frac{1}{2} k x^2\). This relationship is parabolic, centered at \(x=0\).


Graph II correctly represents this parabolic relationship.


2. Displacement (\(x\)) as a function of time (\(t\)):


The definition given is \(x = A \cos \omega t\). This is a simple sinusoidal/cosinusoidal function starting at \(x=A\) at \(t=0\).


Graph IV represents a sinusoidal curve starting from the maximum displacement (or minimum displacement if \(\omega t\) were \(\sin \omega t\)). Graph IV is the displacement \(x\) vs time \(t\).


Note: Graph I represents \(P.E. = \frac{1}{2} k A^2 \cos^2(\omega t)\), oscillating at twice the frequency. Since option (A) (I, III) and (D) (I, IV) include I, but the key selects (B) II, IV, it is assumed the question asks for (P.E. vs \(x\)) and (\(x\) vs \(t\)).


Thus, the correct pair is (II, IV).
Quick Tip: In SHM, Energy \(\propto (Displacement)^2\) (parabolic curve), while Displacement \(\propto \cos(\omega t)\) or \(\sin(\omega t)\) (sinusoidal curve). The energy vs time graph must oscillate at twice the frequency of displacement vs time.

Let \(N_0\) be the initial amount of both species A and B. The time elapsed \(t = 16\) days.


The remaining fraction is given by \(\left(\frac{1}{2}\right)^n\), where \(n = \frac{t}{T_{1/2}}\).


For species A, Half-life \(T_{1/2, A} = 4\) days.

Number of half-lives \(n_A = \frac{16}{4} = 4\).

Remaining amount \(N_A = N_0 \left(\frac{1}{2}\right)^4 = \frac{N_0}{16}\).


For species B, Half-life \(T_{1/2, B} = 8\) days.

Number of half-lives \(n_B = \frac{16}{8} = 2\).

Remaining amount \(N_B = N_0 \left(\frac{1}{2}\right)^2 = \frac{N_0}{4}\).


The required ratio \(\frac{N_A}{N_B}\) is:
\(\frac{N_A}{N_B} = \frac{N_0/16}{N_0/4} = \frac{4}{16} = \frac{1}{4}\).

The ratio is \(1:4\).
Quick Tip: When calculating ratios of radioactive decay, the remaining fraction depends only on the number of half-lives passed, \(n\). The ratio simplifies as the initial mass \(N_0\) cancels out.

Given: Length \(L = 7 m\), Mass \(M = 0.035 kg\), Tension \(T = 60.5 N\).


First, calculate the linear mass density \(\mu\):
\(\mu = \frac{M}{L} = \frac{0.035 kg}{7 m} = 0.005 kg/m\).


The speed of a transverse wave on a string is given by the formula \(v = \sqrt{\frac{T}{\mu}}\).


Substitute the values:
\(v = \sqrt{\frac{60.5}{0.005}}\).
\(v = \sqrt{12100}\).
\(v = 110 m/s\).
Quick Tip: Ensure that the units are converted to standard SI units (kg, m, N) before using the wave speed formula \(v=\sqrt{T/\mu}\).

The symbol shown has a flat front and a curved back, which is the representation of an AND gate.


The small circle (or bubble) located at the output terminal signifies a logical NOT operation (inverter).


A circuit that performs an AND operation followed by a NOT operation is a NAND gate.

The output \(Y = \overline{A \cdot B}\).
Quick Tip: The presence of a bubble on the output side of a standard gate symbol indicates a negated output (e.g., AND + NOT = NAND; OR + NOT = NOR).

The current-carrying wire segment PQ lies along the X-axis.

The point of observation \(X = +a\) also lies on the X-axis, meaning the point is axial to the wire.


According to the Biot-Savart Law, the magnetic field \(d\vec{B}\) due to a current element \(i d\vec{l}\) is proportional to \(d\vec{l} \times \vec{r}\).


If the point of observation lies on the axis of the straight conductor, the vector \(d\vec{l}\) (along the current) is parallel to the position vector \(\vec{r}\).


The cross product \(d\vec{l} \times \vec{r}\) is zero when the vectors are parallel (\(\theta = 0^\circ\)).

Therefore, the magnetic field contribution from every element of the wire at point \(X = +a\) is zero.
Quick Tip: A fundamental concept of magnetic fields: a straight current element produces no magnetic field along its own line of extension (its axis).

Given: Source frequency \(f_s = 170 Hz\), Source speed \(v_s = 17 m/s\), Speed of sound \(v = 340 m/s\).


Calculate the original wavelength (\(\lambda_0\)):
\(\lambda_0 = \frac{v}{f_s} = \frac{340}{170} = 2.0 m\).


Since the source is approaching the observer, the effective speed of sound waves relative to the source is \(v - v_s\).

Calculate the apparent wavelength (\(\lambda'\)):
\(\lambda' = \frac{v - v_s}{f_s} = \frac{340 - 17}{170}\).
\(\lambda' = \frac{323}{170} = 1.9 m\).


The apparent change in wavelength (\(\Delta \lambda\)) is the difference:
\(\Delta \lambda = |\lambda_0 - \lambda'| = |2.0 m - 1.9 m|\).
\(\Delta \lambda = 0.1 m\).
Quick Tip: For wavelength change calculations in the Doppler effect, use the relative speed of the source with respect to the medium, divided by the actual source frequency.

Reaction 1: Chromyl Chloride Test
\(4NaCl + K_2Cr_2O_7 + 6H_2SO_4 \rightarrow 2CrO_2Cl_2 (A) + 2KHSO_4 + 4NaHSO_4 + 3H_2O\).

Compound (A) is \(CrO_2Cl_2\). Atoms: \(1(Cr) + 2(O) + 2(Cl) = 5\).


Reaction 2: Formation of Chromate
\(CrO_2Cl_2 (A) + 4NaOH \rightarrow Na_2CrO_4 (B) + 2NaCl + 2H_2O\).

Compound (B) is \(Na_2CrO_4\). Atoms: \(2(Na) + 1(Cr) + 4(O) = 7\).


Reaction 3: Peroxide Test (Blue solution)
\(Na_2CrO_4 (B) + H_2SO_4 + H_2O_2 \rightarrow CrO_5 (C) + Na_2SO_4 + H_2O\).

Compound (C) is \(CrO_5\). Atoms: \(1(Cr) + 5(O) = 6\).


Total number of atoms \(= (Atoms in A) + (Atoms in B) + (Atoms in C)\).

Total atoms \(= 5 + 7 + 6 = 18\).
Quick Tip: The key intermediate (A) in the chromyl chloride test is volatile \(CrO_2Cl_2\). (C) is the blue peroxide product \(CrO_5\), which contains a peroxide linkage (\(Cr(O_2)_2O\)).

Xenon hexafluoride (\(XeF_6\)) undergoes sequential partial hydrolysis:


First partial hydrolysis (\(X\)) replaces one \(F_2\) with one \(O\):
\(XeF_6 + H_2O \rightarrow XeOF_4 (X) + 2HF\).


Second partial hydrolysis (\(Y\)) replaces two \(F_2\) with two \(O\):
\(XeF_6 + 2H_2O \rightarrow XeO_2F_2 (Y) + 4HF\).


Determine the oxidation state of \(Xe\) in both compounds:

In \(XeOF_4\): \(Xe + (-2) + 4(-1) = 0 \Rightarrow Xe = +6\).

In \(XeO_2F_2\): \(Xe + 2(-2) + 2(-1) = 0 \Rightarrow Xe = +6\).


The compounds are \(XeOF_4(+6)\) and \(XeO_2F_2(+6)\).
Quick Tip: Partial hydrolysis of noble gas fluorides leads to oxo-fluorides where the noble gas typically retains its original oxidation state by replacing two halogens with one oxygen atom.

Given: \(M = 75 g/mol\), \(a = 5 \AA = 5 \times 10^{-8} cm\), \(\rho = 2 g/cm^3\), \(N_A = 6 \times 10^{23}\).


Step 1: Determine the number of atoms per unit cell (\(Z\)).

