JEE MAIN 2020 - 09 JAN S1 2020

$K_1=Ae^{-E_a/{R\times 100}}$ $K_2=A\times e^{\frac{E_a-30}{R\times 500}}$For same rate$K_1 = K_2$ $\therefore E_a=75$

∴ Generally trans is more stable then cis form. Heat of combustion (HOC) $\propto 1/\text{Stability}$Stability : a > b > cHOC : c > b > a

(i) Blue voilet color with Ninhydrine → amino acid derivative. So it cannot be saccharide or sucralose.(ii) Lassaigne extract give +ve test with $AgNO_3$. So Cl is present, -ve test with $Fe_4[Fe(CN)_6]_3$ means N is absent. So it can't be Aspartame or Saccharine or Alitame, so C is sucralose.(iii) Lassaigne solution of B and D given +ve sodium nitroprusside test, so it is having S, so it is Saccharine and Alitame.

$CCl_4$ is molecular solid so does not conduct electricity in liquid & solid state.

A reduces $B O_2$ when temperature is above 1400°C because above 1400°C A has more - ve ΔG° for Δ$O_2$ formation than B to $BO_2$ formation.

$[Pb^{2+}]={300\times 0.134}/400$ $= 1.005 \times 10^{-1} M$ $[Cl^{-}]={100\times 0.4}/400=10^{-1}M$ PbCl_2_(s) ⇌{Pb^{2+}}_{(aq)}+{2Cl^{-1}}_(aq) Q=\[Pb^{2+}]$\times $[Cl^{-}]^2$$= 1.005 × 10^{-3} > k_{sp}$

$[Pd(F)(Cl)(Br)(I)]^{2- }$ have three geometrical isomer so formula for $[Fe(CN)_6]^{n-6}$ is $[Fe(CN)_6]^{3-}$ and CFSE for this complex is $Fe^{3+} \Rightarrow 3d^5 4s^0$Magnetic Moment $= \sqrt3= 1.73 B.M$ CFSE $= [(-0.4\times 5) + (0.6\times 0)] \Delta_0$ $= -2.0 \Delta_0$

$\text{table} ,\text{number of unpaired electron}, \text{magnetic moment};O_2^{-},1 ,1.73 B.M;O^{+}_2,1, 1.73 B.M;O _2,2 ,2.83 BM$

Enthalpy of atomisation of $Br_2(l)$ $\DeltaH_\text{atom}=\DeltaH_\text{vap}+\DeltaH_\text{BE}$ $x=\DeltaH_\text{vap}+y$So, $x > y$

Basic strength order : (i) > (iv) > (ii) > (iii) > (v)

$Be \Rightarrow 1s^2 2s^2$ $B \Rightarrow 1s^2 2s^2 2p^1$ B has a smaller size than Beit is easier to remove 2p electron than 2s electron due to less pentration effect of 2p than 2s. 2p electron of Boron is more shielded from the nucleus by the inner core of electron than the 2s electron of BeB has a smaller size than Be

The character of an oxide generally depends on its constituent element's position in the periodic table. Oxides of non-metals are typically acidic, while oxides of metals are typically basic. Some elements, particularly those near the metal-nonmetal borderline, form amphoteric oxides that can act as both acids and bases.

Let's analyze the correct option:

• $N_2O_3$ (Dinitrogen trioxide) is an oxide of the non-metal nitrogen, making it an acidic oxide.
• $Li_2O$ (Lithium oxide) is an oxide of an alkali metal, making it a strongly basic oxide.
• $Al_2O_3$ (Aluminum oxide) is a classic example of an amphoteric oxide, as it reacts with both strong acids and strong bases.

Therefore, the sequence $N_2O_3, Li_2O, Al_2O_3$ correctly represents acidic, basic, and amphoteric oxides, respectively.

The reaction scheme involves three sequential steps. First, 3-bromoaniline undergoes diazotization with $NaNO_2/HCl$ at low temperatures to form 3-bromobenzenediazonium chloride (X). Next, in a Sandmeyer reaction, compound X is treated with $Cu_2Br_2$ to replace the diazonium group with a bromine atom, yielding 1,3-dibromobenzene (Y). Finally, Y undergoes electrophilic aromatic substitution (nitration) with a mixture of concentrated $HNO_3$ and $H_2SO_4$. The bromine atoms are deactivating but ortho, para-directing. In 1,3-dibromobenzene, the incoming electrophile ($NO_2^+$) is directed to the positions ortho and para to the bromine atoms. The most favorable position is C-4, which is para to one Br and ortho to the other, leading to the major product Z, 1,3-dibromo-4-nitrobenzene (also named 2,4-dibromo-1-nitrobenzene).

