Maths

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

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

Circumcentre (D) $\equiv \left(5,\frac{\alpha }{4}\right)$

${\left(5-\alpha \right)}^2+{\left(\frac{\alpha }{4}+2\right)}^2={\left(5-\alpha \right)}^2+{\left(\frac{\alpha }{4}-6\right)}^2...............\left(i\right)$

${\left(5-\frac{\alpha }{4}\right)}^2+{\left(\frac{\alpha }{4}+2\right)}^2....................\left(ii\right)$

(i) $\frac{\alpha }{4}+2=\pm \left(\frac{\alpha }{4}-6\right)$

(ii) 9 + 16 = 9 + 16

$\oplus \to x$

$\left(-\right)\to \frac{\alpha }{2}=4\Rightarrow \alpha =8$

ar $\left(ABC\right)=24$

2S = 24

R = 5, r = $\frac{\Delta }{s}=\frac{24}{12}=2$

2cos $\left(\frac{x^2+x}{6}\right)=4^x+4^{-x}$

$-2\le LHS\le 2LHS=2\&RHS=2\Rightarrow x=0only\to thenLHS=2also$

RHS $\ge$ 2

$m_1{m}_2=-1,$ for square a,b,c,d let

$A\left(10\left(cos\alpha -sin\alpha \right),10\left(sin\alpha +cos\alpha \right)\right)$

Diagonal : (cosα - sinα)x + (sinα + cosα)y = 10

BD (diagonal)

Dist. Of BD from A is

$\frac{\left|10{\left(cos\alpha -sin\alpha \right)}^2+10{\left(sin\alpha +cos\alpha \right)}^2-10\right|}{\sqrt[]{2}}=\frac{a}{\sqrt[]{2}}$

$\frac{10}{\sqrt[]{2}}=\frac{a}{\sqrt[]{2}}\Rightarrow a=10$

Also, a² + 11a + 3 $\left(m_1^2+m_2^2\right)=220$

210 + 3 $\left(c{m}_1^2+m_2^2\right)=220$

$m_1^2+m_2^2=\frac{10}{3}$

Also, m₁ m₂ = -1

m² + $\frac{1}{m^2}=\frac{10}{3}$

or $-\sqrt[]{3},\frac{1}{\sqrt[]{3}}$

m = $\sqrt[]{3},\frac{-1}{\sqrt[]{3}}$

$m^4-\frac{10}{3}{m}^2+1=0\Rightarrow m^2=\frac{\frac{10}{3}\pm \sqrt[]{\frac{100}{9}-4}}{2}-\frac{\frac{10}{3}\pm \frac{8}{3}}{2}=3,\frac{1}{3}$

m = $\pm \sqrt[]{3},\pm \frac{1}{\sqrt[]{3}}$

Diagonal AC:

$\left(sin\alpha +cos\alpha \right)x-\left(cos\alpha -sin\alpha \right)y$

=10 cos2α - 10cos2α = 0

Slope of AC = $\frac{sin\alpha +cos\alpha }{cos\alpha -sin\alpha }=\frac{tan\alpha +1}{1-tan\alpha }=tan\left(\alpha +\frac{\pi }{4}\right)\alpha =30°$

FIGURE

? = $72\left(\frac{1}{16}+\frac{9}{16}\right)+100-30+13=\frac{720}{16}+83=128$