Class 12th

(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.

Except electron gain enthalpy all the physical properties like sublimation enthalpy, ionization enthalpy and hydration energy affected the value of reduction potential.

Ans. = 3

Clearly, from the graph it could be either Dr. Sharma or Dr. Goyal.

For Dr. Sharma

$=\frac{913.63-750.3}{913.63}*100=24.48$

For Dr. Goyal $=\frac{962.58}{720.93}*100=25.16$

Hence, Dr. Goyal's turnover is maximum.

Electronic configuration of

Gd is ; Gd₆₄ = [Xe]4f⁷ 5d¹ 6s²

Hence $G{d}_{64}^{++}=\left[Xe\right]4f^{7}5d^{1}6s^0$

Total number of electrons in 4f sub-shell = 7

Ans. = 7