$Cr(H_2O)_6Cl_n$if magnetic mement is 3.83 BM then it contain three unpaired electrons. It means chromium in +3 oxidation state so molecular formula is $Cr(H_2O)_6Cl_3$∴ This formula have following isomers (a) $Cr(H_2O)_6Cl_3$ : react with $AgNO_3$ but does not show geometrical isomerism.(b) $[Cr(H_2O)_5Cl]Cl_2 .H_2O$ react with $AgNO_3$ but does not show geometrical isomerism.(c) $[Cr(H_2O)_4Cl_2] Cl. 2H_2O$ react with $AgNO_3$ & show geometrical isomerism.(d) $[Cr(H_2O)_3Cl_3].3H_2O$ does not react with $AgNO_3$ & show geometrical isomerism. $[Cr(H_2O)_4Cl_2]Cl.2H_2O$ react with $AgNO_3$ & show geometrical isomerism and it's IUPAC nomenclature is Tetraaquadichlorido chromium (III) Chloride dihydrate.

(i) $H_2O_2$ act as oxidising agent as well as reducing agent depending on condition.(ii) $H_2SO_3$ act as oxidising agent as well as reducing agent depending on condition.(iii) $HNO_2$ act as oxidising agent as well as reducing agent depending on condition.(iv)$H_3PO_4$ can not act both as oxidising and reducing agent. $H_3PO_4$ can act as only oxidising agent. $H_3PO_4 ⇌3H^{+}+PO_4^{3-}$

2πr = nλ,for n = 1, $r = a_0$n = 4, $r = 16a_0$So, $2\pi \times 16a_0 = 4 \times \lambda$ $\lambda = 8\pi a_0$

$Eu_63\Rightarrow [Xe] 4f^7 5d^0 6s^2$ $Eu^{2⊕} \Rightarrow [Xe] 4f^7$ $Ce_58 \Rightarrow [Xe] 4f^1 5d^1 6s^2$ $Ce^{3⊕} \Rightarrow [Xe] 4f^1$

1 Litre has $10^{-3}$ moles $MgSO_4$So , 1000 litre has 1 mole $MgSO_4$= 1 mole $CaCO_3$= 100 ppm

100 gm soln → 63 gm $HNO_3$ $100/1.4 mL$ → 1 mole $HNO_3$Molarity $=1/{100/1.4\times 1/1000}=14M$

gm eq. of Ag $=108/108=1$gm eq. of $O_2$ (g) = 1Volume of $O_2$ (g) $= 22.7 \times 1/4=5.675$ \litres

M.F. of Histamine is $C_5H_9N_3$Molecular mass of Histamine is 111Now, mass % of nitrogen $=4$2/111×100=37.84%

$\DeltaT_f = i \times m \times K _f$ $0.2=2\times 2\times \frac{w/58.5}{600/1000}$w = 1.755 gm

Let thickness of ice be 'h'.Vol. of ice $=v={4\pi }/3((10+h)^{3-}10^3)$ $\frac{dv}{dt}={4\pi }/3(3(10+h)^2).\frac{dh}{dt}$Given $\frac{dv}{dt}=50cm^3/min$ and h = 5 cm$\Rightarrow 50={4\pi }/3(3(10+5)^2)\frac{dh}{dt}$ $\Rightarrow \frac{dh}{dt}=1/{18\pi }$

No. of five digits numbers = 336kNo. of ways of filling remaining 4 places $=8\times 8\times 7\times 6$ $\therefore k={8\times 8\times 7\times 6}/336=8$

$|\frac{z-i}{z+2i}|=1$ $\Rightarrow |z-i|=|z+2i|$⇒ z lies on perpendicular bisector of (0, 1) and (0, -2).⇒ Imz = $-1/2$Let $z=x-i/2$ $\because |z|=5/2\Rightarrow x^2=6$ $\therefore {|z+3i|}={|x+{5i}/2|}=\sqrt{x^{2+}25/4}$ $=\sqrt{6+25/4}=7/2$

A : Event when card A is drawnB : Event when card B is drawn.$P(A)=P(B)=1/2$Required probability = P(AA or (AB)A or (BA)A or (ABB)A or (BAB)A or (BBA)A)$=1/2\times 1/2+(1/2\times 1/2\times 1/2)\times 2$ $+(1/2\times 1/2\times 1/2\times 1/2)\times 3$ $=11/16$

