Maths NCERT Exemplar Solutions Class 11th Chapter Two

$S_1=1+2+3\dots \dots \dots .n$

$S_2=1+3+5\dots \dots \dots .n$

$S_3=1+4+7\dots \dots \dots .$

$\begin{array}{ll}\left(S_1+S_3\right)=\lambda S_2& S_1=\frac{n}{2}[2a+(n-1\left)1\right]\\ \frac{n}{2}[2a+(n-1\left)1\right]+\frac{n}{2}[2a+(n-1\left)3\right]& S_2=\frac{n}{2}[2a+(n-1\left)2\right]\\ =\lambda \frac{n}{2}[2a+(n-1\left)2\right]& S_3=\frac{n}{2}[2a+(n-1\left)3\right]\end{array}$

$\lambda =2$

$\sim (\sim p\vee \sim q)\to (\sim q\vee r)\Rightarrow p\wedge q\to \sim q\vee r$

$\mathrm{}\mathrm{T}\mathrm{F}\mathrm{T}\Rightarrow \mathrm{F}\to \mathrm{T}\to \mathrm{T}$

$\mathrm{F}\mathrm{T}\mathrm{T}\Rightarrow \mathrm{F}\to \mathrm{T}\to \mathrm{T}$

$\mathrm{T}\mathrm{T}\mathrm{T}\Rightarrow \mathrm{T}\to \mathrm{T}\to \mathrm{T}$

$\mathrm{T}\mathrm{T}\mathrm{F}\Rightarrow \mathrm{T}\to \mathrm{F}\to \mathrm{F}$

${\mathrm{l}\mathrm{i}\mathrm{m}}_{x\to 9^{-}}?\left. [x^2+1\right]+\left. [x^2-1\right]+\left\{x\right\}$

$\Rightarrow \mathrm{}{\mathrm{l}\mathrm{i}\mathrm{m}}_{x\to 9^{-}}?2\left. [x^2\right]+\left\{x\right\}\Rightarrow {\mathrm{l}\mathrm{i}\mathrm{m}}_{h\to 0}?2\left. [(9-x{)}^2\right]+\{9-h\}=160+1=161$

Let $z=x+yi\mathrm{}z-2\bar{z}=1\Rightarrow x=-1\mathrm{}y=0$

$y(ydx+xdy)=x^5\left(\frac{xdy-ydx}{x^2}\right)$

$\Rightarrow d\left(xy\right)=x^5{y}^{-1}\left(\frac{xdy-ydx}{x^2}\right)\Rightarrow d\left(xy\right)=(xy{)}^{\alpha }{\left(\frac{y}{x}\right)}^{\beta }d\left(\frac{y}{x}\right)$

$\alpha -\beta =5\mathrm{}\alpha +\beta =-1$

$2\alpha =4\mathrm{}\alpha =2\mathrm{}\beta =-3$

$\Rightarrow \mathrm{}\left(xy{)}^{-2}d\right(xy)={\left(\frac{y}{x}\right)}^{-3}d\left(\frac{y}{x}\right)\Rightarrow -(xy{)}^{-1}=-\frac{1}{2}{\left(\frac{y}{x}\right)}^{-2}+c$

$\Rightarrow \mathrm{}-(xy{)}^{-1}=-\frac{1}{2}{\left(\frac{y}{x}\right)}^{-2}+c\mathrm{}$ Passing $\left(\mathrm{1,1}\right)$

$\Rightarrow \mathrm{}-1=-\frac{1}{2}+c\Rightarrow c=-\frac{1}{2};-(xy{)}^{-1}=-\frac{1}{2}{\left(\frac{y}{x}\right)}^{-2}-\frac{1}{2}$

Clearly mean will increase by 5 units and variance will be un changed so the sum would be $7+8+5=20$ .7+8+5=20

$(x+y{)}^n={}^n{C}_r{x}^{n-r}{y}^r$

${\left(\frac{3}{2}{x}^2+\frac{2}{x}\right)}^{12}?{}^{12}{C}_r{\left(\frac{3}{2}{x}^2\right)}^{12-r}{\left(\frac{2}{x}\right)}^r$

For independent of X .

$x^{24-2r}{x}^{-r}=x^0;24=3r\Rightarrow r=8$

${}^{12}{C}_8{\left(\frac{3}{2}\right)}^{12-8}(2{)}^8;\frac{\lfloor 12}{\overline{12}44}*\frac{3^4}{2^4}*2^8\Rightarrow \lambda ;\frac{\lambda }{81}\Rightarrow 7920$

$\int {\mathrm{c}\mathrm{o}\mathrm{t}}^{-1}?\left(\frac{1}{1+x}\right)dx$

$=\int {\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}?(1+x)dx$

$=x{\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}?(1+x)-\int \frac{x}{1+(1+x{)}^2}dx;=x{\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}?(1+x)-\int \frac{x}{x^2+2x+2}dx$

$=x{\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}?(1+x)-\int \frac{\frac{1}{2}(2x+2)-1}{x^2+2x+2}dx=x{\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}?(1+x)-\frac{1}{2}\mathrm{l}\mathrm{o}\mathrm{g}?\left(x^2+2x+2\right)+\int \frac{dx}{(x+1{)}^2+1}$

$=x{\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}?(1+x)-\frac{1}{2}\mathrm{l}\mathrm{o}\mathrm{g}?\left(x^2+2x+2\right)+{\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}?(x+1)+c$

Case-I : $a-2>0$

$F(-2)>0$

$4(a-2)+4a+a>0$

$9\mathrm{a}-8>0$

$a>\frac{8}{9}$

$F\left(1\right)>0$

$a-2-2a+a>0$

  $-2>0$  (not possible)

Case-II : $a-2<0;a<2$

$F(-2)<0\mathrm{}\Rightarrow \mathrm{}a<\frac{8}{9}$

$F\left(1\right)<0\mathrm{}\Rightarrow \mathrm{}a\in R$

$-2<\frac{-b}{2a}<1\mathrm{}a<\frac{4}{3}\mathrm{}a\in \left[0,\frac{8}{9}\right)\cup \left\{2\right\}$

$D\ge 0\mathrm{}a\ge 0$

$|\mathrm{A}\mathrm{d}\mathrm{j}?A|=64=|A{|}^{n-1}\Rightarrow |A|=\pm 8$ $\alpha =\pm 8$

$\beta =|\mathrm{a}\mathrm{d}\mathrm{j}?(\mathrm{a}\mathrm{d}\mathrm{j}?A\left)\right|=|A{|}^{(n-1{)}^2}=8^4\mathrm{}\text{Also}\left|\frac{\sqrt[]{\beta }}{\alpha }\right|=\frac{8^2}{8}=8$