JEE Main 24 January 2023 Shift 2 question paper with solutions

Given $\vec{P} \cdot \vec{Q} = 0$. $$\left( \hat{i} + 2m \hat{j} + m \hat{k} \right) \cdot \left( 4 \hat{i} - 2 \hat{j} + m \hat{k} \right) = 0$$ This implies: $$4 - 4m + m^2 = 0$$ Solving the equation: $$(m - 2)^2 = 0 \implies m = 2$$

Given $\($[T^{-1}] = [L^1]^a [M^1 L^{-3}]^b $\left$[ $\frac{MLT^{-2}}{L}$ $\right$]^c$\)$. This implies $\($T^{-1} = M^{b+c} $\cdot$ L^{a-3b} $\cdot$ T^{-2c}$\)$. Given $\($c = $\frac{1}{2}$, b = -$\frac{1}{2}$, a - 3b = 0$\)$. Then $\($a + $\frac{3}{2}$ = 0 $\Rightarrow$ a = -$\frac{3}{2}$$\)$.

Displacement $= \Sigma area = 16 - 8 + 16 - 8 = 16 \, \mathrm{m}$ Distance $= \Sigma |area| = 48 \, \mathrm{m}$ $$\frac{displacement}{Distance} = \frac{1}{3}$$

Natural length = $L_0$ Extension = $x$ $$kx = m(L_0 + x)\omega^2$$ $$\Rightarrow 12.5x = \frac{1}{5}(L_0 + x)25 \Rightarrow 1.5x = L_0$$ $$\Rightarrow \frac{x}{L_0} = \frac{2}{3}$$

Given $\mathbf{F} = t \hat{i} + 3t^2 \hat{j}$. The equation $m \frac{d\mathbf{v}}{dt} = t \hat{i} + 3t^2 \hat{j}$ is given. With $m = 1 kg$, we have: $$\int_0^{\mathbf{v}} dv = \int_0^t t dt \hat{i} + \int_0^t 3t^2 dt \hat{j}$$ This results in: $$\mathbf{v} = \frac{t^2}{2} \hat{i} + t^3 \hat{j}$$ The power is given by: $$Power = \mathbf{F} \cdot \mathbf{V} = \frac{t^3}{2} + 3t^5$$ At $t = 2$, the power is: $$\frac{8}{2} + 3 \times 32$$ This equals $100$.

Given the two cylinders, the moment of inertia for the first cylinder is $$I_1 = \frac{m_1 R^2}{2}$$ and for the second cylinder is $$I_2 = \frac{m_2 \left(\frac{R}{2}\right)^2}{2}.$$ The ratio of the moments of inertia is $$\frac{I_1}{I_2} = \frac{4m_1}{m_2} = \frac{4 \cdot \rho \pi R^2 \ell}{\rho \cdot \frac{\pi R^2}{4} \times \frac{\ell}{2}} \Rightarrow \frac{I_1}{I_2} = 32.$$

Given the equations for gravitational acceleration: $$g' = \frac{g}{\left(1 + \frac{h}{R}\right)^2}$$ $$g' = g \left\{ 1 - \frac{d}{R} \right\}$$ The graph shows the variation of $g$ with respect to $r$ where $r = R$. Statement I is correct and Statement II is incorrect.

The period $T$ is proportional to the inverse of the square root of $g$: $$T \propto \frac{1}{\sqrt{g}}.$$

Given $T^2 \propto R^3$, we have: $$\left( \frac{T_1}{T_2} \right)^2 = \left( \frac{R_1}{R_2} \right)^3$$ Substituting the given values: $$\left( \frac{1}{2.83} \right)^2 = \frac{(1.5 \times 10^6)^3}{R_2}$$ Solving for $R_2$: $$R_2 = \left[ (2.83)^2 \times (1.5 \times 10^6)^3 \right]^{1/3}$$ Calculating the final value: $$= 8^{1/3} \times 1.5 \times 10^6 = 3 \times 10^6 \, km$$

Concept based

When the ball attains terminal velocity $$F_v = (mg - F_B) (\because a = 0)$$ $$= V \sigma_b g - V \rho_\ell g$$ $$= Vg (\sigma_b - \rho_\ell)$$ $$= \frac{4}{3} \pi (10^{-3})^3 \times 9.8 \times (10.5 - 1.5) \times 10^3$$ $$= 3696 \times 10^{-7} \, \mathrm{N}$$ So, $x = 7$

For monoatomic gas $\gamma_1 = \frac{5}{3}$. For diatomic gas at low temperatures $\gamma_2 = \frac{7}{5}$. Therefore, $$\frac{\gamma_1}{\gamma_2} = \frac{\frac{5}{3}}{\frac{7}{5}} = \frac{25}{21}.$$

Given $T = 2\pi \sqrt{\frac{m}{k}} = 1$ and $T' = 2\pi \sqrt{\frac{m+3}{k}} = 2$. The ratio $\frac{T}{T'} = \sqrt{\frac{m}{m+3}} = \frac{1}{2}$. This implies $\frac{m}{m+3} = \frac{1}{4}$. Solving for $m$, we get $m = 1$.

Potential at centre $$V = \frac{(\lambda \cdot \pi R_2)}{4 \pi \varepsilon_0 R_2} + \frac{(\lambda \cdot \pi R_1)}{4 \pi \varepsilon_0 R_1}$$ $$= \frac{\lambda}{2 \varepsilon_0}$$

Given $C_0 = \frac{\varepsilon_0 A}{d} = 15 \, \mathrm{pF}$. Then, $C = \frac{K \varepsilon_0 A}{2d} = \frac{3.5}{2} \times 15 \, \mathrm{pF} = \frac{105}{4} \, \mathrm{pF}$.

