Maths

$f\left(0^{-}\right)=f\left(0\right)=-a$

$f\left(0^{+}\right)=\underset{x\to 0^{+}}{lim}\frac{sin2x\left(\frac{1}{cos2x}-1\right)}{b{x}^3}$ $=\frac{4}{b}$

ab = 4 10 – ab = 14

$S=21+{\displaystyle \sum _{k=1}^{21}{\left(z^k+\frac{1}{z^k}\right)}^3}=\sum {\left(z^k+{\overline{z}}^k\right)}^3=8\sum {\left(Re\left(z^k\right)\right)}^3$

$\left(As\left|z\right|=1\Rightarrow \frac{1}{z}=\overline{z}\right)$

$S=2{\displaystyle \sum _{k=1}^{21}cosk\pi +6}{\displaystyle \sum _{k=1}^{21}cos\frac{k\pi }{3}+21}$

= $2\left(-1\right)+6Re\left(\alpha +{\alpha }^2+{\alpha }^3\right)+21$

Where $\alpha =e^{i\left(2\pi /6\right)}$

$=2\left(-1\right)+6\left(\frac{1}{2}-\frac{1}{2}-1\right)+21$

$=-8+21=13$

$x^2+y^2+{\left(x-1\right)}^2+y^2+x^2+{\left(y-1\right)}^2+{\left(x-1\right)}^2+{\left(y-1\right)}^2=18$

$\Rightarrow x^2+y^2-x-y-\frac{7}{2}=0$

$r^2=\frac{1}{4}+\frac{1}{4}+\frac{7}{2}=4$

(x + y) = 4xy

$\Rightarrow \frac{dy}{dx}=\frac{4xy+4y^2-1}{1-4x^2-4xy},$

$\frac{d^{2}y}{d{x}^2}=\frac{\left(1-4x^2-4xy\right)\left(4y+4xy\text{'}+8yy\text{'}\right)-\left(4xy+4y^2-1\right)\left(-8x-4y-4xy\text{'}\right)}{{\left(1-4x^2-4xy\right)}^2}$

$B\left(1+2\lambda ,3+\lambda ,4+2\lambda \right)$

x – 2y – z = 3 => $\lambda =-6\Rightarrow B\left(-11,-3,-8\right)$

$N\left(1+t,3-2t,4-t\right)$  

$x-2y-z=3\Rightarrow t=2\Rightarrow N\left(3,-1,2\right)$  

Line NB : $\frac{x-3}{7}=\frac{y+1}{1}=\frac{3-2}{5}$

$d^2=P{M}^2=\frac{{\left|P\overrightarrow{Q}*\overrightarrow{R}\right|}^2}{{\left|\overrightarrow{R}\right|}^2}$

${\left|\overrightarrow{R}\right|}^2=75$

$d^2=\frac{1950}{75}=26$  

$LHS={\displaystyle \sum _{r=1}^{15}r\left(r!\right)}$

$=\sum \left(r+1-1\right)r!={\displaystyle \sum _{r=1}^{15}\left(r+1\right)!-r!}$

$=\left(2!-1!\right)+\left(3!-2!\right)+....+\left(16!-15!\right)$

$\begin{array}{l}=16!-1\\ {=}^{16}{P}_{16}-1\end{array}$

Required area

$={\displaystyle \underset{-2}{\overset{4}{\int }}\left(\frac{3x}{2}+6-\frac{3x^2}{4}\right)dx=27}$

$\frac{3x^2}{4}=\frac{3x}{2}+6\Rightarrow x=-2,4$

Equation : 2x² – (6 + k)x + 4 + 3k = 0

D < 0 => 6  $-4\sqrt[]{2}<k<6+4\sqrt[]{2}$

Sum of integral values of K is 1+2+3+.+11 = 66

Side of square = a

Radius of circle = r

Given : 4a + 2pr = 36

S = a² + pr²

^{$=\left(\frac{{\pi }^2}{4}+\pi \right)r^2-9\pi r+81$}

^{$\frac{ds}{dr}=0\Rightarrow r=\frac{18}{\pi +4}$}