JEE Main 30 January 2023 Shift 1 question paper with solutions

The electric field vector is given by $$\vec{E} = \frac{A}{x^2} \hat{i} + \frac{B}{y^3} \hat{j}$$. The dimensional analysis for the term $$\left[ \frac{A}{x^2} \right] = \mathrm{NC^{-1}}$$ implies that $$[A] = \mathrm{Nm^2C^{-1}}$$.

Given $\($ $\frac{dx}{dt}$ = slope $\geq$ 0 $\)$ always increasing. (A - II) $\($ $\frac{dx}{dt}$ 0 $\)$ for first half, $\($ $\frac{dx}{dt}$ < 0 $\)$ for second half. (C - III) $\($ $\frac{dx}{dt}$ = constant $\)$

Given $\left| \frac{d\vec{p}}{dt} \right| = \left| \vec{F} \right|$, it implies $\frac{d\vec{p}}{dt} = Slope of curve$. Max slope (c) min slope (b)

The velocities will be interchanged after collision. Speed of P just before collision = $\sqrt{2gh}$ $$= \sqrt{2 \times 10 \times 0.2} = 2 \, \mathrm{m/s}$$

Given $v_1 = \frac{30}{\sqrt{\frac{2h}{g}}}$, $v_2 = \frac{120}{\sqrt{\frac{2h}{g}}}$. $$(0.01) \, u = (0.2) \frac{30 \sqrt{g}}{\sqrt{2h}} + (0.01) \frac{120 \sqrt{g}}{\sqrt{2h}}$$ $u = 300 + 60 = 360 \, \mathrm{ms^{-1}}$

Given $\($-$\frac{dV}{dr}$ = -$\frac{k}{r^2}$$\)$, we have $\($$\int$_{10}^{V} dV = $\int$_{2}^{3} $\frac{k}{r^2}$ $\,$ dr$\)$. This implies $\($V - 10 = k $\left$[ $\frac{1}{2}$ - $\frac{1}{3}$ $\right$]$\)$. Simplifying, $\($V - 10 = $\frac{k}{6}$ $\Rightarrow$ V = 11$\)$ volts.

Given $$Y = 3\eta (1 + \sigma)$$ $$Y = 3K (1 - \sigma)$$ Therefore, $$2\eta (1 + \sigma) = 3K (1 - 2\sigma)$$ Thus, $$\sigma = \frac{3K - 2\eta}{6K + 2\eta}$$

Given $$h = \frac{2S \cos \theta}{r \rho g}$$ Therefore, $$\frac{h_1}{h_2} = \frac{S_1}{S_2} \frac{\rho_2}{\rho_1}$$ $$\frac{5}{h_2} = \left[ \frac{1}{2} \right] \left[ \frac{2}{1} \right] \implies h_2 = 5 \, \mathrm{cm} = 0.05 \, \mathrm{m}$$ {Info about angle of contact not there so most}

Given $dQ = dU + dW$. Therefore, $dU = nC_V \, dT$. Since $dU = 0$ for isothermal processes, $U$ is constant. Also, $dQ > 0$ (supplied). Hence, $dW > 0$.

Given $Q_1 = \sigma (4 \pi R^2) = 4 \pi R^2 \sigma$ and $Q_2 = \sigma (4 \pi (2R)^2) = 16 \pi R^2 \sigma$. Therefore, $\[$ $\frac{Q_1'}{4 \pi \varepsilon_0 R}$ = $\frac{Q_2'}{4 \pi \varepsilon_0 (2R)}$ $\]$ Thus, $Q_2' = 2Q_1'$. $Q_1' + Q_2' = Q_1 + Q_2$. Therefore, $\[$ $\frac{Q_2'}{2}$ + Q_2' = 20 $\pi$ R^2 $\sigma$ $\]$ $\[$ $\frac{3}{2}$ Q_2' = 20 $\pi$ R^2 $\sigma$ $\]$ Therefore, $\[$ $\frac{Q_2'}{4 \pi (2R)^2}$ = $\frac{2}{3}$ $\cdot$ $\frac{20 \pi R^2 \sigma}{16 \pi R^2}$ $\]$ Thus, $\[$ $\frac{\sigma'}{\sigma}$ = $\frac{5}{6}$ $\]$

Given $Q = \left( \alpha t - \beta t^2 + \gamma t^3 \right)$. The current $i = \frac{dQ}{dt} = \left( \alpha - 2 \beta t + 3 \gamma t^2 \right)$. Differentiating $i$ with respect to $t$, we have: $$\frac{di}{dt} = \left( 3 \gamma t - 2 \beta \right) = 0$$ Solving for $t$, we get: $$t = \frac{\beta}{3 \gamma}$$ Substituting back, we find: $$i = \left( \alpha - 2 \beta t + 3 \gamma t^2 \right) = \left( \alpha - \frac{\beta^2}{3 \gamma} \right)$$

Given $F_m = mg$. Therefore, $ILB = mg$. Thus, $$\left(\frac{V}{R}\right) LB = mg$$ Therefore, $$V = \frac{mgR}{LB}$$ $$= \frac{(1 \times 10^{-3} \, \mathrm{kg})(10 \, \mathrm{m/s^2})(10 \, \Omega)}{(0.1 \, \mathrm{m})(10^3 \times 10^{-4} \, \mathrm{T})} = 10 \, \mathrm{V}$$

Given $M = N I A$. Since $M_A = M_B$, therefore $N_A I_A A_A = N_B I_B A_B$. Thus, $N_A I_A \pi (0.1)^2 = N_B I_B \pi (0.2)^2$. Therefore, $N_A I_A = 4 N_B I_B$.

Given $P = \frac{R}{Z}$, we have $P_1 = \frac{R}{\sqrt{R^2 + X_L^2}} = \frac{R}{R\sqrt{2}}$ (as $X_L = R$). Therefore, $P_1 = \frac{1}{\sqrt{2}}$. For $P_2$, we have $P_2 = \frac{R}{\sqrt{R^2 + (X_L - X_L)^2}} = P_2 = 1$. Thus, $P_1 = \frac{1}{\sqrt{2}}$.

Given $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$, $P = 2D = 2\,\mathrm{m}^{-1}$, $\Rightarrow \frac{1}{f} = \frac{2}{100}\,\mathrm{cm}^{-1}$, $\frac{1}{V} - \left(-\frac{1}{25}\right) = \frac{2}{100}$, $\Rightarrow \frac{1}{V} = \frac{1}{50} - \frac{1}{25}$, $\Rightarrow V = -50\,\mathrm{cm}$.

