Class 12th

By its given condition

$\stackrel{}{}\stackrel{}{}\stackrel{}{}$  $\therefore \overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are linearly independent vectors is $\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]\ne 0$ $\stackrel{}{}\stackrel{}{}\stackrel{}{}$

Now, $\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]$ $\stackrel{}{}\stackrel{}{}\stackrel{}{}$ = $\left|\begin{array}{ccc}\hfill 1+t\hfill & \hfill 1-t\hfill & \hfill 1\hfill \\ \hfill 1-t\hfill & \hfill 1+t\hfill & \hfill 2\hfill \\ \hfill t\hfill & \hfill -t\hfill & \hfill 1\hfill \end{array}\right|\ne 0$ $\begin{array}{ccc}\hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill \end{array}$

$R_1-R_2,R_2+R_3$

$R_1-R_2\Rightarrow \left|\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill -3\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 3\hfill \\ \hfill t\hfill & \hfill -t\hfill & \hfill \mathcal{l}\hfill \end{array}\right|\ne 0\Rightarrow t\ne 0$

T₁ = 3 sec T₂ = 4sec

$T=2\pi \sqrt[]{\frac{I}{\mu B}}$

$\frac{I_1}{I_2}=\frac{3}{2}\Rightarrow \frac{T_1}{T_2}=\sqrt[]{\frac{l_1{\mu }_2}{{\mu }_1{l}_2}}$

$\Rightarrow {\left(\frac{3}{4}\right)}^2=\frac{I_1}{I_2}\left(\frac{{\mu }_2}{{\mu }_1}\right)\Rightarrow \frac{{\mu }_2}{{\mu }_1}=\frac{l_2}{l_1}*\frac{9}{16}=\frac{2}{3}*\frac{9}{16}=\frac{3}{8}$ $\Rightarrow \frac{{\mu }_1}{{\mu }_2}=\frac{8}{3}$

$\left|\frac{x^2+4x+2}{x^2+3}\right|\le 1$

${\left(x^2-4x+2\right)}^2\le {\left(x^2+3\right)}^2$

$\Rightarrow \left(2x^2-4x+5\right)\left(-4x-1\right)\le 0$

$\Rightarrow -4x-1\le 0\to x\ge -\frac{1}{4}$

$\underset{x?0}{lim}\frac{\mathcal{l}n\frac{1+5x}{1+?x}}{x}=10$

$\underset{x?0}{lim}\mathcal{l}n\left\{\frac{1+\left(5??\right)x}{\frac{\left(5??\right)x}{1+?x}?\left(\frac{1+?x}{5??}\right)}\right\}$

5 = 10

= 5

$xdy=\left(\sqrt[]{x^2+y^2}+y\right)dx$

$\frac{xdy-ydx}{x^2}=\sqrt[]{1+\frac{y^2}{x^2}}dx$

$\frac{d\left(\frac{y}{x}\right)}{\sqrt[]{1+{\left(\frac{y}{x}\right)}^2}}=\frac{dx}{x}\Rightarrow \mathcal{l}n\left(\frac{y}{x}+\sqrt[]{{\left(\frac{y}{x}\right)}^2+1}\right)=\mathcal{l}nx+C$

$\alpha =\frac{3}{2}$

$\Delta =0$

$\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 5\hfill & \hfill \alpha \hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \end{array}\right|=0$

15 - 2a + a - 6 – 1 = 0

a = 8

For a = 8, equations are

x + y + 3 = 6

2x + 5y + 8z = b

x + 2y + 3z = 14

$\left(2,5,8\right)=\mathcal{l}\left(1,1,1\right)+m\left(1,2,3\right)$

$\begin{array}{cc}\hfill 2=\mathcal{l}+m\hfill & \hfill 5=\mathcal{l}+2m\hfill \end{array}]\to 3=m,\mathcal{l}=-1$              

8 =  $\mathcal{l}+3m$

$\beta =6\mathcal{l}+14m$

= -6 + 42 = 36

a + b = 8 + 36 = 44

V_AB = 0

$or,\epsilon -i{r}_1=0$

$\Rightarrow \epsilon -\frac{2\epsilon }{r_1+r_2+R}{r}_1=0\Rightarrow 1=\frac{2r_1}{r_1+r_2+R}$

$\Rightarrow R+r_1+r_2=2r_1$

$R=r_1-r_2$

$A=\left[\begin{array}{cc}\hfill -1\hfill & \hfill 2\hfill \\ \hfill 1\hfill & \hfill -1\hfill \end{array}\right]$

$EA=\left[\begin{array}{cc}\hfill a\hfill & \hfill c\hfill \\ \hfill b\hfill & \hfill d\hfill \end{array}\right]\left[\begin{array}{cc}\hfill -1\hfill & \hfill 2\hfill \\ \hfill 1\hfill & \hfill -1\hfill \end{array}\right]$

$=\left[\begin{array}{cc}\hfill -a+c\hfill & \hfill 2a-c\hfill \\ \hfill -b+d\hfill & \hfill 2b-d\hfill \end{array}\right]$

For a = c For $\begin{array}{cc}\hfill -a+c=0\hfill & \hfill 2a-c=1\hfill \end{array}]\to a=1,c=1E=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$  

d = b + 1, d = 1, b = 0

$\begin{array}{cc}\hfill b+d=1\hfill & \hfill 2b-d=-1\hfill \end{array}]\to b=0,d=1R_1\to R_1\to R_2\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

For  $\begin{array}{cc}\hfill -a+c=1\hfill & \hfill 2a-c=1\hfill \end{array}]\to a=0,c=1$

$For\begin{array}{cc}\hfill -a+c=-1\hfill & \hfill 2a-c=2\hfill \end{array}]\to a=1,c=0$

$\begin{array}{cc}\hfill -b+d=-2\hfill & \hfill 2b-d=7\hfill \end{array}]\to b=5,d=3\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 5\hfill & \hfill 3\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

R₂ -> 5R₁ + 3R₂

For $For\begin{array}{cc}\hfill -a+c=-1\hfill & \hfill 2a-c=2\hfill \end{array}]\to a=1,c=1$

$\begin{array}{cc}\hfill -b+d=-1\hfill & \hfill 2b-d=3\hfill \end{array}]\to b=2,d=1$

(A) ® R₁ ® R₁ + R₂

_{(B) ® R2 ® R2 + 2R1 $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 2\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$}

(C) ® R₂ ® 3R₂ + 5R₁

Phenolphthalein is the indicator of weak acid which produces pink colour in basic medium.

Always fresh solution of starch used and recorded immediately after the blue colour appears. Na2S2O3 is used in limited amount.