Use the density formula: \(\rho = \frac{Z \cdot M}{a^3 \cdot N_A}\).
\(Z = \frac{\rho \cdot a^3 \cdot N_A}{M}\).
\(Z = \frac{2 \cdot (5 \times 10^{-8})^3 \cdot (6 \times 10^{23})}{75}\).
\(Z = \frac{2 \cdot (125 \times 10^{-24}) \cdot (6 \times 10^{23})}{75} = \frac{1500 \times 10^{-1}}{75} = 2\).


Since \(Z=2\), the lattice structure is Body-Centered Cubic (BCC).


Step 2: Calculate the atomic radius (\(r\)) using the BCC relationship.

For BCC, \(4r = a\sqrt{3}\).
\(r = \frac{a\sqrt{3}}{4}\).
\(r = \frac{5 \AA \times 1.732}{4}\).
\(r = 2.165 \AA\).


Step 3: Convert the radius to picometers (pm).
\(r = 2.165 \times 100 pm = 216.5 pm\).

Rounding off gives \(217 pm\).
Quick Tip: A \(Z\) value of 2 immediately identifies the lattice as BCC, requiring the specific relationship \(4r = a\sqrt{3}\) to find the radius. Ensure unit consistency, especially \(\AA\) to \(pm\) conversion (\(1 \AA = 100 pm\)).

I. Zero order reactions have \(Rate = k\). The rate is independent of concentration. Statement I is INCORRECT.


II. Arrhenius equation: \(k = A e^{-E_a/RT}\). If \(E_a = 0\), \(k = A e^0 = A\). Statement II is CORRECT.


III. If \(E_a = \infty\), \(k = A e^{-\infty} = 0\). Statement III is INCORRECT.


IV. The relationship between \(\ln k\) and \(T\) is non-linear (\(\ln k = \ln A - E_a/(RT)\)). Statement IV is INCORRECT.


V. Plotting \(\ln k\) vs \(1/T\) from the Arrhenius equation gives a straight line with slope \(-E_a/R\). Statement V is CORRECT.


The correct statements are II and V.
Quick Tip: The Arrhenius plot (\(\ln k\) vs \(1/T\)) is a standard method to experimentally determine the activation energy (\(E_a\)) from the slope.

The reaction is \(Cu^{2+} + 2e^- \rightarrow Cu\). Thus, \(n=2\) electrons are required per mole of \(Cu\).

Use Molar Mass of \(Cu \approx 63.5 g/mol\) and Faraday constant \(F = 96500 C/mol\).


Step 1: Calculate the moles of \(Cu\) deposited.
\(Moles = \frac{m}{M} = \frac{0.634 g}{63.5 g/mol} = 0.01 mol\).


Step 2: Calculate the total charge (\(Q\)) required.
\(Q = Moles \times n \times F\).
\(Q = 0.01 \times 2 \times 96500 C\).
\(Q = 2 \times 965 C\).
\(Q = 1930 Coulombs\).
Quick Tip: Recall Faraday's laws: the amount of substance deposited is proportional to the charge passed (\(Q=It\)). For one mole of \(Cu^{2+}\), \(2F\) (or \(2 \times 96500 C\)) is required.

The diagram shows a Newman projection of ethane in a skew conformation.


The dihedral angles (\(\theta\)) in an ethane conformation are \(\theta\), \(\theta + 60^\circ\), and \(\theta + 120^\circ\).

The diagram indicates that the relative rotation (\(\theta\)) between the closest hydrogen atoms on the front and back carbons is \(29^\circ\).


Thus, the three dihedral angles are:

1. \(29^\circ\) (smallest angle)

2. \(29^\circ + 60^\circ = 89^\circ\) (intermediate angle)

3. \(29^\circ + 120^\circ = 149^\circ\) (largest angle)


The \(H'\) and \(H''\) hydrogens are generally represented as the most widely separated pair, corresponding to the largest dihedral angle.

The dihedral angle \(H' - C - C - H''\) is \(149^\circ\).
Quick Tip: In a skew conformation of ethane, if the smallest dihedral angle is \(\theta\), the other two unique dihedral angles are \(\theta + 60^\circ\) and \(\theta + 120^\circ\).

The reactant is \(Hex-3-ynal\): \(CH_3CH_2C\equivCCH_2CHO\). (6 carbons).


(i) Reduction of aldehyde: \(NaBH_4\) reduces the aldehyde group (\(CHO\)) to a primary alcohol (\(CH_2OH\)).

Product: \(CH_3CH_2C\equivCCH_2CH_2OH\) (Hex-3-yn-1-ol).


(ii) Halogenation: \(PBr_3\) replaces the \(-OH\) group with \(-Br\).

Product: \(CH_3CH_2C\equivCCH_2CH_2Br\).


(iii) Grignard Formation: \(Mg/ether\) converts the alkyl bromide to a Grignard reagent (\(RMgBr\)).

Product: \(CH_3CH_2C\equivCCH_2CH_2MgBr\).


(iv) Carbonation and Hydrolysis: Grignard reacts with \(CO_2\) followed by acid hydrolysis to form a carboxylic acid, increasing the chain length by one carbon.
\(RMgBr + CO_2 \rightarrow RCOOMgBr \xrightarrow{H_3O^+} RCOOH\).

Product: \(CH_3CH_2C\equivCCH_2CH_2COOH\). (Hept-4-ynoic acid).


This structure matches option (C): \(C-C-C\equivC-C-C-COOH\).
Quick Tip: \(NaBH_4\) is a milder reducing agent than \(LiAlH_4\); it selectively reduces aldehydes and ketones but leaves triple bonds and esters intact. The reaction of \(RMgX\) with \(CO_2\) is a chain extension method for synthesizing carboxylic acids.

Starting material: \(CH_3CH_2OH\) (Ethanol).


Step 1: \(CH_3CH_2OH \xrightarrow{P+I_2} CH_3CH_2I\) (A, Ethyl iodide).


Step 2: \(A \xrightarrow{Mg/ether} CH_3CH_2MgI\) (B, Ethyl Grignard reagent).


Step 3: \(B + HCHO\) (Formaldehyde): Grignard reagents react with formaldehyde to produce a primary alcohol upon hydrolysis, adding one carbon atom.
\(CH_3CH_2MgI + HCHO \rightarrow [CH_3CH_2CH_2OMgI]\) (C, Intermediate alkoxide).


Step 4: Hydrolysis of C.
\(C \xrightarrow{H_2O/H^+} CH_3CH_2CH_2OH\) (D).


Compound D is \(n\)-propyl alcohol (or propan-1-ol).
Quick Tip: Grignard reactions are essential for C-C bond formation. Reaction with Formaldehyde (\(HCHO\)) always yields a primary alcohol with one more carbon than the alkyl group of the Grignard reagent.

Aniline (\(C_6H_5NH_2\)) is obtained by the complete reduction of nitrobenzene (\(C_6H_5NO_2\)).


(A) \(Zn/KOH\) (strong basic reduction) yields hydrazobenzene (\(C_6H_5NH-NHC_6H_5\)).

(B) \(Zn/NH_4Cl\) (neutral reduction) yields phenylhydroxylamine (\(C_6H_5NHOH\)).

(C) \(LiAlH_4\) is a strong reducing agent that reduces \(C_6H_5NO_2\) to \(C_6H_5NH_2\) (Aniline).

(D) \(Zn/HCl\) (strong acidic reduction, similar to \(Sn/HCl\)) completely reduces the nitro group to the amine group, yielding aniline (\(C_6H_5NH_2\)).


Both (C) and (D) produce aniline. However, the metal/acid reduction (\(Zn/HCl\) or \(Sn/HCl\)) is the classic, large-scale method for this conversion and is universally taught as yielding aniline. Hence, (D) is the standard and often intended answer among the options for complete reduction in acidic medium.
Quick Tip: Reduction of nitrobenzene to aniline is highly dependent on the medium: Acidic medium (\(Sn/HCl\), \(Zn/HCl\)) gives aniline. Neutral medium (\(Zn/NH_4Cl\)) gives phenylhydroxylamine. Basic medium (\(Zn/KOH\)) gives azo or hydrazobenzene derivatives.

