Ncert Solutions Maths class 12th

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

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

= $100{\cdot }^{18}\text{?}{C}_9$

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 + 24 + 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

$\underset{x\to 10}{lim}{\left(\frac{{\left(x+2cosx\right)}^3+2{\left(x+2cosx\right)}^2+3sin\left(x+2cosx\right)}{{\left(x+2\right)}^3\pm 2{\left(x+2\right)}^2+3sin\left(x+2\right)}\right)}^x$

$e^{\frac{100}{16+3sin2}\left[\frac{12-3\left(4\right)+8\times 1-8+3cos2-3cos2}{1}\right]}$ , using L.H.L Rule, e0 = 1

$2x^2-12x+7=0or3x^2+5x+12=x^2+3x+10$

$x^2+x+1=0$

Sum of roots = 6, no solution.

$f\left(x\right)=0\Rightarrow {\left(x-p\right)}^2-q=0$  

Roots are  $p+\sqrt[]{q},p-\sqrt[]{q}$

Now, $\left|f\left(a_i\right)\right|=500$  

$Let{a}_1,a_2,a_3...ara,a+d,a+2d,a+3d$               

=>  $\frac{9}{4}{d}^2-q=500$           …….(i)

and ${\left|f\left(a_1\right)\right|}^2={\left|f\left(a_2\right)\right|}^2$

From equation (i)

$\frac{9}{4}.\frac{4q}{5}-q=500$     

$\frac{4q}{5}=500$           

and  $2\sqrt[]{q}=2\times \frac{50}{2}=50$

$f\text{'}\left(a\right)=f\text{'}\left(b\right)=f\text{'}\left(c\right)=2$

f’’ (x) is zero for atleast $x_1\in \left(a,b\right)\&x_2\in \left(b,c\right)$ ${}_{}{}_{}$

$RS\equiv \left(\alpha ,-,1\beta \right)$

$DRofPQ\equiv \left(\frac{56}{17}+2,\frac{43}{17}+1,\frac{111}{17}-1\right)$

$\equiv \left(\frac{90}{17},\frac{60}{17},\frac{94}{17}\right)$

$90\alpha +94\beta =60$

$\begin{array}{l}\beta =-15,\alpha =-15\\ {\alpha }^2+{\beta }^2=450\end{array}$

$\lambda =1+\mu$

$\begin{array}{l}\underset{\_}{2\lambda =3\lambda \mu }\\ \lambda =2\to S\left(2,4,6\right)\end{array}$

$\mu =1$

$\overrightarrow{n}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 5\hfill \end{array}\right|=\left(7,-2,-1\right)$

7x – 2y – z + d = 0

(2, 4, 6) Þ d = -14 + 8 + 6

d = 0

7x – 2y – z = 0

7 – 2 = 5

g(1) = ?

    $=f\left(f\left(f\left(1\right)\right)\right)+f\left(f\left(1\right)\right)$            

$f\left(1\right)={\left(2\left(1-\frac{1}{2}\right)\left(2+1\right)\right)}^{\frac{1}{50}}=3^{\frac{1}{50}}$

  $3^{\frac{1}{50}}+13^{\frac{1}{50}}>1^{\frac{1}{50}}$             

3 > 1

$3^{\frac{1}{50}}>2$ $2>3^{\frac{1}{50}}>1$

$3<2^{50}$ $\left[3^{\frac{1}{50}}+1\right]=1+1=2$                     

${\alpha }_n=19^n-12^n$

$\frac{31{\alpha }_9-{\alpha }_{10}}{57.{\alpha }_2}=\frac{31\left(19^9-12^9\right)-\left(19^{10}-12^{10}\right)}{57\left(19^8-12^8\right)}$

$=\frac{19^9\left(31-19\right)-12^9\left(31-12\right)}{57\left(19^8-12^8\right)}$

$=\frac{19^9.12-12^9.19}{57\left(19^8-12^8\right)}=4$