JEE Main 29 January 2023 Shift 2 question paper with solutions

Given $u = \frac{h}{h'} = \frac{5.25}{5.00}$. Least count $= \frac{1}{20} \, \mathrm{cm} - \frac{49}{50} \cdot \frac{1}{20} \, \mathrm{cm}$ $$= \frac{1}{50} \times \frac{1}{20} \, \mathrm{cm} = 0.01 \, \mathrm{mm}$$ $\ln u = \ln h - \ln h'$ $$\frac{du}{u} = \frac{dh}{h} - \frac{dh'}{h'}$$ $$du = \left[ \frac{0.01}{5.25} + \frac{0.01}{5.00} \right] \frac{5.25}{5.00}$$ $$= \frac{41}{10} \times 10^{-3}$$ Ans. = 41

Given $\left( x - At \right)^2 + \left( y - \frac{t}{B} \right)^2 = a^2$. $[At] = A \times \frac{1}{T} = L$. Therefore, $[A] = T^1 L^1$. $\frac{t}{B}$ is in meters. Therefore, $\frac{1}{T[B]} = L$. Thus, $[B] = T^{-1} L^{-1}$. Correct Ans. (2)

The centripetal acceleration is given by $a_c = \frac{v^2}{r}$. Substituting the values, we have $$a_c = \frac{4^2}{2} = \frac{16}{2} = 8 \, \mathrm{m/s^2}.$$ The velocity vector is $\vec{V} = 4\hat{j}$. Therefore, the centripetal acceleration vector is $\vec{a_c} = 8\hat{i}$.

Given $v \frac{dv}{dx} = \frac{v^2}{R}$, we have $\int_{15}^{v} \frac{dv}{v} = \frac{1}{R} \int_{0}^{x} dx$. This gives $v = 15 e^{x/R}$. Then, $\frac{dx}{dt} = 15 e^{x/R}$. Integrating, we have $$\frac{\pi R}{2} \int_{0}^{0} e^{-x/R} dx = 15 \int_{0}^{t_0} dt$$ which results in $$t_0 = 40 \left(1 - e^{-\pi/2}\right).$$

Use $\Delta L = \int_0^t \tau \, dt$ $$L_0 = \int_0^2 mg(v_x t) \, dt$$ $$= mg v_x \frac{t^2}{2} = (0.1)(10)(10\sqrt{2}) \frac{2^2}{2}$$ $$= 20\sqrt{2}$$ $$= \sqrt{800} \, \mathrm{kg} \, \mathrm{m}^2 / \mathrm{s}$$

Given $a_1 = g \sin \theta = \frac{g}{\sqrt{2}}$. $a_2 = g \sin \theta - K g \cos \theta = \frac{g}{\sqrt{2}} - \frac{K g}{\sqrt{2}}$. $t_2 = n t_1$ and $a_1 t_1^2 = a_2 t_2^2$. $$\frac{g}{\sqrt{2}} t_1^2 = \left( \frac{g}{\sqrt{2}} - \frac{K g}{\sqrt{2}} \right) n^2 t_1^2$$ $K = 1 - \frac{1}{n^2}$. Ans. 4

Given the problem, we start with the equation for distance: $$50 = V \times 10$$ Solving for $V$, we get $$V = 5 \, \mathrm{m/s}$$ Using the equation for velocity, $$V = 0 + a \times 20$$ Substituting the value of $V$, $$5 = a \times 20$$ Solving for $a$, we find $$a = \frac{1}{4} \, \mathrm{m/s^2}$$ Using Newton's second law, $$F = ma = 20 \times \frac{1}{4} = 5 \, \mathrm{N}$$

No Solution Available

Given $T^2 \propto R^3$. $$\frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3} \implies \left( \frac{T_1}{T_2} \right)^2 = \left( \frac{R}{\frac{R}{4}} \right)^3$$ Therefore, $$\frac{T_1^2}{T_2^2} = 64$$ Thus, $$T_2^2 = \frac{T_1^2}{64}$$ Therefore, $$T_2 = \frac{24}{8} = 3$$

Given $P_2 A - P_1 A = 5.4 \times 10^5 \times g$. $$P_2 - P_1 = \frac{5.4 \times 10^6}{500} = 5.4 \times 2 \times 10^2 \times 10$$ $$= 10.8 \times 10^3$$ $P_2 + 0 + \frac{1}{2} \rho V_2^2 = P_1 + 0 + \frac{1}{2} \rho V_1^2$ $$P_2 - P_1 = \frac{1}{2} \rho (V_1^2 - V_2^2) = \frac{1}{2} \rho (V_1 - V_2)(V_1 + V_2)$$ $$10.8 \times 10^3 = \frac{1}{2} \times 1.2 (V_1 - V_2) \times 2 \times 3 \times 10^2$$ $$10.8 \times 10 = 3.6 (V_1 - V_2)$$ $$V_1 - V_2 = 30$$ $$\left( \frac{V_1 - V_2}{V} \right) \times 100 = \frac{30}{300} \times 100 = 10\%$$

Given $|F| = \eta A \frac{\Delta v}{\Delta h}$. $$0.1 = 5 \times 10^{-3} \times 0.2 \times \frac{v}{0.25 \times 10^{-3}}$$ $v = 0.025 \, \mathrm{m/s}$ or $v = 25 \times 10^{-3} \, \mathrm{m/s}$

Given $\Delta Q = 184 \times 10^3$. $m = 0.600 \, \mathrm{kg}$ at $-12^\circ \mathrm{C}$. $S = 222.3 \, \mathrm{J/kg}^\circ \mathrm{C}$. $L = 336 \times 10^3 \, \mathrm{J/kg}$. $Q_1 = 0.600 \times 2222.3 \times 12 = 16000.56 \, \mathrm{J}$. Remaining heat $\Delta Q_1 = 184000 - 16000.56 = 167999.44 \, \mathrm{J}$. For meeting at $0^\circ \mathrm{C}$, $\Delta Q_2 = 0.600 \times 336000 = 201600 \, \mathrm{J}$ needed. Therefore, 100$\%$ ice is not melted. Amount of ice melted: $167999.44 = m \times 336000 = 0.4999 \, \mathrm{kg}$. Therefore, mass of water $= 0.4999 \, \mathrm{kg}$. Mass of ice $= 0.1001$. Therefore, Ratio $= \frac{0.1001}{0.4999} \approx 1 : 5$.

Given the equation $$\sqrt{\frac{3RT}{M}} = \sqrt{\frac{\alpha + 5}{\alpha}} \sqrt{\frac{8}{\pi} \frac{RT}{M}}$$. Simplifying, we have $$3 = \frac{\alpha + 5}{\alpha} \frac{8}{\pi}$$. Solving for $\alpha$, we find $$\alpha = 28$$.

Given $\($ $\frac{1}{4}$ a = -25x $\)$, we have $\($ a = -100x $\)$. Since $\($ $\omega$^2 = 100 $\)$, it follows that $\($ $\omega$ = 10 $\)$. Given $\($ $\omega$ A = 4 $\)$, we find $\($ A = $\frac{4}{10}$ = 0.4 \, $\mathrm{m}$ $\)$. Thus, $\($ A = 40 \, $\mathrm{cm}$ $\)$.