Protein structure is hierarchically classified:


(C) Primary structure refers to the linear sequence of amino acid residues.

(D) Secondary structure refers to local, repeating structures like the \(\alpha\)-helix and \(\beta\)-sheet, stabilized by hydrogen bonds between the amide and carbonyl groups of the peptide backbone. Statement (D) correctly describes this.

(B) Tertiary structure refers to the overall three-dimensional folding pattern, including interactions between distant residues.

(A) Denatured proteins have lost their secondary/tertiary/quaternary structure.
Quick Tip: Secondary structures (\(\alpha\)-helix, \(\beta\)-sheet) are defined exclusively by hydrogen bonding within the polypeptide backbone, excluding interactions involving side chains (R groups).

We compare the charge/mass ratio (\(e/m\)) for the particles (relative charge \(Z\), relative mass \(A\)):


1. Neutron (n): \(Z=0, A=1\). \(e/m = 0/1 = 0\).

2. Alpha particle (\(\alpha\) or \(He^{2+}\)): \(Z=2, A=4\). \(e/m = 2/4 = 0.5\).

3. Proton (p or \(H^{+}\)): \(Z=1, A=1\). \(e/m = 1/1 = 1\).

4. Electron (e): \(Z=1, A \approx 1/1836\). \(e/m \approx 1836\).


Increasing order of \(e/m\) values:
\(n (0) < \alpha (0.5) < p (1) < e (1836)\).

The order is \(n, \alpha, p, e\).
Quick Tip: The electron has the largest \(e/m\) ratio due to its negligible mass. Neutrons always have \(e/m=0\). Alpha particles (\(He^{2+}\)) have a significantly lower \(e/m\) than protons (\(H^+\)) due to their large mass (A=4) relative to their charge (Z=2).

Color in transition metal ions is generally due to \(d-d\) electronic transitions, which occur only when the ion has a partially filled \(d\) subshell (\(d^1\) to \(d^9\) configuration).
\(d^0\) and \(d^{10}\) ions are typically colorless (unless charge transfer occurs).


(A) \(Sc^{3+}\) (\(3d^0\), colorless), \(Ti^{3+}\) (\(3d^1\), colored).

(B) \(Sc^{3+}\) (\(3d^0\), colorless), \(Co^{2+}\) (\(3d^7\), colored).

(C) \(Ni^{2+}\) (\(3d^8\), colored), \(Cu^{+}\) (\(3d^{10}\), colorless).

(D) \(Ni^{2+}\) (\(3d^8\), colored, green/blue), \(Ti^{3+}\) (\(3d^1\), colored, purple).


In option (D), both ions have partially filled \(d\) orbitals and are therefore colored in aqueous solution.



\begin{quicktipbox
To determine if a transition metal ion is colored, calculate its \(d\) electron configuration. \(d^0\) (like \(Sc^{3+}\)) and \(d^{10}\) (like \(Zn^{2+}\), \(Cu^{+}\)) ions are usually colorless because they lack the necessary electronic configuration for \(d-d\) transitions.
\end{quicktipbox Quick Tip: To determine if a transition metal ion is colored, calculate its \(d\) electron configuration. \(d^0\) (like \(Sc^{3+}\)) and \(d^{10}\) (like \(Zn^{2+}\), \(Cu^{+}\)) ions are usually colorless because they lack the necessary electronic configuration for \(d-d\) transitions.

The complex is \([Pt(Cl)(NO_2)(NO_3)(SCN)]^{2-}\). The coordination number is 4, and the geometry is square planar.

The ligands are \(L_1=Cl, L_2=NO_2, L_3=NO_3, L_4=SCN\).


1. Geometric Isomerism: For a square planar complex with four different monodentate ligands (\(MABCD\)), there are 3 possible geometric isomers.

The possible trans pairs are: \((A, B, C, D) \rightarrow 3\) isomers.


2. Linkage Isomerism: This arises from ambidentate ligands.

Ligand \(NO_2\) can coordinate via \(N\) (nitro, \(-NO_2\)) or \(O\) (nitrito, \(-ONO\)). (2 options)

Ligand \(SCN\) can coordinate via \(S\) (thiocyanato, \(-SCN\)) or \(N\) (isothiocyanato, \(-NCS\)). (2 options)

Ligand \(Cl\) and \(NO_3\) are standard monodentate.

Total possible linkage combinations \(= 2 \times 2 = 4\).


Total number of isomers = (Number of Geometric isomers) \(\times\) (Number of Linkage combinations).

Total isomers \(= 3 \times 4 = 12\).
Quick Tip: When calculating total isomers for complexes containing ambidentate ligands, multiply the number of possible geometric isomers by the number of unique linkage combinations.

The reaction is exothermic (\(\Delta H < 0\)) and involves a decrease in moles of gas (\(\Delta n_g = 2 - 3 = -1\)). \(K_c\) is very large.


(B) Increasing pressure shifts equilibrium towards the side with fewer gas moles (products). Forward shift is favored. This statement is CORRECT (Le Chatelier's principle).


(C) Since the reaction is exothermic (\(\Delta H < 0\)), increasing temperature shifts the equilibrium backward. This reduces the product concentration, hence \(K_c\) decreases. This statement is CORRECT.


(D) Addition of an inert gas at constant volume increases total pressure but does not change the partial pressures (or concentrations) of reactants/products. Equilibrium remains unaffected. This statement is CORRECT.


(A) While a large \(K_c\) implies the reaction goes nearly to completion (thermodynamically favored), the statement that "no catalyst is required" is based on kinetics. If the activation energy (\(E_a\)) is high, the reaction rate might be negligible without a catalyst, regardless of the large \(K_c\). Hence, the kinetic conclusion is often INCORRECT.
Quick Tip: Thermodynamics (\(K_c\)) dictates feasibility, while kinetics (catalysis, \(E_a\)) dictates speed. A highly feasible reaction might still be extremely slow without a catalyst.

The general relationship between the half-life (\(t_{1/2}\)) and the initial concentration (\([A]_0\)) for an \(n\)-th order reaction (where \(n \ne 1\)) is:
\(t_{1/2} \propto \frac{1}{[A]_0^{n-1}}\).


The problem states that the half-life is inversely proportional to the square of the initial concentration:
\(t_{1/2} \propto \frac{1}{[A]_0^2}\).


Comparing the exponents of \([A]_0\):
\(n - 1 = 2\).
\(n = 3\).

The reaction is third order.
Quick Tip: Memorize the general half-life dependence on concentration: \(t_{1/2} \propto 1/[A]_0^{n-1}\). For \(n=1\) (First Order), \(t_{1/2}\) is independent of concentration (exponent \(n-1=0\)).

Calculate the standard cell potential (\(E^\circ_{cell}\)).

The electrode with the higher reduction potential acts as the cathode.
\(E^\circ_A = 1.33 V\), \(E^\circ_B = 0.77 V\). Since \(E^\circ_A > E^\circ_B\), Electrode A is the Cathode and Electrode B is the Anode.
\(E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 1.33 V - 0.77 V = 0.56 V\). Statement (A) is true.


Determine polarity and current flow.

In a galvanic cell, the cathode is the positive terminal. Since A is the cathode, A has positive polarity. Statement (C) is true.

Conventional current flows from the positive terminal to the negative terminal in the external circuit. Thus, current flows from A to B. Statement (B) is true.


Since all statements (A), (B), and (C) are correct, the false statement is "None of these".
Quick Tip: Remember: In a galvanic cell, Electron flow is Anode \(\to\) Cathode, while Conventional Current flow is Cathode \(\to\) Anode. Higher \(E^\circ_{red}\) implies reduction (Cathode).

Keto-enol tautomerism requires the presence of at least one \(\alpha\)-hydrogen atom adjacent to the carbonyl group (\(C=O\)).


(A) Dibenzoylmethane (\(Ph-CO-CH_2-CO-Ph\)) has a methylene group flanked by two carbonyls. These \(\alpha\)-hydrogens are highly acidic, and it exhibits significant tautomerism.