$I=∫^{2\pi }_{0}\frac{x\sin^8 x}{\sin^8x+\cos^8x}dx$ ....(i) $=[∫^{\pi }_{0}\frac{x\sin^8 x}{\sin^8x+\cos^8x}dx$ $+∫^{\pi }_{0}\frac{(2\pi -x)\sin^8 x}{\sin^8x+\cos^8x}dx]$ $=2\pi ∫^{\pi }_{0}\frac{\sin^8 x}{\sin^8x+\cos^8x}dx$ $I=2\pi [∫^{\pi /2}_{0}\frac{\sin^8 x}{\sin^8x+\cos^8x}dx$ $+∫^{\pi /2}_{0}\frac{ \cos^8 x}{\sin^8x+\cos^8x}dx]$ $=2\pi ∫^{\pi /2}_{0} 2dx=2\pi .\pi /2=\pi ^2$

To solve this, we first simplify the expression inside the inverse tangent. Using trigonometric identities, we can rewrite $\sec x + \tan x$ as:
$\sec x + \tan x = \frac{1}{\cos x} + \frac{\sin x}{\cos x} = \frac{1+\sin x}{\cos x}$
Applying half-angle identities, this becomes $\frac{( \cos(x/2) + \sin(x/2) )^2}{\cos^2(x/2) - \sin^2(x/2)} = \frac{1+\tan(x/2)}{1-\tan(x/2)}$, which is the formula for $\tan(\frac{\pi}{4} + \frac{x}{2})$.
So, $f'(x) = \tan^{-1}(\tan(\frac{\pi}{4} + \frac{x}{2}))$. For the given domain $-\frac{\pi}{2} < x < \frac{\pi}{2}$, the argument $\frac{\pi}{4} + \frac{x}{2}$ lies within the principal range of $\tan^{-1}$, so $f'(x) = \frac{\pi}{4} + \frac{x}{2}$.
To find $f(x)$, we integrate $f'(x)$: $f(x) = \int (\frac{\pi}{4} + \frac{x}{2}) dx = \frac{\pi}{4}x + \frac{x^2}{4} + C$.
Using the initial condition $f(0)=0$, we find $C=0$. Thus, $f(x) = \frac{\pi x + x^2}{4}$.
Finally, evaluating at $x=1$, we get $f(1) = \frac{\pi(1) + 1^2}{4} = \frac{\pi+1}{4}$.

$A=\begin{bmatrix}1 & 1 & 2 \\ 1 & 3 & 4 \\ 1 & -1 & 3\end{bmatrix}$ $\Rightarrow |A|=6$ $\frac{|\text{adj} B|}{|C|} =\frac{|\text{adj}(\text{adjA})|}{|9A|}$ $={|A|}^4/{3^3|A|}={|A|^3}/3^3$ $=6^3/3^3=8$

$e^{4x }+ e^{ 3x }- 4e ^{2x }+ e ^{x} + 1 = 0$ $\Rightarrow e^{2x}+e^x-4+1/e^x+1/e^{2x}=0$ $\Rightarrow (e^{2x}+1/e^{2x})+(e^x+1/e^x)-4=0$ $\Rightarrow (e^{x}+1/e^{x})^{2-}2+(e^x+1/e^x)-4=0$Let $e^{x}+1/e^{x}=t$ $\Rightarrow (e^x-1)^2=0$ $\Rightarrow x=0$∴ Number of real roots = 1

p$= \sqrt5$ is an integer.q : 5 is irrational$~(p \vee q) \equiv ~p \wedge ~q$= $\sqrt5$ is not an integer and 5 is not irrational.

$∑^{10}_{i=1}(x_i-5)=10$ \Rightarrow Mean of observation $x_i-5=1/10 =1/10∑^{3}_{i=1}(x_i-5)=10$ \Rightarrow µ = mean of observation $(x_i - 3)$ = (mean of observation $(x_i- 5)) + 2$ $=1+2=3$ Variance of observation $x_i-5=1/10 ∑^{10}_{i=1}(x_i-5)^2$ $- (\text{Mean} of(x_i-5))^2=3$\Rightarrow λ = variance of observation $(x_i - 3)$= variance of observation $(x_i - 5) = 3$ $\therefore (\mu , \lambda ) = (3, 3)$