Momentum is given by the formula $$Momentum = \frac{Energy}{C}$$ which can also be expressed as $$= \frac{Power \times time}{C}$$ Substituting the given values, we have $$= \frac{(20 \times 10^{-3} \, \mathrm{W})(300 \times 10^{-9} \, \mathrm{s})}{3 \times 10^8 \, \mathrm{m/s}}$$ Simplifying this expression, we get $$= 2 \times 10^{-17} \, \mathrm{kg \cdot m/s}$$

Given $V_n \propto \frac{Z}{n}$. Since $Z = 1$, therefore $V_n \propto \frac{1}{n}$. Thus, $$\frac{V_3}{V_7} = \frac{7}{3}$$ Therefore, $$V_3 = \frac{7}{3} V_7$$ $$= \frac{7}{3} \times 3.6 \times 10^6 \, \mathrm{m/s}$$ $$= 8.4 \times 10^6 \, \mathrm{m/s}$$

Given the expressions: $$(A \cdot A) = A$$ $$\overline{B \cdot B} = \overline{B}$$ $$\overline{\overline{A \cdot B}} = A + B$$ This represents an OR Gate.

Given $A_c + A_m = 120$ and $A_c - A_m = 80$. Therefore, $$A_c = 100$$ $$A_m = 20$$ Modulation index is given by $$\frac{20}{100} = \frac{1}{5}$$ Amplitude of each sideband is $$= \frac{A_c \times (modulation index)}{2}$$ $$= 100 \times \frac{1}{10} = 10 volt$$

Given $t_{AB} = \frac{x}{5 \, \mathrm{m/s}}$. In motion BC, $x = d_1 + d_2$ where $d_1$ and $d_2$ are the distances travelled with $10 \, \mathrm{m/s}$ and $15 \, \mathrm{m/s}$ respectively in equal time intervals $\frac{t}{2}$ each. $$d_1 = \frac{10t}{2}, d_2 = \frac{15t}{2}$$ $$d_1 + d_2 = x = \frac{t}{2}(10 + 15) = \frac{25t}{2}$$ $$\langle v \rangle = \frac{2x}{\frac{x}{5} + \frac{2x}{25}} = \frac{2 \times 25}{5 + 2} = \frac{50}{7} \, \mathrm{m/s}$$ Ans.: 50

($\mathrm{KE}$)_{Rotational} = $\frac{1}{2}$ I $\omega$^2 = E $$E = \frac{1}{2} \frac{m \ell^2}{12} \omega^2$$ $$E = \frac{1}{2} \frac{dA \ell^3}{12} \omega^2$$ $$E = \frac{dA (2)^3}{24} \omega^2$$ $$\sqrt{\frac{3E}{dA}} = \omega$$ $\alpha = 3$ Ans.

Given $x = A \sin(\omega t)$. The potential energy $U_{(x)} = \frac{1}{2} k x^2$, Differentiating with respect to $t$, $$\frac{dU}{dt} = \frac{1}{2} k 2x \frac{dx}{dt}$$ Substituting $x = A \sin(\omega t)$, $$= k A^2 \omega \sin \omega t \cos \omega t \times \frac{2}{2}$$ The maximum rate of change of potential energy is $$\left( \frac{dU}{dt} \right)_{max} = \frac{k A^2 \omega}{2} (\sin 2\omega t)_{max}$$ Solving for $\omega t$, $$2\omega t = \frac{\pi}{2} \Rightarrow t = \frac{\pi}{4} \omega = \frac{T}{8} \Rightarrow \beta = 8$$

Given $C = 900 \, \mu \mathrm{F}$. $Q = CV = 900 \times 10^{-6} \times 100 = 9 \times 10^{-2} = 90 \, \mathrm{MC}$. Now, $$\begin{array}{c} At t = 0 \\ \begin{array}{c|c} 90 \, \mathrm{mc} & -90 \, \mathrm{mc} \\ C & C \\ Q = 0 & \\ \end{array} \end{array}$$ Common potential will be developed across both capacitors by KVL. Total charge on left plates of capacitors should be conserved. $$90 \, \mathrm{mc} + 0 = 2C V_0$$ $$C V_0 = 45 \, \mathrm{mc}$$ $$\begin{array}{c|c} 45 \, \mathrm{mc} & -45 \, \mathrm{mc} \\ C & C \\ \end{array}$$ $$\frac{1}{2} (90 \, \mathrm{mc})^2 + \frac{1}{2} (45 \, \mathrm{mc})^2 = \frac{1}{2} \left[ \frac{Q^2}{C} \right]$$

Junction law at A, $$\frac{x - (y + 5)}{1} + \frac{x - 2}{2} + \frac{x - 0}{2} = 0 \ldots(1)$$ Junction law at B, $$\frac{y + 5 - x}{1} + \frac{y - 0}{1} + \frac{y - 2}{1} = 0 \ldots(2)$$ On solving equation (1) and Equation (2) $$x = 3$$ and $$y = 0$$ At D junction $$I_1 = i_1 + i_2$$ $$I_1 = \frac{y - 0}{1} + \frac{x - 0}{2}$$ $$= \frac{0 - 0}{1} + \frac{3 - 0}{2}$$

Given $\($ $\frac{dI}{dt}$ = -1 $\,$ $\mathrm{A/sec}$ $\)$. $\($ V_A - IR - L $\frac{dI}{dt}$ - 12 = V_B $\)$ $\($ V_A - 2 $\times$ 12 - 6(-1) - 12 = V_B $\)$ $\($ V_A - V_B = 36 - 6 = 30 $\,$ volt $\)$

Given $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$. \[ -\frac{1}{120} - \frac{1}{40} = \frac{1}{f}, \quad f = -30\,\mathrm{cm} \] Now, \[ -\frac{1}{v^2}\,dv - \frac{1}{u^2}\,du = -\frac{1}{f^2}\,df \] Also $(dv = du = \frac{1}{20}\,\mathrm{cm})$ \[ \frac{1}{20}\left(\frac{1}{120}\right)^2 + \frac{1}{20}\left(\frac{1}{40}\right)^2 = \frac{df}{(30)^2} \] On solving \[ df = \frac{1}{32}\,\mathrm{cm} \] Therefore, $k = 32$