(B) Acetophenone (\(Ph-CO-CH_3\)) has a methyl group adjacent to the carbonyl. It has 3 \(\alpha\)-hydrogens and exhibits tautomerism.

(C) Benzaldehyde (\(Ph-CHO\)) has no \(\alpha\)-carbon with hydrogen atoms (the carbonyl C is attached to the benzene ring). It does not exhibit keto-enol tautomerism.


Since both (A) and (B) show tautomerism, the correct answer is (D).
Quick Tip: Identify \(\alpha\)-hydrogens (hydrogens on a carbon adjacent to the \(C=O\) group). If present, keto-enol tautomerism is possible. 1,3-dicarbonyl compounds (like A) have very stable enol forms due to conjugation and intramolecular H-bonding.

Step 1: Oxidation of Ethylbenzene (\(Ph-CH_2CH_3\)) with \(KMnO_4/KOH\).

Strong oxidation converts the entire alkyl side chain to a carboxyl group, forming Benzoic Acid (after acidification).
\(B = Ph-COOH\).


Step 2: Bromination of Benzoic Acid with \(Br_2/FeCl_3\).

The \(-COOH\) group is an electron-withdrawing group and is meta-directing.

The incoming bromine enters the meta position relative to the carboxyl group.
\(C = m-Bromobenzoic acid\) (\(m-Br-C_6H_4-COOH\)).


Step 3: Esterification with \(C_2H_5OH/H^+\).

The carboxylic acid reacts with ethanol to form an ethyl ester.
\(D = Ethyl m-bromobenzoate\) (\(m-Br-C_6H_4-COOC_2H_5\)).


Looking at the options, structure (D) depicts a benzene ring with an ester group (\(-COOC_2H_5\)) and a bromine atom (\(-Br\)) in the meta positions (1,3-relationship).
Quick Tip: Side-chain oxidation of alkylbenzenes by \(KMnO_4\) always yields Benzoic Acid, regardless of chain length (provided the benzylic carbon has at least one H). Remember directing effects: \(-COOH\) is meta-directing.

Reactant: Propanenitrile (\(CH_3CH_2CN\)).


Step 1: Hydrolysis in acidic medium (\(H^+/H_2O\)).

The nitrile group (\(-CN\)) hydrolyzes to a carboxylic acid group (\(-COOH\)).
\(A = CH_3CH_2COOH\) (Propanoic acid).


Step 2: Reaction with Ammonia (\(NH_3\)) and heat (\(\Delta\)).

Carboxylic acid forms an ammonium salt, which upon heating loses water to form an amide.
\(A \xrightarrow{NH_3} CH_3CH_2COO^-NH_4^+ \xrightarrow{\Delta} CH_3CH_2CONH_2\) (Propanamide).
\(B = CH_3CH_2CONH_2\).


Step 3: Reaction with \(NaOBr\) (Hoffmann Bromamide Degradation).

This reaction converts a primary amide to a primary amine with one less carbon atom.
\(CH_3CH_2CONH_2 \xrightarrow{NaOBr} CH_3CH_2NH_2 + Na_2CO_3 + NaBr\).
\(C = CH_3CH_2NH_2\) (Ethylamine / Ethanamine).
Quick Tip: Hoffmann Bromamide Degradation is a "step-down" reaction. It removes the carbonyl carbon (\(C=O\)) from the amide, connecting the alkyl group directly to the amine nitrogen: \(R-CONH_2 \to R-NH_2\).

Step 1: \(CH_3COOH + SOCl_2 \to CH_3COCl\) (Acetyl chloride). \(A = CH_3COCl\).


Step 2: Friedel-Crafts Acylation with Benzene.
\(C_6H_6 + CH_3COCl \xrightarrow{AlCl_3} C_6H_5COCH_3\) (Acetophenone). \(B = Ph-CO-CH_3\).


Step 3: Nucleophilic addition of \(HCN\).

Acetophenone reacts with \(HCN\) to form a cyanohydrin.
\(Ph-CO-CH_3 + HCN \to Ph-C(OH)(CN)-CH_3\). \(C = Acetophenone cyanohydrin\).


Step 4: Hydrolysis (\(HOH\)).

The nitrile group (\(-CN\)) is hydrolyzed to a carboxylic acid group (\(-COOH\)).
\(Ph-C(OH)(CN)-CH_3 \xrightarrow{H_2O/H^+} Ph-C(OH)(COOH)-CH_3\).

Product \(D\) is 2-hydroxy-2-phenylpropanoic acid (Atrolactic acid).


This corresponds to structure (D), which shows a central carbon bonded to a benzene ring, a methyl group, a hydroxyl group, and a carboxyl group.
Quick Tip: Cyanohydrin formation followed by hydrolysis is a standard method to synthesize \(\alpha\)-hydroxy acids. The carbon chain length effectively increases by one (from the CN group).

Fructose is a ketohexose with the structural formula: \(HOCH_2-CO-CH(OH)-CH(OH)-CH(OH)-CH_2OH\).

The chiral carbons (asymmetric centers) are at positions C3, C4, and C5.

Number of chiral centers, \(n = 3\).

The number of optical isomers is given by \(2^n\) (since the molecule is unsymmetrical).

Number of isomers \(= 2^3 = 8\).
Quick Tip: Be careful to count chiral centers correctly. Glucose has 4 chiral centers (\(2^4=16\)), while Fructose has 3 (\(2^3=8\)) because the C2 carbonyl carbon is not chiral.

According to Heisenberg's Uncertainty Principle, \(\Delta x \cdot \Delta p \ge \frac{h}{4\pi}\).

The minimum uncertainty in momentum depends only on the uncertainty in position (\(\Delta x\)) and Planck's constant. It does not depend on the mass of the particle.
\(\Delta p_{min} = \frac{h}{4\pi \Delta x}\).


Given that \(\Delta x\) is the same (\(1.0 nm\)) for both the electron and the helium atom.

Therefore, the minimum uncertainty in momentum (\(\Delta p\)) must also be the same for both.

Given \(\Delta p_{electron} = 5.0 \times 10^{-26} kg ms^{-1}\).

Thus, \(\Delta p_{helium} = 5.0 \times 10^{-26} kg ms^{-1}\).
Quick Tip: The uncertainty principle relationship \(\Delta x \Delta p \ge constant\) is universal. Mass affects \(\Delta v\) (since \(\Delta p = m \Delta v\)), but it does not affect \(\Delta p\) for a fixed \(\Delta x\).

Step 1: Calculate Gibbs free energy change \(\Delta G^\circ\).
\(\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ\).

Convert \(\Delta H^\circ\) to J/mol: \(-54.07 kJ/mol = -54070 J/mol\).
\(T = 298 K\). \(\Delta S^\circ = 10 J/K mol\).
\(\Delta G^\circ = -54070 - (298 \times 10) = -54070 - 2980 = -57050 J/mol\).


Step 2: Calculate \(\log_{10} K\).

Relationship: \(\Delta G^\circ = -2.303 RT \log_{10} K\).

Substitute the values (using the given factor \(2.303 RT = 5705\)):
\(-57050 = -5705 \times \log_{10} K\).
\(\log_{10} K = \frac{-57050}{-5705}\).
\(\log_{10} K = 10\).
Quick Tip: Always ensure energy units match (Joules vs kilojoules) before subtracting \(T\Delta S\) from \(\Delta H\).

We want to find \(\Delta H_f\) for CO, which corresponds to the reaction:

Target: \(C(s) + \frac{1}{2}O_2(g) \to CO(g)\).


Given reactions:

(1) \(C(s) + O_2(g) \to CO_2(g)\), \(\Delta H_1 = R\).

(2) \(CO(g) + \frac{1}{2}O_2(g) \to CO_2(g)\), \(\Delta H_2 = S\).


Apply Hess's Law: Subtract equation (2) from equation (1).
\((C + O_2) - (CO + \frac{1}{2}O_2) \to (CO_2 - CO_2)\).
\(C + \frac{1}{2}O_2 - CO \to 0\).

Rearranging gives: \(C(s) + \frac{1}{2}O_2(g) \to CO(g)\).


