JEE MAIN 2020 - 06 SEP S1 2020

Transition elements are d-block elements, characterized by having partially filled d-orbitals. Let's analyze the atomic numbers ($Z$) in each set based on their position in the periodic table.

The set in option B contains $Z=21$ (Scandium), $Z=25$ (Manganese), $Z=42$ (Molybdenum), and $Z=72$ (Hafnium). All four of these are d-block elements and are thus classified as transition elements.

The other sets contain elements that are not transition metals. For instance, set A includes $Z=32$ (Germanium, a p-block metalloid) and $Z=53$ (Iodine, a p-block halogen). Set C contains only s-block and p-block elements like $Z=9$ (Fluorine) and $Z=38$ (Strontium). Set D includes $Z=37$ (Rubidium, an s-block alkali metal) and $Z=50$ (Tin, a p-block metal).

Thus, only the set in option B consists exclusively of atomic numbers of transition elements.

The stability of the +4 oxidation state in lanthanoids is primarily due to achieving a stable electron configuration, such as an empty ($f^0$) or half-filled ($f^7$) f-orbital. Cerium (Ce) readily forms $Ce^{4+}$ with a stable $f^0$ configuration. Terbium (Tb) forms $Tb^{4+}$, achieving the highly stable half-filled $f^7$ configuration. Dysprosium (Dy) is also known to exhibit a +4 state, forming a less stable $f^8$ ion. Europium (Eu), however, has a very stable +2 oxidation state ($Eu^{2+}$) due to its half-filled $f^7$ shell. The energy required to remove two more electrons to form $Eu^{4+}$ (an unstable $f^5$ configuration) is prohibitively high.

An alloy is a mixture of metals or a metal combined with one or more other elements. Let's examine the composition of each alloy. Bronze is correctly identified as an alloy of copper ($Cu$) and tin ($Sn$). Wrought iron is a purer form of iron manufactured by refining cast iron. German silver is indeed an alloy containing copper ($Cu$), zinc ($Zn$), and nickel ($Ni$). The incorrect statement is B, because brass is an alloy of copper ($Cu$) and zinc ($Zn$). An alloy of copper ($Cu$) and nickel ($Ni$) is known as cupronickel.

Dinitrogen ({\text{N}}_{2}$) has a triple bond between its two nitrogen atoms ({\text{N}}\equiv{\text{N}}$). This bond has a very high bond dissociation enthalpy, making ${\text{N}}_{2}$ exceptionally stable and inert at room temperature. This inert nature allows it to be used as a diluent for reactive chemicals, preventing unwanted reactions by providing a non-reactive atmosphere.

Conversely, liquid ${\text{N}}_{2}$ is commonly used in cryosurgery (A is incorrect). The reaction between ${\text{N}}_{2}$ and ${\text{O}}_{2}$ requires extremely high temperatures, around 2000 K, not $25°{\text{C}}$ (C is incorrect). According to Molecular Orbital Theory, ${\text{N}}_{2}$ has no unpaired electrons and is therefore diamagnetic, not paramagnetic (D is incorrect).

Molarity ($C_2$) is the moles of solute (component 2) per liter of solution. We begin with its definition: $C_2 = \frac{\text{moles of component 2}}{\text{Volume of solution (L)}} = \frac{1000 \cdot n_2}{V_{\text{sol (mL)}}}$. The total mass of the solution is $n_1 M_1 + n_2 M_2$, and its volume is $V_{\text{sol (mL)}} = \frac{\text{Mass}}{d} = \frac{n_1 M_1 + n_2 M_2}{d}$. Substituting the volume into the molarity equation gives $C_2 = \frac{1000 \ d \ n_2}{n_1 M_1 + n_2 M_2}$. To express this in terms of mole fractions, we divide the numerator and denominator by the total moles ($n_1 + n_2$), which gives $C_2 = \frac{1000 \ d \ x_2}{x_1 M_1 + x_2 M_2}$. Finally, by substituting $x_1 = 1 - x_2$, the denominator becomes $(1 - x_2)M_1 + x_2 M_2 = M_1 + x_2(M_2 - M_1)$. Thus, the final expression is $C_2 = \frac{1000 \ d \ x_2}{M_1 + x_2(M_2 - M_1)}$.