Path difference at P be $\Delta x$. $$\Delta x = (\mu_2 - \mu_1)t$$ $$= (1.55 - 1.51) \times 0.1 \, \mathrm{mm}$$ $$= 0.04 \times 10^{-4}$$ $$\Delta x = 4 \times 10^{-6} = 4 \, \mu \mathrm{m}$$ $$y = \frac{\Delta x D}{d} = 4 \times 10^{-6} \frac{D}{d}$$ { $y$ is the distance of central maxima from geometric center } Fringe width $= \frac{\lambda D}{d} = 4 \times 10^{-6} \, \mathrm{m} \frac{D}{d} = 4 \, \mu \mathrm{m} \frac{D}{d}$ Therefore, central bright fringe spot will shift by $x$. Number of shift $= \frac{y}{\beta}$ $$= \frac{4 \times 10^{-6} \, D/d}{4 \times 10^{-7} \, D/d} = 10$$ Ans

Force = $\int$ PdA $\cos$ $\theta$ = $\frac{2I}{C}$ $\int$ dA $\cos$ $\theta$ = $\frac{2I}{C}$ $\pi$ R^2 = 2 $\frac{p_0}{4 \pi R^2}$ $\cdot$ $\frac{\pi R^2}{C}$ = $\frac{p_0}{2C}$ = $\frac{24}{2 \times 3 \times 10^8}$ = 4 $\times$ 10^{-8} $\,$ $\mathrm{N}$ $\;$ ($\mathrm{Ans}$)

Least count = $\frac{\text{Pitch}}{\text{No. of circular divisions}}$ = $\frac{0.5 \, \mathrm{mm}}{100}$ Least count = 5 $\times$ $10^{-3}$ $\mathrm{mm}$ Positive Error = MSR + CSR $\times$ (LC) = 0 $\mathrm{mm}$ + 6 $\times$ (5 $\times$ $10^{-3}$ $\mathrm{mm}$) Reading of Diameter = MSR + CSR $\times$ (LC) Positive zero error = 4 $\times$ 0.5 $\mathrm{mm}$ + (46 $\times$ (5 $\times$ $10^{-3}$)) - 6 $\times$ (5 $\times$ $10^{-3}$) $\mathrm{mm}$

The table shows the atomic numbers and corresponding blocks for different elements. Atomic number 37 (K) belongs to the s-block. Atomic number 78 (Pt) belongs to the d-block. Atomic number 52 (Te) belongs to the p-block. Atomic number 65 (Tb) belongs to the f-block.

Two lone pair one oxygen. Molecule is 'v' shaped. Bond angle is less than $104.5^\circ$ ($102^\circ$). O.S. of 'O' is $+2$.

$\mathrm{IF_7}$ has zero lone pair. $\mathrm{ICl_4^-}$ has two lone pairs. $\mathrm{XeF_6}$ has one lone pair. $\mathrm{XeF_2}$ has three lone pairs.

Due to high hydration energy $\mathrm{Be^{2+}}$ and $\mathrm{Mg^{2+}}$, $\mathrm{BeSO_4}$ and $\mathrm{MgSO_4}$ are readily soluble in water.

The balanced chemical equation is given by: $$8 \mathrm{LiH} + \mathrm{Al_2Cl_6} \rightarrow 2 \mathrm{LiAlH_4} + 6 \mathrm{LiCl}$$

Silica gel prevents water corrosion (rusting) and instrument malfunction by adsorbing moisture from the air.

The acidity of an acid depends upon the stability of its conjugate base. The given structures show the relative stability of the conjugate bases. The stability order is as follows: $$Structure 1 > Structure 2 > Structure 3 > Structure 4$$

The reaction starts with the addition of $\mathrm{H^+}$ to the alkene $\mathrm{H_3C - C = CH_2}$, forming a carbocation $\mathrm{H_3C - \overset{+}{C} - CH_3}$. Under the conditions of $\mathrm{H_2SO_4}$ at $80^\circ \mathrm{C}$, the carbocation rearranges to form $\mathrm{CH_3 - C = CH - CH_3}$ (B). The $\mathrm{O^- - SO_3H}$ group then attacks the carbocation, resulting in the formation of $\mathrm{CH_3 - C - CH_3}$ with $\mathrm{OSO_3H}$ attached (A).

The reactions are as follows: $$\mathrm{NO_{2(g)} \xrightarrow{h\nu} NO_{(g)} + O_{(g)}}$$ This is reaction (A) and (B). $$\mathrm{O_{(g)} + O_{2(g)} \rightleftharpoons O_{3(g)}}$$ This is reaction (B) and (C). $$\mathrm{NO_{(g)} + O_{3(g)} \rightarrow NO_{2(g)} + O_{2(g)}}$$ This is reaction (A) and (C).

The copper ore contains iron, it is mixed with silica before heating in reverberatory furnace. FeO slags off as $\mathrm{FeSiO_3}$. $$\mathrm{FeO} + \mathrm{SiO_2} \rightarrow \mathrm{FeSiO_3}$$

The reaction is given by: $$\mathrm{Cr_2O_7^{2-} + SO_2^{2-} \xrightarrow{H^+} Cr^{3+} + SO_4^{2-}}$$

The increasing order of field strength of ligands (according to spectrochemical series) is $\mathrm{S^{2-} < C_2O_4^{2-} < NH_3 < en < CO}$.

Cis-Platin is used in chemotherapy to inhibit the growth of tumors. (cis[$\mathrm{Pt(NH_3)_2Cl_2}$])

A corresponds to the Wurtz-fitting reaction. B corresponds to the Fitting reaction. C corresponds to the Sandmeyer reaction. D corresponds to the Finkelstein reaction.

Aromatic aldehydes do not give Fehling's test. Both nitrogen and sulfur must be present to obtain blood red colour. Sodium nitroprusside gives blood red colour with S & N.

The reaction of benzyl bromide with silver cyanide $(\mathrm{AgCN})$ gives benzyl isocyanide. The reaction of benzylamine with chloroform $(\mathrm{CHCl_3})$ and aqueous $\mathrm{KOH}$ gives benzyl isocyanide through the carbylamine reaction. The reaction of $\mathrm{N}$-methylbenzylamine with chloroform and aqueous solution gives no reaction. The reaction of benzyl tosylate with potassium cyanide $(\mathrm{KCN})$ gives benzyl cyanide.

Caprolactam is converted to Nylon-6 in the presence of $\mathrm{H_2O}$ and heat $\Delta$. The reaction involves the opening of the lactam ring to form the polymer chain.

