Class 12th

(2 – i) z = (2 + i) $\overline{z}$ $\stackrel{}{}$ , put z = x + iy

$y=\frac{x}{2}........\left(i\right)$

(ii) $\left(2+i\right)z+\left(i-2\right)\overline{z}-4i=0$ $\stackrel{}{}$

x + 2y = 2

(iii) $iz+\overline{z}+1+i=0$ $\stackrel{}{}$

Equation of tangent x – y + 1 = 0

Solving (i) and (ii)

$x=1.y=\frac{1}{2}\Rightarrow centre\left(1,\frac{1}{2}\right)$

Perpendicular distance of point $\left(1,\frac{1}{2}\right)$ $\frac{}{}$ from x – y + 1 = 0 is p = r $\Rightarrow \left|\frac{1-\frac{1}{2}+1}{\sqrt[]{2}}\right|=r$ $\frac{\frac{}{}}{\text{exception 25:}}$

$r=\frac{3}{2\sqrt[]{2}}$

Kindly go throiugh the solution 

 $\lambda =\frac{c}{v}=\frac{Q^3}{W^2}=\frac{I^3{T}^3}{M^2{L}^4{T}^{-4}}\Rightarrow \left[\lambda \right]=\left[M^{-2}{L}^{-4}{T}^7{l}^3\right]$

Let

A : Missile hit the target

B : Missile intercepted

P (B) = $\frac{1}{3}P\left(A/\overline{B}\right)=\frac{3}{4}$ $\frac{}{}\stackrel{}{}\frac{}{}$

$P\left(\overline{B}\right)=\frac{2}{3}$

$\Rightarrow P\left(\overline{B}\cap A\right)=\frac{3}{4}*\frac{2}{3}=\frac{1}{2}$

Required Probability = $\frac{2}{3}*\frac{3}{4}*\frac{2}{3}*\frac{3}{4}*\frac{2}{3}*\frac{3}{4}=\frac{1}{8}$

$\frac{a+20}{2}=3+7r$

a + 20 = 6 + 14r . (i)

b = 2 + 10r. (ii)

a = 18r – 2 . (iii)

Solving (i) and (iii) we get

20 + 18r – 2 = 6 + 14r

r = 3

$$ $\therefore$  a = 14 + 14 (-3) = -56 and b = -2 30 = 32

$\left|a+b\right|\left|-56-32\right|=88$

x – 2y = 1, x – y + kz = -2, ky + 4z = 6

x – 2y + 0. z – 1 = 0

x – y + kz + 2 = 0

0x + ky + 4z – 6 = 0

0x + ky + 4z – 6 = 0

${\Delta }_1=\left|\begin{array}{ccc}\hfill 1\hfill & \hfill -2\hfill & \hfill 0\hfill \\ \hfill -2\hfill & \hfill -1\hfill & \hfill k\hfill \\ \hfill 6\hfill & \hfill k\hfill & \hfill 4\hfill \end{array}\right|=-\left(k+10\right)\left(k+2\right)$

For no solution

$\Delta =0,{\Delta }_1\ne 0$

k = 2

The aim of using displacement current was to allow currrent ot flow inspite of a change in the field. This was not possible in the case of actual current since it lead to some complex situations, due to which scientists had to find a new solution for generating an electric field in such cases.

$l={\displaystyle \underset{1}{\overset{3}{\int }}\left[x^2-2x+1-3\right]dx=}{\displaystyle \underset{1}{\overset{3}{\int }}\left[{\left(x-1\right)}^2-3\right]dx={\displaystyle \underset{1}{\overset{3}{\int }}\left[{\left(x-1\right)}^2\right]dx-3{\displaystyle \underset{1}{\overset{3}{\int }}dx}}}$ .(A)

$l_1={\displaystyle \underset{1}{\overset{3}{\int }}\left[{\left(x-1\right)}^2\right]dxPut{\left(x-1\right)}^2=t}$

$l_1=\frac{1}{2}\left[0{\displaystyle \underset{0}{\overset{1}{\int }}\frac{dt}{\sqrt[]{t}}}{\displaystyle \underset{1}{\overset{2}{\int }}\frac{dt}{\sqrt[]{t}}}+2{\displaystyle \underset{2}{\overset{3}{\int }}\frac{dt}{\sqrt[]{t}}}+3{\displaystyle \underset{3}{\overset{4}{\int }}\frac{dt}{\sqrt[]{t}}}\right]=\frac{1}{2}\left\{{\left|\frac{t^{\frac{-1}{2}+1}}{-\frac{1}{2}+1}\right|}_1^2{\left|+2\frac{t^{\frac{-1}{2}+1}}{-\frac{1}{2}+1}\right|}_2^3{+3\frac{t^{\frac{-1}{2}+1}}{-\frac{1}{2}+1}]}_3^4\right\}$

Hence from (A)

= $5-\sqrt[]{2}-\sqrt[]{3}-\sqrt[]{3}-6=-1-\sqrt[]{2}-\sqrt[]{3}$

2^nd method           

${\displaystyle \underset{1}{\overset{3}{\int }}\left[{\left(x-1\right)}^2\right]dx-6..............\left(A\right)}$

From (A), $l=5-\sqrt[]{2}-\sqrt[]{3}-6=-1-\sqrt[]{2}-\sqrt[]{3}$

Maxwell simply wanted to generate a magnetic field using electric current, which displacement current had the capability to offer. Since both current flows were able to produce an electric field, both of them could be simultaneously used for using in the equation.

    $l={\displaystyle \underset{0}{\overset{2}{\int }}f\left(x\right)dx={\left[xf\left(x\right)\right]}_0^2-{\displaystyle \underset{0}{\overset{2}{\int }}xf\text{'}\left(x\right)dx=2e^2-{\displaystyle \underset{0}{\overset{2}{\int }}xf\text{'}\left(x\right)dx}}}$    .(A)

Put $l_1={\displaystyle \underset{0}{\overset{2}{\int }}xf\text{'}\left(x\right)dx}$               .(i)

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

$l_1={\displaystyle \underset{0}{\overset{2}{\int }}\left(2-x\right)f\text{'}\left(2-x\right)dx={\displaystyle \underset{0}{\overset{2}{\int }}\left(2-x\right)f\text{'}\left(x\right)dx}}$ .(ii)

Adding (i) and (ii) we get

$2l_1=2{\displaystyle \underset{0}{\overset{2}{\int }}f\text{'}\left(x\right)dx\Rightarrow l_1={\left[f\left(x\right)\right]}_0^2}$

f(2) – f(0) = e² – 1

From (A) l = 2e² – e² + 1 = e² + 1

$\widehat{r}-\widehat{i}=2cos\theta \widehat{n}......\left(1\right)$

$\Rightarrow \widehat{r}=\widehat{i}-2\left(\widehat{i}.\widehat{n}\right)\widehat{n}$

$\Rightarrow \overrightarrow{b}=\overrightarrow{a}-2\left(\overrightarrow{a}.\overrightarrow{c}\right)\overrightarrow{c}$