Maths

Let zc be the centre of the circle

$\frac{z_c-2}{z_c+2}=i\Rightarrow z_c=2i$

Radius of the circle =  $\sqrt[]{4+4}=2\sqrt[]{2}$

Square off minimum distance AB =  ${\left(7\sqrt[]{2}\right)}^2=98$

${\left(\frac{x}{4}-\frac{12}{x^2}\right)}^{12}$

According to question, 12 – 3r = 0 r = 4

$\Rightarrow \frac{3^6}{4^4}*55=\frac{3^6}{4^4}*k\Rightarrow k=55$

$?\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+1}{6}=r\left(say\right)$

Let P (1 + 2r, 2 + 3r, 1 + 6r) lies on the plane 2x – y + z = 6

$\therefore r=1\Rightarrow P\left(3,5,5\right)$

Distance between P and Q is PQ =  $\sqrt[]{16+36+9}=\sqrt[]{61}$

$\therefore P{Q}^2=61$

VOWELS

vowels – 2, constants – 4

all the consonants never come together = 6! – 3! 4! =720 – 144 = 576

Since centres C₁ (1, 1) and C₂ (9, 1) lies opposite sides of the line3x + 4y = α

$((3+4-\alpha )((27+4-\alpha )<0\Rightarrow \alpha \in \left(7,31\right)..........\left(i\right)$

Also length of perpendicular from centre of the circle is greater than radius of the circle.

$\frac{\left|3+4-\alpha \right|}{5}\ge 1\Rightarrow \alpha \le 2or\alpha \ge 12............\left(ii\right)$

and $\frac{\left|27+4-\alpha \right|}{5}\ge 2\Rightarrow \alpha \le 21or\alpha \ge 41...........\left(iii\right)$

from (i), (ii) and (iii) we get $$ $\in$ [12, 21]

$\therefore$ sum of all integers = 165 

$l={\displaystyle \underset{-1/2}{\overset{1}{\int }}\left[2x\right]dx+}{\displaystyle \underset{-1/2}{\overset{1}{\int }}\left|2x\right|dx}$

let  $l_1={\displaystyle \underset{-1/2}{\overset{1}{\int }}\left[2x\right]dxput2x=t\Rightarrow dx=\frac{dt}{2}}$

$\therefore l=l_1+l_2=0+\frac{5}{8}=\frac{5}{8}\Rightarrow 8l=5$

$f\left(x\right)=x^2+ax+1\Rightarrow f\text{'}\left(x\right)=2x+a$

for increasing   $f\text{'}\left(x\right)\ge 0$

$\therefore 2x+a\ge 0\Rightarrow a\ge -2x\Rightarrow a\ge -2*2\Rightarrow a\ge -4\therefore R=-4$

And for decreasing   $f\text{'}\left(x\right)\le 0\Rightarrow a\le -2x\Rightarrow a\le -2*1\therefore S=-2$

|R – S| = 2

$\Delta =\left|\begin{array}{ccc}\hfill 2\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill a\hfill \end{array}\right|=2\left(-a-1\right)+\left(1-a\right)+2=-3a+1$

${\Delta }_3=\left|\begin{array}{ccc}\hfill 2\hfill & \hfill 1\hfill & \hfill 5\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 3\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill b\hfill \end{array}\right|=2\left(-b-1\right)+\left(3-b\right)+5*2=-3b+7$

For a = $\frac{1}{3},b\ne \frac{7}{3},$ $\frac{}{}\frac{}{}$ system has no solution

$tan\theta =\frac{3x}{18}=\frac{x}{6}........\left(i\right)$ and

$tan2\theta =\frac{10x}{8}=\frac{5x}{9}..........\left(ii\right)$

Solving (i) and (ii)  $\frac{2tan\theta }{1-ta{n}^2\theta }=\frac{5x}{9}wegetx=\sqrt[]{\frac{72}{5}}$  

Height of pole = 10x =  $12\sqrt[]{10}$

q = 18° Þ 2q + 3q = 90° Þ sin 3q = 1 – sin 2q Þ cos q Þ cosq (4 sin² q + 2sinq - 1) = 0

$?cos\theta \ne 0\therefore 4si{n}^2\theta +2sin\theta -1=0\Rightarrow cose{c}^{2}18°-2cosec18°-4=0$

? x² – 2x – 4 = 0