The work done by the electric field is given by $\omega_E = q \vec{E} \cdot \vec{S}$. Substituting the given values, we have: $$\omega_E = 2 \times 10^{-2} \left[ 30 \hat{i} \cdot (-\hat{i}) \right]$$ This simplifies to: $$= 2 \times 10^{-2} (-30)$$ Further simplifying: $$= -60 \times 10^{-2}$$ Converting to joules: $$= \frac{-60}{100} = -0.6 \, \mathrm{J}$$ Finally, converting to millijoules: $$= -600 \, \mathrm{mJ}$$

Given $$E = -\frac{dV}{dr} = -4ar \equiv \frac{\rho r}{3 \varepsilon_0}$$ (compare). Result inside uniformly charged solid sphere $$\rho = -12a \varepsilon_0$$ $$\lambda = 12$$

Sensitivity of potentiometer wire is inversely proportional to potential gradient.

Given the equation $$\frac{\varepsilon}{r+5} \times 5 = 200x \ldots (1)$$ and $$\frac{\varepsilon \times 15}{r+15} = 300x \ldots (2)$$ solving these equations gives $$\Rightarrow r = 5 \, \Omega$$ Ans. 5

Given $B = \frac{\mu_0 i}{2R} \times 4$. Then $B' = \frac{\mu_0 i}{2R'}$. Given $R' = 4R$, we have $$B' = \frac{\mu_0 i}{8R}$$ Thus, $$\frac{B'}{B} = \frac{1}{16}$$ Finally, $B' = 2T$.

Given $\ell = 50 \, \mathrm{cm}$ and $t = 1 \, \mathrm{sec}$. Therefore, $$V = \frac{0.05}{1} = 0.05 \, \mathrm{m/s}.$$ The current $i$ is given by $$i = \frac{40 \times 0.05 \times 0.05}{10} = 0.01 \, \mathrm{A}.$$ The force $F$ is calculated as $$F = B i \ell = 40 \times 0.01 \times 0.05.$$ Thus, $$F = 0.02 \, \mathrm{N}.$$ Therefore, the work $W$ is $$W = 0.02 \times \ell = 0.02 \times 0.05.$$ Finally, $$W = 1 \times 10^{-3} \, \mathrm{J}.$$

Given a circuit with a resistance of $40 \, \Omega$ and a voltage of $220 \, \mathrm{V}$ at $50 \, \mathrm{Hz}$, the rms current $I_{\mathrm{rms}}$ is calculated as follows: $$I_{\mathrm{rms}} = \frac{220}{40} = 5.5 \, \mathrm{A}$$ In the second circuit, with a resistance of $40 \, \Omega$, an inductance of $50 \, \mathrm{mH}$, and a capacitance of $0.5 \, \mu\mathrm{F}$, $X_L$ is not equal to $X_C$. So rms current in (b) can never be larger than (a).

Given $\frac{1}{2\pi f C}=2\pi f L$. $C=\frac{1}{4\pi^2f^2L}=\frac{1}{4\times\pi^2\times49\times10^6\times2\times10^{-6}}$ $C=\frac{1}{3872}\,\mathrm{F}$ $x=3872$

Statement-I is correct as EMW are neutral. Statement-II is wrong. $$E_0 = \sqrt{\frac{1}{\mu_0 \varepsilon_0}} B_0$$

The formula for P is given by $$P = \frac{2 \mu \sin \theta}{1.22 \lambda}.$$

Given $\mu_1 = \sqrt{2.8 \times 1} = \sqrt{2.8}$. $\mu_2 = \sqrt{6.8 \times 1} = \sqrt{6.8}$. $\mu_1 \sin i = \mu_2 \cos i$ $\tan i = \frac{\mu_2}{\mu_1} = \sqrt{\frac{6.8}{2.8}}$ $\tan i = \left( \frac{2.8 + 4}{2.8} \right)^{1/2}$ $i = \tan^{-1} \left( 1 + \frac{10}{7} \right)^{1/2}$ $\theta = 7$ Ans.

Given $\($ $\frac{\lambda_\alpha}{\lambda_p}$ = $\frac{\frac{h}{\sqrt{2m_\alpha q_\alpha V}}}{\frac{h}{\sqrt{2m_p q_p V}}}$ $\)$. This simplifies to $\($ $\frac{\lambda_\alpha}{\lambda_p}$ = $\sqrt{\frac{1}{8}}$ $\)$ when $\($ m = 8 $\)$. Ans. 3

$(N_0)_A = \frac{320}{16} = 20$ moles $(N_0)_B = \frac{320}{32} = 10$ moles$N_A = \frac{(N_0)_A}{(2)^{2/1}} = \frac{20}{4} = 5$ $N_B = \frac{(N_0)_B}{(2)^{2/0.5}} = \frac{10}{2^4} = 0.625$ Total N = 5.625 No. of atoms = 5.625 $\times$ $6.023 \times 10^{23}$ = $3.38 \times 10^{24}$

The modulation index $\mu$ is given by the formula: $$\mu = \frac{A_{max} - A_{min}}{A_{max} + A_{min}}.$$ Substituting the given values: $$\mu = \frac{14 - 6}{14 + 6} = 0.4.$$

Given $\($ $\frac{R_1 + R_2}{10}$ = $\frac{60}{40}$ = $\frac{3}{2}$ $\Rightarrow$ R_1 + R_2 = 15 $\)$. Now $\($ $\frac{R_1 R_2}{(R_1 + R_2) \times 3}$ = $\frac{40}{60}$ = $\frac{2}{3}$ $\Rightarrow$ R_1 R_2 = 30 $\)$.

The radius $r$ is proportional to $\frac{n^2}{Z}$. $$r_{\mathrm{He}^+} = r_{\mathrm{H}} \times \frac{n^2}{Z}$$ Substituting the values, $$r_{\mathrm{He}^+} = 0.6 \times \frac{(3)^2}{2}$$ $$= 2.7 \, \mathrm{\AA}$$ Therefore, $$r_{\mathrm{He}^+} = 270 \, \mathrm{pm}$$

The first ionization energies (as in NCERT) are as follows: B: $801 \, \mathrm{kJ/mol}$ Al: $577 \, \mathrm{kJ/mol}$ Ga: $579 \, \mathrm{kJ/mol}$ Ga: $[\mathrm{Ar}]3d^{10}4s^24p^1$

Theory based.

