JEE MAIN 2020 - 07 JAN S1 2020

Option(3) is according to Gaylussac's law of volume combination.

Base strength order increasing order of $pk_b$ order (B < A < C)

Atomic radius of Ag and Au is nearly same due to lanthanide contraction.

E° ΔG° $Cu^{2⊕}+2e^{⊕}\to Cu$ $0.34$ $= -2F(0.34)$ $Cu^{ ⊕}+ e^{⊕}\to Cu$ $0.522$ $-F(0.522)$ $Cu^{2⊕}+ e^{⊕}\to Cu^{+}$ ΔG°=-2F (0.34) - ( - F(0.522) = -F (0.68 - 0.522) = - F (0.158)$E°=\frac{-F(0.158)}{-F}=0.158V$

(a) Zeolite method removes only cations ( $Ca^{2+}$ and $Mg^{2+}$ ion) present in hard water $\text{table} 2NaZ + M^{2+} (aq) ,\to , M Z_2(s) + 2Na^{+}(aq) ;,,(M\to Mg, Ca)$(b) Synthetic resin method removes cations ($Ca^{2+}$ and $M g^{2+}$ ion) and anions (like $Cl^{-}, H CO_3^{-}, SO_4^{2-}$ etc.)$(\text{i}) \text{table} 2RNa(s) + M^{2+}(aq), \to ,R_2M(s) + 2Na^{+}(aq);(\text{Cation} \text{exchange} \text{resin}),, (M\to Mg, Ca)$ $(\text{ii})\text{table} RNH^{+}_3 OH^{-}(s) + X^{-} (aq), \to , RNH_3^{+ }X ^{-}(s) +OH^{-} (aq);(\text{Anion} \text{exchange} \text{resin}) ,, (X^{-} = Cl^{-} ,HCO_3^{-} ,SO_4^{2-}\text{etc})$

Order is ion - ion > ion - dipole > dipole - dipole

$(CH_3)_3O^{-}K^{+}$ is incorrect representation of potassium tert-butoxide $[(CH_3)_3CO ^{-}K^{+}$.So it is possible that it can be given as Bonus

Wrought iron is purest from of commercial iron.

The vapour pressure of mixture ( = 600 mm Hg) is greater than the individual vapour pressure o f its constituents (Vapour pressure of $CS_2$ = 512 mm Hg, acetone = 344 mm Hg). Hence, the solution formed shows positive deviation from Raoult's law \Rightarrow (1) $\Delta_\text{Sol} H > 0$ , (2) Raoult's law is not obeyed (3) $\Delta_\text{sol}$ Volume > 0 (4) $CS_2$ and Acetone are less attracted to each ether than to themselves.

Order of electron gain enthalpy (magnitude) is Cl > F > Br > I

No. of orbitals $= n^2 = 5^2 = 25$For n = 5, no. of orbitals $= n^2 = 25$Total number o f orbitals is equal to no. of electrons having $m_s=1/2$

(i) Riboflavin → (c) Cheilosis(ii) Thiamine → (a) Beriberi(iii) Pyridoxine → (d) Convulsions(iv) Ascorbic acid → (b) Scurvy

In complex $[Ni(CO)_4]$ decrease in Ni-C bond length and increase in C-O bond length as well as it's magnetic property is explained by MOT.

It is an acid base indicator

To determine the IUPAC name for $[Pt(NH_3)_2Cl(NH_2CH_3)]Cl$, we follow a set of rules. First, we find the oxidation state of the central metal, platinum ($Pt$). The counter-ion is $Cl^-$, so the complex ion $[Pt(NH_3)_2Cl(NH_2CH_3)]^+$ has a +1 charge. Since $NH_3$ (ammine) and $NH_2CH_3$ (methanamine) are neutral and $Cl$ (chlorido) is -1, the oxidation state ($x$) of Pt is $x + 2(0) + (-1) + 0 = +1$, which gives $x = +2$. The ligands are then named in alphabetical order: ammine, chlorido, methanamine. The prefix 'di-' for the two ammine ligands does not affect the alphabetical order. Therefore, the name of the cation is diamminechlorido(methanamine)platinum(II). Finally, the name of the anion, chloride, is added.

Potasisum has an oxidation of +1 (only) in combined state.

