Class 12th

Seliwanoff's test is used to distinguish carbohydrates while xanthoproteic test is used to distinguish proteins

$\overrightarrow{a}*\left(\overrightarrow{b}*\overrightarrow{c}\right)=3\overrightarrow{b}-\overrightarrow{c}=\overrightarrow{u}$

$\overrightarrow{b}*\left(\overrightarrow{c}*\overrightarrow{a}\right)=\overrightarrow{c}-2\overrightarrow{a}=\overrightarrow{v}$

$\overrightarrow{c}*\left(\overrightarrow{b}*\overrightarrow{a}\right)=3\overrightarrow{b}-2\overrightarrow{a}=\overrightarrow{w}$

$\therefore \overrightarrow{u}+\overrightarrow{v}=\overrightarrow{w}$

so vectors

$\overrightarrow{u},\overrightarrow{v}and\overrightarrow{w}$

are coplanar, hence their Scalar triple product will be zero.

$\frac{dy}{dx}+\frac{2x}{x-1}\cdot y=\frac{1}{{\left(x-1\right)}^2}$

IF $=e^{{\displaystyle \int \frac{2x}{x-1}dx}}$

= $e^{2x}\cdot {\left(x-1\right)}^2$

$y\cdot e^{2x}{\left(x-1\right)}^2=\{e^{2x}\frac{{\left(x-1\right)}^2}{{\left(x-1\right)}^2}dx+C$

y = $\frac{e^{2x}}{2{\left(x-1\right)}^2}+\frac{C}{{\left(x-1\right)}^2}$

y(2) = $\frac{1+e^4}{2e^4},\Rightarrow C=\frac{1}{2}$ $\frac{{}^{}}{{}^{}}\frac{}{}$

y(3) = $\frac{e^{\alpha }+1}{\beta e^{\alpha }}=\frac{e^6+1}{8e^6}$

2, 4-DNP (Brady's Reagent) is used to test carbonyl compound

(a) Na₂CO₃ -> Solvay

(b) Ti -> Van-Arkel

(c) Cl₂ -> Deacon

(d) NaOH -> Castner – kellner

It is an intramolecular aldol condensation

A > B > C > D

Lone pair is localized in (A) while all 3 have delocalized lone pair but they can be compared by 3° > 2° > 1° because methyl group increases the basicity.

Consider the equation of plane,

$P:\left(2x+3y+z+20\right)+\lambda \left(x-3y+5z-8\right)=0$  

$?$  Plane P is perpendicular to 2x + 3y + z + 20 = 0

So,  $4+2\lambda +9-9\lambda +1+5\lambda =0$  

0  $\lambda =7$  

P : 9x – 18y + 36z – 36 = 0

Or P : x – 2y + 4z = 4

If image of

$\left(2,-\frac{1}{2},2\right)$  

In plane P is (a, b, c) then

$\frac{a-2}{1}=\frac{b+\frac{1}{2}}{-2}=\frac{c-2}{4}$  

and  $\left(\frac{a+2}{2}\right)-2\left(\frac{b-\frac{1}{2}}{2}\right)+4\left(\frac{c+2}{2}\right)=4$     

clearly 

$a=\frac{4}{3},b=\frac{5}{6}andc=-\frac{2}{3}$  

So, a : b : c = 8 : 5 : -4

Both (A) and (B) are correct but R is not correct explanation of A. In both oxidation state of metal is +1, also both have similar lattice structure.