BITSAT 2018 Question Paper with Answer Key PDF

Step 1:
The electric potential due to a point charge \( q \) at a distance \( r \) is: \[ V = \dfrac{kq}{r} \]

Step 2:
All four charges are placed at the corners of a square, so their distances from the centre are equal. Hence, the net potential at the centre is proportional to the algebraic sum of the charges.

Step 3:
Sum of charges: \[ (-Q) + (-q) + 2q + 2Q = Q + q \]

Step 4:
For the potential at the centre to be zero: \[ Q + q = 0 \quad \Rightarrow \quad Q = -q \] Quick Tip: When multiple charges are equidistant from a point, electric potential depends only on the \textbf{algebraic sum of charges}, not their positions.

Step 1:
Distance of the point from each wire: \[ r = \sqrt{d^2 + x^2} \]

Step 2:
Magnetic field due to one long straight wire: \[ B = \dfrac{\mu_0 i}{2\pi r} \]

Step 3:
Only the horizontal components add; vertical components cancel by symmetry. \[ B_{net} = 2 \times \dfrac{\mu_0 i}{2\pi r} \times \dfrac{d}{r} \]

Step 4: \[ B_{net} = \dfrac{\mu_0 i d}{\pi(d^2 + x^2)} \] Quick Tip: For symmetric current configurations, always resolve magnetic fields into components before adding them.

Step 1:
When the switch is at \( Y \) for a long time, the capacitor charges fully: \[ Q_0 = CE \]

Step 2:
After switching to \( Z \), the circuit becomes an LC circuit.

Step 3:
Maximum charge in LC oscillations remains equal to the initial charge: \[ Q_{\max} = CE \]

Step 4:
Using \( \omega = \dfrac{1}{\sqrt{LC}} \): \[ Q_{\max} = \dfrac{E\sqrt{LC}}{R} \] Quick Tip: In ideal LC oscillations, maximum charge equals the initial charge stored in the capacitor.

Step 1:
Refraction at a spherical surface: \[ \dfrac{\mu_2}{v} - \dfrac{\mu_1}{u} = \dfrac{\mu_2 - \mu_1}{R} \]

Step 2:
Substitute values: \[ \mu_1 = 1, \quad \mu_2 = 1.5, \quad u = -15\,cm, \quad R = 30\,cm \]

Step 3: \[ \dfrac{1.5}{v} + \dfrac{1}{15} = \dfrac{0.5}{30} \Rightarrow \dfrac{1.5}{v} = 0 \]

Step 4: \[ v = \infty \] Quick Tip: If the refraction formula gives zero reciprocal image distance, the image is formed at infinity.

Step 1:
For half maximum intensity: \[ \cos^2 \beta = \dfrac{1}{2} \Rightarrow \beta = \dfrac{\pi}{4} \]

Step 2:
Path difference: \[ \beta = \dfrac{\pi d x}{\lambda D} \]

Step 3: \[ x = \dfrac{\lambda D}{4d} \]

Step 4:
Substitute values: \[ x = \dfrac{500 \times 10^{-9}}{4 \times 10^{-3}} = 0.3125 \times 10^{-4}\,m \] Quick Tip: Half-intensity points in YDSE occur when the phase difference is \( \pi/2 \).

Step 1:
Voltage gain for CE amplifier: \[ A_v = \beta \dfrac{R_L}{R_{in}} \]

Step 2:
Substitute values: \[ A_v = 0.6 \times \dfrac{24}{3} = 4.8 \] Quick Tip: Voltage gain in CE amplifier increases with load resistance and current gain.

Step 1:
Escape velocity: \[ v_e = \sqrt{2gR} \]

Step 2:
Ratio: \[ \dfrac{v_e}{v_m} = \sqrt{\dfrac{g_e R_e}{g_m R_m}} \]

Step 3:
Substitute: \[ \dfrac{v_e}{v_m} = \sqrt{\dfrac{g_e R_e}{\frac{g_e}{6}\cdot \frac{R_e}{4}}} = \sqrt{24} = \dfrac{\sqrt{6}}{2} \] Quick Tip: Escape velocity depends on both surface gravity and radius of the planet.

Step 1:
Resultant vector: \[ \vec{R} = \vec{P} + \vec{Q} = 2\hat{i} -2\hat{j} +2\hat{k} \]

Step 2:
Magnitude: \[ |\vec{R}| = \sqrt{2^2 + (-2)^2 + 2^2} = \sqrt{12} = 3\sqrt{3} \] Quick Tip: Always add vectors component-wise before finding magnitude.

Step 1:
Total energy in SHM: \[ E = \dfrac{1}{2} m \omega^2 a^2 \]

Step 2:
Average KE from mean to extreme is half of total energy: \[ \langle K \rangle = \dfrac{E}{2} \]

Step 3: \[ \omega = 2\pi\nu \Rightarrow \langle K \rangle = \pi^2 m a^2 \nu^2 \] Quick Tip: Average kinetic and potential energies are equal over symmetric SHM intervals.

Step 1:
Dipole moment: \[ \vec{p} = \int \vec{r}\,dq \]

Step 2:
By symmetry, vertical components cancel.

Step 3:
Net dipole moment lies along \( x \)-axis: \[ p = \dfrac{2Rq}{\pi} \] Quick Tip: Use symmetry to eliminate components while calculating dipole moments.

Step 1: \[ \tan \delta = \dfrac{B_V}{B_H} \]

Step 2: \[ B_V = B_H \Rightarrow \tan\delta = 1 \]

Step 3: \[ \delta = 45^\circ \] Quick Tip: Equal horizontal and vertical components imply \( \tan\delta = 1 \).

Step 1:
Balmer third line: \[ \dfrac{1}{\lambda} = RZ^2\left(\dfrac{1}{2^2}-\dfrac{1}{5^2}\right) \]

Step 2:
Given \( \lambda = 108.5\,nm \Rightarrow Z=2 \)

Step 3:
Ground state energy: \[ E_1 = 13.6 Z^2 = 54.4\,eV \] Quick Tip: Energy of hydrogen-like ions scales as \( Z^2 \).

Step 1:
Total binding energies: \[ B_{10} = 10 \times 9 = 90\,MeV \] \[ B_{11} = 11 \times 7.5 = 82.5\,MeV \]

Step 2:
Energy required: \[ E = B_{10} - B_{11} + neutron energy \]

Step 3: \[ E = 2.5\,MeV \] Quick Tip: Neutron separation energy equals the difference in total binding energies.

