Class 12th

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

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

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

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

$=-8+21=13$

Number of total car owners of TATA = 30 + 20 + 24 + 42 + 23 + 33 + 26 + 50 = 251 lakh

Now, 13% of 251 = 32.63 lakh

Those states which enjoy more than 34.45 Lacs car owners of TATA are Pune and Guwahati.

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

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

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

$d^2=\frac{1950}{75}=26$  

$E_{C{\mathcal{l}}_2/C{\mathcal{l}}^{-}}^0=+1.36V$

$E_{l_2/l^{-}}^0=+0.54V$

$E_{A{g}^{+}/Ag}^0=+0.80V$

$E_{N{a}^{+}/Na}^0=-2.71V$

$E_{L{i}^{+}/Li}^0=-3.05V$

$\therefore E_{C{\mathcal{l}}_2/C{\mathcal{l}}^{-}}^0>E_{A{g}^{+}/Ag}^0>E_{N{a}^{+}/Na}^0>E_{L{i}^{+}/Li}^0$

Required area

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

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

(Sum of the incomes in 2019-20) – (Sum of income in 2018-19)

= 4851.86 – 4139.43

= 712.43

Percentage change

$=\frac{\begin{array}{l}\text{\hspace{0.17em}sumofincomein}2004-05)-\\ \text{\hspace{0.17em}sumofincomein}2005-06\end{array}}{2004-05}*100=14.68\mathrm{\%}$

For AB₃ interhalogen compound, which is T-shaped, only two lone pair electrons are available on central atom.