Maths

  $\frac{2}{-4}=\frac{-1}{2}$

y – 4 = 2 (x – 3)

y = 2x – 2

x² + (2x – 2)² = 25

  $5x^2-8x-21=0$

$z\left(\frac{-7}{5},\frac{-24}{5}\right)$              

x = 2t $A\left(2t,\frac{t^2}{3}\right)$ $\frac{{}^{}}{}$

$\frac{{}^{}}{}$ $y=\frac{t^2}{3}$  S (0, 3)

${\frac{}{}}^{}$  $3y={\left(\frac{x}{2}\right)}^2$ $B\left(0,\lambda \right)$ $$

$3k=\frac{t^2}{3}+3+\lambda =\frac{2t^2}{3}+3-\frac{12t^2}{9-t^2}$

$\underset{t\to 1}{lim}3k=\frac{2}{3}+3-\frac{12}{8}=3-\frac{5}{6}=\frac{13}{6}$

$y^{2}dx+\left(x^2-xy+y^2\right)dy=0$

$\frac{dx}{dy}+\frac{x^2-xy+y^2}{y^2}=0\Rightarrow \frac{dx}{dy}+{\left(\frac{x}{y}\right)}^2-\left(\frac{x}{y}\right)+1=0$

Put x = vy $\Rightarrow v+y\frac{dv}{dy}+v^2-v+1=0$

$\Rightarrow \frac{ydv}{dy}+v^2+1=0$

$\Rightarrow \frac{\pi }{6}+\mathcal{l}n\left|y\right|=\frac{\pi }{4}$

$\mathcal{l}n\left|y\right|=\frac{\pi }{12}$

$\mathrm{(A})\sim \left(p\wedge \left(\sim r\right)\right)\vee q=\left(\sim p\right)\vee \left(r\right)\vee q$

(B) $q\vee \left(-r\vee p\right)$

(C) $\left(\sim p\right)\vee \left(q\vee r\right)$

(D) $\left(\sim p\vee q\right)\vee r$

Using Venn diagram we get B as the correct option.

 $-1\le \frac{x^2-3x+2}{x^2+2x+7}\le 1$

$\Rightarrow 0\le \frac{2x^2-x+9}{x^2+2x+7}\&\frac{-5x-5}{x^2+2x+7}\le 0$

$x\in R\&-1\le x<\infty$

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.

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

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

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

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

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

A(2, 3, 9)

B(5, 2, 1)

C(1, $\lambda$ , 8)

D( $\lambda$ ,2, 3)

$\Delta =\left|\begin{array}{ccc}\hfill 3\hfill & \hfill -1\hfill & \hfill -8\hfill \\ \hfill -4\hfill & \hfill \lambda -2\hfill & \hfill 7\hfill \\ \hfill \lambda -2\hfill & \hfill -1\hfill & \hfill -6\hfill \end{array}\right|$

$\overrightarrow{AB}=\left(3,-1,-8\right)$

$\overrightarrow{AC}=\left(-4,\lambda -2,7\right)$

$\overrightarrow{AD}=\left(\lambda -2,-1,-6\right)$

$\Delta =3\left[-6\lambda +12+7\right]-1\left(7\lambda -14-24\right)-8\left(4-{\left(\lambda -2\right)}^2\right)$

$=57-18\lambda \Rightarrow +38-32+8\left({\lambda }^2-4\lambda +4\right)$

$=95-57\lambda +8{\lambda }^2$

$\Delta =0\Rightarrow {\lambda }_1{\lambda }_2=\frac{95}{8}$

x + 2y + z = 14

${\perp }^{r}linePQ:\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{1}=t$                

Q (1 + t, 2 + 2t, 3 + t)

 x + 2y + z = 14 ⇒ 1 + t + 4 + 4t + 3 + t = 14 ⇒ 6t = 6

t = 1

⇒ Q (2, 4, 4)

PQ = $\sqrt[]{1+4+1}=\sqrt[]{6}$  

$tan60°=\frac{PQ}{QR}\Rightarrow QR=\frac{PQ}{\sqrt[]{3}}=\sqrt[]{2}$

ar (PQR) = $\frac{1}{2}\sqrt[]{6}\cdot \sqrt[]{2}=\sqrt[]{3}$