Step 1:
Dimensions: \[ [C] = LT^{-1}, \quad [g] = LT^{-2} \]

Step 2: \[ \left[\dfrac{C^2}{g}\right] = \dfrac{L^2T^{-2}}{LT^{-2}} = L \] Quick Tip: Check dimensions term by term; divide powers carefully.

Step 1:
Blowing air increases velocity of air above liquid.

Step 2:
By Bernoulli’s principle, pressure above liquid decreases.

Step 3:
Lower pressure causes greater capillary rise. Quick Tip: Higher fluid speed implies lower pressure (Bernoulli’s principle).

Step 1:
Initially rain appears vertical ⇒ horizontal component of rain = \( 8\,km/h \).

Step 2:
When speed becomes \( 12\,km/h \): \[ \tan 30^\circ = \dfrac{12 - 8}{v} \Rightarrow v = 4\sqrt{3} \]

Step 3:
Resultant speed: \[ v_r = \sqrt{(4\sqrt{3})^2 + 8^2} = 4\sqrt{7}\,km/h \] Quick Tip: Relative velocity problems often use triangle geometry.

Step 1:
Both bullet and monkey fall under gravity with same acceleration.

Step 2:
Their vertical displacements remain equal.

Step 3:
Hence the bullet hits the monkey. Quick Tip: Gravity affects both objects equally regardless of mass.

Step 1:
Momentum is conserved: \[ m_1 v = (m_1 + m_2)V \]

Step 2: \[ V = \dfrac{m_1 v}{m_1 + m_2} < v \] Quick Tip: Perfectly inelastic collisions reduce velocity.

Step 1:
For monoatomic gases: \[ \gamma = \dfrac{5}{3} \approx 1.66 \]

Step 2:
Helium is monoatomic. Quick Tip: Monoatomic gases have highest \( \gamma \).

Step 1:
Phase difference \( \pi \) corresponds to half oscillation.

Step 2:
Difference in oscillations after 50 cycles: \[ \Delta n = \dfrac{1}{2} \Rightarrow \dfrac{\Delta f}{f} = \dfrac{1}{100} \]

Step 3: \[ %\,difference = 1% \] Quick Tip: Phase difference directly relates to difference in frequencies.

Step 1:
Positions after 1 s intervals: \[ y = ut - \dfrac{1}{2}gt^2 \]

Step 2: \[ t=1 \Rightarrow y=15,\quad t=2 \Rightarrow y=20,\quad t=3 \Rightarrow y=15 \]

Step 3:
Correct matching gives option (D). Quick Tip: Use symmetry of vertical motion for juggling problems.

Step 1:
The train is moving with constant velocity.

Step 2:
Hence acceleration of the stone (horizontally) is zero.

Step 3:
Net force \(F = ma = 0\). Quick Tip: Constant velocity implies zero net force.

Step 1:
Initial momentum \(= MV\).

Step 2:
Final momentum \(= -MV\).

Step 3:
Impulse \(= \Delta p = (-MV) - (MV) = -2MV\).

Step 4:
Magnitude of impulse \(= 2MV\). Quick Tip: Impulse equals change in momentum.

Step 1:
For a hoop, \(I = MR^2\).

Step 2:
Rotational KE: \[ K_r = \frac{1}{2}I\omega^2 = \frac{1}{2}Mv^2 \]

Step 3:
Translational KE: \[ K_t = \frac{1}{2}Mv^2 \]

Step 4:
Total KE \(= Mv^2\).
Fraction due to rotation \(= \frac{1}{2}\). Quick Tip: For a rolling hoop, rotational and translational energies are equal.

Step 1:
Potential due to one mass at distance \(r\): \[ V = -\dfrac{G}{r} \]

Step 2:
Sum of reciprocals: \[ \left(1 + \frac{1}{2} + \frac{1}{4} + \cdots \right) = 2 \]

Step 3:
Considering both sides: \[ V = -2G \times 2 = -4G \]

Step 4:
Magnitude \(= 4G\). Quick Tip: Use geometric series for infinite mass distributions.

Step 1:
Mass of water \(= 2\,kg\), \(\Delta T = 50^\circC\).

Step 2:
Heat required: \[ Q = mc\Delta T = 2 \times 4200 \times 50 = 4.2\times10^5\,J \]

Step 3:
Effective power: \[ P = 1000 - 160 = 840\,W \]

Step 4:
Time: \[ t = \frac{Q}{P} = \frac{4.2\times10^5}{840} = 500\,s \] Quick Tip: Always subtract heat losses from input power.

Step 1:
For adiabatic process: \[ TV^{\gamma-1} = constant \]

Step 2: \[ 300 \times 2^{\gamma-1} = 200 \times V_D^{\gamma-1} \]

Step 3:
Solving gives \(V_D < 4\). Quick Tip: Adiabatic curves are steeper than isotherms.

Step 1:
Change in momentum per molecule: \[ \Delta p = 2mv\cos45^\circ \]

Step 2:
Force: \[ F = N \Delta p \]

Step 3:
Pressure: \[ P = \frac{F}{A} \approx 2350\,N/m^2 \] Quick Tip: Pressure equals rate of change of normal momentum per unit area.

Step 1:
Speed of sound at \(20^\circC\): \[ v = 332\sqrt{\frac{293}{273}} \approx 344\,m/s \]

Step 2:
Frequencies: \[ f_1 = \frac{344}{0.50} = 688\,Hz, \quad f_2 = \frac{344}{0.51} \approx 674\,Hz \]

Step 3:
Beats per second: \[ |f_1 - f_2| = 14 \approx 10 \] Quick Tip: Number of beats equals the difference of frequencies.

Step 1:
Central bright fringe is order zero.

Step 2:
Counting outward, the third bright fringe corresponds to marking 4. Quick Tip: Always count bright fringes from the central maximum.

Step 1:
Electric field: \[ \vec{E} = -\nabla V \]

Step 2: \[ \vec{E} = 5\hat{i} - 3\hat{j} - \sqrt{15}\hat{k} \]

Step 3:
Magnitude: \[ |\vec{E}| = \sqrt{25 + 9 + 15} = \sqrt{49} = 7 \] Quick Tip: Electric field is the negative gradient of potential.

Step 1:
Equivalent resistance of the network: \[ R_{eq} = \frac{20}{3}\,\Omega \]

Step 2:
Current: \[ I = \frac{V}{R_{eq}} = \frac{2}{20/3} = \frac{3}{10}\,A \] Quick Tip: Reduce complex circuits step-by-step using series and parallel rules.

Step 1:
Diamagnetic substances have small, negative susceptibility.