Only (B) and (C) are correct. (B) $G = H - TS$ At constant $T$ $\Delta G = \Delta H - T\Delta S$ (A) First law is given by $\Delta U = Q + W$ If we apply constant $P$ and reversible work, $\Delta U = Q - P\Delta V$ (C) By definition of entropy change $dS = \dfrac{dq_{\mathrm{rev}}}{T}$ At constant $T$ $\Delta S = \dfrac{q_{\mathrm{rev}}}{T}$ (D) $H = U + PV$ For ideal gas $H = U + nRT$ At constant $T$ $\Delta H = \Delta U + \Delta(nRT)$ $\Delta H = \Delta U + nR\Delta T$ Since $\Delta T = 0$ at constant $T$ $\Delta H = \Delta U$ Choose the most appropriate answer from the options given below:

Given the reactions: $$\mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}, \ K_1 = 4 \times 10^5 ...(i)$$ $$\mathrm{N_2(g) + O_2(g) \rightleftharpoons 2NO(g)}, \ K_2 = 1.6 \times 10^{12} ...(ii)$$ $$\mathrm{H_2(g) + \frac{1}{2}O_2(g) \rightleftharpoons H_2O(g)}, \ K_3 = 1.0 \times 10^{-13} ...(iii)$$ Using the equation $$(ii) + 3 \times (iii) - (i)$$, we get: $$\mathrm{2NH_3(g) + \frac{5}{2}O_2(g) \rightleftharpoons 2NO(g) + 3H_2O(g)}$$ The equilibrium constant is calculated as: $$k_{\mathrm{eq}} = \frac{k_2 \times k_3^3}{k_1} = \frac{1.6 \times 10^{12} \times (10^{-13})^3}{4 \times 10^5}$$ Simplifying gives: $$= \frac{1.6}{4} \times 10^{-32} = 4 \times 10^{-33}$$

Mole of Na $=\dfrac{0.69}{23}=3\times10^{-2}$ $\mathrm{Na + H_2O \rightarrow NaOH + \dfrac{1}{2}H_2}$ By using POAC Moles of NaOH $=3\times10^{-2}$ NaOH reacts with HCl No. of equivalent of NaOH $=$ No. of equivalent of HCl $3\times10^{-2}\times1=\dfrac{73}{36.5}\times V\text{(in L)}\times1$ $V=1.5\times10^{-2}\ \mathrm{L}$ Volume of HCl $=15\ \mathrm{mL}$

The reaction is given by: $$\mathrm{I^- + H_2O_2 \rightarrow I_2 (\mathrm{A}) + H_2O} $$ Then, $\mathrm{I_2}$ reacts with starch as an indicator to form a blue color: $$\mathrm{I_2 + Starch (\mathrm{Indicator}) \rightarrow Blue}$$

The reaction is given by: $$2 \mathrm{LiNO_3} \xrightarrow{\Delta} \mathrm{Li_2O} + 2\mathrm{NO_2} + \frac{1}{2} \mathrm{O_2}$$ Hence three products $\mathrm{Li_2O}$, $\mathrm{NO_2}$, and $\mathrm{O_2}$.

Statement–I is correct. Ni is used in Hydrogenation of unsaturated fat to make edible fats. Statement–II is false as hydride of Silicon is electron precise and neither electron deficient nor electron rich.

The balanced chemical equation is: $$\mathrm{C_xH_y} + \left( x + \frac{y}{4} \right) \mathrm{O_2} \rightarrow x \mathrm{CO_2} + \frac{y}{2} \mathrm{H_2O}$$ Solving the equations: $$x + \frac{y}{4} = 9.5$$ $$\frac{y}{2} = 3$$ Thus, $x = 8$ and $y = 6$.

Let $M$ be the molar mass of the compound (g/mol). The mass of the compound is $0.01 \, M \, \mathrm{gm}$. The mass of carbon is $0.01 \, M \times \frac{60}{100}$. The moles of carbon are given by: $$\frac{0.01 \, M}{12} \times \frac{60}{100}$$ The moles of $\mathrm{CO_2}$ from combustion are: $$\frac{4.4}{44} = moles of carbon$$ Equating the moles of carbon: $$\frac{0.01 \, M}{12} \times \frac{60}{100} = \frac{4.4}{44}$$ Solving for $M$: $$M = \frac{4.4}{44} \times \frac{100}{60} \times \frac{12}{0.01} = 200 \, \mathrm{gm/mol}$$

The growth of fish gets inhibited if the concentration of dissolved Oxygen in water is less than 6 ppm and Biochemical Oxygen demand in clean water should be less than 5 ppm.

One unit cell of hcp contains 18 voids. No. of voids in 0.02 mol of hcp $$= \frac{18}{6} \times 6.02 \times 10^{23} \times 0.02$$ $$\approx 3.6 \times 10^{22}$$ $$\approx 36 \times 10^{21}$$

$(A)$ van't Hoff factor, $i$ $$i = \frac{\text{Normal molar mass}}{\text{Abnormal molar mass}}$$ $(B)$ $k_f =$ Cryoscopic constant $(C)$ Solutions with same osmotic pressure are known as isotonic solutions. $(D)$ Solutions with same composition of vapour over them are called Azeotrope.

$Zn(s)+Sn^{2+}(aq)\rightleftharpoons Zn^{2+}(aq)+Sn(s)$ $\Delta G^\circ=-2.303RT\log_{10}K_{eq}$ $-nFE^\circ_{cell}=-2.303RT\log_{10}K_{eq}$ $E^\circ_{Zn/Zn^{2+}}+E^\circ_{Sn^{2+}/Sn} =\dfrac{0.059}{2}\log_{10}K_{eq}$ $0.76+E^\circ_{Sn^{2+}/Sn} =\dfrac{0.059}{2}\log_{10}10^{20}$ $0.76+E^\circ_{Sn^{2+}/Sn} =\dfrac{0.059\times20}{2}$ $E^\circ_{Sn^{2+}/Sn} =0.59-0.76=-0.17$ $\left|E^\circ_{Sn^{2+}/Sn}\right| =17\times10^{-2}\,V$ Ans. $=17$

As unit of rate constant is $(conc.)^{1-n} time^{-1}$ $$\Rightarrow (\mathrm{L\ mol^{-1}}) \Rightarrow 1-n = -1$$ $$n = 2$$

A. Osmosis III B. Reverse osmosis I C. Electro osmosis IV D. Electrophoresis II

Calamine: $\mathrm{ZnCO_3}$ Siderite: $\mathrm{FeCO_3}$ Sphalerite: $\mathrm{ZnS}$ Malachite: $\mathrm{CuCO_3} \cdot \mathrm{Cu(OH)_2}$

Acidic oxides are $\mathrm{N_2O_3}$, $\mathrm{NO_2}$, $\mathrm{Cl_2O_7}$, $\mathrm{SO_2}$.

A solution of $\mathrm{CrO_5}$ in amyl alcohol has a blue colour. So, option (4) is correct.

(i), (ii) and (iv) correct. Manganese exhibits $+7$ oxidation state in its oxide ($\mathrm{Mn_2O_7}$). Ru and Os from $\mathrm{RuO_4}$ and $\mathrm{OsO_4}$ oxide in $+8$ oxidation state. Cr in $+6$ oxidation act is oxidizing. Sc does not show $+4$ oxidation state.

For $[\mathrm{FeF}_6]^{3-}$: $\mathrm{Fe}^{3+} = 3d^5$, $\Delta_o P$. Number of unpaired $e^- = 0$. Therefore, $\mu = 0 \, \mathrm{BM}$.