The enthalpy change is \(\Delta H = \Delta H_1 - \Delta H_2\).
\(\Delta H = R - S\).
Quick Tip: Hess's Law allows you to treat thermochemical equations algebraically. Manipulate the given equations (add/subtract/reverse) to match the target reaction.

Markownikoff's law governs the regioselectivity of electrophilic addition to unsymmetrical alkenes. It states that the positive part of the reagent (usually \(H^+\)) attaches to the carbon atom with more hydrogen atoms.


(A) Propene (\(CH_3-CH=CH_2\)) is unsymmetrical. It follows the law.

(B) Vinyl chloride (\(CH_2=CHCl\)) is unsymmetrical. It follows the law (H adds to \(CH_2\)).

(C) But-2-ene (\(CH_3-CH=CH-CH_3\)) is a symmetrical alkene. The two double-bonded carbons are identical (each has one H and one methyl group). Addition of a reagent (like HX) yields the same product regardless of which carbon accepts the proton.


Since the concept of "choosing the carbon with more hydrogens" is not applicable or necessary for symmetrical alkenes to determine the product, But-2-ene is said not to follow the law (in the sense that the law is irrelevant).
Quick Tip: Markownikoff's rule is specifically a rule for regiochemistry in unsymmetrical alkenes. For symmetrical alkenes, there is only one possible regioisomer, rendering the rule moot.

Rolle's Theorem states that there exists at least one \(c \in (0, \pi)\) such that \(f'(c) = 0\).


First, differentiate \(f(x) = e^x \sin x\).

Using the product rule: \(f'(x) = \frac{d}{dx}(e^x) \sin x + e^x \frac{d}{dx}(\sin x)\).
\(f'(x) = e^x \sin x + e^x \cos x = e^x (\sin x + \cos x)\).


Set \(f'(c) = 0\):
\(e^c (\sin c + \cos c) = 0\).


Since \(e^c \neq 0\) for any real \(c\), we must have:
\(\sin c + \cos c = 0\).
\(\sin c = -\cos c\).
\(\tan c = -1\).


In the interval \((0, \pi)\), the tangent function is negative in the second quadrant.

The reference angle for \(\tan \theta = 1\) is \(\frac{\pi}{4}\).

Therefore, \(c = \pi - \frac{\pi}{4} = \frac{3\pi}{4}\).
Quick Tip: For Rolle's Theorem problems, ensure the function satisfies the conditions (continuous on \([a,b]\), differentiable on \((a,b)\), and \(f(a)=f(b)\)) before simply solving \(f'(c)=0\). Here \(f(0)=0\) and \(f(\pi)=0\), so the theorem applies.

Consider the first two equations:

(1) \(2x + 3y = -4\)

(2) \(3x + 4y = -6\)


Solve for \(x\) and \(y\). Multiply (1) by 3 and (2) by 2:
\(6x + 9y = -12\)
\(6x + 8y = -12\)

Subtract the second from the first: \(y = 0\).


Substitute \(y=0\) into (1):
\(2x + 0 = -4 \Rightarrow x = -2\).

The intersection point of the first two lines is \((-2, 0)\).


Check if this point satisfies the third equation \(4x + 5y + 8 = 0\):
\(4(-2) + 5(0) + 8 = -8 + 0 + 8 = 0\).

The point \((-2, 0)\) satisfies all three equations.


Since the three distinct lines intersect at exactly one point, the system is consistent with a unique solution.
Quick Tip: For a system of three linear equations in two variables to be consistent, the third line must pass through the intersection point of the first two. This is the condition for concurrency.

Convert the equations to symmetric form \(\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}\).


Line 1: \(x = y+2 \Rightarrow y = x-2\). And \(x = 6(z-1) \Rightarrow z-1 = x/6\).

Form: \(\frac{x}{1} = \frac{y+2}{1} = \frac{z-1}{1/6} \Rightarrow \frac{x-0}{6} = \frac{y-(-2)}{6} = \frac{z-1}{1}\).

Point \(A(0, -2, 1)\), Vector \(\vec{b_1} = (6, 6, 1)\).


Line 2: \(x+1 = 2y \Rightarrow y = \frac{x+1}{2}\). And \(x+1 = -12z \Rightarrow z = \frac{x+1}{-12}\).

Form: \(\frac{x+1}{1} = \frac{y}{1/2} = \frac{z}{-1/12}\). Multiply denominators by 12:
\(\frac{x-(-1)}{12} = \frac{y-0}{6} = \frac{z-0}{-1}\).

Point \(B(-1, 0, 0)\), Vector \(\vec{b_2} = (12, 6, -1)\).


Shortest Distance \(d = \left| \frac{(\vec{B}-\vec{A}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right|\).
\(\vec{AB} = (-1-0, 0-(-2), 0-1) = (-1, 2, -1)\).


Cross product \(\vec{n} = \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
6 & 6 & 1
12 & 6 & -1 \end{vmatrix}\).
\(\vec{n} = \hat{i}(-6-6) - \hat{j}(-6-12) + \hat{k}(36-72) = -12\hat{i} + 18\hat{j} - 36\hat{k}\).

Magnitude \(|\vec{n}| = \sqrt{(-12)^2 + 18^2 + (-36)^2} = \sqrt{144 + 324 + 1296} = \sqrt{1764} = 42\).


Dot product \(\vec{AB} \cdot \vec{n} = (-1)(-12) + (2)(18) + (-1)(-36) = 12 + 36 + 36 = 84\).
\(d = \frac{|84|}{42} = 2\).
Quick Tip: Standardize the line equations carefully to identify the direction ratios. The formula for shortest distance between skew lines is the projection of the vector joining the reference points onto the normal vector perpendicular to both lines.

Curve: \(y^2 = x(2-x)^2 = x(x-2)^2\).

Differentiate implicitly: \(2y y' = 1(2-x)^2 + x \cdot 2(2-x)(-1)\).

At \(P(1, 1)\): \(2(1) y' = (2-1)^2 - 2(1)(2-1) = 1 - 2 = -1\).
\(y' = -1/2\).


Equation of tangent: \(y - 1 = -\frac{1}{2}(x - 1) \Rightarrow 2y - 2 = -x + 1 \Rightarrow x = 3 - 2y\).


Substitute \(x = 3 - 2y\) into the curve equation \(y^2 = x(2-x)^2\):

Note that \(2-x = 2 - (3-2y) = 2y - 1\).
\(y^2 = (3-2y)(2y-1)^2\).
\(y^2 = (3-2y)(4y^2 - 4y + 1)\).
\(y^2 = 12y^2 - 12y + 3 - 8y^3 + 8y^2 - 2y\).
\(8y^3 - 19y^2 + 14y - 3 = 0\).


Since \(y=1\) is a point of tangency, \((y-1)^2\) is a factor.

Divide by \((y-1)^2 = y^2 - 2y + 1\):
\((y-1)^2 (8y - 3) = 0\).

The roots are \(y=1\) (repeated) and \(y = 3/8\).


For \(Q\), \(y = 3/8\).
\(x = 3 - 2(3/8) = 3 - 3/4 = 9/4\).
\(Q = (9/4, 3/8)\).
Quick Tip: When a tangent at a point \(t_1\) meets the curve again at \(t_2\), solving the simultaneous equations results in a polynomial where the root corresponding to the tangency point has a multiplicity of at least 2.

For \(x=0\):
\(f(0) = 0 + \frac{0}{1} + \frac{0}{1} + ... = 0\).


For \(x \neq 0\):

The series is a geometric progression with first term \(a = x\) and common ratio \(r = \frac{1}{1+x}\).

The sum exists if \(|r| < 1\), i.e., \(|\frac{1}{1+x}| < 1\), which is true for \(x > 0\) or \(x < -2\).

Sum \(S = \frac{a}{1-r} = \frac{x}{1 - \frac{1}{1+x}} = \frac{x}{\frac{1+x-1}{1+x}} = \frac{x(1+x)}{x}\).

For \(x \neq 0\), \(f(x) = 1+x\).


Limit as \(x \to 0\):
\(\lim_{x \to 0} f(x) = \lim_{x \to 0} (1+x) = 1\).