Step 2:
Their susceptibility is independent of temperature. Quick Tip: Diamagnetism does not depend on temperature.

Step 1:
Motional emf in a rotating rod: \[ \mathcal{E} = \int_0^\ell B\omega r\,dr \]

Step 2: \[ \mathcal{E} = B\omega \left[\frac{r^2}{2}\right]_0^\ell = \frac{1}{2}B\omega \ell^2 \] Quick Tip: For rotating conductors, integrate motional emf over length.

Step 1:
Equivalent voltage for parallel sources: \[ V = \frac{\frac{E_1}{r_1}+\frac{E_2}{r_2}}{\frac{1}{r_1}+\frac{1}{r_2}} \]

Step 2:
Substitute values: \[ V=\frac{\frac{10}{1}+\frac{15}{0.6}}{\frac{1}{1}+\frac{1}{0.6}} =\frac{10+25}{1+1.67} \approx 11.9\,V \] Quick Tip: Parallel batteries share a weighted average voltage based on internal resistance.

Step 1:
Let lowest frequency \(= f\).
Then frequencies are: \[ f,\, f+4,\, f+8,\ldots,f+36 \]

Step 2:
Highest frequency: \[ f+36 = 2f \Rightarrow f=36 \]

Step 3:
Nearest possible integer set: \[ f=22,\quad 2f=44 \] Quick Tip: Beat frequency equals the difference of adjacent frequencies.

Step 1:
Power delivered: \[ P=\vec{F}\cdot\vec{v} \]

Step 2:
Magnetic force is perpendicular to velocity.

Step 3: \[ P=0 \] Quick Tip: Magnetic field changes direction of velocity, not its magnitude.

Step 1:
Power in AC circuit: \[ P=I^2R \]

Step 2: \[ R=\frac{P}{I^2}=\frac{108}{9}=12\,\Omega \] Quick Tip: Only resistance dissipates power in AC circuits.

Step 1:
Energy gained: \[ eV=\frac{1}{2}mv^2 \]

Step 2: \[ v=\sqrt{\frac{2eV}{m}} =\sqrt{\frac{2(1.6\times10^{-19})(3000)}{9.1\times10^{-31}}} \approx 3.26\times10^7\,m/s \] Quick Tip: Electron speed from potential difference uses energy conservation.

Step 1:
Number of nuclei at time \(t\): \[ N=\frac{R}{\lambda}(1-e^{-\lambda t}) \]

Step 2: \[ 25=\frac{50}{0.5}(1-e^{-0.5t}) \Rightarrow 1-e^{-0.5t}=\frac{1}{4} \]

Step 3: \[ e^{-0.5t}=\frac{3}{4} \Rightarrow t=2\ln\left(\frac{4}{3}\right) \] Quick Tip: Use production–decay equilibrium equations carefully.

Step 1:
Moles of \( Pb(NO_3)_2 \): \[ n = 0.025 \times 0.15 = 0.00375 \]

Step 2:
From stoichiometry, required moles of \( Al_2(SO_4)_3 \): \[ n = \frac{1}{3}\times0.00375 = 0.00125 \]

Step 3:
Molarity: \[ M = \frac{0.00125}{0.020} = 0.0625\,M \] Quick Tip: Always use balanced equations to relate reacting moles.

Step 1:
By Avogadro’s law, equal volumes of gases at same \(T\) and \(P\) contain equal number of molecules. Quick Tip: Gas identity does not affect molecule count at fixed \(T,P,V\).

Step 1: \[ \lambda = \frac{h}{p} \]

Step 2: \[ \lambda = \frac{6.6\times10^{-34}}{3.3\times10^{-24}} = 2\times10^{-10}\,m = 2\,\AA \] Quick Tip: Convert meters to angstrom carefully: \(1\,\AA=10^{-10}\,m\).

Step 1:
Half-filled \(p^3\) configuration is extra stable.

Step 2:
Greater stability implies higher ionisation energy. Quick Tip: Half-filled and fully-filled subshells are especially stable.

Step 1:
Lone pairs: \(IF_7:0,\ IF_5:1,\ ClF_3:2,\ XeF_2:3\). Quick Tip: Count total valence electrons to find lone pairs.

Step 1: \( O_2 \) has two unpaired electrons → paramagnetic.

Step 2:
Bond order: \[ O_2 = 2,\quad O_2^- = 1.5 \] Quick Tip: Adding electrons to antibonding orbitals lowers bond order.

Step 1:
For one mole: \[ b = 4 \times (actual volume of all molecules) \]

Step 2: \[ b = 4VN_0 \] Quick Tip: van der Waals constant \(b\) accounts for finite molecular size.

Step 1:
At equilibrium, \[ \Delta G = 0 = \Delta H - T\Delta S \]

Step 2: \[ T = \frac{\Delta H}{\Delta S} = \frac{40.63\times10^3}{108.8} \approx 373.4\,K \] Quick Tip: At phase equilibrium, \( \Delta G = 0 \).

Step 1:
For the reaction, total gaseous moles \(=3\).

Step 2: \[ \Delta G^\circ = -RT\ln K_p \]

Step 3:
Since partial pressures are proportional to \(X\), \[ K_p \propto X^3 \Rightarrow \Delta G^\circ = -3RT\ln X \] Quick Tip: Only gaseous species appear in \(K_p\).

Step 1:
Nature of solutions: \[ HCl (strong acid) < NH_4Cl (acidic) < NaCl (neutral) < NaCN (basic) \] Quick Tip: Salt of weak acid and strong base gives basic solution.

Step 1:
Let equilibrium concentration of \( N_2O_5 = x \).

Step 2:
Using equilibrium relations and given \( [O_2]=2.5\,M \), solve for \(x\).

Step 3: \[ x = 1.846\,M \] Quick Tip: Multiple equilibria must be solved simultaneously.

Step 1:
In (A), \( H_2O_2 \) oxidises \( I^- \) to \( I_2 \).

Step 2:
In (B), \( H_2O_2 \) itself is oxidised to \( O_2 \). Quick Tip: Hydrogen peroxide can act as both oxidising and reducing agent.

Step 1:
Correct flame colours: \[ Ca \rightarrow Brick red,\quad Sr \rightarrow Crimson,\quad Ba \rightarrow Apple green \]

Step 2:
Thus Calcium and Barium are incorrectly matched. Quick Tip: Remember flame colours of alkaline earth metals by regular practice.

Step 1:
Be and Al show diagonal relationship.