The equivalent resistance is calculated as follows: $$R_{eq} = \frac{400 \times 100}{500} + 100$$ $$= 180 \, \Omega$$ The current $i$ is given by: $$i = \frac{90}{180} = \frac{1}{2} \, A$$ The reading is calculated as: $$Reading = \frac{1}{2} \times \frac{400}{500} \times 100$$ $$= 40 \, volt$$

As volume is constant, so resistance is proportional to (length)$^2$. $$\Rightarrow % change in resistance = 20 + 20 + \frac{400}{100} = 44\%$$

Given $B = \mu_0 n I$. $$= 4 \pi \times 10^{-7} \times 70 \times 10^2 \times 2$$ $$= 56 \pi \times 10^{-4} \, \mathrm{T}$$ $$= 176 \times 10^{-4} \, \mathrm{T}$$

Force on 5 cm side is $|\vec{F}| = ILB \sin \theta$. $$= (2)(5 \times 10^{-2}) \times \frac{3}{4} \times \frac{12}{13} = \frac{9}{130} \, \mathrm{N}$$ So, $x = 9$

The integral of the electromotive force is given by $$\int d\varepsilon = \int B(\omega x) \, dx$$. Evaluating the integral, we have $$\varepsilon = B \omega \int_0^L x \, dx = \frac{B \omega L^2}{2}.$$

After long time an inductor behaves as a resistance-less path. So current through cell $$I = \frac{12}{R/3} = 3 \, \mathrm{A} \{\because R = 12 \, \Omega\}$$

Given the equation for C: $$C = \frac{\omega}{k} = \frac{E_0}{B_0}$$

Based on fact.

Given $\($ $\frac{1}{f_{\mathrm{H_2O}}}$ = $\left$( $\frac{\mu_g}{\mu_{\mathrm{H_2O}}}$ - 1 $\right$) $\left$( $\frac{2}{R}$ $\right$) $\)$ $\($ = $\frac{1}{8}$ $\left$( $\frac{2}{R}$ $\right$) $\)$ $\($ = $\frac{1}{(4f_{\mathrm{air}})}$ $\)$ So, $\($ f_{$\mathrm{H_2O}$} = 4f_{$\mathrm{air}$} = 72 \, $\mathrm{cm}$ $\)$ So change in focal length $\($ = 72 - 18 = 54 \, $\mathrm{cm}$ $\)$

Given $\lambda_D = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$. Therefore, $\lambda \propto \frac{1}{\sqrt{m}}$. Since $m_\alpha > m_p > m_e$, it follows that $\lambda_e > \lambda_p > \lambda_\alpha$.

Given $( \frac{hc}{\lambda} = \left[ 1 - \frac{1}{16} \right] (13.6\ \mathrm{eV}) )$. So, $( \lambda = 94.1\ \mathrm{nm} )$.

Number of moles is given by $$\frac{120}{240} = \frac{1}{2}$$. The number of molecules is $$\frac{1}{2} \times N_A$$. The energy released is $$\frac{1}{2} \times 6 \times 10^{23} \times 200$$ which equals $$6 \times 10^{25} \, MeV$$.

The truth table shows the values for $A_1$, $B_1$, and $Y$. The expression for $Y$ is given by: $$Y = \overline{A_1 \odot B_1} NAND$$

AM Broadcast $540 - 1600 \, \mathrm{KHz}$ FM Broadcast $\rightarrow 88 - 108 \, \mathrm{MHz}$ Television $\rightarrow 54 - 890 \, \mathrm{MHz}$ Satellite communication $\rightarrow 3.7 - 4.2 \, \mathrm{GHz}$ Therefore: A-II, B-I, C-IV, D-III

The spectrum of Black body radiation is explained using quantization of energy. With increase in temperature, peak of spectrum shifts to shorter wavelength or higher frequency. For above graph $\rightarrow T_1 > T_2 > T_3 > T_4$.

For $\mathrm{N_2}$: $$\sigma 1s^2 \, \sigma^* 1s^2 \, \sigma 2s^2 \, \sigma^* 2s^2 \, \pi 2p_x^2 = \pi 2p_y^2$$ $$\frac{\sigma 2p_z^2}{HOMO}$$ For $\mathrm{N_2^+}$: $$\sigma 1s^2 \, \sigma^* 1s^2 \, \sigma 2s^2 \, \sigma^* 2s^2 \, \pi 2p_x^2 = \pi 2p_y^2$$ $$\frac{\sigma 2p_z^1}{HOMO}$$ For $\mathrm{O_2}$: $$\sigma 1s^2 \, \sigma^* 1s^2 \, \sigma 2s^2 \, \sigma^* 2s^2 \, \sigma 2p_z^2$$ $$\pi 2p_x^2 = \pi 2p_y^2$$ $$\pi^* 2p_x^1 = \pi^* 2p_y^1 (HOMO)$$ For $\mathrm{O_2^+}$: $$\sigma 1s^2 \, \sigma^* 1s^2 \, \sigma 2s^2 \, \sigma^* 2s^2 \, \sigma 2p_z \, \pi 2p_x = \pi 2p_y^2$$ $$\pi^* 2p_x^1 = \pi^* 2p_y^0 (HOMO)$$ $\mathrm{N_2} \Rightarrow 0$ unpaired $e^-$ in HOMO $\mathrm{N_2^+} \Rightarrow 1$ unpaired $e^-$ in HOMO $\mathrm{O_2} \Rightarrow 2$ unpaired $e^-$ in HOMO $\mathrm{O_2^+} \Rightarrow 1$ unpaired $e^-$ in HOMO

At $(a) \rightarrow \mathrm{CO_2}$ exists as gas. $(b) \rightarrow$ liquefaction of $\mathrm{CO_2}$ starts. $(c) \rightarrow$ liquefaction ends. $(d) \rightarrow \mathrm{CO_2}$ exists as liquid. Between $(b)$ and $(c) \rightarrow$ liquid and gaseous $\mathrm{CO_2}$ co-exist. As volume changes from $(b)$ to $(c)$, gas decreases and liquid increases. $(A), (C) \rightarrow$ correct.