1. Ranitidine: Antacid 2. Meprobamate: Tranquilizer 3. Terfenadine: Antihistamine 4. Brompheniramine: Antihistamine

Zn^{2+}, Co^{2+}, Ni^{2+} = $\mathrm{IV^{th}}$ Group Fe^{3+} = $\mathrm{III^{rd}}$ Group

Seliwanoff's test is a differentiating test for ketose and aldose. This test relies on the principle that the keto hexose are more rapidly dehydrated to form 5-hydroxy methyl furfural when heated in acidic medium which on condensation with resorcinol, cherry red or brown red coloured complex is formed rapidly indicating a positive test.

For one photon $E = h \nu$. For one mole photon, $$E = 6.023 \times 10^{23} \times 6.626 \times 10^{-34} \times 2 \times 10^{12}$$ $$= 798.16 \, \mathrm{J}$$ $$\approx 798 \, \mathrm{J}$$

For ideal gas $U = f(T)$ and for isothermal process, $\Delta U = 0$

Total millimoles of $\mathrm{H}^+$ = $(600 \times 0.01) + (400 \times 0.01 \times 2)$ $$= 14$$ $$[\mathrm{H}^+] = \frac{14}{1000} = 14 \times 10^{-3}$$ pH = $3 - \log 14$ $$= 1.86$$ $$= 186 \times 10^{-2}$$

Mole of $\mathrm{CO_2} = 0.2 \, \mathrm{M} \times (300 \times 10^{-3}) \, \mathrm{L}$ $$= 0.06 \, Mole$$ Volume of $0.06$ mole $\mathrm{CO_2}$ at S.T.P $$= 0.06 \times 22.7$$ $$= 1.362 \, \mathrm{L}$$

Given $\Delta T_b = 373.52 - 373$. This equals $0.52$. We have $\Delta T_b = K_b \cdot m$. Thus, $0.52 = 0.52 \times \frac{2}{Molar Mass} \times \frac{1}{20 \times 10^{-3}}$. Therefore, the Molar Mass is $100 \, \mathrm{g/mol}$.

Molarity = $\frac{mole}{volume}$ = $\frac{x / 85}{0.67141}$. 2.6 $\times$ 10^{-3} = $\frac{x / 85}{0.67141}$. x = 0.148 $\,$ g. conc. Fo DCM in ppm = $\frac{0.148}{1.49 \times 671.141}$ $\times$ 10^6 = 148 $\,$ ppm

$Pt(s)|H_2(g,1\,atm)|H^+(aq,1M)\parallel Fe^{3+}(aq),Fe^{2+}(aq)|Pt(s)$ At anode $H_2\rightarrow2H^++2e^-$ At cathode $Fe^{3+}+e^-\rightarrow Fe^{2+}$ $E^\circ=E^\circ_{H_2|H^+}+E^\circ_{Fe^{3+}|Fe^{2+}}=0.771\,V$ $E=E^\circ-\dfrac{0.06}{1}\log\left(\dfrac{Fe^{2+}}{Fe^{3+}}\right)$ $0.712=(0+0.771)-\dfrac{0.06}{1}\log\left(\dfrac{Fe^{2+}}{Fe^{3+}}\right)$ $\log\left(\dfrac{Fe^{2+}}{Fe^{3+}}\right)=\dfrac{0.059}{0.06}\approx1$ $\therefore\ \dfrac{Fe^{2+}}{Fe^{3+}}=10$

The reaction is given by $\mathrm{A + B \rightarrow P}$. Initially, at $t = 0$, the concentration is $7 \, \mathrm{g}$, and at $t = t$, the concentration is $2 \, \mathrm{g}$. At constant volume, the time $t$ is calculated as follows: $$t = \frac{2.303}{K} \log \frac{[\mathrm{A}]_0}{[\mathrm{A}]_t}$$ Substituting the given values: $$= \frac{2 \cdot 303}{2 \cdot 011 \times 10^{-3}} \log \frac{7}{2}$$ $$= \frac{2 \cdot 303 \times 0 \cdot 544}{2 \cdot 011 \times 10^{-3}}$$ $$= 622.989$$ Therefore, $t \approx 623$.

The balanced redox reaction is given by: $$\mathrm{MnO_4^- + 4H^+ + 3e^- \rightarrow MnO_2 + 2H_2O}$$

The compound with the structure $\mathrm{C_{10}H_{12}O_2}$ can have the following groups present: $\mathrm{CH=O}$ and $\mathrm{C_3H_7}$ (both groups can be present). Alternatively, the compound can have $\mathrm{CH_2OH}$ and $\mathrm{C=CH_3}$ (both groups can be present). The reaction with $\mathrm{NaOH}$ and $\mathrm{CH_3I}$ converts the compound to have $\mathrm{OCH_3}$, resulting in $\mathrm{CH=O + C_3H_7}$.

Given $\log_{\cos x} \cot x + 4 \log_{\sin x} \tan x = 1$. This implies $$\frac{\ln \cos x - \ln \sin x}{\ln \cos x} + 4 \frac{\ln \sin x - \ln \cos x}{\ln \sin x} = 1$$ which simplifies to $$\left( \ln \sin x \right)^2 - 4 \left( \ln \sin x \right) \left( \ln \cos x \right) + 4 \left( \ln \cos x \right)^2 =$$ $$\ln \sin x = 2 \ln \cos x$$ Therefore, $$\sin^2 x + \sin x - 1 = 0 \implies \sin x = \frac{-1 + \sqrt{5}}{2}$$ Thus, $\alpha + \beta = 4$.

If $a_n = \frac{-2}{4n^2 - 16n + 15}$, then $a_1 + a_2 + \ldots + a_{25}$. $$\Rightarrow \sum_{n=1}^{25} a_n = \sum \frac{-2}{4n^2 - 16n + 15}$$ $$= \sum \frac{-2}{4n^2 - 6n - 10n + 15}$$ $$= \sum \frac{-2}{2n(2n-3) - 5(2n-3)}$$ $$= \sum \frac{-2}{(2n-3)(2n-5)}$$ $$= \sum \left( \frac{1}{2n-3} - \frac{1}{2n-5} \right)$$ $$= \frac{1}{47} - \frac{1}{(-3)}$$ $$= \frac{50}{141}$$