Copper tartarate complex. Denticity = 2

The reaction starts with $\mathrm{CH_3 - CH_2 - CH(CH_3) - CH_2 - Br}$ which undergoes a reaction to form $\mathrm{C - C - C = C}$ with a $\mathrm{CH_3}$ group attached. This is labeled as (1). Next, $\mathrm{CH_3 - CH - CH_3}$ with a $\mathrm{Br}$ group undergoes a reaction to form $\mathrm{CH_3CH=CH_2}$, also labeled as (1). Then, $\mathrm{CH_3 - CH_2 - CH_2 - CH(CH_3) - CH_3}$ with a $\mathrm{Br}$ group reacts to form $\mathrm{C - C - C - C - C = C}$ and $\mathrm{C - C - C = C - C}$, with cis and trans isomers, labeled as (3). Finally, $\mathrm{C - C - C - C - C}$ with a $\mathrm{CH_3}$ and $\mathrm{Br}$ group reacts to form $\mathrm{C - C - C - C = C}$ with a $\mathrm{C}$ group attached, labeled as (1).

The reaction starts with the compound reacting with $NaCN$ to form a cyanohydrin. This intermediate is then treated with $EtOH, H_3O^+$ to form an ester. The ester undergoes a reaction with $MeMgBr$ to form a tertiary alcohol. Further reaction with $MeMgBr$ leads to the formation of a diol.

The reaction shown is the Hoffmann Bromamide reaction. It involves the conversion of an amide to an amine with one less carbon atom. The given amide is converted to ethylamine using $\mathrm{Br_2/KOH}$. This reaction results in the removal of the carbonyl group and the formation of an amine.

Neoprene: Elastomer Polyester: Fibre Polystyrene: Thermoplastic Urea–Formaldehyde Resin: Thermosetting polymer

Theory based.

Hydrolysis of the given tetrapeptide will give the following: Phenylalanine (F), Leucine (L), Aspartic acid (D), Tyrosine (Y).

Then $\alpha_1 \alpha_2 - \alpha_3 \alpha_4 + \alpha_5 \alpha_6$ is equal to ______. Given equation can be rearranged as $$x(x^6 + 3x^4 - 13x^2 - 15) = 0$$ clearly $x = 0$ is one of the root and other part can be observed by replacing $x^2 = t$ from which we have $$t^3 + 3t^2 - 13t - 15 = 0$$ $$\Rightarrow (t - 3)(t^2 + 6t + 5) = 0$$ So, $t = 3$, $t = -1$, $t = -5$. Now we are getting $x^2 = 3$, $x^2 = -1$, $x^2 = -5$. $$\Rightarrow x = \pm \sqrt{3}, x = \pm i, x = \pm \sqrt{5}i$$ From the given condition $|\alpha_1| \geq |\alpha_2| \geq \ldots \geq |\alpha_7|$ We can clearly say that $|\alpha_7| = 0$ and $|\alpha_6| = \sqrt{5} = |\alpha_5|$ $|\alpha_4| = \sqrt{3} = |\alpha_3|$ and $|\alpha_2| = 1 = |\alpha_1|$ So we can have, $\alpha_1 = \sqrt{5}i$, $\alpha_2 = -\sqrt{5}i$, $\alpha_3 = \sqrt{3}i$, $\alpha_4 = -\sqrt{3}$, $\alpha_5 = i$, $\alpha_6 = -i$. Hence $$\alpha_1 \alpha_2 - \alpha_3 \alpha_4 + \alpha_5 \alpha_6$$ $$= 1 - (-3) + 5 = 9$$

Given $\alpha = 8 - 14i$ and $z = x + iy$. Then, $$az = (8x + 14y) + i(-14x + 8y)$$ $$z + \bar{z} = 2x z - \bar{z} = 2iy$$ Set A: $$\frac{2i(-14x + 8y)}{i(4xy - 112)} = 1$$ $$(x - 4)(y + 7) = 0$$ $$x = 4 or y = -7$$ Set B: $$x^2 + (y + 3)^2 = 16$$ When $x = 4$, $y = -3$ When $y = -7$, $x = 0$ Therefore, $A \cap B = \{4 - 3i, 0 - 7i\}$ So, $$\sum_{z \in A \cap B} (Rez - Imz) = 4 - (-3) + (0 - (-7)) = 14$$

Let's arrange the letters of OUGHT in alphabetical order. G, H, O, T, U Words starting with G$\_$$\_$$\_$$\_$ $\rightarrow$ 4! H$\_$$\_$$\_$$\_$ $\rightarrow$ 4! O$\_$$\_$$\_$$\_$ $\rightarrow$ 4! TG$\_$$\_$$\_$ $\rightarrow$ 3! TH$\_$$\_$$\_$ $\rightarrow$ 3! TOG$\_$$\_$ $\rightarrow$ 2! TOH$\_$$\_$ $\rightarrow$ 2! TOUGH $\rightarrow$ 1! Total = 89

N should be divisible by 2 but not by 3. N = (Numbers divisible by 2) - (Numbers divisible by 6) $$N = \frac{9000}{2} - \frac{9000}{6} = 4500 - 1500 = 3000$$

As, $S = \frac{a_1}{2} + \frac{a_2}{2^2} + \ldots + \frac{a_9}{2^9} + \frac{a_{10}}{2^{10}}$. Therefore, $$\frac{S}{2} = \frac{b_1}{2^1} + \frac{b_2}{2^2} + \ldots + \frac{b_9}{2^{10}} + \frac{b_{10}}{2^{11}}.$$ Subtracting, $$S - \frac{S}{2} = \frac{a_1}{2} + \frac{a_2}{2^2} + \ldots + \frac{a_9}{2^9} - \frac{b_{10}}{2^{11}}.$$ Therefore, $$\frac{S}{2} = b_1 - \frac{b_{10}}{2^{10}} + \left( \frac{a_1}{2^2} + \frac{a_2}{2^3} + \ldots + \frac{a_9}{2^9} \right).$$ $$S = b_1 - \frac{b_{10}}{2^{11}} + \left( \frac{a_1}{2} + \frac{a_2}{2^2} + \ldots + \frac{a_9}{2^{10}} \right).$$ Subtracting, $$S = \frac{b_1}{2} - \frac{b_{10}}{2^{11}} + \frac{a_1}{2^2} + \frac{a_9}{2^{10}} + \left( \frac{1}{2^2} + \frac{2}{2^3} + \ldots + \frac{8}{2^9} \right).$$ Therefore, $$S = \frac{a_1 + b_1}{2} - \frac{(b_{10} + 2a_9)}{2^{11}} + \frac{T}{4}.$$ Therefore, $$2S = 2(a_1 + b_1) - \frac{(b_{10} + 2a_9)}{2^9} + T.$$ $$2^7 (2S - T) = 2^8 (a_1 + b_1) - \frac{(b_{10} + 2a_9)}{4}.$$ Given $a_n - a_{n-1} = n - 1$, therefore, $$a_2 - a_1 = 1,$$ $$a_3 - a_2 = 2,$$ $$\vdots$$ $$a_9 - a_8 = 8.$$ Therefore, $$a_9 - a_1 = 1 + 2 + \ldots + 8 = 36.$$ Therefore, $$a_9 = 37 (a_1 = 1).$$ Also, $$b_n - b_{n-1} = a_n - a_{n-1}.$$ Therefore, $$b_{10} - b_1 = a_1 + a_2 + \ldots + a_9$$ $$= 1 + 2 + 4 + 7 + 11 + 16 + 22 + 29 + 37.$$ Therefore, $2^7(2S-T)=2^8(a_1+b_1)-\frac{b_{10}+2a_9}{4}$ $=2^9-\frac{204}{4}$ $=512-51$ $=461$.

