Maths

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

${\left|2\overrightarrow{a}+3\overrightarrow{b}\right|}^2={\left|3\overrightarrow{a}+\overrightarrow{b}\right|}^2\Rightarrow 4{\left|\overrightarrow{a}\right|}^2+9{\left|\overrightarrow{b}\right|}^2+12\overrightarrow{a}.\overrightarrow{b}=9{\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2+6\overrightarrow{a}.\overrightarrow{b}$

$\Rightarrow -5{\left|\overrightarrow{a}\right|}^2+8{\left|\overrightarrow{b}\right|}^2+6\overrightarrow{a}.\overrightarrow{b}=0.........\left(i\right)$

$cos60°=\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}=\frac{1}{2}\therefore \overrightarrow{a}.\overrightarrow{b}=4\left|\overrightarrow{b}\right|and\left|\overrightarrow{a}\right|=8$

from (i) we get $\left|\overrightarrow{b}\right|=5$

Applying Leibniz theorem,  

$\sqrt[]{1-{\left(f\text{'}\left(x\right)\right)}^2}=f\left(x\right)\Rightarrow \frac{f\text{'}\left(x\right)}{\sqrt[]{1-{\left(f\left(x\right)\right)}^2}}=1$ on integrating both sides, we get

$f\left(x\right)=sinx+C$ put x = 0 and f (0) = 0 we get C = 0

$Now\underset{x\to 0}{lim}\frac{{\displaystyle \underset{0}{\overset{x}{\int }}f\left(t\right)dt}}{x^2},\left(\frac{0}{0}\right)$ by L' Hospital rule $\underset{x\to 0}{lim}\frac{f\left(x\right)}{2x}=\underset{x\to 0}{lim}\frac{sinx}{2x}=\frac{1}{2}$

$\frac{dy}{dx}=\frac{2^{x+y}-2^x}{2^y}\Rightarrow {\displaystyle \int \frac{2^{y}dy}{2^y-1}={\displaystyle \int 2^{x}dx\Rightarrow put{2}^y-1=t\Rightarrow 2^{y}ln2dy=dt}}$

$\frac{1}{ln2}{\displaystyle \int \frac{dt}{t}={\displaystyle \int 2^{x}dx\Rightarrow \frac{1}{ln2}lnt=\frac{2^x}{ln2}+\frac{C}{ln2}}}$    put x = 0 and y = 1 we get C = -1

$\therefore ln\left(2^y-1\right)=2^x-1$    put x = 1 and we get y = log₂ (1 + e)