This is a two-step reaction sequence. The first step involves an E2 elimination reaction of an alkyl mesylate. Potassium tert-butoxide ($KO^tBu$) is a strong, bulky base, which favors the abstraction of the less sterically hindered proton. This leads to the formation of the less substituted alkene, the Hofmann product. Therefore, 3-methylbut-1-ene ($(CH_3)_2CH-CH=CH_2$) is the major intermediate.

The second step is the oxidative ozonolysis of this alkene using $O_3$ followed by $H_2O_2$. This reaction cleaves the double bond and oxidizes the resulting fragments. The internal carbon of the double bond is oxidized to a carboxylic acid, forming 2-methylpropanoic acid ($(CH_3)_2CHCOOH$). The terminal $=CH_2$ group is first converted to formaldehyde, which is then further oxidized by $H_2O_2$ to formic acid ($HCOOH$). Thus, the major products are 2-methylpropanoic acid and formic acid.

The Kraft temperature ($T_k$) is a critical property of surfactants. Below the Kraft temperature, the solubility of the surfactant is lower than its Critical Micelle Concentration (CMC). This means that even if you add enough surfactant, it won't form micelles and will instead precipitate out of the solution. As you increase the temperature, the solubility of the surfactant increases. The Kraft temperature is the specific point at which the surfactant's solubility equals the CMC. Only at temperatures above $T_k$ is the surfactant soluble enough to reach the CMC and begin forming micelles.

The Assertion (A) states that ethene polymerized using a Ziegler-Natta catalyst ($TiCl_4$ + $Al(C_2H_5)_3$) is used to make buckets and dustbins. This process yields High-Density Polyethene (HDP). HDP is known for its high strength and rigidity, making it the ideal material for manufacturing such durable goods. Thus, the Assertion (A) is correct.

The Reason (R) states that high-density polymers are closely packed and chemically inert. This is also correct. HDP consists of linear polymer chains that can pack together efficiently, resulting in a high density and crystalline structure, which imparts strength and hardness. Its saturated hydrocarbon nature makes it chemically inert.

The properties mentioned in (R)-close packing (leading to strength and rigidity) and chemical inertness-are precisely why HDP is suitable for making buckets and dustbins as mentioned in (A). Therefore, (R) is the correct explanation for (A).

The spin-only magnetic moment is calculated using the formula $\mu_s = \sqrt{n(n+2)} {\text{ BM}}$, where $n$ is the number of unpaired electrons. A magnetic moment of $5.9 {\text{ BM}}$ corresponds to $n=5$ unpaired electrons, since $\sqrt{5(5+2)} = \sqrt{35} \approx 5.92 {\text{ BM}}$. We must identify the complex with five unpaired electrons. In the complex $[{\text{MnBr}}_{4}]^{2-}$, the oxidation state of manganese is $+2$. The electronic configuration of ${\text{Mn}}^{2+}$ is $[{\text{Ar}}] 3d^5$. Since ${\text{Br}}^-$ is a weak field ligand and the geometry is tetrahedral, the complex is high-spin. This results in all five $3d$ electrons being unpaired ($n=5$), matching the given magnetic moment.

This reaction is the acid-catalyzed hydration of an alkyne, which proceeds via Markovnikov addition. The regioselectivity is determined by the electronic nature of the substituents. The methoxy group ($-OCH_3$) is electron-donating ($+M$ effect), stabilizing a positive charge on the adjacent alkyne carbon. In contrast, the nitro group ($-NO_2$) is electron-withdrawing ($-M$ effect) and destabilizes a positive charge. Therefore, the nucleophile ($H_2O$) preferentially attacks the carbon atom bonded to the methoxy-substituted phenyl ring. The initially formed enol then tautomerizes to the more stable ketone, with the carbonyl group adjacent to the 4-methoxyphenyl ring.