Given that $c_k = a_k + b_k$ and $a_1 = b_1 = 4$. Also $a_2 = 4r_1$, $a_3 = 4r_1^2$, $b_2 = 4r_2$, $b_3 = 4r_2^2$. Now $c_2 = a_2 + b_2 = 5$ and $c_3 = a_3 + b_3 = \frac{13}{4}$. Therefore, $r_1 + r_2 = \frac{5}{4}$ and $r_1^2 + r_2^2 = \frac{13}{16}$. Hence, $r_1 r_2 = \frac{3}{8}$ which gives $r_1 = \frac{1}{2}$ and $r_2 = \frac{3}{4}$. $$\sum_{k=1}^{\infty} c_k - (12a_6 + 8b_4)$$ $$= \frac{4}{1 - r_1} + \frac{4}{1 - r_2} - \left(\frac{48}{32} + \frac{27}{2}\right)$$ $$= 24 - 15 = 9$$

Total 3 digit number = 900 Divisible by 3 = 300 (Using $\frac{900}{3} = 300$) Divisible by 4 = 225 (Using $\frac{900}{4} = 225$) Divisible by 3 $\&$ 4 = 108, .... (Using $\frac{900}{12} = 75$) Number divisible by either 3 or 4 = 300 + 225 - 75 = 450 We have to remove divisible by 48, 144, 192, ....., 18 terms Required number of numbers = 450 - 18 = 432

In the expansion of $(1+x)^{99}$ $=C_{0}+C_{1}x+C_{2}x^{2}+\cdots+C_{99}x^{99}$ $K=C_{1}+C_{3}+\cdots+C_{99}$ $=2^{98}$ Let $a=$ middle term in the expansion of $\left(2+\frac{1}{\sqrt{2}}\right)^{200}$ $T_{200/2+1}$ $={}^{200}C_{100}(2)^{100}\left(\frac{1}{\sqrt{2}}\right)^{100}$ $={}^{200}C_{100}2^{50}$ So, $\frac{{}^{200}C_{99}\times2^{98}}{{}^{200}C_{100}\times2^{50}}$ $=\frac{100\times2^{48}}{101}$ So,$\frac{25}{m}\times2^{50}=\frac{n}{2^{k}}$ $\therefore$ $m,n$ are odd. $(\ell,n)$ become $(50,101)$.

Given $\lambda = \cos^2 x - 2 \sin^4 x - 2 \cos^2 x$. Convert all in to $\cos x$. $$\lambda = (2 \cos^2 x - 1)^2 - 2(1 - \cos^2 x)^2 - 2 \cos^2 x$$ $$= 4 \cos^4 x - 4 \cos^2 x + 1 - 2(1 - 2 \cos^2 x + \cos^4 x) - 2 \cos^2 x$$ $$= 2 \cos^4 x - 2 \cos^2 x + 1 - 2$$ $$= 2 \cos^4 x - 2 \cos^2 x - 1$$ $$= 2 \left[ \cos^4 x - \cos^2 x - \frac{1}{2} \right]$$ $$= 2 \left[ \left( \cos^2 x - \frac{1}{2} \right)^2 - \frac{3}{4} \right]$$ $$\lambda_{\max} = 2 \left[ \frac{1}{4} - \frac{3}{4} \right] = 2 \times \left( -\frac{2}{4} \right) = -1 (max Value)$$ $$\lambda_{\min} = 2 \left[ 0 - \frac{3}{4} \right] = -\frac{3}{2} (Minimum Value)$$ So, Range = $\left$[ -$\frac{3}{2}$, -1 $\right$].

The given line is polar or $P(2, \beta)$ with respect to the given circle $$x^2 + y^2 - 4x - 6y - 3 = 0$$ Chord or contact $$\alpha x + \beta y - 2(x + \alpha) - 3(y + \beta) - 3 = 0$$ $$\Rightarrow (\alpha - 2)x + (\beta - 3)y - (2\alpha + 3\beta + 3) = 0 \ldots (i)$$ But the equation of chord of contact is given as $$x + y - 3 = 0 \ldots (ii)$$ Comparing the coefficients $$\frac{\alpha - 2}{1} = \frac{\beta - 3}{1} = -\left(\frac{2\alpha + 3\beta + 3}{-3}\right)$$ On solving $\alpha = -6$ $\beta = -5$ Now $$4\alpha - 7\beta = 11$$

Given $S_1 : y^2 = 2x$ and $S_2 : x^2 + y^2 = 4x$. The point $P(2,2)$ is a common point on $S_1$ and $S_2$. $T_1$ is tangent to $S_1$ at $P$, therefore $T_1 : y \cdot 2 = x + 2$. This implies $T_1 : x - 2y + 2 = 0$. $T_2$ is tangent to $S_2$ at $P$, therefore $T_2 : x \cdot 2 + y \cdot 2 = 2(x+2)$. This implies $T_2 : y = 2$. And $L_3 : x + y + 2 = 0$ is the third line. For the triangle $\triangle PQR$, $$PQ = a = \sqrt{20}$$ $$QR = b = \sqrt{8}$$ $$RP = c = 6$$ The area of $\triangle PQR = \Delta = \frac{1}{2} \times 6 \times 2 = 6$. Therefore, $\frac{abc}{4\Delta} = \frac{\sqrt{160}}{4} = \sqrt{10} \implies r^2 = 10$.

Given $y^2 = 3x$. Tangent $P(x_1, y_1)$ is parallel to $x + 2y = 1$. Then slope at $P = -\frac{1}{2}$. $$2y \frac{dy}{dx} = 3$$ $$\Rightarrow \frac{dy}{dx} = \frac{3}{2y} = -\frac{1}{2}$$ $$\Rightarrow y_1 = -3$$ Coordinates of $P(3, -3)$. Similarly $Q \left( \frac{4}{\sqrt{3}}, \frac{1}{\sqrt{5}} \right)$, $R \left( -\frac{4}{\sqrt{5}}, -\frac{1}{\sqrt{5}} \right)$. Area of $\triangle APQR$ $$= \frac{1}{2} \begin{vmatrix} 3 & -3 & 1 \\ \frac{4}{\sqrt{5}} & \frac{1}{\sqrt{5}} & 1 \\ -\frac{4}{\sqrt{5}} & \frac{1}{\sqrt{5}} & 1 \end{vmatrix}$$ $$= \frac{1}{2} \left[ 3 \left( \frac{2}{\sqrt{5}} \right) + 3 \left( \frac{8}{\sqrt{5}} \right) + 0 \right] = \frac{30}{2\sqrt{5}} = 3\sqrt{5}$$