$1 \rightarrow 2 \Rightarrow$ Isobaric process $2 \rightarrow 3 \Rightarrow$ Isochoric process $3 \rightarrow 1 \Rightarrow$ Isothermal process $W = W_{1\to2} + W_{2\to3} + W_{3\to1}$ $= \left[-P(V_2-V_1)+0+\left(-P_1V_1\ln\left(\frac{V_2}{V_1}\right)\right)\right]$ $= \left[-1\times(40-20)+0+\left(-1\times20\ln\left(\frac{20}{40}\right)\right)\right]$ $= -20 + 20\ln2$ $= -20 + 20\times2.3\times0.3$ $= -6.2\ \text{bar L}$ $|W| = 6.2\ \text{bar L} = 620\ \text{J}$

Concentration of calcium lactate $=0.005\ \mathrm{M}$. Concentration of lactate ion $=(2\times0.005)\ \mathrm{M}$. Calcium lactate is a salt of weak acid + strong base. $\therefore$ Salt hydrolysis will take place. $\mathrm{pH}=7+\frac{1}{2}\left(\mathrm{p}K_a+\log C\right)$ $=7+\frac{1}{2}\left(5+\log(2\times0.005)\right)$ $=7+\frac{1}{2}\left(5-2\log10\right)$ $=7+\frac{1}{2}\times3$ $=8.5=85\times10^{-1}$

The reaction is given by $\mathrm{C_6H_5COOH} + \mathrm{NaOH} \rightarrow \mathrm{C_6H_5COONa} + \mathrm{H_2O}$. (A) $\rightarrow$ (B) Free $\mathrm{H^+}$ ions are replaced by $\mathrm{Na^+}$ which decreases conductance. (B) $\rightarrow$ $(C)$ Un-dissociated benzoic acid reacts with $\mathrm{NaOH}$ and forms salt which increases ions and conductance increases. $(C)$ $\rightarrow$ (D) After equivalence point at (3), $\mathrm{NaOH}$ added further increases $\mathrm{Na^+}$ and $\mathrm{OH^-}$ ions which further increases the conductance.

The reaction is given by: $$\mathrm{HOCl} + \mathrm{H_2O_2} \rightarrow \mathrm{H_3O^+} + \mathrm{Cl^-} + \mathrm{O_2}$$ In this reaction, $\mathrm{HOCl}$ acts as the oxidizing agent (O.A) and $\mathrm{H_2O_2}$ acts as the reducing agent (R.A). The $\mathrm{Cl}$ is reduced and the $\mathrm{O}$ is oxidized.

(1) $\mathrm{Na} > \mathrm{Cs} > \mathrm{Li}$ – true (If considered with sign) The low solubility of $\mathrm{CsI}$ is due to smaller hydration enthalpy of its two ions. $\mathrm{Li_2CO_3}$ is highly stable to heat – false In Conc. $\mathrm{NH_3}$, $\mathrm{K}$ formed blue solution – true

Be has less negative value compared to other AEM. However, its reducing nature is due to large hydration energy associated with the small size of $\mathrm{Be^{2+}}$ ion and relatively large value of the atomization enthalpy of metal.

Given $Z = 55$ for $[\mathrm{Cs}]$, the configuration is $[\mathrm{Xe}] \, 6s^1$. For $[\mathrm{Cs}^+]$, the configuration is $[\mathrm{Xe}]$, i.e., up to $5s$ count $e^-$ of the $s$-subshell. This means $1s, 2s, 3s, 4s, 5s$ contribute to 10 electrons.

Most easily deprotonation. (More resonance stabilised)

For S-I, the reaction with $\mathrm{Zn-Hg}$ and conc. $\mathrm{HCl}$ does not lead to the reduction with hydrolysis of amide at high temperature only. Therefore, it is false. For S-II, the reaction with $\mathrm{NH_2NH_2}$ and $\mathrm{OH^-}$/Glycol does not lead to reduction with elimination. Therefore, it is false.

Assertion – A: Benzene is more stable than cyclohexatriene (True) Reason – R: Delocalised $\pi$–e cloud lies B.M.O so more attracted by nuclei of carbon atom. (True & Correct Explanation)

Let V.P. of pure A be $P_A^0$. Let V.P of pure B be $P_B^0$. When $X_A = 0.7$ and $X_B = 0.3$, $P_s = 350$. $$\Rightarrow P_A^0 \times 0.7 + P_B^0 \times 0.3 = 350 (i)$$ When $X_A = 0.2$ and $X_B = 0.8$, $P_s = 410$. $$\Rightarrow P_A^0 \times 0.2 + P_B^0 \times 0.8 = 410 (ii)$$ Solving (i) and (ii), $P_A^0 = 314 \, \mathrm{mm \, Hg}$, $P_B^0 = 434 \, \mathrm{mm \, Hg}$.

Mass percent, mole fraction, molarity, ppm, molality are used for measuring concentration terms.

Given $t_{1/2} \propto (P_0)^{1-n}$. $$\frac{\left(t_{1/2}\right)_1}{\left(t_{1/2}\right)_2} = \frac{\left(P_0\right)_1^{1-n}}{\left(P_0\right)_2^{1-n}}$$ $$\Rightarrow \left(\frac{4}{2}\right) = \left(\frac{50}{100}\right)^{1-n}$$ $$\Rightarrow 2 = \left(\frac{1}{2}\right)^{1-n}$$ $$\Rightarrow 2 = (2)^{n-1}$$ $$\Rightarrow n-1 = 1$$ $$\Rightarrow n = 2$$

For physisorptions (a) Decreases with increase in temperature (b) No appreciable activation energy is required

Ag. $$4\mathrm{Ag} + 8\mathrm{CN}^- + \mathrm{O}_2 + 2\mathrm{H}_2\mathrm{O} \rightarrow 4[\mathrm{Ag(CN)}_2]^- + 4\mathrm{OH}^-$$ $$2[\mathrm{Ag(CN)}_2]^- + \mathrm{Zn} \rightarrow 2\mathrm{Ag} \downarrow + [\mathrm{Zn(CN)}_4]^{2-}$$

$\mathrm{Ce^{+4}}$ and $\mathrm{Tb^{+4}}$ act as oxidising agent

The reaction is given by: $$3\mathrm{SO_2} + \mathrm{Cr_2O_7^{2-}} + 2\mathrm{H^+} \rightarrow 3\mathrm{SO_4^{2-}} + 2\mathrm{Cr^{3+}} + \mathrm{H_2O}$$ The color of the solution is green.