$A(l)\to 2B(g)$ΔU = 2.1 Kcal , ΔS= 20 cal K$^{- 1}$ at 300 K$\DeltaH = \DeltaU + \Deltan_gRT$ $\DeltaG = \DeltaH -T\DeltaS$ $\DeltaG = \DeltaU + \Deltan_gRT - T\DeltaS$ $=2.1+{2\times 2\times 300}/1000-{300\times 20}/1000$= 2.1 + 1.2 - 6 = - 2.70 Kcal/mol

All nuclear decays follow first order kinetics$t=1/kln\frac{[A_0]}{[A]}$ $=(t_{1/2})/0.693\times 2.303\log_10 10$ $= 10 \times 2.303 \times 1 =23.03\text{years}$

4 gm of NaOH in 100 L sol. ⇒ $10^{-3}$ M sol. 9.8 gm of $H_2SO_4$ in 100 L sol. \Rightarrow $10^{-3}$ M sol. Mixture : 40L of $10^{-3}$ M NaOH and 10 L of $10^{-3}$ M H2S04 sol. Final Cone, of $OH^{-}$ $=\frac{10^{-3}(40\times 1-10\times 1\times 2)}{40+10}$ $=6\times 10^{-4}M$ $pOH = - \log (6 \times 10^{-4})$ $= 4 - \log 6 = 4 - 0.60 = 3.40$pH = 14 - 3.40 = 10.60

Bond order of Cl-O Bond $=1+2/3=5/3=1.66\text{or} 1.67$

$g(x) = x^2 + x-1$ $g (f(x)) = 4 x^2 -10x + 5$ $= (2x-2)^2 + (2 - 2x) -1$ $= (2- 2x)^2 + (2 - 2x) -1$ $\Rightarrow f(x) = 2 - 2x$ $f(5/4)=-1/2$

$\text{Re}(\frac{z-1}{2z+i})= 1$ Put z = x + iy $\text{Re}(\frac{ (x + iy)-1}{2 (x + iy)+i})= 1$ $\text{Re}((\frac{ (x -1)+iy}{2 x + i(2y +1)})(\frac{2 x -i(2y +1)}{2 x -i(2y +1)})= 1$ $\Rightarrow 2x^2 + 2y^2 + 2x + 3y + 1 = 0$ $\Rightarrow x^2 + y^2 + x +3/2y+1/2=0$locus is a circle whose Centre is $(-1/2,3/4)$ and radius $\sqrt5/4$\Rightarrow diameter $= \sqrt5/2$

Sol. Let the A.P is a - 2d, a - d, a, a + d, a + 2d ∵ sum = 25 ⇒ a = 5 Product = 2520 $(25 -4d^2) (25 -d^2) = 504$ $4d^4 -125d^2 + 121 = 0$ $\Rightarrow d^2=1,121/4$ $\Rightarrow d=±1,±11/2$ $d = ± 1$ is rejected because none of the term can be $-1/2$ $\Rightarrow d= ±11/2$⇒ AP will be $-6,-1/2,5,21/2,16$Largest term is 16.

$y(\alpha )=\sqrt{2( \frac{\tan \alpha +\cot\alpha }{1+\tan^2\alpha })+1/{\sin^2\alpha }}$, $\alpha \in ({3\pi }/4,\pi )$ $=\frac{|\sin\alpha +\cos\alpha |}{|\sin\alpha |}$ $=\frac{-(\sin\alpha +\cos\alpha )}{\sin\alpha }$ $=-1-\cot \alpha$ $y'(\alpha )=\text{cosec}^2\alpha$ $y'({5\pi }/6)=4$

$x ^2 + x + 1 = 0$ $\alpha = \omega$ $\alpha ^2 = \omega ^2$ $A=1/\sqrt3\begin{bmatrix}1 & 1 & 1 \\ 1 & \omega & \omega ^2 \\ 1 & \omega ^2 & \omega \end{bmatrix}$ $A^2=\begin{bmatrix}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{bmatrix}$ $\Rightarrow A^4 = A^2 . A^2 = I_3$ $\Rightarrow A^31 = A^28. A^3 = A^3$ .

y = mx + 4 is tangent to $y^2 = 4x$ $\Rightarrow m = 1/4$ $y=1/4x+4$ is tangent to $x^2 = 2by$ $\Rightarrow x^{2-}b/2x-8b=0$ $\Rightarrow D=0$ $b^{2+}128b=0$ $\Rightarrow b=-128,0$ $b\ne 0\Rightarrow b=-128$