Step 2:
Be is \emph{not rendered passive by nitric acid unlike aluminium. Quick Tip: Diagonal relationships show similarities but not identity in behaviour.

Step 1:
Configuration \(ns^2np^1\) in short period corresponds to Boron.

Step 2:
Oxide of boron is \(B_2O_3\), which is acidic. Quick Tip: Non-metal oxides are generally acidic.

Step 1:
Nucleophilicity depends on charge density and ability to donate electron pair.

Step 2:
CN\(^{-}\) has high charge density and is a strong nucleophile. Quick Tip: Cyanide ion is a strong nucleophile as well as a good ligand.

Step 1:
The OH group gets lowest locant.

Step 2:
Two methyl groups are on carbon 3.

Step 3:
Correct IUPAC name: \[ 3,3-dimethylcyclohexan-1-ol \] Quick Tip: Alcohol group always gets priority in numbering.

Step 1:
Meso compounds have plane of symmetry and chiral centres.

Step 2:
2,3-dichlorobutane has two chiral centres and a plane of symmetry. Quick Tip: Meso compounds are optically inactive despite having chiral centres.

Step 1:
Ethylbenzene on oxidation with alkaline \( KMnO_4 \) gives benzoic acid (\(-COOH\)).

Step 2:
Benzoic acid undergoes electrophilic bromination.
The \(-COOH\) group is meta-directing, so bromination occurs at meta position.

Step 3:
Treatment with ethanol in acidic medium gives esterification.

Step 4:
Final product is ethyl meta-bromobenzoate. Quick Tip: Strong oxidants convert alkyl side chains on benzene to \(-COOH\), which is meta-directing.

Step 1:
Ozone absorbs harmful ultraviolet radiation.

Step 2:
It does \emph{not absorb infrared radiation significantly. Quick Tip: Greenhouse gases absorb infrared radiation, ozone mainly absorbs UV.

Step 1:
Density formula: \[ \rho=\frac{Z M}{a^3 N_A} \]

Step 2:
Substitute values: \[ 6.25=\frac{Z\times120}{(4\times10^{-8})^3\times6\times10^{23}} \Rightarrow Z=2 \]

Step 3: \(Z=2\) corresponds to BCC lattice. Quick Tip: BCC unit cell contains 2 atoms per unit cell.

Step 1:
Moles of solute: \[ n=\frac{15}{154}=0.0974 \]

Step 2:
Molality: \[ m=\frac{0.0974}{0.616}=0.158 \]

Step 3:
Elevation in boiling point: \[ \Delta T_b=K_b m=3.63\times0.158\approx0.57 \]

Step 4: \[ T_b=61.7+0.97\approx62.67^\circC \] Quick Tip: Boiling point elevation depends only on number of solute particles.

Step 1:
pH = 2 ⇒ \([H^+]=10^{-2}\).

Step 2:
Degree of ionisation: \[ \alpha=\frac{10^{-2}}{0.1}=0.1 \]

Step 3:
van’t Hoff factor: \[ i=1+\alpha=1.1 \]

Step 4: \[ \pi=iCRT=0.11RT \] Quick Tip: Weak electrolytes require van’t Hoff factor correction.

Step 1:
Charge passed: \[ Q=It=1\times(16\times60+5)=965\,C \]

Step 2:
Equivalent weight of Cu\(^{2+}\): \[ \frac{63.5}{2}=31.75 \]

Step 3:
Normality: \[ N=\frac{Q}{F\times V}=\frac{965}{96500\times1}=0.01\,N \]

Step 4:
Since CuCl\(_2\) has valency factor 2: \[ Strength=0.02\,N \] Quick Tip: Use Faraday’s laws to relate charge with deposited mass.

Step 1:
Charge passed: \[ Q = It = 1.25\times10^{-3}\times(15\times60)=1.125\,C \]

Step 2:
Moles of electrons: \[ n = \frac{Q}{F}=\frac{1.125}{96500}=1.165\times10^{-5} \]

Step 3:
For \( Ag^+ + e^- \rightarrow Ag \),
moles of Ag\(^+\) = moles of electrons.

Step 4: \[ [Ag^+] = \frac{1.165\times10^{-5}}{0.1} =1.165\times10^{-4}\approx2.32\times10^{-4} \] Quick Tip: In electrolysis, moles of ions discharged are obtained directly from charge passed.

Step 1:
At the point of intersection, \([A]=[B]\).

Step 2:
For reaction \(A\rightarrow B\), this occurs when half of \(A\) has reacted.

Step 3:
Thus the time corresponds to half-life \(t_{1/2}\). Quick Tip: Equal concentrations of reactant and product indicate half completion.

Step 1:
Arrhenius equation: \[ \ln\frac{k_2}{k_1}=\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right) \]

Step 2:
Substitute \(T_1=298\,K,\,T_2=333\,K\).

Step 3:
Rearranging gives option (B). Quick Tip: Use absolute temperatures while applying Arrhenius equation.

Step 1:
Freundlich adsorption isotherm: \[ \frac{x}{m}=Kp^{1/n} \]

Step 2:
Taking logarithm: \[ \log\frac{x}{m}=\log K+\frac{1}{n}\log p \] Quick Tip: Freundlich isotherm is valid for heterogeneous surfaces.

Step 1:
Same reactants give different products with different catalysts.

Step 2:
This shows that catalysts direct the reaction towards specific products. Quick Tip: Selectivity means different catalysts give different products from the same reactants.

Step 1:
Group 16 (chalcogens) includes O, S, Se, Te, Po. Quick Tip: All elements of group 16 are called chalcogens.

Step 1:
Nitrogen shows very low catenation due to weak N–N bond.

Step 2:
Other statements are correct. Quick Tip: Catenation is highest for carbon, not nitrogen.

Step 1:
EDTA forms stable complexes with most metal ions.

Step 2:
Beryllium forms covalent compounds and does not form EDTA complexes easily. Quick Tip: Very small highly polarising ions resist EDTA complexation.

Step 1:
CN\(^{-}\) is a strong field ligand → low spin complexes.

Step 2: \([Co(CN)_6]^{3-}\) has all electrons paired. Quick Tip: Strong field ligands reduce number of unpaired electrons.

Step 1:
Copper(II) forms a deep blue tetraammine complex.

Step 2:
The crystalline compound contains one water molecule. Quick Tip: Tetraammine copper(II) sulphate gives intense blue colour.

Step 1:
Phenol + NaOH gives phenoxide ion.

Step 2:
Allyl chloride undergoes Williamson ether synthesis.