$2^{1/4}. 4^{1/16}.8^{1/48}. 16^{1/128}........\infty$ $=2^{1/4}.2^{2/16}.2^{3/48}. 2^{4/128}........\infty$ $=2^{1/4}.2^{1/8}.2^{1/16}. 2^{1/32}........\infty$ $=2^{1/2}$

Equation of family of circle touching y-axis at (0, 4) is given by $(x - 0)^2 + (y - 4)^2 + \lambda x = 0$.It passes through (2, 0)$\Rightarrow \lambda = -10.$⇒ Required circle is $(x - 0)^2 + (y - 4)^2 - 10x = 0$ $\Rightarrow x^2 + y^2 -10x - 8y + 16 = 0$center of circle = (5,4) and radius = 5distance of 4x + 3y - 8 = 0 from (5, 4) $=|24/5|\ne \text{radius}$

$e_1=\sqrt{1-4/18}=\sqrt7/3$ $e_2=\sqrt{1+4/9}=\sqrt13/3$ $\because (e_1,e_2)$ lies on $15x^2 + 3y^2 = k$ $\Rightarrow 15e_1^{2+}3e_2^2=k$ $\Rightarrow k=16$

it is clear from graph that $m_1 > m_2$ $\Rightarrow \frac{f(c)-f(a)}{c-a} > \frac{f(b)-f(c)}{b-c}$ $\Rightarrow \frac{f(c)-f(a)}{f(b)-f(c)} > \frac{c-a}{b-c}$

For planes to intersect on a line⇒ there should be infinite solution of the given system of equations for infinite solutions$\Delta=\begin{vmatrix}1 & 4 & -2 \\ 1 & 7 & -5 \\ 1 & 5 & \alpha \end{vmatrix}=0$ $\Rightarrow 3\alpha +9=0\Rightarrow \alpha =-3$ $\Delta_z=\begin{vmatrix}1 & 4 & 1 \\ 1 & 7 & \beta \\ 1 & 5 & 5\end{vmatrix}=0$ $\Rightarrow 13-\beta =0\Rightarrow \beta =13$ $\therefore \alpha +\beta =-3+13=10$

$I=∫\frac{dx}{(x+4)^{8/7}(x-3)^{6/7}}$ $=∫\frac{dx}{(\frac{x+4}{x-3})^{8/7}(x-3)^2}$Let $\frac{x+4}{x-3}=t$ $\Rightarrow \frac{dx}{(x-3)^2}=-1/7dt$ $\Rightarrow I=-1/7∫{dt}/(t^{8/7})=t^{-1/7}+C$ $=(\frac{x-3}{x+4})^{1/7}+C$

Centroid of Δ = (2, 2)line passing through intersection ofx + 3y - 1 = 0 and3x - y + 1 = 0, be given by(x + 3y - 1) + A,(3x - y + 1) = 0∵ It passes through (2, 2)$\Rightarrow 7+5\lambda =0\Rightarrow \lambda =-7/5$∴ Required line is 8 x - 11 y + 6 = 0∵ (-9, -6) satisfies this equation.

$\lim_{x\to 0^{-}}f(x)=\lim_{x\to 0}({\sin(a+2)x}/x+{\sin x}/x)=a+3$ $\lim_{x\to 0^{+}}f(x)=\lim_{x\to 0} {(x+3x^2)^{1/3}-x^{1/3}}/x^{4/3}$ $=\lim_{x\to 0}{(1+3x)^{1/3}-1}/x=1$ $f(0)=b$ for continuity at x = 0$\lim_{x\to 0^{-}}f(x)=f(0)=\lim_{x\to 0^{+}}f(x)$⇒ a + 3 = b = 1⇒ a = -2, b = 1⇒ a + 2b = 0

$\cos^3(\pi /8 )\cos({3\pi }/8) +\sin^3(\pi /8)\sin({3\pi }/8)$ $=\sin \pi /8.\cos \pi /8 =1/2\sin \pi /4=1/{2\sqrt2}$

$f(x) = a + bx + cx^2$ ∫^{1}_{0}f(x)dx=\[ax+{bx^2}/2+{cx^3}/3]^1_0$$=a+b/2+c/3=1/6[6a+3b+c]$$=1/6{f(0)+f(1)+4f(1/2)}$

$(1 + x + x^2)^10$ $=^10C_0+^10C_1x(1+x)+^10C_2x^2(1+x)^2$ +^10C_3 x^3(1+x)^{3+}^10C_4 x^4(1+x)^{4+}....Coeff. of $x^4=^10C_2+^10C_3\times ^3C_1+^10C_4$ = 615