Linear Programming

$\Delta -\left|\begin{array}{ccc}\hfill 3\hfill & \hfill -2\hfill & \hfill -k\hfill \\ \hfill 2\hfill & \hfill -4\hfill & \hfill -2\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill -1\hfill \end{array}\right|=\left|\begin{array}{ccc}\hfill 3\hfill & \hfill -2\hfill & \hfill -k\hfill \\ \hfill 0\hfill & \hfill -8\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill -1\hfill \end{array}\right|,\left[R_2-R_1-2R_3\right]$

= -8 (-3 + k)

For inconsistent $\Delta =0\Rightarrow k=3$  

${\Delta }_x=\left|\begin{array}{ccc}\hfill 10\hfill & \hfill -2\hfill & \hfill -k\hfill \\ \hfill 6\hfill & \hfill -4\hfill & \hfill -2\hfill \\ \hfill 5m\hfill & \hfill 2\hfill & \hfill -1\hfill \end{array}\right|=32-40m\ne 0\Rightarrow m\ne \frac{4}{5}$              

  $l={\displaystyle {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\left(\left[x\right]+\left[-sinx\right]\right)dx}$ . (i)

$I={\displaystyle {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\left(\left[-x\right]+\left[sinx\right]\right)dx}$ . (ii)

by using property $\left({\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx=}{\displaystyle \underset{a}{\overset{b}{\int }}f\left(a+b-x\right)dx}\right)$

Adding (i) and (ii) we get 2l = ${\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\left(-2\right)dx=-2\pi \Rightarrow l=-\pi }$

$l_{n,m}={\displaystyle \underset{0}{\overset{\frac{1}{2}}{\int }}\frac{x^n}{x^m-1}dx,m,n\in N,n>m}$

$l_{6+i,3}-l_{3+i,3}$

$A=\frac{1}{2^5}\left[\begin{array}{ccc}\hfill \frac{1}{5}\hfill & \hfill \frac{1}{5}\hfill & \hfill \frac{1}{5}\hfill \\ \hfill 0\hfill & \hfill \frac{1}{12}\hfill & \hfill \frac{1}{12}\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill \frac{1}{28}\hfill \end{array}\right]$ $=\frac{B}{32}$

$\left|A\right|={\left(\frac{1}{32}\right)}^3\left|B\right|=\frac{1}{105.2^{19}}$

$\alpha +\beta +\gamma =2\pi$

$\Delta =\left|\begin{array}{ccc}\hfill 1\hfill & \hfill cos\gamma \hfill & \hfill cos\beta \hfill \\ \hfill cos\gamma \hfill & \hfill 1\hfill & \hfill cos\alpha \hfill \\ \hfill cos\beta \hfill & \hfill cos\alpha \hfill & \hfill 1\hfill \end{array}\right|$

$=1-co{s}^2\alpha -co{s}^2\beta -co{s}^2\gamma +2cos\alpha .cos\beta .cos\gamma$

$=-cos\gamma cos\left(\alpha -\beta \right)+cos\gamma cos\left(\alpha -\beta \right)=0$

Given 2x + y – z = 3         . (i)

x – y – z = α        . (ii)

3x + 3y + βz = 3                . (iii)

(i) x 2 – (ii) – (iii) – (1 + β) z = 3 - α

For infinite solution 1 + β = 0 = 3 - α

=> α = 3, β = -1

So, α + β - αβ = 5

  $S_1=\left\{z,\in c:\lfloor z_1-3\rfloor =\frac{1}{2}\right\}and{S}_2=\left\{z_2\in c:\left|z_2-\left|z_2+1\right|\right|=\left|z_2+\left|z_2-1\right|\right|\right\}$

Then, for $z_1\in S_1$  and $z_2\in S_2,$  the least value of |z₂ – z₁|

${\left|z_2+\left|z_2-1\right|\right|}^2={\left|z_2-\left|z_2+1\right|\right|}^2$      

$\Rightarrow \left(z_2+{\overline{z}}_2\right)\left(\left|z_2-1\right|+\left|z_2+1\right|-2\right)=0$      

$\therefore z_2+{\overline{z}}_2=0or\left|z_2-1\right|+\left|z_2+1\right|-2=0$          

$\therefore$ z₂ lies on imaginary axis or on real axis with in [-1, 1]

also $\left|z_1-3\right|=\frac{1}{2}$  lie on circle having centre 3 and radius   $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

Clearly $\left|z_1-z_2\right|min=\frac{5}{2}-1=\frac{3}{2}$

cota = 1        & secβ =   $\frac{-5}{3}$

< < $\frac{}{}$       $\frac{3\pi }{2}$  secβ =   $\frac{-5}{3}$

$\alpha =\left(\pi +\frac{\pi }{4}\right)$  cosβ = $\frac{-3}{5}=cos\left(180-53\right)$  

tanb =  $\frac{-4}{3}$  

tan(α + β) =   $\frac{tan\alpha +tan\beta }{1-tan\alpha tan\beta }$

$\therefore A-\frac{1}{4}\&4thquadrant.$

$=\frac{1-\frac{4}{3}}{1+\frac{4}{3}}=\frac{-1}{7}$

$\therefore -\frac{1}{4}\&4thquadrant.$

Case – I $\Delta \equiv \nabla \equiv \wedge$

$\left(p\wedge q\right)\to \left(\left(p\wedge q\right)\wedge r\right)$

it can be false if r is false,

so not a tautology

Case – II If $\Delta \equiv \nabla \equiv \vee$

$\left(p\vee q\right)\text{?}\to \left(\left(p\vee q\right)\vee r\right)\equiv$ tautology

then $\left(p\vee q\right)\vee r\equiv \left(p\Delta r\right)\vee q$

Case – III If $\Delta \equiv v,\nabla \equiv \wedge$

then $\left(p\wedge q\right)\to \left\{\left(p\vee q\right)\wedge r\right\}$

Not a tautology

Case – IV If $\Delta \equiv \wedge ,\nabla \equiv \vee$

$\left(p\wedge q\right)\to \left\{\left(p\wedge q\right)\vee r\right\}$

Not a tautology

Clearly r must be equal to $\sim$ $$ p

$?\sim p\vee \sim p=\sim p$

$and\left(p\wedge q\right)\vee \sim p=p$

$$ $\therefore \sim p\Rightarrow p=$  tautology.