\begin{tabular}{|l|l|l|l|} \hline p & q & (p $\rightarrow$ q) & (p $\lor$ q) & (p $\rightarrow$ q) $\land$ (p $\lor$ q) \\ \hline T & T & F & T & T \\ \hline T & F & F & F & T \\ \hline F & T & T & T & T \\ \hline F & F & T & T & T \\ \hline \end{tabular} \begin{tabular}{|l|l|l|l|} \hline A $\Rightarrow$ B & $\neg$ A $\Rightarrow$ B & B $\Rightarrow$ (A $\Rightarrow$ B) & A $\Rightarrow$ (($\neg$ A) $\Rightarrow$ B) & B $\Rightarrow$ (($\neg$ A) $\Rightarrow$ B) \\ \hline T & T & T & T & T \\ \hline F & T & T & T & T \\ \hline T & T & T & T & T \\ \hline T & F & T & T & T \\ \hline \end{tabular}

Given $\bar{x} = \frac{1}{31} \sum_{i=11}^{41} i = \frac{11 + 41}{2} = 26$ (31 elements) and $\bar{y} = \frac{1}{31} \sum_{j=61}^{91} j = \frac{61 + 91}{2} = 76$ (31 elements). Combined mean, $\mu = \frac{31 \times 26 + 31 \times 76}{31 + 31} = \frac{26 + 76}{2} = 51$. $$\sigma^2 = \frac{1}{62} \left( \sum_{i=1}^{31} (x_i - \mu)^2 + \sum_{i=1}^{31} (y_i - \mu)^2 \right) = 705$$ Since, $x_i \in X$ are in A.P. with 31 elements & common difference 1, same is $y_i \in y$, when written in increasing order. $$\therefore \sum_{i=1}^{31} (x_i - \mu)^2 = \sum_{i=1}^{31} (y_i - \mu)^2 = 10^2 + 11^2 + \ldots + 40^2$$ $$= \frac{40 \times 41 \times 81}{6} - \frac{9 \times 10 \times 19}{6} = 21855$$ $$\therefore \left| \bar{x} + \bar{y} - \sigma^2 \right| = \left| 26 + 76 - 705 \right| = 603$$

Given $a \, R \, a \Rightarrow 5a$ is multiple of $5$. So reflexive. $a \, R \, b \Rightarrow 2a + 3b = 5\alpha$. Now $b \, R \, a$. $$2b + 3a = 2b + \left(\frac{5\alpha - 3b}{2}\right) \cdot 3$$ $$= \frac{15}{2}\alpha - \frac{5}{2}b = \frac{5}{2}(3\alpha - b)$$ $$= \frac{5}{2}(2a + 2b - 2\alpha)$$ $$= 5(a + b - \alpha)$$ Hence symmetric. $a \, R \, b \Rightarrow 2a + 3b = 5\alpha$. $b \, R \, c \Rightarrow 2b + 3c = 5\beta$. Now $$2a + 5b + 3c = 5(\alpha + \beta)$$ $$\Rightarrow 2a + 5b + 3c = 5(\alpha + \beta)$$ $$\Rightarrow 2a + 3c = 5(\alpha + \beta - b)$$ $$\Rightarrow a \, R \, c$$ Hence relation is equivalence relation.

If it is invertible, then determinant value $\neq 0$. So, $$\begin{vmatrix} e^t & e^{-t}(\sin t - 2 \cos t) & e^{-t}(-2 \sin t - \cos t) \\ e^t & e^{-t}(2 \sin t + \cos t) & e^{-t}(\sin t - 2 \cos t) \\ e^t & e^{-t} \cos t & e^{-t} \sin t \end{vmatrix} \neq 0$$ $$\Rightarrow e^t \cdot e^{-t} \cdot e^{-t} \begin{vmatrix} 1 & \sin t - 2 \cos t & -2 \sin t - \cos t \\ 1 & 2 \sin t + \cos t & \sin t - 2 \cos t \\ 1 & \cos t & \sin t \end{vmatrix} \neq 0$$ Applying, $R_1 \rightarrow R_1 - R_2$ then $R_2 \rightarrow R_2 - R_3$, we get $$e^{-t} \begin{vmatrix} 0 & -\sin t - \cos t & -3 \sin t + \cos t \\ 0 & 2 \sin t & -2 \cos t \\ 1 & \cos t & \sin t \end{vmatrix} \neq 0$$ By expanding we have, $$e^{-t} \times 1 (2 \sin t \cos t + 6 \cos^2 t + 6 \sin^2 t - 2 \sin t \cos t) \neq 0$$ $$\Rightarrow e^{-t} \times 6 \neq 0$$ for $\forall \ t \in \mathbb{R}$

Now $ac - b^2 = 2$ and $2a + b = 1$ and $2b + c = 2$. Solving all these above equations we get $$\frac{1-b}{2} \times \frac{(2-2b)}{1} - b^2 = 2$$ $$\Rightarrow (1-b)^2 - b^2 = 2$$ $$\Rightarrow 1 - 2b = 2$$ $$\Rightarrow b = -\frac{1}{2} and a = \frac{3}{4} and c = 3$$ Hence $\alpha = 3a + \frac{3b}{2} = \frac{9}{4} - \frac{3}{4} = \frac{3}{2}$ and $\beta = 3b + \frac{3c}{2} = -\frac{3}{2} + \frac{9}{2} = 3$. Also $s = a + c = \frac{15}{4}$. Therefore, $$\frac{\beta s}{\alpha^2} = \frac{3 \times 15}{4 \times \frac{9}{4}} = 5$$

Given for $x \geq 2$ $f(1) + 2f(2) + \ldots + xf(x) = x(x+1)f(x)$ Replace $x$ by $x+1$ $\[$ $\Rightarrow$ x(x+1)f(x) + (x+1)f(x+1) = (x+1)(x+2)f(x+1) $\]$ $\[$ $\Rightarrow$ $\frac{x}{f(x+1)}$ + $\frac{1}{f(x)}$ = $\frac{(x+2)}{f(x)}$ $\]$ $\[$ $\Rightarrow$ x f(x) = (x+1)f(x+1) = $\frac{1}{2}$, x $\geq$ 2 $\]$ $f(2) = \frac{1}{4}, f(3) = \frac{1}{6}$ Now $f(2022) = \frac{1}{4044}$ $f(2028) = \frac{1}{4056}$ So, $\[$ $\frac{1}{f(2022)}$ + $\frac{1}{f(2028)}$ = 4044 + 4056 = 8100 $\]$