Crystal field theory introduces the spectrochemical series based upon the experimental values of $\Delta$ but can't explain its order. While other three points are explained by CFT. Especially when the CFSE increases, thermodynamic stability of the complex increases.

The complex $[\mathrm{Co(NH_3)_6}]^{3+}$ has a $d^2sp^3$ hybridization and is diamagnetic.

The reaction of the given compound with $\mathrm{Cl_2/hv}$ leads to the formation of isomers. The first isomer is formed by the substitution of a chlorine atom at the terminal carbon, resulting in: $$\begin{array}{c} \mathrm{CH_3} \\ \mathrm{|} \\ \mathrm{CH_2 - C - CH_2 - CH_2 - C - CH_3} \\ \mathrm{|} \\ \mathrm{Cl} \\ \mathrm{CH_3} \end{array}$$ The second isomer is formed by the substitution of a chlorine atom at the chiral center, resulting in: $$\begin{array}{c} \mathrm{CH_3} \\ \mathrm{|} \\ \mathrm{CH_3 - C^* - CH - CH_2 - C - CH_3} \\ \mathrm{|} \\ \mathrm{CH_3} \\ \mathrm{Cl} \end{array}$$ This isomer is optically active, indicated by $(\pm)$. The total number of isomers is 3.

The reaction involves the addition of water to an alkene. Using $BH_3, THF$ followed by $H_2O_2, HO^-$ results in an anti-Markovnikov product, which is compound (A). Using $Hg(OAc)_2, H_2O$ followed by $NaBH_4$ results in a Markovnikov product, which is compound (B).

Statement 1 is (True) Pure aniline is colourless liquid. Statement 2 is (False) Aniline becomes dark brown due to action of air and light (oxidation).

The reaction involves the coupling of a naphthylamine derivative with a diazonium salt. The diazonium salt is represented as $\mathrm{N=N-O-C-CH_3}$ with a sulfonic acid group $\mathrm{SO_3H}$. The reaction produces an azo compound with a red color, indicated by the presence of $\mathrm{SO_3H (Red)}$, and acetic acid $\mathrm{CH_3COOH}$ as a byproduct.

Theoretical.

Theoretical, NCERT based.

The number of possible tripeptides with Val and Pro is $2^3$. (1) val – val – val (2) pro – pro – pro (3) val – pro – pro (4) pro – val – pro (5) val – val – pro (6) val – pro – val (7) pro – pro – val (8) pro – val – val

Given the equation: $$3 \left( x^2 + \frac{1}{x^2} \right) - 2 \left( x + \frac{1}{x} \right) + 5 = 0$$ Rewriting it as: $$3 \left[ \left( x + \frac{1}{x} \right)^2 - 2 \right] - 2 \left( x + \frac{1}{x} \right) + 5 = 0$$ Let $x + \frac{1}{x} = t$. Then the equation becomes: $$3t^2 - 2t - 1 = 0$$ Solving the quadratic equation: $$3t^2 - 3t + t - 1 = 0$$ Factorizing: $$3t(t - 1) + 1(t - 1) = 0$$ $$(t - 1)(3t + 1) = 0$$ Thus, $t = 1, -\frac{1}{3}$. For $x + \frac{1}{x} = 1, -\frac{1}{3}$, there is no solution.

Let $\sin \frac{2\pi}{9} + i \cos \frac{2\pi}{9} = z$. $$\left( \frac{1+z}{1+\frac{1}{z}} \right)^3 = \left( \frac{1+z}{1+\frac{1}{z}} \right)^3 = z^3$$ $$\Rightarrow \left( i \left( \cos \frac{2\pi}{9} - i \sin \frac{2\pi}{9} \right) \right)^3$$ $$= -i \left( \cos \frac{2\pi}{3} - i \sin \frac{2\pi}{3} \right) = -i \left( -\frac{1}{2} - i \frac{\sqrt{3}}{2} \right)$$ $$\Rightarrow \frac{-1}{2} (\sqrt{3} - i).$$

Four digit numbers greater than 7000 $$= 2 \times 4 \times 3 \times 2 = 48$$ Five digit number $$= 5! = 120$$ Total number greater than 7000 $$= 120 + 48 = 168$$

$1^3+2^3+3^3+\cdots+n^3=\left(\frac{n(n+1)}{2}\right)^2$ Now, $1\cdot3+2\cdot5+3\cdot7+\cdots+n$ terms $=\sum_{r=1}^{n}r(2r+1)$ $=\sum_{r=1}^{n}(2r^2+r)$ $=\frac{2\cdot n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}$ $=\frac{n(n+1)}{6}\Big(2(2n+1)+3\Big)$ $=\frac{n(n+1)}{6}(4n+5)$ If $\frac{n^2(n+1)^2}{4}=\frac{9}{5}\cdot\frac{n(n+1)}{6}(4n+5)$ then $\frac{n(n+1)}{2}=\frac{9(4n+5)}{15}$ $\Rightarrow 5n(n+1)=\frac{9(4n+5)}{3}$ $\Rightarrow 15n(n+1)=18(4n+5)$ $\Rightarrow 15n^2+15n=72n+90$ $\Rightarrow 15n^2-57n-90=0$ $\Rightarrow 5n^2-19n-30=0$ $\Rightarrow (n-5)(5n+6)=0$ $\Rightarrow n=5$ or $n=-\frac{6}{5}$ Since $n$ is positive, $\Rightarrow n=5$.

S = 0 $\cdot$ $({}^{30}C_{0})^2$ + 1 $\cdot$ $({}^{30}C_{1})^2$ + 2 $\cdot$ $({}^{30}C_{2})^2$ + $\ldots$ + 30 $\cdot$ $({}^{30}C_{30})^2$. S = 30 $\cdot$ $({}^{30}C_{0})^2$ + 29 $\cdot$ $({}^{30}C_{1})^2$ + 28 $\cdot$ $({}^{30}C_{2})^2$ + $\ldots$ + 0 $\cdot$ $({}^{30}C_{30})^2$. 2S = 30 $\cdot$ $\left( ({}^{30}C_{0})^2 + ({}^{30}C_{1})^2 + \ldots + ({}^{30}C_{30})^2 \right)$. S = 15 $\cdot$ ${}^{60}C_{30}$ = 15 $\cdot$ $\frac{60!}{(30!)^2}$. $\frac{15 \cdot 60!}{(30!)^2}$ = $\frac{\alpha \cdot 60!}{(30!)^2}$. $\Rightarrow$ $\alpha$ = 15.

