Class 12th

$n\left(n+1\right)x^2+2\left(2n+1\right)x+4$

$=n\left(n+1\right)x^2+\left\{\left(2n+2\right)+2n\right\}x+4$

${\displaystyle \sum _{n=1}^9\frac{x}{\left(nx+2\right)\left(\left(n+1\right)x+2\right)}}$

$={\displaystyle \sum _{n=1}^9\left[\frac{1}{nx+2}-\frac{1}{\left(n+1\right)x+2}\right]}$

$\left[\left(\frac{1}{x+2}-\frac{1}{2x+2}\right)+\left(\frac{1}{2x+2}-\frac{1}{3x+2}\right)+....+\left(\frac{1}{9x+2}-\frac{1}{10x+2}\right)\right]$

$\therefore \underset{x\to 2}{Lt}\frac{9x}{\left(10x+2\right)\left(x+2\right)}=\frac{9}{44}$              

${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left[\frac{1+si{n}^{2}x}{1+{\pi }^{sinx}}+\frac{1+si{n}^{2}x}{1+{\pi }^{-sinx}}\right]dx}$

$={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(1+si{n}^{2}x\right)dx=\frac{\pi }{2}+\frac{1}{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(1-cos2x\right)dx}}$

$=\frac{\pi }{2}+\frac{1}{2}{\left[x-\frac{sin2x}{2}\right]}_0^{\frac{\pi }{2}}$
$=\frac{\pi }{2}+\frac{\pi }{4}=\frac{3\pi }{4}$

FeCl₃. 3H₂O i.e. [Fe (H₂O)₃], K₃ [Fe (CN]₆ and [Co (NH₃)₆]Cl₃ the last two complex are inner-orbital complex due to presence of strong ligand.

$\therefore {\Delta }_{0}of{K}_3\left[Fe{\left(CN\right)}_6\right]>{\Delta }_{0}of\left[Co{\left(N{H}_3\right)}_6\right]C{l}_3$

Because CN^- is strong ligand

$\therefore {\Delta }_0\propto \frac{1}{\lambda }$

More ${\Delta }_0$ smaller value of absorbed $\lambda$

$K_3\left[Fe{\left(CN\right)}_6\right]\to F{e}^{+3}=4s^{0}3d^{5}4p^0$

$\therefore u.e=1$

$\therefore \mu =\sqrt[]{1*\left(1+2\right)}B.M=\sqrt[]{3}B.M$

= 1.732 B.M

$\approx$ 2

$A^2=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)$

$A^3=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)$

$A^{2025}-A^{2020}$

$=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 2024\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)-\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 2019\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 5\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)=A^6-A$

Number of replaceable H is 2.

Average the highest marks

$\frac{85+80+75}{3}=\frac{240}{3}=80$

Therefore, the correct answer is option (d)

Percentage Increase $\frac{55-46}{46}*100=20\mathrm{\%}$

The difference is 9. Therefore the correct answer is option (a)

$Average=\frac{52+66+64+75+58}{5}=\frac{315}{5}=63$

Please find the answer below