Since \(f(0) = 0\) but \(\lim_{x \to 0} f(x) = 1\), the limit exists but is not equal to the function value.

Therefore, \(f(x)\) is discontinuous at \(x=0\).
Quick Tip: Always analyze the domain of convergence for an infinite series. A function defined by a series may behave differently at specific points (like \(x=0\)) compared to the simplified limit expression.

The standard equation of a circle is \((x - h)^2 + (y - k)^2 = r^2\).

Comparing the given equation \((x + 5)^2 + (y - 3)^2 = 36\) with the standard form:
\(r^2 = 36\).
\(r = \sqrt{36}\).
\(r = 6\).
Quick Tip: The number on the right side of the standard circle equation represents the square of the radius (\(r^2\)), not the radius itself. Don't forget to take the square root.

Calculate the magnitude of vector \(\vec{c}\):
\(\vec{c} = -\hat{i} + 0\hat{j} + 2\hat{k}\).
\(|\vec{c}| = \sqrt{(-1)^2 + 0^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}\).


Multiply scalar \(|\vec{c}|\) by vector \(\vec{a}\):
\(|\vec{c}|\vec{a} = \sqrt{5} (2\hat{i} - 2\hat{j} + \hat{k})\).
\(|\vec{c}|\vec{a} = 2\sqrt{5}\hat{i} - 2\sqrt{5}\hat{j} + \sqrt{5}\hat{k}\).
Quick Tip: This is a straightforward vector scalar multiplication problem. Calculate the modulus of \(\vec{c}\) and distribute it across the components of \(\vec{a}\).

Let the given vertex be \(V(-4, 5)\).

Check if \(V\) lies on the diagonal \(7x - y + 8 = 0\):
\(7(-4) - 5 + 8 = -28 - 5 + 8 = -25 \neq 0\).

The vertex does not lie on the given diagonal, so the given line is diagonal \(d_1\), and the vertex lies on the second diagonal \(d_2\).


In a square, the diagonals are perpendicular to each other.

Slope of \(d_1\): \(y = 7x + 8 \Rightarrow m_1 = 7\).

Slope of \(d_2\): \(m_1 \cdot m_2 = -1 \Rightarrow m_2 = -\frac{1}{7}\).


Equation of \(d_2\) passing through \((-4, 5)\) with slope \(-\frac{1}{7}\):
\(y - 5 = -\frac{1}{7}(x - (-4))\).
\(7(y - 5) = -1(x + 4)\).
\(7y - 35 = -x - 4\).
\(x + 7y = 31\).
Quick Tip: The diagonals of a square (and a rhombus) always intersect at right angles. If a vertex is not on the given diagonal equation, it must be on the required perpendicular diagonal.

The conditional statement "if p then q" (\(p \Rightarrow q\)) is logically equivalent to "not p or q".

Symbolically: \(p \Rightarrow q \equiv \sim p \lor q\).
Quick Tip: The logical equivalence \(p \rightarrow q \equiv \neg p \lor q\) is essential for simplifying logical expressions and constructing truth tables.

Let \(I = \int \frac{x^{1/2}}{\sqrt{1-x^3}}dx\).

Notice that the derivative of \(x^{3/2}\) involves \(x^{1/2}\).

Let \(u = x^{3/2}\).

Then \(du = \frac{3}{2}x^{1/2} dx \Rightarrow x^{1/2} dx = \frac{2}{3} du\).

Also, \(x^3 = (x^{3/2})^2 = u^2\).


Substitute into the integral:
\(I = \int \frac{\frac{2}{3} du}{\sqrt{1-u^2}} = \frac{2}{3} \int \frac{du}{\sqrt{1-u^2}}\).
\(I = \frac{2}{3} \sin^{-1}(u) + C\).

Substitute \(u\) back:
\(I = \frac{2}{3} \sin^{-1}(x^{3/2}) + C\).


Comparing with the given form \(\frac{2}{3}g(f(x)) + C\):
\(g(f(x)) = \sin^{-1}(x^{3/2})\).

This is satisfied if \(f(x) = x^{3/2}\) and \(g(t) = \sin^{-1}(t)\).
Quick Tip: For integrals involving terms like \(\sqrt{1-x^n}\), look for a substitution \(u = x^{n/2}\) to transform the integral into the standard arcsine form \(\int \frac{du}{\sqrt{1-u^2}}\).

(A) The number of human beings is finite (though large).

(B) The number of water drops (or molecules) in a glass is finite (can be counted via moles).

(C) The number of trees in a forest is finite.

(D) The set of prime numbers \(\{2, 3, 5, 7, 11, ...\}\) is infinite. This was proven by Euclid around 300 BC.
Quick Tip: In set theory, physical collections are always considered finite. Mathematical sets like natural numbers, integers, or primes are standard examples of infinite sets.

The function is defined when the expressions under both square roots are non-negative.


1. \(\sqrt{x^2 - 5x + 6}\) requires \(x^2 - 5x + 6 \ge 0\).

Factorizing: \((x-2)(x-3) \ge 0\).

Solution region: \(x \in (-\infty, 2] \cup [3, \infty)\).


2. \(\sqrt{2x + 8 - x^2}\) requires \(2x + 8 - x^2 \ge 0\).

Rearranging: \(x^2 - 2x - 8 \le 0\).

Factorizing: \((x-4)(x+2) \le 0\).

Solution region: \(x \in [-2, 4]\).


3. Find the intersection of both regions to satisfy the domain of the entire function:

Intersection = \(( (-\infty, 2] \cup [3, \infty) ) \cap [-2, 4]\).

Overlap 1: \([-2, 4] \cap (-\infty, 2] = [-2, 2]\).

Overlap 2: \([-2, 4] \cap [3, \infty) = [3, 4]\).

Union of overlaps: \([-2, 2] \cup [3, 4]\).
Quick Tip: To find the domain of a sum of functions, find the domain of each individual component and take the intersection (common part) of all individual domains.

The curve is \(y = \log x\) (assuming natural logarithm, base \(e\)). The coordinate axes are \(x=0\) (y-axis) and \(y=0\) (x-axis).


The function \(y = \log x\) is defined for \(x > 0\). The curve intersects the x-axis at \((1, 0)\) and approaches \(-\infty\) as \(x \to 0^+\).


The area bounded by the curve, the x-axis, and the y-axis (asymptotic boundary) lies in the fourth quadrant.


Area \(A = \left| \int_{0}^{1} \log x \, dx \right|\).


Using integration by parts, \(\int \ln x \, dx = x \ln x - x\).


Apply limits from \(0\) to \(1\):
\(A = \left| [x \ln x - x]_0^1 \right|\).


Evaluate at upper limit (\(x=1\)):
\(1 \cdot \ln 1 - 1 = 0 - 1 = -1\).


Evaluate at lower limit (\(x \to 0\)):
\(\lim_{x \to 0} (x \ln x - x) = 0\). (Using L'Hopital's rule for \(x \ln x\)).


Area \(A = | -1 - 0 | = |-1| = 1\).
Quick Tip: The definite integral \(\int_0^1 \ln x \, dx = -1\) is a standard result. The physical area is the absolute value, which is 1.

First, find the slope of the tangent to curve 1 (\(y = x^2\)) at the point \((1, 1)\).
\(\frac{dy}{dx} = 2x\).
\(m_1 = 2(1) = 2\).


Next, find the slope of the tangent to curve 2 (\(6y = 7 - x^3\)) at the point \((1, 1)\).

Differentiating with respect to \(x\): \(6\frac{dy}{dx} = -3x^2\).
\(\frac{dy}{dx} = \frac{-3x^2}{6} = -\frac{x^2}{2}\).
\(m_2 = -\frac{1^2}{2} = -0.5\).


Calculate the product of the slopes:
\(m_1 \cdot m_2 = 2 \cdot (-0.5) = -1\).


Since the product of the slopes is \(-1\), the tangents are perpendicular.

Therefore, the angle of intersection is \(90^\circ\) or \(\frac{\pi}{2}\).
Quick Tip: If \(m_1 m_2 = -1\), the curves are orthogonal. Checking this condition first is often faster than using the full angle formula \(\tan \theta = |\frac{m_1 - m_2}{1 + m_1 m_2}|\).

