Class 11th

3-Ethylhex-3-ene will not show stereo isomerism

Methylhex-1-ene will show steroisomerism (optical isomerism)

Since it has chiral carbon.

3, 4-Dimethylhex-3-ene will show stereoisomerism (Geometrical isomerism)

4-Methylhex-1-ene will show stereoisomerism (optical) since it has one chiral carbon

$2NaB{H}_4+l_2\to 2Nal+H_2+B_2{H}_6$

Diborane is produced when NaBH₄ is reacted with I₂

Both boron atoms are sp³ hybridised & molecule is non-planar. Diborane as two bridged 3 centre-2-electron bonds.

${\left(\sqrt[]{50}\right)}^2={\left(\sqrt[]{45}\right)}^2+{\left(\sqrt[]{5}\right)}^2$

$\therefore \angle B=90°$ circum centre = $O\left(\frac{1}{2},\frac{11}{2}\right)$

Mid point of BC = $D\left(2,\frac{17}{2}\right)$

Equation of OD is y = 2x + $\frac{9}{2}.$ This line passes through

$\left(0,\frac{\alpha }{2}\right)\therefore \alpha =9$

  $C{H}_3-C{H}_2-C{H}_2-C-N{H}_2\to \pi -\mathcal{l}pconjugation$  

              $C{H}_3-C{H}_2-CH=CH-C{H}_2-N{H}_2\to$ No conjugation

          $C{H}_3-C{H}_2-O-CH=C{H}_2\to \pi -\mathcal{l}p$     conjugation

$a_{n+1}=\frac{a_{n+2}-a_n}{2}letp={\displaystyle \sum _{n=1}^{\infty }\frac{a_n}{8^n}}$

$\Rightarrow 64\frac{a_{n+2}}{8^{n+2}}=\frac{16a_{n+1}}{8^{n+1}}+\frac{a_n}{8^n}$

$\Rightarrow 64{\displaystyle \sum _{n=1}^{\infty }\frac{a_{n+2}}{8^{n+2}}=}{\displaystyle \sum _{n=1}^{\infty }\frac{16a_{n+1}}{8^{n+1}}+}{\displaystyle \sum _{n=1}^{\infty }\frac{a_n}{8^n}}\Rightarrow 64\left(p-\frac{a_1}{8}-\frac{a_2}{8^2}\right)=16\left(p-\frac{a_1}{8}\right)+p$

$\Rightarrow 64\left(p-\frac{1}{8}-\frac{1}{8^2}\right)=16\left(p-\frac{1}{8}\right)+p\Rightarrow 64p-8-1=16p-2+p$

Only tritium is the isotope of hydrogen which is radioactive in nature that emits low energy b^- particles since its $\frac{n}{p}$ ratio is above stability belt.

$?f\left(x\right)=ln\left(x+\sqrt[]{1+x^2}\right)\therefore f\left(-x\right)=-f\left(x\right)$ .

Hence f(x) is an odd function.

$Nowg\left(t\right)={\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}cos\left(\frac{\pi }{4}t+f\left(x\right)\right)dx}$

Put t = 0, g(0) = ${\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}cos\left(f\left(x\right)\right)dx=2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}cos\left(f\left(x\right)\right)dx}...............\left(i\right)$

where cos(f(x)) is an even function.

Now again put t = 1,

$\therefore g\left(1\right)=\frac{1}{\sqrt[]{2}}*g\left(0\right),from\left(i\right)$

$\therefore g\left(0\right)=\sqrt[]{2}g\left(1\right)$

$f\left(x\right)=\frac{5x+3}{6x-\alpha }=y.....................(i)$

$\Rightarrow x=\frac{\alpha y+3}{6y-5}\Rightarrow f^{-1}\left(x\right)=\frac{\alpha x+3}{6x-5}..............(ii)$

According to question, $f\left(x\right)=f^{-1}\left(x\right)\therefore$  from (i) and (ii) we get $\propto =5$