Class 12th

Let angle of dip =

Then B cos 45° = B' cos 60°- (i)

B' sin 60 = B sin - (ii)

$\left(i\right)÷\left(ii\right)\Rightarrow cost60=cot\alpha cos45°$

$cot\alpha =\frac{cot60°}{cos45°}=\frac{1\sqrt[]{2}}{\sqrt[]{3}x}=\sqrt[]{\frac{2}{3}}$

$tan\alpha =\sqrt[]{\frac{3}{2}}$

$\alpha =ta{n}^{-1}\sqrt[]{\frac{3}{2}}$

a₀ = 0, a₁ = 0

a_n+2 = 3a_n+1 – 2a_n + 1

a₂₅ a₂₃ – 2a₂₅a₂₂ – a₂₃a₂₄ + 4a₂₂a₂₄ =?

a₂ = 3a₁ – 2a₀ + 1

a₃ = 3a₂ – 2a₁ + 1

a₄ = 3a₃ – 2a₂ + 1

a₅ = 3a₄ – 2a₃ + 1

a_n+2  = 3a_n+1 – 2a_n + 1

$\left(+\right)\Rightarrow \left(a_2+a_3+a_4+....+a_{n+1}+a_{n+2}\right)$

-> a_n+2 = 2 (a₂ + a₃ + …. + a_n + a_n+1) –2 (a₁ + a₂ + ….+ a_n) + n + 1

a_n+2 = 2a_n+1 + n + 1

a₂₅ a₂₃ -2a₂₅ a₂₂ -a₂₃ a₂₄ + 4a₂₂ a₂₄

= a₂₅ (a₂₃ – 2a₂₂) -2a₂₄ (a₂₃ – 2a₂₂)

As a_n+2 = 2a_n+1 + n + 1

-> a_n+2 – 2a_n+1 = n + 1

-> a_n+1 -2a_n = n

-> 24 * 22 = 528

cos^-1 x = 2 sin^-1 x = cos^-1 2x

$co{s}^{-1}x-2\left(\frac{\pi }{2}-co{s}^{-1}x\right)=co{s}^{-1}2x$          

$cos\left(3co{s}^{-1}x\right)=cos\left(\pi +co{s}^{-1}2x\right)$    

$4x^3-3x=-2x$        

$4x^3=x\Rightarrow x=0\pm \frac{1}{2}$      

All satisfy the original equation

Sum =  $-\frac{1}{2}+0+\frac{1}{2}=0$

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

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

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

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

$R=r_1-r_2$