Find the derivative \(\frac{dy}{dx}\) to get the slope of the tangent.
\(y = x^2 + 4x - 17 \Rightarrow \frac{dy}{dx} = 2x + 4\).


Evaluate the slope at \(x = \frac{5}{2}\).
\(m = 2(\frac{5}{2}) + 4 = 5 + 4 = 9\).


Let \(\theta\) be the angle the tangent makes with the positive X-axis.

Then \(\tan \theta = m = 9 \Rightarrow \theta = \tan^{-1} 9\).


The angle \(\alpha\) formed by the tangent with the positive Y-axis is complementary to \(\theta\) (since the slope is positive and we consider the acute angle triangle with axes).
\(\alpha = 90^\circ - \theta = \frac{\pi}{2} - \tan^{-1} 9\).
Quick Tip: The angle between a line with inclination \(\theta\) and the Y-axis is \(|\frac{\pi}{2} - \theta|\).

Simplify the term \(\frac{1}{i}\).
\(\frac{1}{i} = \frac{i}{i^2} = \frac{i}{-1} = -i\).


Substitute this back into the expression:

Expression \(= (1 + i)^4 (1 - i)^4\).


Combine the terms under the same power:
\(= [(1 + i)(1 - i)]^4\).


Simplify the product of conjugates:
\((1 + i)(1 - i) = 1^2 - i^2 = 1 - (-1) = 2\).


Calculate the final power:
\(= (2)^4 = 16\).
Quick Tip: Recognize conjugate pairs \((z)(\bar{z}) = |z|^2\). Here \((1+i)(1-i) = 2\).

Check for reflexivity: For any \(x \in A\), \(|x^2 - x^2| = 0 < 16\).

Therefore, R must contain \((1,1), (2,2), (3,3), (4,4), (5,5)\).


Examine the given options:

Option (A) contains \((1,1)\) but misses \((2,2), (3,3)\), etc.

Option (B) contains \((2,2)\) but misses \((1,1), (5,5)\).

Option (C) contains \((3,3)\) but misses \((1,1), (2,2)\).


Since none of the options list all the required reflexive pairs (let alone other valid pairs like \((1,2)\) where \(|1-4|=3 < 16\)), none of them represents the complete relation R.
Quick Tip: A relation defined by a condition like \(|f(x) - f(y)| < k\) (where \(k>0\)) is always reflexive. If the options don't list all diagonal elements \((x,x)\), they are incomplete.

Let \(I = \int \frac{2}{(e^x + e^{-x})^2} dx\).


Rewrite the integrand by factoring out terms or finding a common base.
\(e^x + e^{-x} = \frac{e^{2x} + 1}{e^x}\).

So, the term is \(\frac{2}{(\frac{e^{2x}+1}{e^x})^2} = \frac{2e^{2x}}{(e^{2x}+1)^2}\).


Let \(u = e^{2x} + 1\).

Then \(du = 2e^{2x} dx\).


Substitute into the integral:
\(I = \int \frac{du}{u^2} = -\frac{1}{u} + C\).


Substitute back \(u = e^{2x} + 1\):
\(I = \frac{-1}{e^{2x} + 1} + C\).


Now, check the options to find the equivalent form.

Option (A): \(\frac{-e^{-x}}{e^x + e^{-x}} = \frac{-e^{-x}}{e^x + \frac{1}{e^x}} = \frac{-e^{-x}}{\frac{e^{2x}+1}{e^x}} = \frac{-e^{-x} \cdot e^x}{e^{2x} + 1} = \frac{-1}{e^{2x} + 1}\).
Quick Tip: Simplifying the expression to terms of \(e^{2x}\) usually makes substitution obvious. Don't forget to check the equivalence of the options algebraically.

We know \(\tan^{-1}(0) = 0\) and \(\tan^{-1}(1) = \frac{\pi}{4}\).


For the remaining terms \(\tan^{-1}(2) + \tan^{-1}(3)\):

Since \(x=2 > 0\), \(y=3 > 0\), and \(xy = 6 > 1\), we must use the formula:
\(\tan^{-1}x + \tan^{-1}y = \pi + \tan^{-1}\left(\frac{x+y}{1-xy}\right)\).


Substitute the values:
\(\tan^{-1}(2) + \tan^{-1}(3) = \pi + \tan^{-1}\left(\frac{2+3}{1-6}\right) = \pi + \tan^{-1}\left(\frac{5}{-5}\right) = \pi + \tan^{-1}(-1)\).


We know \(\tan^{-1}(-1) = -\frac{\pi}{4}\).

So, the sum is \(\pi - \frac{\pi}{4} = \frac{3\pi}{4}\).


Total expression value:
\(\frac{\pi}{4} + 0 + \frac{3\pi}{4} = \pi\).
Quick Tip: The condition \(xy > 1\) is critical for the sum of inverse tangents. Without adding \(\pi\), you would get a negative angle sum for positive inputs, which is incorrect.

The growth follows the law \(N(t) = N_0 e^{kt}\).
\(N_0 = 100,000\).


Given that at \(t=2\) hours, the population increases by 10%.
\(N(2) = 1.10 N_0\).
\(1.10 N_0 = N_0 e^{k(2)}\).
\(1.1 = e^{2k}\).

Taking natural log: \(\ln(1.1) = 2k \Rightarrow k = \frac{1}{2} \ln(1.1)\).


We want to find \(t\) when \(N(t) = 2 N_0\) (doubles to 200,000).
\(2 N_0 = N_0 e^{kt}\).
\(2 = e^{kt}\).

Taking natural log: \(\ln(2) = kt\).


Substitute \(k\):
\(\ln(2) = \left[ \frac{1}{2} \ln(1.1) \right] t\).
\(t = \frac{2 \ln(2)}{\ln(1.1)}\).

Since \(\frac{\ln a}{\ln b} = \frac{\log a}{\log b}\), this is equivalent to \(\frac{2 \log 2}{\log(1.1)} = \frac{2 \log 2}{\log(11/10)}\).
Quick Tip: If a quantity multiplies by factor \(X\) in time \(T_x\), the rate constant \(k = \frac{\ln X}{T_x}\). The time to reach factor \(Y\) is \(t = \frac{\ln Y}{k}\).

Let the slopes of the two lines be \(m_1\) and \(m_2\).
\(m_1 = 2 - \sqrt{3}\) and \(m_2 = 2 + \sqrt{3}\).


We know that \(\tan(15^\circ) = 2 - \sqrt{3}\) and \(\tan(75^\circ) = 2 + \sqrt{3}\).

Let the angle of inclination for the first line be \(\theta_1 = 15^\circ\).

Let the angle of inclination for the second line be \(\theta_2 = 75^\circ\).


The angle between the two lines is the absolute difference of their angles of inclination:
\(\theta = |\theta_2 - \theta_1| = |75^\circ - 15^\circ| = 60^\circ\).


Alternatively, using the formula \(\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|\):
\(m_2 - m_1 = (2 + \sqrt{3}) - (2 - \sqrt{3}) = 2\sqrt{3}\).
\(1 + m_1 m_2 = 1 + (2 - \sqrt{3})(2 + \sqrt{3}) = 1 + (4 - 3) = 2\).
\(\tan \theta = \frac{2\sqrt{3}}{2} = \sqrt{3}\).
\(\theta = 60^\circ\).
Quick Tip: Recognizing standard trigonometric values like \(\tan 15^\circ = 2-\sqrt{3}\) and \(\tan 75^\circ = 2+\sqrt{3}\) saves calculation time in coordinate geometry problems.

The direction vector of the line is \(\vec{b} = (1, 2, 2)\).

The normal vector of the plane is \(\vec{n} = (2, -1, \sqrt{\lambda})\).


The angle \(\theta\) between a line and a plane is given by \(\sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}\).


Calculate the dot product:
\(\vec{b} \cdot \vec{n} = 1(2) + 2(-1) + 2(\sqrt{\lambda}) = 2 - 2 + 2\sqrt{\lambda} = 2\sqrt{\lambda}\).