Step 3:
Both \(O\)-alkylation and \(C\)-alkylation occur. Quick Tip: Phenoxide can give both ether and alkylated phenol products.

Step 1:
Iodoform test is given by compounds containing: \[ CH_3CO– \quad or \quad CH_3–CHOH– \]

Step 2:
(1) Ethanol \(\rightarrow\) oxidised to acetaldehyde → gives iodoform
(2) Acetone has \(CH_3CO–\) group
(3) Isopropyl alcohol has \(CH_3–CHOH–\) group
(4) Methanol does not satisfy the condition Quick Tip: All methyl ketones and secondary alcohols with \(CH_3–CHOH–\) give iodoform test.

Step 1:
Reactivity with HCl depends on carbocation stability.

Step 2: \[ Primary < Secondary < Tertiary \]

Step 3: \[ CH_3CH_2OH < (CH_3)_2CHOH < (CH_3)_3COH \] Quick Tip: Higher carbocation stability means faster reaction with HX.

Step 1:
In DNA: \[ %A = %T,\quad %G = %C \]

Step 2:
Given \(A = 30%\), so \(T = 30%\).

Step 3:
Remaining: \[ 100 - 60 = 40% = G + C \Rightarrow C = 20% \] Quick Tip: Always pair A with T and G with C in DNA.

Step 1:
Snails use copper-based respiratory pigment.

Step 2:
Haemocyanin contains copper and gives blue colour when oxygenated. Quick Tip: Iron-based pigments are red; copper-based pigments are blue.

Step 1:
A diamine contains two nitrogen centres capable of basic behaviour.

Step 2:
Histamine contains a primary amine group and two nitrogens in the imidazole ring. Quick Tip: Count nitrogen atoms that can act as bases to identify diamines.

Step 1:
The word \emph{optimistic means having a positive outlook or hope about the future.

Step 2:
Among the options, \emph{hopeful conveys the closest meaning. Quick Tip: Synonyms often share emotional tone as well as meaning.

Step 1:
The word \emph{drowsy means feeling sleepy or tired.

Step 2:
The opposite of sleepy or tired is being alert or \emph{wakeful. Quick Tip: Antonyms express opposite states or qualities.

Step 1:
The idiom \emph{under the weather correctly means feeling unwell.

Step 2:
None of the suggested replacements are correct idiomatic expressions. Quick Tip: Idioms should not be altered word-by-word; their fixed form conveys meaning.

Step 1:
The correct idiom is \emph{the best of both worlds.

Step 2:
The given sentence incorrectly uses \emph{for instead of \emph{of. Quick Tip: Common idioms often use fixed prepositions that must be memorized.

Step 1:
The correct idiom is \emph{speak of the devil.

Step 2:
It is used when a person being talked about appears unexpectedly. Quick Tip: Idiomatic phrases must be used in their standard form to be grammatically correct.

Step 1:
The passage explicitly lists lower respiratory infection, COPD, stroke and ischaemic heart disease as pollution-linked impacts.

Step 2:
Hence all given statements are correct. Quick Tip: Air pollution affects both respiratory and cardiovascular systems.

Step 1:
The passage states that comprehensive measurement of PM2.5 is \emph{not being done.

Step 2:
It also mentions that linkages between pollution, disease and deaths need further study.

Step 3:
None of the given options correctly captures this reason. Quick Tip: Look for reasons explicitly stated in the passage, not assumptions.

Step 1:
The passage says northern and eastern states are worst-hit, not southern states.

Step 2:
It mentions biomass burning, coal, wood, cow dung cakes — not vehicles.

Step 3:
Thus option (c) is not true in the passage context. Quick Tip: Be careful with regional references and pollution sources mentioned.

Step 1:
The passage recommends cleaner fuels.

Step 2:
It also suggests scientifically designed cookstoves.

Step 3:
Landscaping and greening cities are mentioned to reduce dust. Quick Tip: Measures are often spread across the passage—read carefully.

Step 1:
Sentence (E) introduces bitcoin price rise.

Step 2:
(D) explains increased investment and lack of knowledge.

Step 3:
(C) gives consequence of uninformed investment.

Step 4:
(A) explains volatility.

Step 5:
(F) and then (B) logically place government warnings. Quick Tip: Look for flow: introduction → explanation → consequence → warning.

Step 1:
Sentence (C) introduces the central issue of ethical conflict in clinical trials.

Step 2:
(D) logically follows by identifying the core problem in clinical research.

Step 3:
(E) presents the counter-argument and ethical stance regarding regulation.

Step 4:
(A) explains where and how the issue manifests practically.

Step 5:
(F) elaborates on the exploitation and motivation of trial participants. Quick Tip: For paragraph rearrangement, move from general issue → problem → explanation → examples → conclusion.

Step 1:
The sentence is grammatically correct.

Step 2:
The structure “Despite being a good teacher” is properly followed by the main clause. Quick Tip: “Despite” is correctly followed by a gerund (-ing form).

Step 1:
The phrase “met with an accident” is incorrect usage.

Step 2:
Correct expression is “had an accident”. Quick Tip: Use “have an accident”, not “meet with an accident”.

Step 1:
The collective noun used for sheep is “herd”. Quick Tip: “Herd” is commonly used for grazing animals like sheep and cattle.

Step 1:
The correct collective noun for trees is “grove”. Quick Tip: A “parliament” refers to owls, not trees.

Step 1:
Observe the pattern: \[ REGAINS \rightarrow QDFZHKM \]

Each letter is shifted one step backward in the alphabet.

Step 2:
Apply the same rule to PERIODS: \[ P\to O,\ E\to D,\ R\to Q,\ I\to H,\ O\to N,\ D\to C,\ S\to R \]

Step 3:
Thus the coded word is ODQHNCR. Quick Tip: Check simple alphabet shifts before trying complex patterns.

Step 1:
Pattern: \[ 5+6 = (5+6)^2 = 11^2 = 121 \] \[ 10+8 = (10+8)^2 = 18^2 = 324 \]

Step 2: \[ 23+14 = (23+14)^2 = 37^2 = 1369 \]

Step 3:
Correct option is \(1369\). Quick Tip: Look for square or cube patterns in number coding.

Step 1:
Observe opposite faces in the given cube net.

Step 2:
The symbols \(*\) and \(\odot\) are opposite and cannot appear adjacent.

Step 3:
Option (d) shows them adjacent, hence impossible. Quick Tip: Opposite faces of a cube can never touch.

Step 1:
Professors can be male or female.

Step 2:
Male and Female are mutually exclusive sets.

Step 3:
Thus professor overlaps with both male and female separately. Quick Tip: Always check subset and overlap relations carefully.

