Class 12th

$f\left(x\right)=4\left|2x+3\right|+9\left[x+\frac{1}{2}\right]-12\left[x+20\right],-20<x<20$ doubtful points for differentiability :   $x=\frac{-3}{2},$

$f\left(x\right)=4\left(2x+3\right)+9\left(-1\right)-12\left(-2\right)-240=8x-213forx=\frac{-3}{2}+h$         

= $-8x-230$  for x = $\frac{-3}{2}-h$  

Not diff. at x = $\frac{-3}{2}$  

other doubtful points : $x+\frac{1}{2}=integer$  

$-20+\frac{1}{2}<x+\frac{1}{2}<20+\frac{1}{2}$

$x+\frac{1}{2}=-19,-18,....,19,20$

$x=-19.5,-18.5,-17.5,.......,18.5,19.5\to$ total 40 numbers.

 No. of number = 19.5 – $\left(-19.5\right)+1=40\left(-1.5\right)included$  

  $-20<x<90\Rightarrow x=-19,-18,.....,18,15\to 39points$

No. of number = 19 - (-19) + 1 = 39

Total : 40 + 39 = 79

$r{\cdot }^n\text{?}{C}_r=n{\cdot }^{n-1}\text{?}{C}_{r-1}$

${\displaystyle \sum _{k=1}^{10}{\left(k{\cdot }^{10}\text{?}{C}_k\right)}^2}={\displaystyle \sum _{k=1}^{10}{\left(10{\cdot }^9\text{?}{C}_{K-1}\right)}^2}$

= $100{\cdot }^{18}\text{?}{C}_9$  

22000L =   $100{\cdot }^{18}\text{?}{C}_9$

 L =

$18!=2^{16}\cdot 3^8\cdot 5^3\cdot 7^2\cdot 11^1\cdot 13\cdot 17$

$9+4+2+1$ $9!=2^7\cdot 3^4\cdot 5^1\cdot 7^1$

6 + 2                    4 + 2 + 1   $\frac{18!}{{\left(9!\right)}^2}=2^2\cdot 5\cdot 11\cdot 13\cdot 17$            

3 +                       3 + 1

 = 221

$A=\left[\begin{array}{ccc}\hfill -1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 6\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]$

$A^2=\left[\begin{array}{ccc}\hfill -1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 6\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill -1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 6\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]$

$=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 6\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=I+B,B=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 6\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$

$A^4=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 6\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 6\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]B^2\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]=0$

$A^{2n}={\left(I+B\right)}^n$

I + nB + 0 + 0 +……….

$A^{2n}=I+\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 6n\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 6n\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

$X\text{'}{A}^{k}x=\left[33\right]$ $=\left[111\right]A^k\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right]$

$\left[111\right]A^{2n}\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right]\left(k=2n\right)$

$\left[1+1+6n+1\right]=\left[6n+3\right]=\left[33\right]n=5$

k = 10

np + npq = 82.5 Þ np (1 + q) = 82.5

$np\cdot npq=1350\Rightarrow {\left(np\right)}^{2}q=1350$

n = ?

$\Rightarrow \frac{{\left(1+q\right)}^2}{q}=\frac{82.5*82.5}{1350}$

$q^2+2q+1=\frac{121}{24}q$

-> 24q² – 73q + 24 = 0

q $=\frac{73\pm \sqrt[]{73^2-4*576}}{48}$  

$=\frac{73\pm 55}{48}=\frac{128}{48},\frac{18}{48}=\frac{8}{3},\frac{3}{8}$

$p=\frac{5}{8},q=\frac{3}{8}$

                $n\cdot \frac{5}{8}\cdot \frac{11}{8}=\frac{165}{2}$

n = 96

Structure of Bithinol is;

Total 3 streo- isomers

Data contradiction.

Common tangents

$T_1:y=mx\pm \sqrt[]{4m^2+9}$      

$T_2:y=mx\pm \sqrt[]{42m^2-13}$    

So, 4m² + 9 = 42 m² – 13 Þ m =   $\pm 2$

  $\Rightarrow c=\pm 5$              

So tangent  $y=\pm 2x\pm 5$

y = 2x + 5 does not pass through 4^th quadrant compare this tangent with T = 0 to get pt. of intersection

$\frac{x{x}_2}{42}-\frac{y{y}_2}{143}=1\Rightarrow x_2=\frac{-84}{5}$

Mole of polyhydric alcohol =  $\frac{1.84*10^{-3}}{92}=2.0*10^{-5}mole.$

Mole of H₂ gas produced = $\frac{1.344}{22400}=6.0*10^{-5}mole.$  

No of -OH gp present =  $\frac{6.0*10^{-5}}{2.0*10^{-5}}=3$

$f\left(x\right)=\frac{2}{\sqrt[]{3}}{\displaystyle {\int }_0^{\sqrt[]{3}}f\left(\frac{{\lambda }^{2}x}{3}\right)d\lambda }$

Substitute $\frac{{\lambda }^{2}x}{3}=t$ $\frac{{}^{}}{}$

$f\left(x\right)=\frac{1}{\sqrt[]{x}}{\displaystyle {\int }_0^x\frac{f\left(t\right)}{\sqrt[]{t}}dt}$

differentiate using leibneitz rule

$\sqrt[]{x}.f\text{'}\left(x\right)+f\left(x\right).\frac{1}{2\sqrt[]{x}}=\frac{f\left(x\right)}{\sqrt[]{x}}$

f (1) = $\sqrt[]{3}\Rightarrow f\left(x\right)=\sqrt[]{3x}$