Class 12th

P (H) = x . P (T) = 1 – x

P (4H. 1T) = P (5H)           

6x = 5 = 0             $x=\frac{5}{6}$  

P (atmost 2H)

$P\left(OH,5T\right)+P\left(1H,4T\right)+P\left(2H,3T\right)$  

$=\frac{1}{6^5}\left(1+25+250\right)=\frac{276}{6^5}=\frac{4^6}{6^4}$  

Given a > b

Area common to x² + y² $\le a^{2}and\frac{x^2}{a^2}+\frac{y^2}{b^2}\le 1$  

is $\pi a^2-\pi ab=30\pi ..............\left(i\right)$  

Similarly $\pi ab-\pi b^2=18\pi .................\left(ii\right)$  

Equation (i) and equation (ii)  $\frac{a}{b}=\frac{5}{3}$  

Equation (i) + equation (ii) $a^2-b^2=48$  

a² = 75, b² = 27

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.