$K _{ P }= K _{ C }( RT )^{\Delta n _{ g }}$ $= K _{ C }( RT )^{1-\;\frac{3}{2}}$ $= K _{ C }( RT )^{-\;\frac{1}{2}}$ $\Rightarrow K _{ C }= K _{ p }( RT )^{1 / 2}$

(A) 10^{-2} MHCl \Rightarrow \[ H ^{+}]$=10^{-2} M \Rightarrow pH =2 \Rightarrow pOH =14-2=12$(B)$10^{-2} M NaOH \Rightarrow $[ OH ^{-}]$=10^{-2} M \Rightarrow pOH =2$(C)$10^{-2} M CH 3 COO ^{-} Na ^{+} \Rightarrow $[ OH ^{+}]$>10^{-7} \Rightarrow pOH <7$(D)$10^{-2} M NaCl \Rightarrow $Neutral$pOH =7$Order of$pOH$value$A > D > C > B$

Fluoride ions in drinking water at a concentration of about $1$ ppm are beneficial for dental health. This concentration helps prevent tooth decay, a process also known as dental caries. The fluoride ion ($F^−$) converts the hydroxyapatite ($Ca_5(PO_4)_3OH$), the main component of tooth enamel, into the much harder and more acid-resistant fluorapatite ($Ca_5(PO_4)_3F$). This strengthens the enamel and protects it from acid produced by bacteria. Concentrations above $1.5-2$ ppm can become harmful, causing a condition called dental fluorosis (mottling of teeth). Thus, $1$ ppm is considered a safe and effective level.

Based on the reactions, we can deduce the structures of the ozonolysis products 'B' and 'C'. 'B' gives a positive Tollens' test (silver mirror), indicating it's an aldehyde, and a positive iodoform test (yellow ppt), which for an aldehyde is specific to acetaldehyde ($CH_3CHO$). 'C' does not give an iodoform test, so it's not a methyl ketone. Reduction of 'C' with $LiAlH_4$ gives 'D', which reacts with the Lucas reagent (Anhydrous $ZnCl_2$/conc. $HCl$) to form turbidity in 5 minutes, identifying 'D' as a secondary alcohol. This means 'C' must be a ketone that is not a methyl ketone.

The starting alkene 'A' ($C_7H_{14}$) breaks down into 'B' ($C_2H_4O$) and 'C'. By carbon conservation, 'C' must be a 5-carbon ketone. The only 5-carbon ketone that is not a methyl ketone is pentan-3-one ($CH_3CH_2COCH_2CH_3$). Reversing the ozonolysis by combining the carbonyl carbons of acetaldehyde ($CH_3CHO$) and pentan-3-one ($(CH_3CH_2)_2CO$) gives the structure of 'A' as 3-ethylpent-2-ene, $CH_3-CH=C(CH_2CH_3)_2$.

The ${\text{pK}}_{{\text{b}}}$ value is inversely proportional to the basicity of a compound ({\text{pK}}{{\text{b}}} = -\log {\text{K}}{{\text{b}}}$). Basicity depends on the availability of the lone pair of electrons on the nitrogen atom. Electron-donating groups (EDGs) increase basicity (decrease${\text{pK}}{{\text{b}}}$), while electron-withdrawing groups (EWGs) decrease basicity (increase${\text{pK}}{{\text{b}}}$).

In compound I, the -OCH$_3$ group at the para position exerts a strong electron-donating resonance (+R) effect, making it the most basic. Compound II (N,N-dimethylaniline) is the reference. In compound IV, the -OH group at the meta position has a weak electron-withdrawing inductive (-I) effect. In compound III, the -CN group at the meta position has a strong electron-withdrawing inductive (-I) effect, making it the least basic. The order of basicity is therefore I > II > IV > III. This corresponds to the increasing order of ${\text{pK}}_{{\text{b}}}$ values as I < II < IV < III.

The solubility of alkaline earth metal sulfates in water is determined by the interplay between lattice enthalpy and hydration enthalpy. For a salt to dissolve, the hydration enthalpy released must overcome the lattice enthalpy required to break the crystal structure.

