Class 12th

${\left(x+2\sqrt[]{3}\right)}^2+y^2=4^2$                

$y^2=8\left(x+\frac{1}{2}\right)$

Point of intersection are (0, 2) and (0, -2)

  $2{\displaystyle \underset{0}{\overset{2}{\int }}\left[\left(\sqrt[]{16-y^2}-2\sqrt[]{3}\right)-\frac{\left(y^2-4\right)}{8}\right]dy}$              

  $=\frac{1}{3}\left(8\pi +4-12\sqrt[]{3}\right)$

Let f(x) = 8 sin x - sin 2x

$f\text{'}\left(x\right)=8cosx-2cos2x$              

$f\text{'}\text{'}\left(x\right)=-8sinx+4sin2x$               

= -8 (sin x) (1 – cos x)

f''(x) < 0, f'(x) is a decreasing function

  ^{$f\text{'}\left(\frac{\pi }{3}\right)<f\text{'}\left(x\right)<f\text{'}\left(\frac{\pi }{a}\right)$}               

  $5<f\text{'}\left(x\right)<\frac{8}{\sqrt[]{2}}$

${\displaystyle \underset{\pi /4}{\overset{\pi /3}{\int }}dx<{\displaystyle \underset{\pi /4}{\overset{\pi /3}{\int }}\frac{f\left(x\right)}{x}<{\displaystyle \underset{\pi /4}{\overset{\pi /3}{\int }}\frac{8}{\sqrt[]{2}}}}}$

$5\pi 12<l<\frac{8}{\sqrt[]{2}}.\frac{\pi }{\sqrt[]{2}}$             

Find RHL, x = -1 + h

$\underset{h\to 0}{lim}a.sin\left(\frac{\pi \left[-1+h\right]}{2}\right)+\left[2+1-h\right]=-a+2$               

 Find LHL, x = -1 – h

$\underset{h\to 0}{lim}asin\frac{\left(\pi \left[-1-h\right]\right)}{2}=\left[2+1+h\right]=3$               

${\displaystyle \underset{0}{\overset{4}{\int }}f\left(x\right)dx={\displaystyle \underset{0}{\overset{1}{\int }}1dx+{\displaystyle \underset{1}{\overset{2}{\int }}\left(-1\right)dx+{\displaystyle \underset{2}{\overset{3}{\int }}\left(-1\right)dx+{\displaystyle \underset{3}{\overset{4}{\int }}\left(-1\right)dx=-2}}}}}$               

R_w = 2R_y

$\Rightarrow \rho \left(\frac{2x}{\raisebox{1ex}{$A$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)=\frac{2\rho \left(1-x\right)}{A}$

$\frac{4\rho x}{A}=\frac{2\rho \left(1-x\right)}{A}$

4x = 2 – 2x 6x = 2

x = $\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$6$}\right.=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$ $\therefore \frac{L_x}{L_y}=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{1-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=\frac{1}{3}*\frac{3}{2}$

$=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

Kindly go through the solution

Use characteristic equation = 0

$V={\left|z\right|}^2+{\left|z-3\right|}^2+{\left|z-6i\right|}^2$  

Put z = x + iy

$V=3\left[{\left(x-1\right)}^2+{\left(y-2\right)}^2+10\right]$                

For minimum x = 1, y = 2, V₀ = 30

So, z₀ = 1 + 2i

Based on theoretical data.

f (a) = a, a is max of powers of prime P such that P^a divides a.

f (2) = 1; g (2) = 3

f (3) = 1; g (3) = 4

f (4) = 2; g (4) = 5

f (5) = 1; g (5) = 6

f (2) + g (2) = 4

f (3) + g (3) = 5

f (4) + g (4) = 7

f (5) + g (5) = 7

So, f (x) + g (x) is many-one-and an into function. We won't get 'I' in the range.

V = $\frac{q_{\left(capacitance\right)}}{C}=\frac{q_1-q_2}{2c}$

For sphere 'C' after contacting with 'A'. q_A = q_C = $\frac{q}{2}.$

offer contacting with 'B'. q_B = q_C = $\frac{3q}{4}$

$F_{Net}=\left|F_1-F_2\right|$

F₂ = $\frac{K\cdot 3q^2*4}{r^{2}8}=\frac{3}{2}\frac{k{q}^2}{r^2}$

$F_1=\frac{9k{q}^2*4}{16r^2}=\frac{9k{q}^2}{4r^2}=\frac{9}{4}F$

$\therefore F_{Net}=\left|\frac{9}{4}F-\frac{3}{2}F\right|$

= $\frac{9-6}{4}F=\frac{3}{4}F$