Given Binomial $\left( x - \frac{3}{x^2} \right)^n$, $x \neq 0$, $n \in \mathbb{N}$, Sum of coefficients of first three terms $$^nC_0 - ^nC_1 \cdot 3 + ^nC_2 \cdot 3^2 = 376$$ $$\Rightarrow 3n^2 - 5n - 250 = 0$$ $$\Rightarrow (n - 10)(3n + 25) = 0$$ $$\Rightarrow n = 10$$ Now general term $$^{10}C_r x^{10-r} \left( \frac{-3}{x^2} \right)^r$$ $$= ^{10}C_r x^{10-r} (-3)^r \cdot x^{-2r}$$ $$= ^{10}C_r (-3)^r \cdot x^{10-3r}$$ Coefficient of $x^4$ $\Rightarrow 10 - 3r = 4$ $$\Rightarrow r = 2$$ $$^{10}C_2 (-3)^2 = 405$$

Given $\tan(\pi \cos \theta) + \tan(\pi \sin \theta) = 0$. This implies $\tan(\pi \cos \theta) = -\tan(\pi \sin \theta)$. Therefore, $\tan(\pi \cos \theta) = \tan(-\pi \sin \theta)$. Thus, $\pi \cos \theta = n\pi - \pi \sin \theta$. We have $\sin \theta + \cos \theta = n$ where $n \in \mathbb{I}$. Possible values are $n = 0, 1$ and $-1$ because $$-\sqrt{2} \leq \sin \theta + \cos \theta \leq \sqrt{2}$$ Now it gives $\theta \in \left\{ 0, \frac{\pi}{2}, \frac{3\pi}{4}, \frac{7\pi}{4}, \frac{3\pi}{2}, \pi \right\}$. So, $$\sum_{\theta \in S} \sin^2 \left( \theta + \frac{\pi}{4} \right) = 2(0) + 4 \left( \frac{1}{2} \right) = 2$$

Assume $B(\alpha, -2\alpha)$ and $C(\beta + 3, \beta)$. $$\frac{\alpha + \beta + 3 + 1}{3} = 2$$ also $$\frac{-2\alpha - 2 + \beta}{3} = a$$

In $\triangle CPB$, $\($ $\cos$ $\frac{\theta}{2}$ = $\frac{PC}{2}$ $\Rightarrow$ PC = 2 $\cos$ $\frac{\theta}{2}$ $\)$ $\($ $\Rightarrow$ (h - 4)^2 + (k - 5)^2 = 4 $\cos$^2 $\frac{\theta}{2}$ $\)$ Now $\($ (x - 4)^2 + (y - 5)^2 = $\left$( 2 $\cos$ $\frac{\theta}{2}$ $\right$)^2 $\)$ $\($ $\Rightarrow$ r_1 = 2 $\cos$ $\frac{\pi}{6}$ = $\sqrt{3}$ $\)$ $\($ r_2 = 2 $\cos$ $\frac{\theta}{2}$ $\)$ $\($ r_3 = 2 $\cos$ $\frac{\pi}{3}$ = 1 $\)$ $\($ $\Rightarrow$ 5 = 3 + $\cos$ $\frac{\theta}{2}$ + 1 $\)$

AB: ($\lambda$ + 1)x + $\lambda$ y = 4 AC: $\lambda$ x + (1 - $\lambda$)y + $\lambda$ = 0 Vertex A is on y-axis $\Rightarrow$ x = 0 So y = $\frac{4}{\lambda}$, y = $\frac{\lambda}{\lambda - 1}$ $\Rightarrow$ $\frac{4}{\lambda}$ = $\frac{\lambda}{\lambda - 1}$ $\Rightarrow$ $\lambda$ = 2 AB: 3x + 2y = 4 AC: 2x - y + 2 = 0 $\Rightarrow$ A(0, 2) Let C($\alpha$, 2$\alpha$ + 2) Now (Slope of Altitude through C) $\left$(-$\frac{3}{2}$$\right$) = -1 $\left$($\frac{2\alpha}{\alpha - 1}$$\right$)$\left$(-$\frac{3}{2}$$\right$) = -1 $\Rightarrow$ $\alpha$ = -$\frac{1}{2}$ So C$\left$(-$\frac{1}{2}$, 1$\right$) Let Equation of tangent be y = mx + $\frac{\gamma}{2m}$

Given $$\lim_{x \to a} ([x - 5] - [2x + 2]) = 0$$ $$\lim_{x \to a} ([x] - 5 - [2x] - 2) = 0$$ $$\lim_{x \to a} ([x] - [2x]) = 7$$ Let $[a] - [2a] = 7$. If $a \in \mathbb{I}$, then $a = -7$. If $a \notin \mathbb{I}$, then $a = I + f$. Now, $[a] - [2a] = 7$ implies $$-I - [2f] = 7$$ Case-I: $f \in \left(0, \frac{1}{2}\right)$ Then $2f \in (0, 1)$ $$-I = 7$$ Thus, $I = -7 \implies a \in (-7, -6.5)$ Case-II: $f \in \left(\frac{1}{2}, 1\right)$ Then $2f \in (1, 2)$ $$-I - 1 = 7$$ Thus, $I = -8 \implies a \in (-7.5, -7)$ Hence, $a \in (-7.5, -6.5)$.

We start with the expression $\sim (p \land (p \to \sim q))$. This is equivalent to $\sim p \lor \sim (\sim p \lor \sim q)$. Simplifying further, we get $\sim p \lor (p \land q)$. This can be rewritten as $(\sim p \lor p) \land (\sim p \lor q)$. Since $\sim p \lor p$ is a tautology, it simplifies to $t \land (\sim p \lor q)$. Finally, this reduces to $\sim p \lor q$.

