Class 12th

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

Consider the image below

There is no mirror image of each other, having some configuration.

Consider the image below

Consider the image below

Given : $\widehat{a}\cdot \widehat{b}=\widehat{b}\cdot \widehat{c}=\widehat{c}\cdot \widehat{a}=cos\theta \left(say\right)$

$\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\left|\overrightarrow{c}\right|=14$

$\left(\overrightarrow{a}*\overrightarrow{b}\right)\cdot \left(\overrightarrow{b}*\overrightarrow{c}\right)$

$=\overrightarrow{a}\cdot \left[\left(\overrightarrow{b}\cdot \overrightarrow{c}\right)\overrightarrow{b}-\left(\overrightarrow{b}\cdot \overrightarrow{b}\right)\overrightarrow{c}\right]$

$=\left(\overrightarrow{a}\cdot \overrightarrow{b}\right)\left(\overrightarrow{b}\cdot \overrightarrow{c}\right)-{\left|\overrightarrow{b}\right|}^2\overrightarrow{a}\cdot \overrightarrow{c}$

$=\left|\overrightarrow{a}\right|{\left|\overrightarrow{b}\right|}^2\left|\overrightarrow{c}\right|\left(co{s}^2\theta -cos\theta \right)=14\left|\overrightarrow{b}\right|\left(co{s}^2\theta -cos\theta \right)$

Similarly, $\left(\overrightarrow{b}*\overrightarrow{c}\right)\cdot \left(\overrightarrow{c}*\overrightarrow{a}\right)$

= $\left|\overrightarrow{b}\right|{\left|\overrightarrow{c}\right|}^2\left|\overrightarrow{a}\right|\left(co{s}^2\theta -sin\theta \right)=14\left|\overrightarrow{c}\right|\left(cos2\theta -sin\theta \right)\&\left(\overrightarrow{c}*\overrightarrow{a}\right)\cdot \left(\overrightarrow{a}*\overrightarrow{b}\right)$

$=\left|\overrightarrow{c}\right|{\left|\overrightarrow{a}\right|}^2\left|\overrightarrow{b}\right|\left(co{s}^2\theta -sin\theta \right)=14\left|\overrightarrow{a}\right|\left(co{s}^2\theta -sin\theta \right)$

Given : 14 $\left(co{s}^2\theta -cos\theta \right)\left(\left|\overrightarrow{a}\right|+\left|\overrightarrow{b}\right|+\left|\overrightarrow{c}\right|\right)=168$

$\left|\overrightarrow{a}\right|+\left|\overrightarrow{b}\right|+\left|\overrightarrow{c}\right|=\frac{12}{co{s}^2\theta -cos\theta }=\frac{12}{\frac{1}{4}-\left(-\frac{1}{2}\right)}=\frac{12}{\frac{3}{4}}=16$

Given : $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are coplanar & pair wise equal angle.

$\left|z-1+i\right|\ge \left|z\right|$

$\left|z+i\right|=\left|z-1\right|$

W = (2x, y) = (α, y)

Let S represent the line segment AB

For 'B'

x² + y² = 4

x = y

x² = 2

$x=\pm \sqrt[]{2}$

$B\left(-\sqrt[]{2},\sqrt[]{2}\right)$

$A\left(\frac{1}{2},-\frac{1}{2}\right)$

W (2x, y) lies on AB

$-\sqrt[]{2}<2x\le \frac{1}{2}$

$\left(\left|z\right|<2\right)$

$-\frac{1}{\sqrt[]{2}}<x\le \frac{1}{4}$

 $\begin{array}{ccc}\hfill 3R\hfill & \hfill 2R\hfill & \hfill 4B\hfill \\ \hfill 5B\hfill & \hfill 3W\hfill & \hfill 2W\hfill \end{array}$

I        II

Let

E₁ : a red ball is transferred from I to II

E₂ : a black is transferred from I to II

E₃ :a white transferred from I to II

E : a black ball is drawn from 2nd bag after a ball from I to II was transferred.