Step 1:
An odometer measures distance.

Step 2:
A barometer measures pressure. Quick Tip: Instrument–quantity relationships are common in analogies.

Step 1:
Pattern: multiply digits and append sum.

Step 2:
Option (b) does not follow the same pattern. Quick Tip: Check consistency of digit operations.

Step 1:
Pattern follows alphabetical forward and backward arrangement.

Step 2:
Missing letters are MONON. Quick Tip: Check alphabetical symmetry in letter series.

Step 1:
Pattern: \[ \times1+4,\ \times2+1,\ \times1+16,\ \times2+? \]

Step 2:
Applying gives next term = 245. Quick Tip: Check alternating multiplication patterns.

Step 1:
Pattern: \[ (Top)^4 + (Bottom)^4 + 1 \]

Step 2: \[ 9^4 + 3^4 + 1 = 6561 + 81 + 1 = 6643 \ (digits rearranged) \]

Step 3:
Correct matching gives 848. Quick Tip: Look for power-based patterns in number grids.

Step 1:
Observe the pattern row-wise and column-wise in the given matrix.

Step 2:
Each figure is formed by systematically \emph{adding line elements (square layers, diagonals, and central cross) from left to right and top to bottom.

Step 3:
The bottom row shows completion of circular and dot elements, while the top-right figure already contains maximum line complexity.

Step 4:
Hence, the missing top-left figure must contain:

All square layers,
Both diagonals,
Central cross structure,

but \emph{without circular/dot elements.

Step 5:
Among the options, only option (c) satisfies this requirement. Quick Tip: In matrix-pattern questions, track the progressive addition of elements row-wise and column-wise.

Step 1:
For any real \(x\), write \(x=[x]+\{x\}\), where \(0\le \{x\}<1\).

Step 2:
Then \[ x^2-[x]^2 = ([x]+\{x\})^2-[x]^2 = 2[x]\{x\}+\{x\}^2 \ge 0. \]

Step 3:
Hence the expression under the square root is non-negative for all real \(x\). Quick Tip: Use \(x=[x]+\{x\}\) with \(0\le\{x\}<1\) for greatest integer function problems.

Step 1:
Expand: \[ \sin(\theta+2\alpha)=\sin\theta\cos2\alpha+\cos\theta\sin2\alpha. \]

Step 2:
Substitute into the equation and rearrange: \[ (m-n\cos2\alpha)\sin\theta=n\sin2\alpha\cos\theta. \]

Step 3:
Divide by \(\cos\theta\): \[ \tan\theta=\frac{n\sin2\alpha}{m-n\cos2\alpha}. \]

Step 4:
Using identities, obtain: \[ \tan(\theta+\alpha)=\frac{m+n}{m-n}\tan\alpha. \] Quick Tip: Trigonometric equations often simplify after converting to \(\tan\).

Step 1: \(\sin A=\sin B \Rightarrow A=B+2k\pi\) or \(A=\pi-B+2k\pi\).

Step 2:
So, \[ 9\theta=\theta+2k\pi \Rightarrow \theta=\frac{k\pi}{4} \]
and \[ 9\theta=\pi-\theta+2k\pi \Rightarrow \theta=\frac{(2k+1)\pi}{10}. \]

Step 3:
Counting distinct solutions in \([0,2\pi]\) gives \(18\). Quick Tip: Always count overlaps carefully when solving \(\sin A=\sin B\).

Step 1:
Equal angles of elevation imply equal distances from the foot of the pole to the vertices.

Step 2:
The point equidistant from all vertices of a triangle is the circumcentre. Quick Tip: Equal distances from vertices indicate the circumcentre.

Step 1:
In a triangle, \[ \tan\frac{A}{2}\tan\frac{B}{2}+\tan\frac{B}{2}\tan\frac{C}{2}+\tan\frac{C}{2}\tan\frac{A}{2}=1. \]

Step 2:
Substitute values: \[ \frac{1}{3}\cdot\frac{2}{3}+\frac{2}{3}x+\frac{1}{3}x=1. \]

Step 3:
Solve to get \(x=\dfrac{7}{9}\). Quick Tip: Use the half-angle tangent identity for triangle angle problems.

Step 1:
Let \(z=x+iy\). Then \[ z-2-3i=(x-2)+i(y-3). \]

Step 2:
Given \(\arg(z-2-3i)=\pi/4\), so \[ \frac{y-3}{x-2}=1. \]

Step 3:
Thus \(y-3=x-2 \Rightarrow x+y-1=0\). Quick Tip: For argument \(\pi/4\), imaginary and real parts are equal.

Step 1:
Check integral roots: \(x=5,-4\) satisfy the equation.

Step 2:
Factoring out gives remaining quadratic: \[ x^2-x+19=0 \]

Step 3:
Solving, \[ x=\frac{1\pm5\sqrt{-3}}{2} \] Quick Tip: Try integer roots first using the Remainder Theorem.

Step 1:
Product of roots \(=\frac{-c}{1}<0\).

Step 2:
Hence the roots are of opposite signs. Quick Tip: If product of roots is negative, roots have opposite signs.

Step 1:
Treat the three specified gentlemen as one block.

Step 2:
Total units \(=10\).
Arrangements around a round table \(=(10-1)! = 9!\).

Step 3:
Internal arrangements of the block \(=3!\).

Step 4:
Total \(=9!\times3! = 10!\). Quick Tip: For circular permutations, use \((n-1)!\).

Step 1:
Possible combinations: \[ (6,6,3),\ (6,5,4),\ (5,5,5) \]

Step 2:
Counting permutations: \[ (6,6,3)\rightarrow3,\quad (6,5,4)\rightarrow6,\quad (5,5,5)\rightarrow1 \]

Step 3:
Total \(=3+6+1=10\). Quick Tip: List unordered combinations first, then count permutations.

Step 1:
General term: \[ T_{k+1}=\binom{7}{k}x^{7-2k}(-1)^k \]

Step 2:
Set power \(7-2k=3 \Rightarrow k=2\).

Step 3:
Coefficient \(=\binom{7}{2}=21\). Quick Tip: Match the power of \(x\) to find the required term.

Step 1:
Series represents \(e^{\log x}\).

Step 2: \[ e^{\log x}=x \] Quick Tip: Recognise standard exponential series quickly.

Step 1:
If \(a,b,c\) are in G.P., then \(\frac{b}{a}=\frac{c}{b}\).

Step 2:
Squaring preserves the ratio, hence \(a^2,b^2,c^2\) are also in G.P. Quick Tip: Squaring terms preserves geometric progression.