Given $a_1 + a_3 = 10 = a_1 + d \Rightarrow d = 5$. $a_1 + a_2 + a_3 + a_4 + a_5 + a_6 = 57$. Therefore, $$\frac{6}{2} [a_1 + a_6] = 57$$ This implies $a_1 + a_6 = 19$. Thus, $2a_1 + 5d = 19$ and $a_1 + d = 5$. Solving these, we get $a_1 = 2$, $d = 3$. The numbers are: $2, 5, 8, 11, 14, 17$. Variance $= \sigma^2 = mean of squares - square of mean$. $$= \frac{2^2 + 5^2 + 8^2 + (11)^2 + (14)^2 + (17)^2}{6} - \left(\frac{19}{2}\right)^2$$ $$= \frac{699}{6} - \frac{361}{4} = \frac{105}{4}$$ Finally, $8\sigma^2 = 210$.

Given $R = \{(a, b), (b, c), (b, d)\}$. In order to make it equivalence relation as per given set, $R$ must be $$\{(a, a), (b, b), (c, c), (d, d), (a, b), (b, a), (b, c), (c, b), (b, d), (d, b), (a, c), (a, d), (c, d), (d, c), (c, a), (d, a)\}$$ There already given so 13 more to be added.

In each row and each column exactly one is to be placed. Therefore, the number of such materials is $5 \times 4 \times 3 \times 2 \times 1 = 120$. Alternate: $$\begin{bmatrix} 0 & 0 & 1 & 0 & 0 \end{bmatrix} \rightarrow 5 ways$$ $$\begin{bmatrix} 0 & 1 & 0 & 0 & 0 \end{bmatrix} \rightarrow 4 ways$$ $$\begin{bmatrix} 0 & 0 & 0 & 1 & 0 \end{bmatrix} \rightarrow 3 ways$$ $$\begin{bmatrix} 0 & 0 & 0 & 0 & 1 \end{bmatrix} \rightarrow 2 ways$$ $$\begin{bmatrix} 1 & 0 & 0 & 0 & 0 \end{bmatrix} \rightarrow 1 way$$ Step-1: Select any 1 place for 1's in row 1. Automatically some column will get filled with 0's. Step-2: From next now select 1 place for 1's. Automatically some column will get filled with 0's. Each time one less place will be available for putting 1's. Repeat step-2 till last row. Required ways $= 5 \times 4 \times 3 \times 2 \times 1 = 120$ $3 \sqrt{3}$

Given $|adj (adj (adj A))| = 12^4$. $$\Rightarrow |A|^{(n-1)^3} = 12^4$$ Given $n = 3$ $$\Rightarrow |A|^8 = 12^4$$ $$\Rightarrow |A|^2 = 12$$ $$|A| = 2\sqrt{3}$$ We are asked $$|A^{-1} \cdot adj A|$$ $$= |A^{-1}| \cdot |adj A|$$ $$= \frac{1}{|A|} \cdot |A|^{3-1}$$ $$= |A| = 2\sqrt{3}$$

Given the equations: $$x + 2y + 3z = 3 \ldots (i)$$ $$4x + 3y - 4z = 4 \ldots (ii)$$ $$8x + 4y - \lambda z = 9 + \mu \ldots (iii)$$ Multiply equation (i) by 4 and subtract equation (ii): $$5y + 16z = 8 \ldots (iv)$$ Multiply equation (ii) by 2 and subtract equation (iii): $$2y + (\lambda - 8)z = -1 - \mu \ldots (v)$$ Multiply equation (iv) by 2 and subtract equation (iii) multiplied by 5: $$(32 - 5(\lambda - 8))z = 16 - 5(-1 - \mu)$$ For infinite solutions: $$72 - 5\lambda = 0 \implies \lambda = \frac{72}{5}$$ Solve for $\mu$: $$21 + 5\mu = 0 \implies \mu = \frac{-21}{5}$$ Thus, $$\Rightarrow (\lambda, \mu) \equiv \left( \frac{72}{5}, \frac{-21}{5} \right)$$

Given $f(x) = \frac{4^x}{4^x + 2}$. $f(x) + f(1-x) = \frac{4^x}{4^x + 2} + \frac{4^{1-x}}{4^{1-x} + 2}$ $$= \frac{4^x}{4^x + 2} + \frac{4}{4 + 2(4^x)}$$ $$= \frac{4^x}{4^x + 2} + \frac{2}{2 + 4^x}$$ $$= 1$$ Thus, $f(x) + f(1-x) = 1$. Now $f\left(\frac{1}{2023}\right) + f\left(\frac{2}{2023}\right) + f\left(\frac{3}{2023}\right) + \ldots + \ldots + f\left(1 - \frac{3}{2023}\right) + f\left(1 - \frac{2}{2023}\right) + f\left(1 - \frac{1}{2023}\right)$ Now the sum of terms equidistant from the beginning and end is 1. Sum $= 1 + 1 + 1 + \ldots + 1 \ (1011 times)$ $$= 1011$$

Given $f(x + y) = f(x) \cdot f(y)$ for all $x, y \in \mathbb{N}$, $f(1) = 3$. $f(2) = f^2(1) = 3^2$ $f(3) = f(1) f(2) = 3^3$ $f(4) = 3^4$ $f(k) = 3^k$ $$\sum_{k=1}^{n} f(k) = 3279$$ $f(1) + f(2) + f(3) + \ldots + f(k) = 3279$ $3 + 3^2 + 3^3 + \ldots + 3^k = 3279$ $$\frac{3(3^k - 1)}{3 - 1} = 3279$$ $$\frac{3^k - 1}{2} = 1093$$ $$3^k - 1 = 2186$$ $$3^k = 2187$$ $k = 7$