Option (2) Coefficient of $x^{15}$ in $\left( ax^3 + \frac{1}{bx^{1/3}} \right)^{15}$ $$T_{r+1} = \binom{15}{r} (ax^3)^{15-r} \left( \frac{1}{bx^{1/3}} \right)^r$$ $$45 - 3r - \frac{r}{3} = 15$$ $$30 = \frac{10r}{3}$$ $$r = 9$$ Coefficient of $x^{15} = \binom{15}{9} a^6 b^{-9}$ Coefficient of $x^{-15}$ in $\left( ax^{1/3} - \frac{1}{bx^3} \right)^{15}$ $$T_{r+1} = \binom{15}{r} (ax^{1/3})^{15-r} \left( -\frac{1}{bx^3} \right)^r$$ $$5 - \frac{r}{3} - 3r = -15$$ $$\frac{10r}{3} = 20$$ $$r = 6$$

$(1+x)^{500} +x(1+x)^{499} +x^{2}(1+x)^{498} +\cdots +x^{500}$ $=(1+x)^{500} \left[ 1+\frac{x}{1+x} +\left(\frac{x}{1+x}\right)^{2} +\cdots +\left(\frac{x}{1+x}\right)^{500} \right]$ $=(1+x)^{500} \left[ \frac{1-\left(\frac{x}{1+x}\right)^{501}} {1-\frac{x}{1+x}} \right]$ $=(1+x)^{500} \left[ \frac{(1+x)^{501}-x^{501}} {(1+x)^{501}} \right] (1+x)$ $=(1+x)^{501}-x^{501}$ Coefficient of $x^{301}$ in $(1+x)^{501}-x^{501}$ is ${}^{501}C_{301}$ $={}^{501}C_{200}$

Option (1) $$\tan 15^\circ = 2 - \sqrt{3}$$ $$\frac{1}{\tan 75^\circ} = \cot 75^\circ = 2 - \sqrt{3}$$ $$\frac{1}{\tan 105^\circ} = \cot(105^\circ) = -\cot 75^\circ = \sqrt{3} - 2$$ $$\tan 195^\circ = \tan 15^\circ = 2 - \sqrt{3}$$ Therefore, $$2(2 - \sqrt{3}) = 2a \implies a = 2 - \sqrt{3}$$ Thus, $$a + \frac{1}{a} = 4$$

Equation of straight line: $\($ $\frac{x}{a}$ + $\frac{y}{b}$ = 1 $\)$ Or $\($ x $\cos$ $\frac{\pi}{3}$ + y $\sin$ $\frac{\pi}{3}$ = p $\)$ $\[$ $\frac{x}{2}$ + $\frac{y \sqrt{3}}{2}$ = p $\]$ $\[$ $\frac{x}{3p}$ + $\frac{y}{2p}$ = 1 $\]$ Comparing both: $\($ a = 2p, b = $\frac{2p}{\sqrt{3}}$ $\)$ Now area of $\($ $\triangle$ OAB = $\frac{1}{2}$ $\cdot$ ab = $\frac{98}{3}$ $\cdot$ $\sqrt{3}$ $\)$ $\[$ $\frac{1}{2}$ $\cdot$ 2p $\cdot$ $\frac{2p}{\sqrt{3}}$ = $\frac{98}{3}$ $\cdot$ $\sqrt{3}$ $\]$ $\[$ p^2 = 49 $\]$ $\[$ $\frac{p^2 \cdot 49}{2}$ = 2 $\]$

Given $L_1: y = x + 2$, $L_2: 4y = 3x + 6$, $L_3: 3y = 4x + 1$. Bisector of lines $L_2$ and $L_3$ is given by: $$\frac{4x - 3y + 1}{5} = \pm \left( \frac{3x - 4y + 6}{5} \right)$$ For the positive case: $$4x - 3y + 1 = 3x - 4y + 6$$ Simplifying gives: $$x + y = 5$$ The center lies on the bisector of $4x - 3y + 1 = 0$ and $$3x - 4y + 6 = 0$$ Therefore, $h + k = 5$.

Equation of normal $y = -tx + 2at + at^3$ $y = -tx + \frac{2}{16} t + \frac{1}{16} t^3$ It passes through $(0, 33)$ $$33 = \frac{t}{8} + \frac{t^3}{16}$$ $$t^3 + 2t - 528 = 0$$ $$(t - 8)(t^2 + 8t + 66) = 0$$ $t = 8$ $P(at^2, 2at) = \left( \frac{1}{16} \times 64, 2 \times \frac{1}{16} \times 8 \right) = (4, 1)$ Parabola: $y^2 = 4(x + y)$ $\Rightarrow y^2 - 4y = 4x$ $\Rightarrow (y - 2)^2 = 4(x + 1)$ Equation of directix: $x + 1 = -1$ $x = -2$ Distance of point = 6

For $S_1 \equiv ((p \lor q) \Rightarrow r) \Leftrightarrow (p \Rightarrow r)$, the truth table is as follows: \begin{tabular}{|l|l|l|l|l|l|} \hline p & q & r & (p \lor q) & (p \lor q) \Rightarrow r & ((p \lor q) \Rightarrow r) \Leftrightarrow (p \Rightarrow r) \\ \hline T & T & T & T & T & T \\ \hline T & T & F & T & F & F \\ \hline T & F & T & T & T & T \\ \hline T & F & F & T & F & F \\ \hline F & T & T & T & T & T \\ \hline F & T & F & T & F & T \\ \hline F & F & T & F & T & T \\ \hline F & F & F & F & T & T \\ \hline \end{tabular} $_$ For $S_2 \equiv (p \lor q) \Rightarrow r \Leftrightarrow ((p \Rightarrow r) \lor (q \Rightarrow r))$, the truth table is as follows: \begin{tabular}{|l|l|l|l|l|l|} \hline p & q & r & (p \lor q) & (p \lor q) \Rightarrow r & (p \Rightarrow r) \lor (q \Rightarrow r) & S_2 \\ \hline T & T & T & T & T & T & T \\ \hline T & T & F & T & F & T & F \\ \hline T & F & T & T & T & T & T \\ \hline T & F & F & T & F & T & F \\ \hline F & T & T & T & T & T & T \\ \hline F & T & F & T & F & T & F \\ \hline F & F & T & F & T & T & T \\ \hline F & F & F & F & T & T & T \\ \hline \end{tabular}

For symmetric $(a, b), (b, c) \in R$ implies $(b, a), (c, b) \in R$. For transitive $(a, b), (b, c) \in R$ implies $(a, c) \in R$. Now 1. Symmetric $\therefore (a, c) \in R \implies (c, a) \in R$ 2. Transitive $\therefore (a, b), (b, a) \in R$ implies $(a, a) \in R \& (b, c), (c, b) \in R$ implies $(b, b) \& (c, c) \in R$ $\therefore$ Elements to be added $$\left\{ (b, a), (c, b), (a, c), (c, a), (a, a), (b, b), (c, c) \right\}$$ Number of elements to be added $= 7$

