Class 12th

Here A is CH₃ - CH₂ – CH₂ – COO – CH₂ – CH₃

The given sequence of reaction is:

$C{H}_3-C{H}_2-C{H}_2-C{H}_2-OH\underset{}{\overset{Oxidation}{\to }}C{H}_3-C{H}_2-C{H}_2-COOH$

Ce, Pr, Nd, Tb and Dy are the only lanthanoids which shows + 4 O.S, so can form MO₂, but Yb does not shows +4 O.S so can't form MO₂.

Only 1° aromatic amines give stable diazonium salt on reaction with nitrous acid

Here 1° aromatic amine is  which gives most stable diazonium salt

Now equation of line OA be

$\frac{x-1}{4}=\frac{y-3}{-5}=\frac{z-5}{2}=\lambda$

direction cosines of plane are 4, -5, 2

Equation of any point on OA be

$O\left(4\lambda +1,-5\lambda +3,2\lambda +5\right)$

Since O lies on given plane so

$4\left(4\lambda +1\right)-5\left(-5\lambda +3\right)+2\left(2\lambda +5\right)=8$

So, O (9/5,2,27/5). Hence by mid-point formula

B $\left(\frac{13}{5},1,\frac{29}{5}\right)\Rightarrow 5\left(\alpha +\beta +\gamma \right)=47$

$\underset{x\to a}{lim}\frac{xf\left(a\right)-af\left(x\right)}{x-a}\left(\frac{0}{0}\right)$

By L'hospital Rule

$=\underset{x\to a}{lim}\frac{f\left(a\right)-af\text{'}\left(x\right)}{1}=f\left(a\right)-af\text{'}\left(a\right)$

$=4-2a$

Now equation of line OA be

3, 4, 5, 5

In remaining six places you have to arrange

3, 4, 5,5

So no. of ways $=\frac{6!}{2!2!2!}$

Total no. of seven digits nos. = $\frac{7!}{2!3!2!}*1$

Hence Req. prob. $\frac{\frac{6!}{2!2!2!}}{\frac{7!}{2!3!2!}}=\frac{6!3!}{2!7!}=\frac{3}{7}$

Electrophilic addition of bromine to an alkene is anti-addition, in which cis-alkene gives two enantiomers and trans – alkene gives meso form

Here; trans-but-2-ene will give meso products

Calamine is the ore of zinc i.e. ZnCO₃.

Malachite is the ore of copper i.e. CuCO₃.Cu (OH)₂.

  $g\left(2\right)=\underset{x\to 2}{lim}g\left(x\right)=\underset{x\to 2}{lim}\frac{x^2-x-2}{2x^2-x-6}=\underset{x\to 2}{lim}\frac{\left(x-2\right)\left(x+1\right)}{2x\left(x-2\right)+3\left(x-2\right)}$         

$Nowfog=f\left(g\left(x\right)\right)=si{n}^{-1}g\left(x\right)=si{n}^{-1}\left(\frac{x^2-x-2}{2x^2-x-6}\right)$

$\frac{3x^2-2x-8}{2x^2-x-6}\ge 0\&\frac{-x^2+4}{2x^2-x-6}\le 0$

On solving we get  $x\in \left(-\infty ,-2\right)\cup \left[\frac{-4}{3},\infty \right)$

As x = 2 also lies in domain since g(2) $=\underset{x\to 2}{lim}g\left(x\right)$