Given $f(x) = x^3 - x^2 f'(1) + x f''(2) - f'''(3)$, $x \in \mathbb{R}$. Let $f'(1) = a$, $f''(2) = b$, $f'''(3) = c$. Then, $$f(x) = x^3 - ax^2 + bx - c$$ Differentiating, $$f'(x) = 3x^2 - 2ax + b$$ $$f''(x) = 6x - 2a$$ $$f'''(x) = 6$$ Given $c = 6$, $a = 3$, $b = 6$, we have $$f(x) = x^3 - 3x^2 + 6x - 6$$ Calculating values, $$f(1) = -2, \ f(2) = 2, \ f(3) = 12, \ f(0) = -6$$ Finally, $$2f(0) - f(1) + f(3) = 2 = f(2)$$

Given $$\int_{\frac{3\sqrt{2}}{4}}^{\frac{3\sqrt{3}}{4}} \frac{48}{\sqrt{9 - 4x^2}} \, dx$$. We have $$\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1} \frac{x}{a} + C$$. Hence, $$\int_{\frac{3\sqrt{2}}{4}}^{\frac{3\sqrt{3}}{4}} \frac{48}{\sqrt{9 - 4x^2}} \, dx = \frac{48}{2} \left[ \sin^{-1} \frac{2x}{3} \right]_{\frac{3\sqrt{2}}{4}}^{\frac{3\sqrt{3}}{4}}$$ $$= 24 \times \left[ \sin^{-1} \left( \frac{2}{3} \times \frac{3\sqrt{3}}{4} \right) - \sin^{-1} \left( \frac{2}{3} \times \frac{3\sqrt{2}}{4} \right) \right]$$ $$= 24 \times \left[ \sin^{-1} \frac{\sqrt{3}}{2} - \sin^{-1} \frac{1}{\sqrt{2}} \right]$$ $$= 24 \times \left( \frac{\pi}{3} - \frac{\pi}{4} \right)$$ $$= 24 \times \frac{\pi}{12} = 2\pi$$

Given $$f(x) + \int_0^x f(t) \sqrt{1 - (\log_e f(t))^2} \, dt = e$$ This implies $$f(0) = e$$ Differentiating, we have $$f'(x) + f(x) \sqrt{1 - (\ln f(x))^2} = 0$$ Let $f(x) = y$. Then $$\frac{dy}{dx} = -y \sqrt{1 - (\ln y)^2}$$ Integrating both sides, $$\int \frac{dy}{y \sqrt{1 - (\ln y)^2}} = -\int dx$$ Put $\ln y = t$, then $$\int \frac{dt}{\sqrt{1 - t^2}} = -x + C$$ This gives $$\sin^{-1} t = -x + C \implies \sin^{-1} (\ln y) = -x + C$$ Thus, $$\sin^{-1} (\ln f(x)) = -x + C$$ Given $f(0) = e$, we have $$\frac{\pi}{2} = C$$ Therefore, $$\sin^{-1} (\ln f(x)) = -x + \frac{\pi}{2}$$ Evaluating at $x = \frac{\pi}{6}$, $$\sin^{-1} \left( \ln f \left( \frac{\pi}{6} \right) \right) = -\frac{\pi}{6} + \frac{\pi}{2}$$ This simplifies to $$\sin^{-1} \left( \ln f \left( \frac{\pi}{6} \right) \right) = \frac{\pi}{3}$$

Given $y^2 - 2y = -x$. This implies $y^2 - 2y + 1 = -x + 1$. Therefore, $(y - 1)^2 = -(x - 1)$. We have $y = -x$. Points of intersection are found by solving $x^2 + 2x = -x$, which gives $x^2 + 3x = 0$. Thus, $x = 0, -3$. The area $A$ is given by: $$A = \int_{0}^{3} (-y^2 + 2y + y) \, dy$$ Evaluating the integral: $$= \frac{3y^2}{2} - \frac{y^3}{3} \bigg|_{0}^{3} = \frac{9}{2}$$ Finally, $8A = 36$.

$(x^2 - 3y^2) \, dx + 3xy \, dy = 0$ $\frac{dy}{dx} = \frac{3y^2 - x^2}{3xy}$ $\implies \frac{dy}{dx} = \frac{y}{x} - \frac{1}{3} \frac{x}{y}$ ...(1) Put $y = vx$ $\frac{dy}{dx} = v + x \frac{dv}{dx}$ (1) $\implies v + x \frac{dv}{dx} = v - \frac{1}{3} \frac{1}{v}$ $\implies v \, dv = -\frac{1}{3x} \, dx$ Integrating both sides $\frac{v^2}{2} = -\frac{1}{3} \ln x + c$ $\implies \frac{y^2}{2x^2} = -\frac{1}{3} \ln x + c$ $y(1) = 1 \implies \frac{1}{2} = c$ $\implies \frac{y^2}{2x^2} = -\frac{1}{3} \ln x + \frac{1}{2}$ $y^2(x^2) = \frac{2}{3} x^2 \ln x^2 - e^2$

Let $\vec{\beta}_1 = \lambda \vec{\alpha}$. Now $\vec{\beta}_2 = \vec{\beta} - \vec{\beta}_1$. $$= (\hat{i} + 2\hat{j} - 4\hat{k}) - \lambda (4\hat{i} + 3\hat{j} + 5\hat{k})$$ $$= (1 - 4\lambda) \hat{i} + (2 - 3\lambda) \hat{j} - (5\lambda + 4) \hat{k}$$ $\vec{\beta}_2 \cdot \vec{\alpha} = 0$. $$\Rightarrow 4(1 - 4\lambda) + 3(2 - 3\lambda) - 5(5\lambda + 4) = 0$$ $$\Rightarrow 4 - 16\alpha + 6 - 9\lambda - 25\lambda - 20 = 0$$ $$\Rightarrow 50\lambda = -10$$ $$\Rightarrow \lambda = \frac{-1}{5}$$ $$\vec{\beta}_2 = \left( 1 + \frac{4}{5} \right) \hat{i} + \left( 2 + \frac{3}{5} \right) \hat{j} - (-1 + 4) \hat{k}$$ $$\vec{\beta}_2 = \frac{9}{5} \hat{i} + \frac{13}{5} \hat{j} - 3\hat{k}$$ $$5\vec{\beta}_2 = 9\hat{i} + 13\hat{j} - 15\hat{k}$$ $$5\vec{\beta}_2 \cdot (\hat{i} + \hat{j} + \hat{k}) = 9 + 13 - 15 = 7$$