Given $$A = \begin{bmatrix} m & n \\ p & q \end{bmatrix}, |A - d(adj A)| = 0$$ We have $$|A - d(adj A)| = \left| \begin{bmatrix} m & n \\ p & q \end{bmatrix} - d \begin{bmatrix} q & -n \\ -p & m \end{bmatrix} \right|$$ This simplifies to $$= \left| \begin{bmatrix} m - qd & n(1 + d) \\ p(1 + d) & q - md \end{bmatrix} \right| = 0$$ Thus, $$(m - qd)(q - md) - np(1 + d)^2 = 0$$ Expanding, we get $$mq - m^2d - q^2d + mqd^2 - np(1 + d)^2 = 0$$ Rearranging terms, $$(mq - np) + d^2(mq - np) - d(m^2 + q^2 + 2np) = 0$$ This leads to $$d + d^3 - d((m + q)^2 - 2d) = 0$$ Simplifying further, $$1 + d^2 = (m + q)^2 - 2d$$ Finally, $$(1 + d)^2 = (m + q)^2$$ Therefore, Option (1) is correct.

Given the matrix equation: $$\begin{vmatrix} 1 & 1 & k \\ 2 & 3 & -1 \\ 3 & 4 & 2 \end{vmatrix} = 0$$ We have: $$1(10) - 1(7) + k(-1) - 0$$ This simplifies to: $$k = 3$$ For $k = 3$, the second system is: $$4x + 5y = 7 .....(1)$$ and $$7x + 8y = 10 .....(2)$$ Clearly, they have a unique solution. Subtracting equation (1) from equation (2): $$(2) - (1) \Rightarrow 3x + 3y = 3$$ This implies: $$x + y = 1$$

Given $5f(x+y) = f(x) \cdot f(y)$. $5f(0) = f(0)^2 \Rightarrow f(0) = 5$. $5f(x+1) = f(x) \cdot f(1)$. Therefore, $$\frac{f(x+1)}{f(x)} = \frac{f(1)}{5}$$ Thus, $$\frac{f(1)}{f(0)} \cdot \frac{f(2)}{f(1)} \cdot \frac{f(3)}{f(2)} = \left(\frac{f(1)}{5}\right)^3$$ This implies $$\frac{320}{5} = \frac{(f(1))^3}{5^3} \Rightarrow f(1) = 20$$ Therefore, $5f(x+1) = 20 \cdot f(x) \Rightarrow f(x+1) = 4f(x)$. The sum is $$\sum_{n=0}^{5} f(n) = 5 + 5 \cdot 4 + 5 \cdot 4^2 + 5 \cdot 4^3 + 5 \cdot 4^4 + 5 \cdot 4^5$$ This equals $$= \frac{5 \left[4^6 - 1\right]}{3} = 6825$$

Normal of line is parallel to line $x + 90y + 2 = 0$ $$m_N = -\frac{1}{90}$$ $$-\left(\frac{dx}{dy}\right)_{(x_1, y_1)} = -\frac{1}{90} \implies \left(\frac{dy}{dx}\right)_{(x_1, y_1)} = 90$$ Now, $$\frac{dy}{dx} = 270x^4 - 540x^3 - 210x^2 + 360x + 210 = 90$$ $$\implies x = 1, 2, -\frac{2}{3}, -\frac{1}{3}$$ (4) normals

Given the integral $$\int_1^2 x^2 e^{[x^3]+1} \, dx$$. Let $x^3 = t$, then $3x^2 \, dx = dt$. The integral becomes $$= \frac{e}{3} \int_1^8 e^{[t]} \, dt$$. This can be expanded as $$= \frac{e}{3} \left( \int_1^2 e \, dt + \int_2^3 e^2 \, dt + \ldots + \int_7^8 e^7 \, dt \right)$$. Evaluating the integrals, we have $$= \frac{e}{3} (e + e^2 + \ldots + e^7)$$. Simplifying further, $$= \frac{e^2}{3} (1 + e + \ldots + e^6) = \frac{e^2 (e^7 - 1)}{3 (e - 1)}$$. Therefore, $$\frac{3(e-1)}{e} \int_1^2 x^2 \times e^{[x]+[x^3]} \, dx = \frac{3}{e} (e-1) \times \frac{e^2 (e^7 - 1)}{3 (e-1)}$$. This simplifies to $$= e(e^7 - 1)$$ which equals $$= e^8 - e$$.

Given $\frac{dy}{dx} + \left( -\frac{3x^5 \tan^{-1} x^3}{(1+x^6)^{3/2}} \right) y = 2e^{\left[ \frac{x - \tan x}{\sqrt{1+x^6}} \right]}$. The integrating factor (I.F.) is $e^{\int \frac{-3x^5 \tan^{-1} x^3}{(1+x^6)^{3/2}} \, dx}$. $\text{I.F.} = e^{\frac{\tan^{-1} x^3 - x^3}{\sqrt{1+x^6}}} = e^{\frac{\tan^{-1} x^3 - x^3}{\sqrt{1+x^6}}}$. Solution of the differential equation: \[ y \cdot e^{\frac{\tan^{-1} x^3 - x^3}{\sqrt{1+x^6}}} = \int 2xe^{\left( \frac{x^3 - \tan^{-1} x^3}{\sqrt{1+x^6}} \right)} \cdot e^{\left( \frac{\tan^{-1} (x^3) - x^3}{\sqrt{1+x^6}} \right)} \, dx \] \[ = \int 2x \, dx + c \] $y \cdot e^{\frac{\tan^{-1} x^3 - x^3}{\sqrt{1+x^6}}} = x^2 + c$ Also, it passes through the origin, so $c = 0$. $y(1) \cdot e^{\frac{\tan^{-1} (1) - 1}{\sqrt{2}}} = 1$ $y(1) \cdot e^{\frac{-1}{4\sqrt{2}}} = 1$ $y(1) \cdot e^{\frac{-\pi - 4}{4\sqrt{2}}} = 1$ $y(1) = \frac{1}{e^{\frac{-\pi - 4}{4\sqrt{2}}}} = e^{\frac{4 - \pi}{4\sqrt{2}}}$

Given $\hat{n} \perp \vec{c}$ and $\vec{a} = \alpha \vec{b} - \vec{n}$. We have $\vec{b} \cdot \vec{c} = 12$. Then, $\vec{a} \cdot \vec{c} = \alpha (\vec{b} \cdot \vec{c}) - \vec{n} \cdot \vec{c}$. Since $\vec{n} \perp \vec{c}$, $\vec{n} \cdot \vec{c} = 0$. Therefore, $\vec{a} \cdot \vec{c} = \alpha (\vec{b} \cdot \vec{c})$. Now, $|\vec{c} \times (\vec{a} \times \vec{b})| = |(\vec{c} \cdot \vec{b}) \vec{a} - (\vec{c} \cdot \vec{a}) \vec{b}|$. Substituting, we get $|(\vec{c} \cdot \vec{b}) \vec{a} - \alpha (\vec{b} \cdot \vec{c}) \vec{b}|$. This simplifies to $|(\vec{c} \cdot \vec{b})||\vec{a} - \alpha \vec{b}|$. Therefore, $= 12 \times (|\hat{n}|) = 12 \times 1 = 12$.