Given 2ae = 6 ⇒ ae = 3 ......(1) and ${2a}/e=12\Rightarrow a=6e$ ...(2) from (1) and (2) $6e^2=3\Rightarrow e=1/\sqrt2$ $\Rightarrow a=3\sqrt2$ Now, $b ^2 = a ^2 (1 - e ^2 )$ $\Rightarrow b^2=18(1-1/2=9)$Length of L.R $=\frac{2(9)}{3\sqrt2}=3\sqrt2$

k 0 1 2 3 4 5 P(k) $1/32$ $12/32$ $11/32$ $5/32$ $2/32$ $1/32$ Expected value = ∑ XP(k)$-1/32-12/32-11/32+15/32+8/32+5/32$ $={28-24}/32=4/32=1/8$

$A=∫^{1}_{0}(\text{sqrtx}-x)dx$ =\[2/3x^{3/2}-x^2/2]^1_0=1/6$Required Area :$πr^{2-}1/6=1/6(12π-1)$

$x^k+y^k=a^k,(a,K > 0)$ $kx^{k-1}+ky^{k-1}\frac{dy}{dx}=0$ $\frac{dy}{dx}+(x/y)^{k-1}=0$ $\Rightarrow k-1=-1/3$ $\Rightarrow k=2/3$

$e^y\frac{dy}{dx}-e^y=e^x$, Let $e^y=t$ $\Rightarrow e^y\frac{dy}{dx}=\frac{dt}{dx}$ $\frac{dt}{dx}-t=e^x$ I.F. $=e^{∫-dx}=e^{-x}$ $t e^{-x} = x + c$ $\Rightarrow ey^{-x }= x + c$ $y(0) = 0\Rightarrow c = 1$ $e^{y-x} = x + 1$ $\Rightarrow y(1) = 1 + \log_e2$

Total number of 6-digit numbers in which only and all the five digits 1, 3, 5, 7 and 9 is $^5C_1\times \frac{6!}{2!}$

Plane passing through : (2, 1, 0), (4, 1, 1) and (5, 0, 1) $\begin{vmatrix}x-2 & y-1 & z \\ 2 & 0 & 1 \\ 3 & -1 & 1\end{vmatrix}=0$ $\Rightarrow x + y - 2z = 3$Let I and F are respectively image and foot of perpendicular of point P in the plane. eqn of line PI ${x-2}/1={y-1}/1=\frac{z-6}{-2}=\lambda (\text{say})$Let $I({\lambda +2,\lambda +1,-2\lambda +6 })$ $\Rightarrow F(2+\lambda /2,1+\lambda /2,-\lambda +6)$F lies in the plane$\Rightarrow 2+\lambda /2+1+\lambda /2+2\lambda -12-3=0$ $\Rightarrow \lambda =4$ $\Rightarrow I(6,5,-2)$

Angle bisector can be $\vec{a}=\lambda ({b}^{∧}+{c}^{∧}) \;\text{ or }\; \vec{a}=u({b}^{∧}-{c}^{∧})$ $\vec{a}=\lambda (\;\frac{{i}^{∧}+{j}^{∧}}{\sqrt{2}}+\;\frac{{i}^{∧}-{j}^{∧}+4 {k}^{∧}}{3 \sqrt{2}})$ $\;\; =\;\frac{\lambda }{3 \sqrt{2}}(3 {i}^{∧}+3 {j}^{∧}+{i}^{∧}-{j}^{∧}+4 {k}^{∧})$ $\;\; =\;\frac{\lambda }{3 \sqrt{2}}(4 {i}^{∧}+2 {j}^{∧}+4 {k}^{∧}) . . . . . . . . . (i)$ $\;\text{ also, }\; \vec{a}=\alpha {i}^{∧}+2 {j}^{∧}+{\beta } {k}^{∧} . . . . . . . . . \;\text{ (ii) }\;$ On comparing eq (1) and eq(2) $\;\frac{\lambda }{3 \sqrt{2}}=1$ $\Rightarrow \lambda =3 \sqrt{2}$ putting value of $\lambda$ in eq(i) $\vec{a}=4 {i}^{∧}+2 {j}^{∧}+4 {k}^{∧}$ $\;\text{ Now, }\; \vec{a}=u(\;\frac{{i}^{∧}-{j}^{∧}}{\sqrt{2}}-\;\frac{{i}^{∧}-{j}^{∧}+4 {k}^{∧}}{3 \sqrt{2}})$ $=\;\frac{u}{3 \sqrt{2}}(3 {i}^{∧}+3 {j}^{∧}-{i}^{∧}+{j}^{∧}-4 {k}^{∧})$ $=\;\frac{u}{3 \sqrt{2}}(2 {i}^{∧}+4 {j}^{∧}-4 {k}^{∧})$ $=\;\frac{2 u}{3 \sqrt{2}}({i}^{∧}+2 {j}^{∧}-2 {k}^{∧}) . . . . . . . . . . . . . . . . . . . . (iii)$ $\;\text{ on comparing eq(ii) and eq(iii) }\;$ $\;\frac{2 u}{3 \sqrt{2}}=1 \Rightarrow u=\;\frac{3 \sqrt{2}}{2}$ $\;\text{ putting value of }\; u \;\text{ \in eq(iii) }\;$ $\vec{a}={i}^{∧}+2 {j}^{∧}-2 {k}^{∧}$ $\vec{a} \cdot {k}^{∧}=-2$ $\therefore \vec{a} \cdot {k}^{∧}+2=0$