The cations $\text{Be}^{2+}$ and $\text{Mg}^{2+}$ are very small and have a high charge density. This leads to a very high hydration enthalpy upon dissolution. Although their lattice enthalpies are also high, the magnitude of the hydration enthalpy is significantly greater. This makes the overall enthalpy of solution ($\Delta H_{\text{sol}}$) negative, favoring dissolution. Therefore, both $\text{BeSO}_4$ and $\text{MgSO}_4$ are highly soluble in water. As we move down the group, the cation size increases, causing the hydration enthalpy to decrease more rapidly than the lattice enthalpy, leading to decreased solubility.

This reaction is an electrophilic addition of two equivalents of HBr to the polarized aromatic ring of p-nitrotoluene. The electron-donating methyl group ($-\text{CH}_3$) and the electron-withdrawing nitro group ($-\text{NO}_2$) are in a para "push-pull" arrangement.

The first molecule of HBr adds via protonation at the carbon ortho to the activating methyl group. This generates the most stable carbocation intermediate, a tertiary carbocation at C1, which is stabilized by the methyl group. The bromide ion ($Br^−$) then attacks this carbocation, forming a geminal bromo-methyl group at the C1 position.

This step disrupts aromaticity, yielding a cyclohexadiene intermediate. The second molecule of HBr adds to this diene. Under the influence of the strongly electron-withdrawing nitro group at C4, the second bromine atom adds to this position, C4. This results in the formation of a second geminal center. The overall reaction gives 1,4-dibromo-1-methyl-4-nitrocyclohexane as the major product.

$\ln [\;\frac{k_{2}}{k_{1}}]=\;\frac{\Delta H^{\circ}}{R}\{\;\frac{1}{T_{1}}-\;\frac{1}{T_{2}}\}$ $\Rightarrow \ln (10)=\;\frac{\Delta H^{\circ}}{R}\{\;\frac{1}{298}-\;\frac{1}{373}\}$ $\Rightarrow \Delta H^{\circ}=28.37 kJ / mol$ $\Delta G^{\circ}=-R T \ln K$ $T_{1}=25^{\circ} C K _{1}=10$ $\Delta G^{\circ}$ at $T_{1}=-8.314 \times 298 \times 2.303 \times \log 10$ $=-5.71 kJ / mol$ $\Delta G^{\circ}$ at $T_{2}=-8.314 \times 373 \times 2.303 \times \log (100)$ $=-14.29 kJ / mol$

The rate law for a reaction is given by ${\text{rate}} = k[{\text{conc}}]^n$, where $n$ is the order of the reaction. Taking the logarithm of this equation gives us ${\text{log(rate)}} = n \cdot {\text{log(conc)}} + {\text{log(k)}}$. This equation is in the form of a straight line, $y = mx + c$.

When we plot ${\text{log(rate)}}$ (y-axis) vs. ${\text{log(conc)}}$ (x-axis), the slope ($m$) of the line is equal to the order of the reaction ($n$). A steeper line indicates a larger slope and, therefore, a higher reaction order.

By observing the graph, we can compare the steepness (slope) of the four lines. The order of steepness is: Line [D] > Line [B] > Line [A] > Line [C]. Since the slopes correspond to the reaction orders d, b, a, and c respectively, the correct sequence for the order of the reactions is $d > b > a > c$.

Geometrical isomerism, or cis-trans isomerism, arises in alkenes when there is a carbon-carbon double bond ($C=C$) and each of the two carbon atoms in the double bond is attached to two different groups. Let's examine 4-methylpent-2-ene, which has the structure $CH_3-CH=CH-CH(CH_3)_2$. The double bond is between carbon-2 and carbon-3. Carbon-2 is bonded to a hydrogen atom (-H) and a methyl group (-$CH_3$), which are different. Carbon-3 is bonded to a hydrogen atom (-H) and an isopropyl group (-$CH(CH_3)_2$), which are also different. Since both carbons of the double bond have two distinct groups, this compound exhibits geometrical isomerism. The other options fail this condition; for example, in 2-methylpent-2-ene, C2 is bonded to two identical methyl groups.