Given $\mathbf{a} = \hat{i} + 2\hat{j} + \lambda \hat{k}$, $\mathbf{b} = 3\hat{i} - 5\hat{j} - \lambda \hat{k}$, $\mathbf{a} \cdot \mathbf{c} = 7$. $\mathbf{a} \times \mathbf{c} - \mathbf{b} \times \mathbf{c} = \mathbf{0}$, $$(\mathbf{a} - \mathbf{b}) \times \mathbf{c} = \mathbf{0} \Rightarrow (\mathbf{a} - \mathbf{b}) is paralleled to \mathbf{c}$$ $\mathbf{a} - \mathbf{b} = \mu \mathbf{c}$, where $\mu$ is a scalar $$-2\hat{i} + 7\hat{j} + 2\lambda \hat{k} = \mu \cdot \mathbf{c}$$ Now $\mathbf{a} \cdot \mathbf{c} = 7$ gives $2\lambda^2 + 12 = 7\mu$ And $\mathbf{b} \cdot \mathbf{c} = -\frac{43}{2}$ gives $4\lambda^2 + 82 = 43\mu$ $\mu = 2$ and $\lambda^2 = 1$ $$|\mathbf{a} \cdot \mathbf{b}| = 8$$

Equation of plane passing through point of intersection of P1 and P2 $$P = P1 + kP2$$ $$\left( x + (\lambda + 4)y + z - 1 \right) + k \left( 2x + y + z - 2 \right) = 0$$ Passing through $(0, 1, 0)$ and $(1, 0, 1)$ $$(\lambda + 4 - 1) + k(1 - 2) = 0$$ $$(\lambda + 3) - k = 0 ....(1)$$ Also passing $(1, 0, 1)$ $$(1 + 1 - 1) + k(2 + 1 - 2) = 0$$ $$1 + k = 0$$ $$k = -1$$ put in (1) $$\lambda + 3 + 1 = 0$$ $$\lambda = -4$$ Then point $(2 \lambda, \lambda, -\lambda)$ $$(-8, -4, 4)$$ $$d = \frac{| -16 - 4 + 4 - 2 |}{\sqrt{6}}$$ $$d = \frac{18}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}} = 3 \sqrt{6}$$

Direction ratio of line is $\($ $\begin{vmatrix}$ $\hat{i}$ & $\hat{j}$ & $\hat{k}$ $\\$ 1 & 2 & 1 $\\$ 0 & 3 & -1 $\end{vmatrix}$ $\)$. $\($ = $\hat{i}$(-5) - $\hat{j}$(-1) + $\hat{k}$(3) $\)$ $\($ = -5$\hat{i}$ + $\hat{j}$ + 3$\hat{k}$ $\)$ Point $\($ P(1, 9, 7) $\)$ $\($ M $\left$( -5$\lambda$ + 3, $\lambda$ + 2, 3$\lambda$ + 1 $\right$) $\)$ $\($ $\overline{\mathrm{PM}}$ $\perp$ (-5$\hat{i}$ + $\hat{j}$ + 3$\hat{k}$) $\)$ $\($ -5(-5$\lambda$ + 2) + ($\lambda$ - 7) + 3(3$\lambda$ - 6) = 0 $\)$ $\($ $\Rightarrow$ 25$\lambda$ + $\lambda$ + 9$\lambda$ - 10 - 7 - 18 = 0 $\)$ $\($ $\Rightarrow$ $\lambda$ = 1 $\)$ Point $\($ M = (-2, 3, 4) = ($\alpha$, $\beta$, $\gamma$) $\)$ $\($ $\alpha$ + $\beta$ + $\gamma$ = 5 $\)$

Shortest distance between the lines $$\frac{x + \sqrt{6}}{2} = \frac{y - \sqrt{6}}{3} = \frac{z - \sqrt{6}}{4}$$ $$\frac{x - \lambda}{3} = \frac{y - 2\sqrt{6}}{4} = \frac{2 + 2\sqrt{6}}{5}$$ is 6. Vector along line of shortest distance $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix}, \Rightarrow -\hat{i} + 2\hat{j} - \hat{k} (its magnitude is \sqrt{6} )$$ Now $$\frac{1}{\sqrt{6}} \begin{vmatrix} \sqrt{6} + \lambda & \sqrt{6} & -3\sqrt{6} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix} = \pm 6$$ $$\Rightarrow \lambda = -2\sqrt{6}, 10\sqrt{6}$$ So, square of sum of these values is 384.

Given $m^2 + 2m - 3 = 0$. Solving for $m$, we get $m = 1, -3$. So the tangent which touches in the first quadrant at $T$ is $$T \equiv \left( \frac{a}{m^2}, \frac{2a}{m} \right)$$ $$\equiv \left( \frac{3}{2}, 3 \right)$$ Therefore, $CT = \sqrt{4 + 4} = 2\sqrt{2}$. Consider the urns: Urn A: Red 4, Black 6 Urn B: Red 5, Black 5 Urn C: Red $\lambda$, Black 4 The probability is given by: $$P\left( \frac{C}{R} \right) = \frac{P(C)P\left( \frac{R}{C} \right)}{P(A)P\left( \frac{R}{A} \right) + P(B)P\left( \frac{R}{B} \right) + P(C)P\left( \frac{R}{C} \right)}$$ Substituting the values: $$0.4 = \frac{\frac{1}{3} \times \frac{\lambda}{\lambda + 4}}{\frac{1}{3} \times \frac{4}{10} + \frac{1}{3} \times \frac{5}{10} + \frac{1}{3} \times \frac{\lambda}{\lambda + 4}}$$ Solving, we find $\lambda = 6$. For the tangent line, $\tan 30^\circ = 3t = \frac{3}{2}t^2$. Thus, $$\frac{1}{\sqrt{3}} = \frac{2}{t}$$