Line $\ell$, is given by $$L_1: \frac{x-2}{2} = \frac{y-6}{1} = \frac{z-2}{-2}$$ Given, $$L_2: \frac{x+1}{2} = \frac{y+4}{-3} = \frac{z}{2}$$ Shortest distance $= \frac{|\overline{AB} \cdot \overline{MN}|}{|\overline{MN}|}$ $$\overline{AB} = 3\hat{i} + 10\hat{j} + 2\hat{k}$$ $$\overline{MN} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 2 & -3 & 2 \end{vmatrix} = -4\hat{i} - 8\hat{j} - 8\hat{k}$$ $$|\overline{MN}| = \sqrt{16 + 64 + 64} = 12$$ Therefore, Shortest distance $= \frac{| -12 - 80 - 16 |}{12} = 9$ Thus, Option (4) is correct.

Equation of plane: $$\begin{vmatrix} x-1 & y-2 & z-3 \\ 1 & 1 & 1 \\ 0 & 3 & 4 \end{vmatrix} = 0$$ $\($ $\Rightarrow$ [x-1] - 4[y-2] + 3[z-3] = 0 $\)$ $\($ $\Rightarrow$ x - 4y + 3z = 2 $\)$ D.R's of normal of plane $\($ $\langle$ 1, -4, 3 $\rangle$ $\)$ D.C's of $$\left\langle \pm \frac{1}{\sqrt{26}}, \pm \frac{4}{\sqrt{26}}, \pm \frac{3}{\sqrt{26}} \right\rangle$$ $\($ $\cos$ $\beta$ = $\frac{4}{\sqrt{26}}$ $\)$ $\($ $\cos$ $\alpha$ = $\frac{-1}{\sqrt{26}}$ $\)$ $\($ $\frac{\pi}{2}$ < $\alpha$ < $\pi$ $\)$ $\($ $\cos$ $\gamma$ = $\frac{-3}{\sqrt{26}}$ $\)$ $\($ $\frac{\pi}{2}$ < $\gamma$ < $\pi$ $\)$ Ans.: (1)

Either all outcomes are positive or any two are negative. Now, $p = P(positive) = \frac{3}{6} = \frac{1}{2}$ $q = p(negative) = \frac{2}{6} = \frac{1}{3}$ Required probability $$= \binom{5}{5} \left( \frac{1}{2} \right)^5 + \binom{5}{2} \left( \frac{1}{3} \right)^2 \left( \frac{1}{2} \right)^3 + \binom{5}{4} \left( \frac{1}{3} \right)^4 \left( \frac{1}{2} \right)^1$$ $$= \frac{521}{2592}$$ Therefore, Option (2) is correct.

Given $z = 1 + i$. $$z_1 = \frac{1 + i \bar{z}}{\bar{z}(1-z) + \frac{1}{z}}$$ Substituting $z = 1 + i$, we have: $$z_1 = \frac{1 + i(1-i)}{(1-i)(1-1-i) + \frac{1}{1+i}}$$ Simplifying further: $$= \frac{1 + i - i^2}{(1-i)(-i) + \frac{1-i}{2}}$$ $$= \frac{2 + i}{-3i - 1} = \frac{4 + 2i}{-3i - 1}$$ Multiplying numerator and denominator by the conjugate: $$= \frac{-(4 + 2i)(3i - 1)}{(3i)^2 - (1)^2}$$ The argument of $z_1$ is: $$Arg(z_1) = \frac{3\pi}{4}$$ Therefore: $$\frac{12}{\pi} arg(z_1) = \frac{12}{\pi} \times \frac{3\pi}{4} = 9$$

For a number to be divisible by 15, the last digit should be 5 and the sum of digits must be divisible by 3. Possible combinations are 1 2 1 5 Numbers = 3 2 2 3 5 Numbers = 3 3 3 1 5 Numbers = 3 1 1 5 5 Numbers = 3 2 3 5 5 Numbers = 6 3 5 5 5 Numbers = 3 Total Numbers = 21

The given expression is $$\sum_{n=0}^{\infty} \frac{n^3 \left( (2n)! \right) + (2n-1)(n!)}{(n!)((2n)!)}$$ which simplifies to $$\sum_{n=0}^{\infty} \frac{1}{(n-3)!} + \sum_{n=0}^{\infty} \frac{3}{(n-2)!} + \sum_{n=0}^{\infty} \frac{1}{(n-1)!} + \sum_{n=0}^{\infty} \frac{1}{(2n-1)!} - \sum_{n=0}^{\infty} \frac{1}{(2n)!}.$$ This further simplifies to $$= e + 3e + e + \frac{1}{2} \left( e - \frac{1}{e} \right) - \frac{1}{2} \left( e + \frac{1}{e} \right).$$ Therefore, $$= 5e - \frac{1}{e}.$$ Finally, $$a^2 - b + c = 26.$$

Given $\($ $\frac{x_1 + x_2 + \ldots + x_7}{7}$ = 8 $\)$. $\($ $\frac{x_1 + x_2 + x_3 + \ldots + x_6 + 14}{7}$ = 8 $\)$ This implies $\($ x_1 + x_2 + $\ldots$ + x_6 = 42 $\)$. Therefore, $\($ $\frac{x_1 + x_2 + \ldots + x_6}{6}$ = $\frac{42}{6}$ = 7 = a $\)$. $\($ $\frac{\Sigma x_i^2}{7}$ - 8^2 = 16 $\)$ $\($ $\Sigma$ x_i^2 = 560 $\)$ $\($ $\Rightarrow$ x_1^2 + x_2^2 + $\ldots$ + x_6^2 = 364 $\)$ $\($ b = $\frac{x_1^2 + x_2^2 + \ldots + x_6^2}{6}$ - 7^2 $\)$ $\($ = $\frac{364}{6}$ - 49 $\)$ $\($ b = $\frac{70}{6}$ $\)$ $\($ a + 3b - 5 = 7 + 3 $\times$ $\frac{70}{6}$ - 5 $\)$ $\($ = 37 $\)$