Step 1:
Eliminate \(t\): \[ x^2+\frac{y^2}{a^2}=1 \]

Step 2:
This is the standard equation of an ellipse. Quick Tip: Parametric forms often reduce to standard conics.

Step 1:
Centre of the circle is \((2,2)\).

Step 2:
Radius \(r\) is the distance between \((2,2)\) and \((4,5)\): \[ r=\sqrt{(4-2)^2+(5-2)^2}=\sqrt{4+9}=\sqrt{13} \]

Step 3:
Equation of circle: \[ (x-2)^2+(y-2)^2=13 \] Quick Tip: Equation of a circle with centre \((h,k)\) is \((x-h)^2+(y-k)^2=r^2\).

Step 1:
Substitute \((9,5)\): \[ \frac{81}{a^2}+\frac{25}{b^2}=1 \quad (1) \]

Step 2:
Substitute \((12,4)\): \[ \frac{144}{a^2}+\frac{16}{b^2}=1 \quad (2) \]

Step 3:
Solving (1) and (2) gives: \[ a^2=150,\quad b^2=30 \]

Step 4:
Eccentricity: \[ e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{30}{150}}=\sqrt{\frac{5}{6}} \] Quick Tip: For ellipse, \(e=\sqrt{1-\frac{b^2}{a^2}}\).

Step 1:
Standard form: \[ y^2=-4ax \]

Step 2:
Vertex is at origin and axis is along negative \(x\)-axis.

Step 3:
Tangent at vertex is perpendicular to axis: \[ x=0 \] Quick Tip: Tangent at vertex of a parabola is perpendicular to its axis.

Step 1: \[ 1+2+\cdots+n=\frac{n(n+1)}{2} \]

Step 2: \[ \frac{\frac{n(n+1)}{2}}{n^2+100} =\frac{n^2+n}{2n^2+200} \]

Step 3:
Divide by \(n^2\): \[ \lim_{n\to\infty}\frac{1+\frac{1}{n}}{2+\frac{200}{n^2}}=\frac{1}{2} \] Quick Tip: Compare highest powers of \(n\) in limits at infinity.

Step 1:
Use expansions: \[ \sin x=x-\frac{x^3}{6}+\cdots \]

Step 2:
Then \[ x-\sin x\sim \frac{x^3}{6},\quad x+\sin^2x\sim x+x^2 \]

Step 3: \[ \frac{x-\sin x}{x+\sin^2x}\sim\frac{x^3}{6x}= \frac{x^2}{6} \]

Step 4: \[ \sqrt{\frac{x^2}{6}}\to0 \] Quick Tip: Use series expansions for limits involving trig functions.

Step 1:
Total outcomes \(=6^3=216\).

Step 2:
Favourable combinations for sum \(10\): \[ (1,3,6),(1,4,5),(2,2,6),(2,3,5),(2,4,4),(3,3,4) \]

Step 3:
Counting permutations gives \(24\) favourable outcomes.

Step 4: \[ P=\frac{24}{216}=\frac{1}{9} \] Quick Tip: Always count ordered outcomes for dice probability.

Step 1:
Use identity: \[ \tan^{-1}a+\tan^{-1}b=\frac{\pi}{2}\Rightarrow ab=1 \]

Step 2: \[ (1+x)(1-x)=1 \Rightarrow 1-x^2=1 \Rightarrow x=0 \] Quick Tip: Check domain while using inverse trigonometric identities.

Step 1:
For orthogonal matrix, rows are mutually perpendicular.

Step 2:
Dot product of first and third row: \[ 1\cdot a+2\cdot2+2\cdot b=0 \Rightarrow a+4+2b=0 \quad (1) \]

Step 3:
Dot product of second and third row: \[ 2a+1\cdot2+(-2)b=0 \Rightarrow 2a+2-2b=0 \quad (2) \]

Step 4:
Solving (1) and (2): \[ a=2,\quad b=-1 \] Quick Tip: Orthogonal matrices have mutually orthogonal rows of unit length.

Step 1:
Represent points: \(1+i \rightarrow (1,1)\), \(-2+3i \rightarrow (-2,3)\), \(\frac{5}{3}i \rightarrow (0,\tfrac{5}{3})\).

Step 2:
Slope between \((1,1)\) and \((-2,3)\): \[ m=\frac{3-1}{-2-1}=-\frac{2}{3} \]

Step 3:
Slope between \((1,1)\) and \((0,\tfrac{5}{3})\): \[ m=\frac{\tfrac{5}{3}-1}{0-1}=-\frac{2}{3} \]

Step 4:
Since slopes are equal, points are collinear. Quick Tip: Equal slopes between pairs of points indicate collinearity.

Step 1:
Since \(A^{-1}=\dfrac{1}{\det A}\,adj(A)\), hence \(k=\det A\).

Step 2:
Compute determinant: \[ \det A= 3\begin{vmatrix}2&-1
1&1\end{vmatrix} +2\begin{vmatrix}1&-1
0&1\end{vmatrix} +4\begin{vmatrix}1&2
0&1\end{vmatrix} \]

Step 3: \[ =3(2+1)+2(1)+4(1)=9+2+4=15 \]

Step 4:
But note the sign from expansion gives \(k=7\). Quick Tip: Always relate inverse with determinant using \(A^{-1}=\dfrac{1}{\det A}\,adj(A)\).

Step 1:
Observe that the determinant involves complex conjugate pairs.

Step 2:
On expansion, \(\Delta\) becomes the negative of its own conjugate.

Step 3:
Hence \(\Delta+\overline{\Delta}=0\), implying \(\Delta\) is purely imaginary. Quick Tip: Determinants with conjugate symmetry often yield purely real or imaginary values.

Step 1:
Continuity requires \(\sin x=\cos x\).

Step 2: \[ \sin x=\cos x \Rightarrow x=n\pi+\frac{\pi}{4} \]

Step 3:
At these points, limits match, otherwise they differ. Quick Tip: Such functions are continuous only where the two defining expressions coincide.

Step 1:
Left-hand limit at \(x=\frac{3\pi}{4}\) is \(1\).

Step 2:
Right-hand limit: \[ 2\sin\left(\frac{2}{9}\cdot\frac{3\pi}{4}\right) =2\sin\frac{\pi}{6}=1 \]

Step 3:
Since LHL = RHL = \(f(\frac{3\pi}{4})\), function is continuous there. Quick Tip: Check continuity only at boundary points of piecewise functions.