Mass of organic compound $=1.6 {g}$ Mass of $A g B r=1.88 g$ Moles of ${Br}=$ Moles of ${AgBr}=\;\frac{1.88}{188}=0.01$ Mass of ${Br}=0.01 \times 80=0.80 {g}$ $\%$ of ${Br}=\;\frac{0.80 \times 100}{ {I} .60}=50 \%$ Alternate Method: $\%$ of Br $=\;\frac{\;\text{ Wt. of AgBr }\;}{\;\text{ Wt of O.C. }\;} \times \;\frac{\;\text{ Molar mass of Br }\;}{\;\text{ AgBr }\;} \times 100$ $=\;\frac{1.88}{1.6} \times \;\frac{80}{188} \times 100=50 \%$

The elevation of boiling point is a colligative property calculated by $\Delta T_b = i K_b m$, where $i$ is the van't Hoff factor. For the $0.05 \text{ m } \text{CaCl}_2$ solution, the salt dissociates into three ions (\text{Ca}^{2+}$and$2\text{Cl}^-$), so its van't Hoff factor,$i_{\text{CaCl}2}$, is 3. Its elevation of boiling point is$\Delta T{b, \text{CaCl}_2} = 3 × K_b × 0.05 = 0.15 K_b$.

The chromium complex has a coordination number of 6. Since all 'x' $\text{NH}_3$ molecules are inside the coordination sphere, the complex formula is $[\text{Cr}(\text{NH}_3)_x\text{Cl}_{6-x}]\text{Cl}_{x-3}$. Upon $100\%$ ionization, it produces one complex ion and $(x-3)$ chloride ions, for a total of $1 + (x-3) = x-2$ particles. Thus, its van't Hoff factor is $i_{\text{complex}} = x-2$.

The elevation of boiling point for the complex solution is $\Delta T_{b, \text{complex}} = (x-2) \times K_b \times 0.10$. We are given that $\Delta T_{b, \text{complex}} = 2 \times \Delta T_{b, \text{CaCl}_2}$.
Therefore, $(x-2) \times 0.10 K_b = 2 \times (0.15 K_b)$.
This simplifies to $0.10(x-2) = 0.30$, which gives $x-2 = 3$, so $x = 5$.

For gas when temperature and number of moles are constant then we can apply Boyle's law. According to Boyle's law, ${P}_{1} {V}_{1}= {P}_{2} {V}_{2}$ $\;\text{ Given, }\; {P}_{1}=48 \times 10^{-3} {bar}$ ${V}_{1}=\;\frac{4}{3} \pi (3)^{3}$ ${V}_{2}=\;\frac{4}{3} \pi (12)^{3}$ ${P}_{2}=?$ $\therefore {P}_{2}=\;\frac{P_{1} V_{1}}{V_{2}}$ $=\;\frac{48 \times 10^{-3} \times (3)^{3}}{(12)^{3}}$ $=7.5 \times 10^{-4}=750 \times 10^{-6}$

To determine the number of ${\text{Cl}}={\text{O}}$ bonds in perchloric acid ({\text{HClO}}_4$), we must draw its most stable Lewis structure. The central atom is chlorine (Cl), bonded to four oxygen (O) atoms, with the hydrogen (H) attached to one of the oxygens. To minimize the formal charges on all atoms, the chlorine atom expands its octet. This results \in a structure where Cl forms a single bond with the hydroxyl ({\text{-OH}}$) group and double bonds with the remaining three oxygen atoms. Therefore, perchloric acid contains three ${\text{Cl}}={\text{O}}$ double bonds.