$P\left(\frac{E_1}{E}\right)=\frac{P\left(E_1\cap E\right)}{P\left(E\right)}$

$P\left(E\right)=P\left(E_1\cap E\right)+P\left(E_2\cap E\right)+P\left(E_3\cap E\right)$

= $P\left(E_1\right)\cdot P\left(\frac{E}{E_1}\right)+....+.....$

= $\frac{3}{10}\cdot \frac{5}{10}+\frac{4}{10}\cdot \frac{6}{10}+\frac{3}{10}\cdot \frac{5}{10}=\frac{54}{100}$

$P\left(E_1/E\right)=\frac{15/100}{54/100}=\frac{5}{18}$

 $\frac{dy}{dx}+\left(\frac{2x^2+11x+13}{x^3+6x^2+11x+6}\right)y=\frac{x+3}{x+1},x>-1$

IF = $e^{{\displaystyle \int pdx}}=\frac{{\left(x+1\right)}^2\left(x+2\right)}{x+3}$

${\displaystyle \int Pdx={\displaystyle \int \frac{2x^2+11x+13}{x^3+6x^2+11x+6}dn={\displaystyle \int \left(\frac{2}{x+1}+\frac{1}{x+2}-\frac{1}{x+3}\right)dx}}}$

$=\mathcal{l}n\left({\left(x+1\right)}^2\cdot \left(x+2\right)/\left(x+3\right)\right)$

$\frac{2x^2+11x+13}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{x+3}$

$2x^2+11x+13=A\left(x+2\right)\left(x+3\right)+B\left(x+1\right)\left(x+3\right)+C\left(x+1\right)\left(x+2\right)$

x = -1

⇒ 4 = 2A ⇒ A = 2

x = -2

⇒ -1 = -B Þ B = 1

x₂ – 3 ⇒ -2 = 2c

c = -1

$y\cdot \frac{{\left(x+1\right)}^2\left(x+2\right)}{x+3}={\displaystyle \int \frac{x+3}{x+1}\cdot \frac{{\left(x+1\right)}^2\left(x+2\right)}{x+3}dx}$

= ${\displaystyle \int \left(x+1\right)\left(x+2\right)dx}$

= $\frac{x^3}{3}+\frac{3x^2}{2}+2x+c$

$\left(0,1\right)\Rightarrow 1\cdot \frac{2}{3}=c$

x = 1 $y\cdot \left(3\right)=\frac{1}{3}+\frac{3}{2}+2+\frac{2}{3}=\frac{3}{2}+3=\frac{9}{2}$

y = $\frac{3}{2}$

$\frac{dy}{dx}=\frac{x+y-2}{x-y}$

Let x – 1 = X, y – 1 = Y

then DE: $\frac{dY}{dX}=\frac{X+Y}{X-Y}=\frac{1+\frac{Y}{X}}{1-\frac{Y}{X}}$

Put y = vx

then $\frac{dY}{dX}=V+X\frac{dV}{dX}$

V + $X\frac{dV}{dX}=\frac{1+V}{1-V}$

$X\frac{dV}{dX}=\frac{1+V}{1-V}-V$

$=\frac{1+V^2}{1-V}$

$\frac{1-V}{1+V^2}dV=\frac{dX}{X}$

$\frac{V-1}{V^2+1}dV+\frac{dX}{X}=0$

$\frac{1}{2}\mathcal{l}n\left|V^2+1\right|-ta{n}^{-1}V+\mathcal{l}n\left|X\right|=c$

$\mathcal{l}n\sqrt[]{V^2+1}\cdot X-ta{n}^{-1}V=c$

$\mathcal{l}n\left(\sqrt[]{1+{\left(\frac{Y-1}{X-}\right)}^2}\cdot \left|X-1\right|\right)ta{n}^{-1}\frac{Y-1}{X-1}=c$

$\left(2,1\right)\Rightarrow \mathcal{l}n\left(\sqrt[]{1+0}\cdot 1\right)-0=c$

c = 0

$\mathcal{l}n\sqrt[]{{\left(X-1\right)}^2+{\left(Y-1\right)}^2}=ta{n}^{-1}\frac{Y-1}{X-1}$

point (k + 1, 2) $\mathcal{l}n\sqrt[]{k^2+1}=ta{n}^{-1}\frac{1}{k}$

$\Rightarrow \frac{1}{2}\mathcal{l}n\left(k^2+1\right)=ta{n}^{-1}\frac{1}{k}$