Given $$f''(x) = g''(x) + 6x (1)$$ $$f'(1) = 4g'(1) - 3 = 9 (2)$$ $$f(2) = 3g(2) = 12 (3)$$ By integrating (1) $$f'(x) = g'(x) + 6 \frac{x^2}{2} + C$$ At $x = 1$, $$f'(1) = g'(1) + 3 + C$$ $$\Rightarrow 9 = 4 + 3 + C \Rightarrow C = 3$$ $$\therefore \ f'(x) = g'(x) + 3x^2 + 3$$ Again by integrating, $$f(x) = g(x) + \frac{3x^3}{3} + 3x + D$$ At $x = 2$, $$f(2) = g(2) + 8 + 3(2) + D$$ $$\Rightarrow 12 = 4 + 8 + 6 + D \Rightarrow D = -6$$ So, $$f(x) = g(x) + x^3 + 3x - 6$$ $$\Rightarrow f(x) - g(x) = x^3 + 3x - 6$$ At $x = -2$, $$\Rightarrow g(-2) - f(-2) = 20 (Option (1) is true)$$ Now, for $-1 < x < 2$, $$h(x) = f(x) - g(x) = x^3 + 3x - 6$$ $$\Rightarrow h'(x) = 3x^2 + 3$$ $$\Rightarrow h(x) \uparrow$$ So, $h(-1) < h(x) < h(2)$ $$\Rightarrow -10 < h(x) < 8$$ $$\Rightarrow |h(x)| < 10 (option (2) is NOT true)$$

Given $\($ y = $\frac{x-a}{(x+b)(x-2)}$ $\)$. At point $\($(1, -3)$\)$, $\[$ -3 = $\frac{1-9}{(1+b)(1-2)}$ $\]$ $\($ $\Rightarrow$ 1-a = 3(1+b) $\)$ $\($ $\ldots$ $\)$ (1) Now, $\($ y = $\frac{x-a}{(x+b)(x-2)}$ $\)$ $\[$ $\Rightarrow$ $\frac{dy}{dx}$ = $\frac{(x+b)(x-2)(1)-(x-a)(2x+b-2)}{(x+b)^2(x-2)^2}$ $\]$ At $\($(1, -3)$\)$ slope of normal is $\($ $\frac{1}{4}$ $\)$ hence $\($ $\frac{dy}{dx}$ = -4 $\)$, So, $\[$ -4 = $\frac{(1+b)(-1)-(1-a)b}{(1+b)^2(-1)^2}$ $\]$ Using equation (1) $\[$ $\Rightarrow$ -4 = $\frac{(1+b)(-1)-3(b+1)b}{(1+b)^2}$ $\]$ $\[$ $\Rightarrow$ -4 = $\frac{(-1)-3b}{(1+b)}$ (b $\neq$ -1) $\]$ $\($ $\Rightarrow$ b = -3 $\)$ So, $\($ a = 7 $\)$ Hence, $\($ a+b = 7-3 = 4 $\)$

Given $$I = \int_{1}^{2} \left( \frac{t^4 + 1}{t^6 + 1} \right) \, dt$$ This can be rewritten as $$= \int_{1}^{2} \left( \frac{t^4 + 1 - t^2 + t^2}{(t^2 + 1)(t^4 - t^2 + 1)} \right) \, dt$$ Simplifying, we have $$= \int_{1}^{2} \left( \frac{1}{t^2 + 1} + \frac{t^2}{t^6 + 1} \right) \, dt$$ Further simplifying, $$= \int_{1}^{2} \left( \frac{1}{t^2 + 1} + \frac{1}{3} \frac{3t^2}{(t^3)^2 + 1} \right) \, dt$$ This results in $$= \tan^{-1}(t) + \frac{1}{3} \tan^{-1}(t^3) \bigg|_{1}^{2}$$ Evaluating the integral, $$= (\tan^{-1}(2) - \tan^{-1}(1)) + \frac{1}{3} (\tan^{-1}(2^3) - \tan^{-1}(1^3))$$ Finally, we have $$= \tan^{-1}(2) + \frac{1}{3} \tan^{-1}(8) - \frac{\pi}{3}$$

Given $$I = \int_{1/2}^{2} \frac{\tan^{-1} x}{x} \, dx ...... (i)$$ Put $x = \frac{1}{t}$, $dx = -\frac{1}{t^2} \, dt$ $$I = -\int_{2}^{1/2} \frac{\tan^{-1} \frac{1}{t}}{\frac{1}{t}} \cdot \frac{1}{t^2} \, dt = -\int_{2}^{1/2} \frac{\tan^{-1} \frac{1}{t}}{t} \, dt$$ $$I = \int_{1/2}^{2} \frac{\cot^{-1} t}{t} \, dt = \int_{1/2}^{2} \frac{\cot^{-1} x}{x} \, dx ...... (ii)$$ Add both equations $$2I = \int_{1/2}^{2} \frac{\tan^{-1} x + \cot^{-1} x}{x} \, dx = \frac{\pi}{2} \int_{1/2}^{2} \frac{dx}{x} = \frac{\pi}{2} (\ln 2)_{1/2}^{2}$$ $$= \frac{\pi}{2} \left( \ln 2 - \ln \frac{1}{2} \right) = \pi \ln 2$$ $$I = \frac{\pi}{2} \ln 2$$

Given $|\cos x - \sin x| \leq y \leq \sin x$. Intersection point of $\cos x - \sin x = \sin x$ implies $\tan x = \frac{1}{2}$. Let $\psi = \tan^{-1} \frac{1}{2}$. So, $\tan \psi = \frac{1}{2}$, $\sin \psi = \frac{1}{\sqrt{5}}$, $\cos \psi = \frac{2}{\sqrt{5}}$. Area $= \int_{\psi}^{\pi/2} (\sin x - |\cos x - \sin x|) \, dx$ $= \int_{\psi}^{\pi/4} (\sin x - (\cos x - \sin x)) \, dx$ $+ \int_{\pi/4}^{\pi/2} (\sin x - (\sin x - \cos x)) \, dx$ $= \int_{\psi}^{\pi/4} (2 \sin x - \cos x) \, dx + \int_{\pi/4}^{\pi/2} \cos x \, dx$ $= [-2 \cos x - \sin x]_{\psi}^{\pi/4} + [\sin x]_{\pi/4}^{\pi/2}$ $= -\sqrt{2} - \frac{1}{\sqrt{2}} + 2 \cos \psi + \sin \psi + \left(1 - \frac{1}{\sqrt{2}}\right)$