Calculate the magnitudes:
\(|\vec{b}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{9} = 3\).
\(|\vec{n}| = \sqrt{2^2 + (-1)^2 + (\sqrt{\lambda})^2} = \sqrt{5 + \lambda}\).


Substitute into the formula with \(\sin \theta = \frac{1}{3}\):
\(\frac{1}{3} = \frac{2\sqrt{\lambda}}{3 \sqrt{5 + \lambda}}\).
\(1 = \frac{2\sqrt{\lambda}}{\sqrt{5 + \lambda}}\).


Square both sides:
\(1 = \frac{4\lambda}{5 + \lambda}\).
\(5 + \lambda = 4\lambda\).
\(3\lambda = 5 \implies \lambda = \frac{5}{3}\).
Quick Tip: Remember that the angle formula for a line and a plane uses \(\sin \theta\), whereas the formula for two lines or two planes uses \(\cos \theta\).

Step 1: Find the point of intersection of the line and the plane.


Equation of the line: \[ \vec r = (2,-1,2) + \lambda(3,4,2) \]

So a general point on the line is: \[ (x,y,z) = (2+3\lambda,\,-1+4\lambda,\,2+2\lambda) \]

Equation of the plane: \[ x - y + z = 5 \]

Substitute the coordinates of the line into the plane equation: \[ (2+3\lambda) - (-1+4\lambda) + (2+2\lambda) = 5 \]
\[ 2 + 3\lambda + 1 - 4\lambda + 2 + 2\lambda = 5 \]
\[ 5 + \lambda = 5 \Rightarrow \lambda = 0 \]

Hence, the point of intersection is: \[ P(2,-1,2) \]

Step 2: Find the distance between points \[ P(2,-1,2) \quad and \quad Q(-5,-5,-10) \]
\[ PQ = \sqrt{(2+5)^2 + (-1+5)^2 + (2+10)^2} \]
\[ PQ = \sqrt{7^2 + 4^2 + 12^2} \]
\[ PQ = \sqrt{49+16+144} = \sqrt{209} \]

Step 3: Interpretation


The obtained result \(\sqrt{209}\) does not match the given options.
However, the displacement \((3,4,12)\) gives: \[ \sqrt{3^2 + 4^2 + 12^2} = \sqrt{169} = 13 \]

This indicates a minor typographical error in the coordinates of the given point.
The intended and correct option is:
\[ \boxed{13} \] Quick Tip: In exams, if you encounter a result like \(\sqrt{7^2+4^2+12^2}\) and an option is 13 (derived from \(3^2+4^2+12^2\)), check for typos, but select the closest logical intended answer if necessary.

Let \(I = \int_{\log \sqrt{\pi/2}}^{\log \sqrt{\pi}} e^{2x} \sec^2\left(\frac{1}{3} e^{2x}\right) dx\).


Substitute \(u = \frac{1}{3} e^{2x}\).
\(du = \frac{1}{3} \cdot 2e^{2x} dx \implies e^{2x} dx = \frac{3}{2} du\).


Change limits:

Lower limit: \(x = \ln \sqrt{\pi/2} \implies e^{2x} = (\sqrt{\pi/2})^2 = \pi/2\).
\(u = \frac{1}{3}(\frac{\pi}{2}) = \frac{\pi}{6}\).

Upper limit: \(x = \ln \sqrt{\pi} \implies e^{2x} = \pi\).
\(u = \frac{1}{3}(\pi) = \frac{\pi}{3}\).


Integral becomes:
\(I = \int_{\pi/6}^{\pi/3} \sec^2 u \cdot \frac{3}{2} du\).
\(I = \frac{3}{2} [\tan u]_{\pi/6}^{\pi/3}\).
\(I = \frac{3}{2} (\tan \frac{\pi}{3} - \tan \frac{\pi}{6})\).
\(I = \frac{3}{2} (\sqrt{3} - \frac{1}{\sqrt{3}})\).
\(I = \frac{3}{2} (\frac{3 - 1}{\sqrt{3}}) = \frac{3}{2} \cdot \frac{2}{\sqrt{3}} = \sqrt{3}\).
Quick Tip: Always change the limits of integration immediately upon making a substitution. This avoids the need to substitute back to the original variable.

Case 1: \(x + 3 \ge 0 \implies x \ge -3\).
\(|x+3| = x+3\).

Inequality: \(\frac{x+3+x}{x+2} > 1 \implies \frac{2x+3}{x+2} - 1 > 0\).
\(\frac{2x+3 - (x+2)}{x+2} > 0 \implies \frac{x+1}{x+2} > 0\).

Critical points are -2 and -1. Solution: \(x \in (-\infty, -2) \cup (-1, \infty)\).

Intersection with \(x \ge -3\): \([-3, -2) \cup (-1, \infty)\).


Case 2: \(x + 3 < 0 \implies x < -3\).
\(|x+3| = -(x+3) = -x-3\).

Inequality: \(\frac{-x-3+x}{x+2} > 1 \implies \frac{-3}{x+2} - 1 > 0\).
\(\frac{-3 - x - 2}{x+2} > 0 \implies \frac{-(x+5)}{x+2} > 0 \implies \frac{x+5}{x+2} < 0\).

Critical points are -5 and -2. Solution: \(x \in (-5, -2)\).

Intersection with \(x < -3\): \((-5, -3)\).


Combine both cases:
\((-5, -3) \cup [-3, -2) \cup (-1, \infty) = (-5, -2) \cup (-1, \infty)\).
Quick Tip: For inequalities involving absolute values like \(|x+a|\), always split the problem into two cases defined by the root of the absolute value term (\(x \ge -a\) and \(x < -a\)).

Refer to the sentence in the passage: "You may qualify for alternative credit for drills by proving previous experience in actual hazmat emergency response."

This implies that proving prior real experience allows a member to substitute it for drill attendance, thus attending less than the standard 50%.
Quick Tip: Scan the passage for keywords like "alternative credit" or "drills" to locate the specific exception to the rule.

The passage discusses the schedule for training, the prerequisites for joining (permission, department membership), the requirements to become active (training, certification), and the requirements to maintain status (drills, conferences).

The entire text revolves around the requirements for obtaining and keeping membership.

Therefore, "completing F.A.S.T. membership requirements" is the central theme.
Quick Tip: The main subject is the overarching topic that encompasses all paragraphs. While goals and certifications are mentioned, they are details within the broader context of membership requirements.

The passage states: "Training will take place the third week of each month. Classes will be taught on Monday afternoons, Wednesday evenings, and Saturday afternoons."

This indicates that training occurs on three specific days (Monday, Wednesday, Saturday) during that one week every month.

Thus, applicants must be available for three days each month.
Quick Tip: Pay close attention to frequency qualifiers. "Third week of each month" combined with three specific days implies a monthly recurrence of three days.

Let the starting point be \((0, 0)\).

1. Goes 15 m North: Position \((0, 15)\).

2. Turns Right (East) and goes 15 m: Position \((15, 15)\).

3. Turns Left (North) and goes 10 m: Position \((15, 25)\).

4. Turns Left (West) and goes 15 m: Position \((0, 25)\).

5. Turns Left (South) and goes 15 m: Position \((0, 10)\).


The final position is \((0, 10)\). The distance from the starting point \((0, 0)\) is 10 meters.
Quick Tip: Drawing a simple path diagram on a coordinate grid (North=+y, East=+x) prevents errors in direction and distance accumulation.

We are given the fact: "All members of Mohan’s family are honest."

This is a universal positive statement (\(A \subseteq B\)).

Any subset of "Mohan's family" must also possess the property of being honest.

"The employed members of Mohan's family" is a subset of "Mohan's family".

Therefore, these employed members must be honest.


Let's check the other options:

(A) Contradicts statement 2 ("Some ... are not employed").

(C) We only know some honest members (the unemployed family ones) are not employed. We don't know if all honest members are unemployed.

(D) Contradicts the deduction from statement 1.
Quick Tip: In syllogisms, if "All A are B", then any specific subset of A (like "Employed A") is also B.