First, calculate the moles of ${\text{KClO}}_{3}$ to be produced: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{10 {\text{ g}}}{122 {\text{ g/mol}}} \approx 0.082 {\text{ mol}}$.
From the balanced half-reaction, producing 1 mole of ${\text{ClO}}_{3}^{-}$ requires 6 moles of electrons, so the n-factor is 6.
The total charge ($Q$) required is given by Faraday's law: $Q = n \times (\text{n-factor}) \times F$.
$Q = \left(\frac{10}{122}\right) \times 6 \times 96500 {\text{ C}} \approx 47459 {\text{ C}}$.
Since the current efficiency is $60\%$, the effective current used is $I_{\text{eff}} = 2 {\text{ A}} \times 0.60 = 1.2 {\text{ A}}$.
The time required is calculated using $Q = I_{\text{eff}} \times t$, so $t = \frac{Q}{I_{\text{eff}}} = \frac{47459 {\text{ C}}}{1.2 {\text{ A}}} \approx 39549 {\text{ s}}$.
Converting seconds to hours gives $\frac{39549 {\text{ s}}}{3600 {\text{ s/hr}}} \approx 10.986$ hours, which rounds to $11.0$ hours.

Let foot of perpendicular is (h, k) $\frac{x^{2}}{4}+\frac{y^{2}}{2}=1$ (Given) ${a}=2, {b}=\sqrt{2}$, ${e}=\sqrt{1-\frac{2}{4}}=\frac{1}{\sqrt{2}}$ ∴ Focus $({ae}, 0)=(\sqrt{2}, 0)$ Equation of tangent, $y=m x+\sqrt{a^{2} m^{2}+b^{2}}$ $y=m x+\sqrt{4 m^{2}+2}$ Passes through (h, k) $({k}-{mh})^{2}=4 {m}^{2}+2$ ....(1) line perpendicular to tangent will have slope $-1/m$ $y-0=-\frac{1}{m}(x-\sqrt{2})$ $m y=-x+\sqrt{2}$ $({h}+{mk})^{2}=2$.....(2) Add equation (1) and (2) $k^{2}(1+m^{2})+h^{2}(1+m^{2})=4(1+m^{2})$ ${h}^{2}+{k}^{2}=4$ ${x}^{2}+{y}^{2}=4$ (Auxiliary circle) $\therefore (-1,\sqrt3)$ lies on the locus.

Total numbers in three families $=3+3+4=10$ so total arrangement $=10 !$ Favourable cases $={3!}_{\text{Arrangment} \text{of} 3\text{Families}}\;\; {3!\times 3!\times 4!}_{\text{Interval} \text{Arrangment} \text{of} \text{families} \text{members}}$ ∴ Probability of same family memebers are together $=\frac{3 ! 3 ! 3 ! 4 !}{10 !}=\frac{1}{700}$

\lim _{x \to 1}({{∫ _{0} ^{(x-1)^{2}}t\cos(t^2)\text{dt}}/{(x-1) \sin (x-1)})(0/0) Apply L Hopital Rule $=\lim _{x \to 1} \frac{2(x-1) .(x-1)^{2} \cos (x-1)^{4}-0}{(x-1) .\cos (x-1)+\sin (x-1)}(\frac{0}{0})$ $=\lim _{x \to 1} \frac{2(x-1)^{3} .\cos (x-1)^{4}}{(x-1)[\cos (x-1)+\frac{\sin (x-1)}{(x-1)}]}$ $=\lim _{x \to 1} \frac{2(x-1)^{2} \cos (x-1)^{4}}{(x-1)[\cos (x-1)+\frac{\sin (x-1)}{(x-1)}]}$ $=\lim _{x \to 1} \frac{2(x-1)^{2} \cos (x-1)^{4}}{\cos (x-1)+\frac{\sin (x-1)}{(x-1)}}$on taking limit $=0/{1+1}=0$

$\{\frac{3^{200}}{8}\}=\{\frac{(3^{2})^{100}}{8}\}$ $=\{\frac{(1+8)^{100}}{8}\}$ $=\{\frac{1+{ }^{100} {C}_{1} . 8+{ }^{100} {C}_{2} . 8^{2}+...+{ }^{100} {C}_{100} 8^{100}}{8}\}$ $=\{\frac{1+8 {m}}{8}\}=1/8$

