Class 12th

$\frac{dy}{dx}=\frac{\left(x+1\right)\left(x^2-x+1\right)+\sqrt[]{\left(1-x\right)\left(1+x\right)}}{\left(x-1\right)\left(x+1\right)}$

$\frac{dy}{dx}=\frac{x\left(x-1\right)+1}{\left(x-1\right)}+\sqrt[]{\frac{\left(1-x\right)\left(1+x\right)}{{\left(x-1\right)}^2{\left(x+1\right)}^2}}$

$\frac{dy}{dx}=x+\frac{1}{x-1}+\frac{1}{\sqrt[]{\left(1-x\right)\left(1+x\right)}}$

$dy=xd+\frac{1}{\left(x-1\right)}dx+\frac{dx}{\sqrt[]{1-x^2}}$

$y=\frac{x^2}{2}+ln\left|x-1\right|+si{n}^{-1}x+c$

at x = 0, y = 2 2 = c

$y=\frac{x^2}{2}+ln\left|x-1\right|+si{n}^{-1}x+2$

$y\left(\frac{1}{2}\right)=\frac{17}{8}+\frac{\pi }{6}-ln2$

-> $-1\le \left|\begin{array}{cc}\hfill \overline{2-\left|x\right|}\hfill & \hfill 4\hfill \end{array}\right|\le 1$

-> $\left|\frac{2-\left|x\right|}{4}\right|\le 1$

$-1\le \frac{2-\left|x\right|}{4}\le 1$

–4 £ 2 – |x| £ 4

–6 £ – |x| £ 2

–2 £ |x| £ 6

|x| £ 6

->x Î [–6, 6]              …(1)

Now, 3 – x ¹ 1

And x ¹ 2                    …(2)

and 3 – x > 0

x < 3                            (3)

From (1), (2) and (3)

->x Î [–6, 3] – {2}

a = 6

b = 3

g = 2

a + b + g = 11

$f\text{'}\left(x\right)=\frac{g\text{'}\left(x\right)-g\text{'}\left(2-x\right)}{2},f\text{'}\left(\frac{3}{2}\right)=\frac{g\text{'}\left(\frac{3}{2}\right)-g\text{'}\left(\frac{1}{2}\right)}{2}=0$  

Also  $f\text{'}\left(\frac{1}{2}\right)=\frac{g\text{'}\left(\frac{1}{2}\right)-g\text{'}\left(\frac{3}{2}\right)}{2}=0$ , f' (1) = 0

-> $f\text{'}\left(\frac{3}{2}\right)=f\text{'}\left(\frac{1}{2}\right)=0$  

->roots in $\left(\frac{1}{2},1\right)$ and $\left(1,\frac{3}{2}\right)$  

->f" (x) is zero at least twice in $\left(\frac{1}{2},\frac{3}{2}\right)$

Area of ?

$=\frac{1}{2}\left|\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill x\hfill & \hfill y\hfill & \hfill 1\hfill \\ \hfill -x\hfill & \hfill y\hfill & \hfill 1\hfill \end{array}\right|$

-> $\left|\frac{1}{2}\left(xy+xy\right)\right|=\left|xy\right|$

->Area (D) = |xy| = |x (– 2x² + 54x)|

$\frac{d\left(\Delta \right)}{dx}=\left|\left(-6x^2+108x\right)\right|\Rightarrow \frac{d\Delta }{dx}=0$  at x = 0 and 18

->at x = 0, minima

and at x = 18 maxima

Area (D) = |18 (– 2 (18)² + 54 * 18)| = 5832

(a – 1) * 2 + (b – 2) * 5 + (g – 3) * 1 = 0

2a + 5b + g – 15 = 0

Also, P lie on line

a + 1 = 2λ

b – 2 = 5λ

g – 4 = λ

2 (2λ – 1) + 5 (5λ + 2) + λ + 4 – 15 = 0

4λ + 25λ + λ – 2 + 10 + 4 – 15 = 0

30λ – 3 = 0

$\lambda =\frac{1}{10}$  

a + b + g = (2λ – 1) + (5λ + 2) + (λ + 4)

$=8\lambda +5=\frac{8}{10}+5=5.8$

Group number = 11 (Atomic number = 111)

Coca alkaloid or cocaine is obtained from Erythroxylum coca.

Functional group should get the lowest possible number.

Please find the solution below:

The boiling point of alcohol increases with increase in the number of carbon atoms. So, hexanol has maximum boiling point.