Given $x \log_e x \frac{dy}{dx} + y = x^2 \log_e x$, $(x > 1)$. $$\Rightarrow \frac{dy}{dx} + \frac{y}{x \ln x} = x$$ Linear differential equation I.F. $= e^{\int \frac{1}{x \ln x} \, dx} = |\ln x|$ Therefore, solution of differential equation $$y |\ln x| = \int x |\ln x| \, dx$$ $$= |\ln x| \frac{x^2}{2} - \int \frac{1}{x} \cdot \frac{x^2}{2} \, dx$$ $$\Rightarrow y |\ln x| = |\ln x| \left( \frac{x^2}{2} \right) - \frac{x^2}{4} + c$$ For constant $y(2) = 2 \Rightarrow c = 1$ So, $y(x) = \frac{x^2}{2} - \frac{x^2}{4 |\ln x|} + \frac{1}{|\ln x|}$ Hence, $y(e) = \frac{e^2}{2} - \frac{e^2}{4} + 1 = 1 + \frac{e^2}{4}$

$$\vec{r}\times\vec{b}-\vec{c}\times\vec{b}=0$$ $$\Rightarrow (\vec{r}-\vec{c})\times\vec{b}=0$$ $$\Rightarrow \vec{r}-\vec{c}=\lambda\vec{b}$$ $$\Rightarrow \vec{r}=\vec{c}+\lambda\vec{b}$$ And given that $$\vec{r}\cdot\vec{a}=0$$ $$\Rightarrow (\vec{c}+\lambda\vec{b})\cdot\vec{a}=0$$ $$\Rightarrow \vec{c}\cdot\vec{a}+\lambda\,\vec{b}\cdot\vec{a}=0$$ $$\Rightarrow \lambda=\frac{-\,\vec{c}\cdot\vec{a}}{\vec{b}\cdot\vec{a}}$$ Now $$\vec{r}\cdot\vec{c}=(\vec{c}+\lambda\vec{b})\cdot\vec{c}$$ $$=\left(\vec{c}-\frac{\vec{c}\cdot\vec{a}}{\vec{b}\cdot\vec{a}}\vec{b}\right)\cdot\vec{c}$$ $$=|\vec{c}|^{2}-\left(\frac{\vec{c}\cdot\vec{a}}{\vec{b}\cdot\vec{a}}\right)(\vec{b}\cdot\vec{c})$$ $$=74-\left(\frac{15}{3}\right)8$$ $$=74-40=34$$

Given $\vec{a} \times \vec{b} = 15\hat{i} - 20\hat{j} - 25\hat{k}$. Let $\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}$. Therefore, $15x - 20y - 25z + 25 = 0$. This simplifies to $3x - 4y - 5z = -5$. Also, $x + y + z = 4$. And $\frac{\vec{c} \cdot \vec{a}}{|\vec{a}|} = 1$ implies $4x + 3y = 5$. Therefore, $\vec{c} = 2\hat{i} - \hat{j} + 3\hat{k}$. The projection of $\vec{c}$ or $\vec{b}$ is $\frac{25}{5\sqrt{2}} = \frac{5}{\sqrt{2}}$.

Given $\($ $\frac{x-1}{2}$ = $\frac{y+8}{-7}$ = $\frac{z-4}{5}$ $\)$ and $\($ $\frac{x-1}{2}$ = $\frac{y-2}{1}$ = $\frac{z-6}{-3}$ $\)$. $\($ $\mathbf{p}$ = 2$\hat{i}$ - 7$\hat{j}$ + 5$\hat{k}$, $\mathbf{q}$ = 2$\hat{i}$ + $\hat{j}$ - 3$\hat{k}$ $\)$. $\($ $\mathbf{a}$ = $\hat{i}$ - 8$\hat{j}$ + 4$\hat{k}$ $\)$ and $\($ $\mathbf{b}$ = $\hat{i}$ + 2$\hat{j}$ + 6$\hat{k}$ $\)$. Calculate $\($ $\mathbf{p}$ $\times$ $\mathbf{q}$ $\)$: $$ \mathbf{p} \times \mathbf{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -7 & 5 \\ 2 & 1 & -3 \end{vmatrix} $$ $$ = \hat{i}(16) - \hat{j}(-16) + \hat{k}(16) $$ $$ = 16(\hat{i} + \hat{j} + \hat{k}) $$ Calculate $\($ d = $\frac{|(\mathbf{a} - \mathbf{b}) \cdot (\mathbf{p} \times \mathbf{q})|}{|\mathbf{p} \times \mathbf{q}|}$ $\)$: $$ d = \frac{|(-10\hat{j} - 2\hat{k}) \cdot 16(\hat{i} + \hat{j} + \hat{k})|}{16\sqrt{3}} $$ $$ = \frac{|-12|}{\sqrt{3}} = 4\sqrt{3} $$

Given points A(a, -2, 4), B(2, b, -3). The ratio AC : CB = 2 : 1. Therefore, $$C \equiv \left( \frac{a+4}{3}, \frac{2b-2}{3}, \frac{-2}{3} \right)$$ C lies on the line $2x - y + 2 = 4$. Thus, $$\frac{2a+8}{3} - \frac{2b-2}{3} - \frac{2}{3} = 4$$ This implies $a - b = 2 \ldots (1)$ Also, $OC = \sqrt{5}$ Therefore, $$\left( \frac{a+4}{3} \right)^2 + \left( \frac{2b-2}{3} \right)^2 + \frac{4}{9} = 5 \ldots (2)$$ Solving equations (1) and (2), $$(b+6)^2 + (2b-2)^2 = 41$$ This implies $$5b^2 + 4b - 1 = 0$$ Thus, $b = -1$ or $\frac{1}{5}$ Therefore, $a = 1$ or $\frac{11}{5}$ But $ab < 0$ implies $(a, b) = (1, -1)$ Thus, $$C \equiv \left( \frac{5}{3}, -\frac{4}{3}, -\frac{2}{3} \right), \ P \equiv (2, -1, -3)$$ Finally, $$CP^2 = \frac{1}{9} + \frac{1}{9} + \frac{49}{9} = \frac{51}{9} = \frac{17}{3}$$

Point on $L_1 \equiv (\lambda + 1, 2\lambda + 2, \lambda - 3)$ Point on $L_2 \equiv (2\mu + a, 3\mu - 2, \mu + 3)$ $\lambda - 3 = \mu + 3 \Rightarrow \lambda = \mu + 6 \ldots (1)$ $2\lambda + 2 = 3\mu - 2 \Rightarrow 2\lambda = 3\mu - 4 \ldots (2)$ Solving, (1) and (2) $\Rightarrow \lambda = 22 and \mu = 16$ $\Rightarrow P \equiv (23, 46, 19)$ $\Rightarrow a = -9$

Let $P(w_1)=\lambda$. Then $P(w_2)=\frac{\lambda}{2}, \ldots, P(w_n)=\frac{\lambda}{2^{n-1}}$. As $$ \sum_{k=1}^{\infty} P(w_k)=1 \implies \frac{\lambda}{1-\frac{1}{2}}=1 \implies \lambda=\frac{1}{2}. $$ So, $$ P(w_n)=\frac{1}{2^n}. $$ Also, $$ A=\{2k+3\ell;\,k,\ell\in\mathbb{N}\} =\{5,7,8,9,10,\ldots\}. $$ Hence, $$ B=\{w_n:n\in A\} =\{w_5,w_7,w_8,w_9,w_{10},w_{11},\ldots\}. $$ Observe that $$ A=\mathbb{N}\setminus\{1,2,3,4,6\}. $$ Therefore, $$ P(B) =1-\left[P(w_1)+P(w_2)+P(w_3)+P(w_4)+P(w_6)\right] $$ $$ =1-\left[ \frac{1}{2} +\frac{1}{4} +\frac{1}{8} +\frac{1}{16} +\frac{1}{64} \right] $$ $$ =1-\frac{32+16+8+4+1}{64} =\frac{3}{64}. $$