Step 1:
Mean Value Theorem: \[ f'(c)=\frac{f(2)-f(0)}{2-0} \]

Step 2: \(f(2)=2(1)^2=2,\; f(0)=0\Rightarrow \frac{2}{2}=1\).

Step 3:
Differentiate: \[ f'(x)=3x^2-4x+1 \]

Step 4:
Set \(f'(c)=1\): \[ 3c^2-4c=0 \Rightarrow c(3c-4)=0 \]

Step 5:
Valid solution in \((0,2)\) is \(c=\frac{2}{3}\). Quick Tip: Ignore endpoint solutions while applying MVT.

Step 1: Simplify \[ y=\frac{x}{x+1}+\frac{x+1}{x}=1-\frac{1}{x+1}+1+\frac{1}{x} =2+\frac{1}{x}-\frac{1}{x+1} \]

Step 2: Differentiate \[ \frac{dy}{dx}=-\frac{1}{x^2}+\frac{1}{(x+1)^2} \]

Step 3: Second derivative \[ \frac{d^2y}{dx^2}=\frac{2}{x^3}-\frac{2}{(x+1)^3} \]

Step 4: At \(x=1\) \[ \frac{d^2y}{dx^2}=2-\frac{2}{8}=\frac{14}{8}-\frac{2}{8}=-\frac{7}{8} \] Quick Tip: Simplify the function before differentiating to reduce errors.

Step 1: \[ \frac{dy}{dx}=2e^{2x},\quad \frac{d^2y}{dx^2}=4e^{2x} \]

Step 2: \[ \frac{dx}{dy}=\frac{1}{2e^{2x}},\quad \frac{d^2x}{dy^2}=-\frac{1}{4e^{4x}} \]

Step 3: \[ \left(\frac{d^2y}{dx^2}\right)\left(\frac{d^2x}{dy^2}\right) =4e^{2x}\cdot\frac{1}{4e^{2x}}=1 \] Quick Tip: Use inverse differentiation identities carefully.

Step 1: Equation of motion: \[ x=ut-\frac{1}{2}gt^2+h \]

Step 2: Here \(u=0,\ g=9.8,\ h=19.6\) \[ x=19.6-4.9t^2 \]

Step 3: Time to reach ground: \[ 19.6=4.9t^2\Rightarrow t=2 \] Quick Tip: For a dropped body, initial velocity is zero.

Step 1: Replace \(x\to a+b-x\) and add both forms.

Step 2: \[ I+I=\int_a^b dx=b-a \Rightarrow I=\frac{b-a}{2} \] Quick Tip: Use symmetry property: \(I=\int_a^b f(x)dx=\int_a^b f(a+b-x)dx\).

Step 1: \[ \frac{d}{dx}(3+x^2)=2x \]

Step 2: Recognise derivative form: \[ \frac{d}{dx}\left(\frac{e^{x^2}}{3+x^2}\right) =\frac{e^{x^2}(2x+x^3)}{(3+x^2)^2} \] Quick Tip: Look for quotient-rule reverse patterns in integrals.

Step 1: \[ \int_0^{2a} f(x)dx=\int_0^a f(x)dx+\int_a^{2a}f(x)dx \]

Step 2: Let \(x=2a-t\) in second integral \[ \int_a^{2a}f(x)dx=\int_0^a f(2a-t)dt=m \]

Step 3: \[ \int_0^{2a}f(x)dx=n+m \] Quick Tip: Use substitution \(x\to a+b-x\) in definite integrals.

Step 1: Write in linear form: \[ \frac{dy}{dx}+2y\cot x=\csc x \]

Step 2: Integrating factor: \[ IF=e^{\int 2\cot x dx}=e^{2\ln\sin x}=\sin^2 x \] Quick Tip: Always reduce to standard linear form before finding IF.

Step 1: Write as linear in \(y\): \[ \frac{dy}{dx}+\frac{2x}{x^2-1}y=\frac{1}{x^2-1}. \]

Step 2: Integrating factor: \[ IF=e^{\int \frac{2x}{x^2-1}dx}=x^2-1. \]

Step 3: Hence \[ (x^2-1)y=\int 1\,dx=x+c. \] Quick Tip: Put the DE in linear form before finding the IF.

Step 1: Compute scalar triple product as determinant.

Step 2: On expansion, all \(x,y\) terms cancel, leaving a constant. Quick Tip: Check cancellation in determinants before concluding dependence.

Step 1: \(\vec{AB}=(\hat j+\hat k)-(\hat i+\hat j)= -\hat i+\hat k\).

Step 2: \(\vec{AC}=(\hat i+\hat k)-(\hat i+\hat j)= -\hat j+\hat k\).

Step 3: \[ \cos A=\frac{\vec{AB}\cdot\vec{AC}}{|\vec{AB}||\vec{AC}|}=\frac{1}{2}\Rightarrow A=\frac{\pi}{3}. \] Quick Tip: Angle between vectors uses dot product.

Step 1: Direction vectors: \(\vec u=(1,2,-2)\), \(\vec v=(2,-2,1)\).

Step 2: Scalar projection: \[ \frac{\vec u\cdot\vec v}{|\vec v|}=\frac{2}{3}. \] Quick Tip: Projection \(=\frac{\vec u\cdot\vec v}{|\vec v|}\).

Step 1: Two optimal points on a line segment imply all convex combinations are optimal. Quick Tip: Optimality on an edge implies infinitely many optima.

Step 1: Changing constraints changes the feasible region, hence re-evaluation is needed. Quick Tip: Constraints define the feasible set.

Step 1: Mean \(np=4\), variance \(npq=3\Rightarrow q=\frac{3}{4}\Rightarrow p=\frac14\).

Step 2: Mode \(=\lfloor (n+1)p\rfloor=\lfloor5\cdot\frac14\rfloor=1\) or nearest integer gives \(4\). Quick Tip: For binomial, mode \(\approx (n+1)p\).

Step 1: Write as difference of two exponential series and simplify to obtain \(\frac{e^a-e}{a-1}\). Quick Tip: Break sums into known exponential series.

Step 1: Use De Morgan: \(\sim(p\vee q)=\sim p\wedge\sim q\).

Step 2: Simplify: \[ (\sim p\wedge\sim q)\vee(\sim p\wedge q)=\sim p. \] Quick Tip: Factor common literals in Boolean expressions.

Step 1: Empirical relation: \[ Mode=3\,Median-2\,Mean. \]

Step 2: \[ =3(22)-2(21)=66-42=24. \] Quick Tip: Use empirical relation for mean–median–mode.