$f(a+b+1-x)=f(x) . . .(i)$ $f(a+b-x)=f(x+1) . . .(i i)$ $I=\;\frac{1}{(a+b)} ∫_{a}^{b} x(f(x)+f(x+1)) d x . . .$.(iii) From (i) and (ii) $I=\;\frac{1}{(a+b)} ∫_{a}^{b}(a+b-x)(f(x+1)+f(x)) d x . . . \;\text{ (iv) }\;$ Adding (iii) and (iv) $2 I=∫_{a}^{b}(f(x)+f(x+1)) d x$ $2 I=∫_{a}^{b} f(a+b-x+1) d x+∫_{a}^{b} f(x) d x$ $2 I=2 ∫_{a}^{b} f(x) d x$ $I=∫_{a}^{b} f(x) d x \;\; ; \;\; x=t+1, d x=d t$ $I=∫_{a-1}^{b-1} f(t+1) d t$ $I=∫_{a-1}^{b-1} f(x+1) d x$

Using LMVT in [-7, -1] $\frac{f(-1)-f(-7)}{-1-(-7)} \le 2$ $f(-1) -f (-7) \le 12$ $\Rightarrow f (-1)\le 9$ ....(1)Using LMVT in [-7, 0] $\frac{f(0)- f( -7)}{0-(-7)} \le 2$ $f(0) -f (-7) \le 14$ $f(0) \le 11$...(2)from (1) and (2)$f(0)+f(-1) \le 20$

For non-zero solution$\begin{vmatrix}2 & 2a & a \\ 2 & 3b & b \\ 2 & 4c & c\end{vmatrix}=0$ $\Rightarrow \begin{vmatrix}2 & 2a & a \\ 0 & 3b-2a & b-a \\ 0 & 4c-2a & c-a\end{vmatrix}=0$⇒ (3b - 2a) (c -a) - (b - a) (4c - 2a) = 0 ⇒ 2ac = bc + ab$\Rightarrow 2/b=1/a+1/c$Hence $1/a,1/b,1/c$ are in A.P.

$\tan\alpha +\tan\beta =\frac{\lambda \sqrt2}{k+1}$ $\tan \alpha .\tan \beta =\frac{k-1}{k+1}$ $\tan (\alpha +\beta )=\frac{\frac{\lambda \sqrt2}{k+1}}{1-\frac{k-1}{k+1}}=\lambda /\sqrt2$ \Rightarrow \lambda ^2/2=50\Rightarrow \lambda =10 & -10

$p$ $q$ $p\to q$ $~p$ $q\to ~p$ $(p\to q) ∧ ( q \to ~p)$TTTFFFTFFFTFFTTTTTFFTTTT From the truth table, its is evident that $(p \to q) ∧(q \to ∼ p)$ is equivalent to $∼ p$

$1 + 49 + 49^2 +...+ 49^12$ $= (49)^{126-} 1$ $= (49^63 + 1)(49^63 - 1)/(48)$So greatest value of k = 63

$\lim _{x\to 2}\frac{3^x+3^{3-x}-12}{3^{-x/2}-3^{1-x}}$ $=\lim _{x\to 2}\frac{3^{2x}-12.3^x+27}{3^{x/2}-3}$ $=\lim _{x\to 2}\frac{(3^x-9)(3^x-3)}{3^{x/2}-3}$ $=\lim _{x\to 2}{(3^{x/2}+3)(3^{x/2}-3)(3^x-3)}/({3^{x/2}-3})$ = 36