For infinite many solutions ${\text{D}}={\text{D}}_{1}={\text{D}}_{2}={\text{D}}_{3}=0$ Now, $\text{D}=\begin{vmatrix}1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & \lambda\end{vmatrix}=0$ $1 .(2 \lambda-9)-1 .(\lambda-3)+1 .(3-2)=0$ $\therefore \lambda=5$ Now, $\text{D}_1=\begin{vmatrix}2 & 1 & 1 \\ 5 & 2 & 3 \\ \mu & 3 & 5\end{vmatrix}=0$ $2(10-9)-1(25-3 \mu)$ $+1(15-2 \mu)=0$ $\therefore \mu=8$

$|x|+|y| \le 1$ $2 y^{2} \ge |x|$ For point of intersection $x+y=1 \Rightarrow x=1-y$ ${y}^{2}=\frac{{x}}{2}$ $\Rightarrow 2 {y}^{2}={x}$ $2 {y}^{2}=1-{y}$ $\Rightarrow 2 {y}^{2}+{y}-1=0$ $(2 {y}-1)({y}+1)=0$ ${y}=\frac{1}{2}$ or $-1$ Now Area of $\Delta{\text{OAB}}=\frac{1}{2} \times 1 \times 1=\frac{1}{2}$ Area of Region ${\text{R}}_{1}=\frac{1}{2} \times \frac{1}{2}\times \frac{1}{2}=\frac{1}{8}$ Area of Region ${\text{R}}_{2}=\frac{1}{\sqrt{2}} ∫^{1/2}_{0}\sqrt{{x}} {d} {x}=\frac{1}{6}$ Now area of shaded region in first quadrant $=$ Area of $\Delta{\text{OAB}}-{\text{R}}_{1}-{\text{R}}_{2}$ $=1/2-(1/6)-(1/8)=5/24$ So required area $=4(\frac{5}{24})=\frac{5}{6}$

Out of 11 consecutive natural numbers either 6 even and 5 odd numbers or 5 even and 6 odd numbers when 3 numbers are selected at random then total cases $={ }^{11} {C}_{3}$ since these 3 numbers are in A.P. Let no's are ${a}, {b}, {c}$ 2b \Rightarrow even number $a+c\Rightarrow (\text{table} \text{even} + \text{even};\text{odd}+\text{odd})$ so favourable cases $={ }^{6} {C}_{2}+{ }^{5} {C}_{2}$ $=15+10=25$ ${\text{P}}(3 \text{numbers} \text{are} \in {\text{A}.\text{P}.}={25}/{^{11} {\text{C}}_{3}}=\frac{25}{165}=\frac{5}{33})$

${\text{S}} . {\text{D}}=\sqrt{\frac{\sum _{i=1} ^{n}(x_{i}-a)}{n}-(\frac{\sum _{i=1} ^{n}(x_{i}-a)}{n})^{2}}$ $=\sqrt{\frac{{na}}{{n}}-(\frac{{n}}{{n}})^{2}}$ $\{ \text{Given} \sum _{i=1} ^{n}(x_{i}-a)=n \sum _{i=1} ^{n}(x_{i}-a)^{2}=n a\}=\sqrt{a-1}$

${y}^{2}=4({x}+1)$ equation of tangent $y=m(x+1)+\frac{1}{m}$ ${y}={mx}+{m}+\frac{1}{{m}}$ ${y}^{2}=8({x}+2)$ equation of tangent $y=m'(x+2)+{2}{m'}$ ${y}={m}'{x}+2({m} +\frac{1}{{m}'})$ since lines intersect at right angles $\therefore {mm}'=-1$ Now ${y}={mx}+{m}+\frac{1}{{m}}$ ...(1) ${y}={m}'{x}+2({m}'+\frac{1}{{m} })$ ${y}=-\frac{1}{{m}} {x}+2(-\frac{1}{{m}}-{m})$ ...(2) From equation (1) and (2) ${mx}+{m}+\frac{1}{{m}}=-\frac{1}{{m}} {x}-2({m}+\frac{1}{{m}})$ $({m}+\frac{1}{{m}}) {x}+3({m}+\frac{1}{{m}})=0$ $\therefore x+3=0$