Let $S = \{1, 2, 3, 4, 5, 6\}$, then the number of one-one functions, $f: S \to P(S)$, where $P(S)$ denotes the power set of $S$, such that $f(n) < f(m)$ where $n < m$ is $$n(S) = 6$$ $$P(S) = \left\{ \emptyset, \{1\}, \ldots, \{6\}, \{1, 2\}, \ldots, \{5, 6\}, \ldots, \{1, 2, 3, 4, 5, 6\} \right\}$$ - 64 elements case - 1 $f(6) = S$ i.e. 1 option, $f(5) = any 5 element subset A of S$ i.e. 6 options, $f(4) = any 4 element subset B of A$ i.e. 5 options, $f(3) = any 3 element subset C of B$ i.e. 4 options, $f(2) = any 2 element subset D of C$ i.e. 3 options, $f(1) = any 1 element subset E of D or empty subset i.e. 3 options,$ Total functions = 1080 Case - 2 $f(6) = any 5 element subset A of S$ i.e. 6 options, $f(5) = any 4 element subset B of A$ i.e. 5 options, $f'(4) = any 3 element subset C of B$ i.e. 4 options, $f(3) = any 2 element subset D of C$ i.e. 3 options, $f'(2) = any 1 element subset E of D$ i.e. 2 options, $f(1) = empty subset i.e. 1 option$ Total functions = 720 Case - 3 $f(6) = S$ $f(5) = any 4 element subset A of S$ i.e. 15 options, $f(4) = any 3 element subset B of A$ i.e. 4 options, $f(3) = any 2 element subset C of B$ i.e. 3 options, $f(2) = any 1 element subset D of C$ i.e. 2 options, $f(1) = empty subset i.e. 1 option$ Total functions = 360 Case - 4 $f(6) = S$ $f(5) = any 5 element subset A of S$ i.e. 6 options, $f(4) = any 3 element subset B of A$ i.e. 10 options, $f(3) = any 2 element subset C of B$ i.e. 3 options, $f(2) = any 1 element subset D of C$ i.e. 2 options, $f(1) = empty subset i.e. 1 option$

Given $$f^1(x) = \frac{3x+2}{2x+3}$$ Then $$f^2(x) = \frac{13x+12}{12x+13}$$ Then $$f^3(x) = \frac{63x+62}{62x+63}$$ Therefore, $$f^5(x) = \frac{1563x+1562}{1562x+1563}$$ Finally, $$a + b = 3125$$

Given the limit expression: $$48 \lim_{x \to 0} \frac{\int_0^x \frac{t^3}{t^6 + 1} \, dt}{x^4}$$ This is an indeterminate form $\($ $\frac{0}{0}$ $\)$. Applying L'Hôpital's Rule: $$48 \lim_{x \to 0} \frac{x^3}{x^6 + 1} \times \frac{1}{4x^3}$$ The result is: $$= 12$$

Given $y = x$ and $y^2 = 8x$. Solving it, $x^2 = 8x$. Therefore, $x = 0, 8$. Thus, $y = 0, 8$. The intersection at $x = 2$ will occur at $y^2 = 16 \Rightarrow y = \pm 4$. Therefore, the area of the shaded region is $$\int_{2}^{8} (\sqrt{8x - x}) \, dx = \int_{2}^{8} (2\sqrt{2} \sqrt{x} - x) \, dx$$ $$= \left[ 2\sqrt{2} \cdot \frac{x^{3/2}}{3/2} - \frac{x^2}{2} \right]_{0}^{8}$$ $$= \left( \frac{4\sqrt{2}}{2} \cdot 2^{9/2} - 32 \right) - \left( \frac{4\sqrt{2}}{2} \cdot 2^{9/2} - 2 \right)$$ Therefore, $3A = 22$.

Given the equations of the planes: $$x - 3y + 2z - 1 = 0$$ $$4x - y + z = 0$$ The cross product of the normals $\vec{n}_1 \times \vec{n}_2$ is: $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ 4 & -1 & 1 \end{vmatrix}$$ This results in: $$=-\hat{i} + 7\hat{j} + 11\hat{k}$$ The direction ratios of the normal to the plane are $-1, 7, 11$. Equation of the plane: $$-1(x - 1) + 7(y - 1) + 11(z - 2) = 0$$ Simplifying gives: $$-x + 7y + 11z = 28$$ Dividing throughout by 28: $$\frac{-1}{28}x + \frac{7y}{28} + \frac{11z}{28} = 1$$ This is in the form $Ax + By + Cz = 1$. Calculating $140(C - B + A)$: $$140 \left( \frac{11}{28} - \frac{7}{28} - \frac{1}{28} \right)$$ This simplifies to: $$= 140 \times \frac{3}{28} = 15$$

Given $\mathbf{P}_1 = \mathbf{r} \cdot (3\hat{i} - 5\hat{j} + \hat{k}) = 7$ and $\mathbf{P}_2 = \mathbf{r} \cdot (\lambda \hat{i} + \hat{j} - 3\hat{k}) = 9$. $\theta = \sin^{-1}\left(\frac{2\sqrt{6}}{5}\right)$ Therefore, $\sin \theta = \frac{2\sqrt{6}}{5}$. Thus, $\cos \theta = \frac{1}{5}$. $\cos \theta = \frac{\mathbf{r} \cdot \mathbf{r}}{|\mathbf{r}_1||\mathbf{r}_2|}$ $$= \frac{(3\hat{i} - 5\hat{j} + \hat{k})(\lambda \hat{i} + \hat{j} - 3\hat{k})}{\sqrt{35} \cdot \sqrt{\lambda^2 + 10}}$$ $$\frac{1}{5} = \left|\frac{3\lambda - 8}{\sqrt{35} \cdot \sqrt{\lambda^2 + 10}}\right|$$ Square both sides: $$\Rightarrow \frac{1}{25} = \frac{9\lambda^2 + 64 - 48\lambda}{35(\lambda^2 + 10)}$$ $$\Rightarrow 19\lambda^2 - 120\lambda + 125 = 0$$ $$19\lambda^2 - 95\lambda - 25\lambda + 125 = 0$$