Class 12th

$R=\frac{v_2}{P}=\frac{10000}{50}\Rightarrow R=200\Omega$

$V=\sqrt[]{V_C^2+V_R^2}$

$\Rightarrow 200=\sqrt[]{V_C^2+{\left(100\right)}^2}$

$x_c=\frac{1}{{\omega }_c}=\frac{\pi \sqrt[]{x}}{100\pi *50}=\frac{\sqrt[]{x}}{5000*10^{-6}}=\frac{\sqrt[]{x}*10^3}{5}$

$=200\sqrt[]{x}$

$V_C=i{x}_C$

$\Rightarrow \sqrt[]{3}.\sqrt[]{1+x}=2\sqrt[]{x}$

$\Rightarrow 3\left(1+x\right)=4x\Rightarrow x=3$

 ${?}_A=\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.,{?}_B=\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$3$}\right.$

$I_A=I_1+I_2+2\sqrt[]{l_1{l}_2}cos?=l+4l+2\sqrt[]{l*4l}cos\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.=5l$

$I_B=I+4I+2\sqrt[]{l.4l}cos\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$3$}\right.=5l+2l=7l$

$\Delta l=7l-5l=2l\Rightarrow x=2$

$\frac{30}{70}=\frac{S}{5.6k\Omega }\Rightarrow S=\frac{3}{7}*5.6k\Omega =3*0.8k\Omega$

= 2.4k $\Omega$ = 2400 $\Omega$

The CBSE Class 12 compartment exams for 2025 are scheduled to be held in July. Specifically, the exams are expected to take place from July 15, 2025. The practical exams are scheduled from July 10 to July 15, 2025. The admit cards for these exams are likely to be released in late June. The results for the compartment exams are expected to be announced in August. 

$?x=sin\left(2ta{n}^{-1}\alpha \right)=\frac{2\alpha }{1+{\alpha }^2}$ ……. (i)

and

$y=sin\left(\frac{1}{2}ta{n}^{-1}\frac{4}{3}\right)=sin\left(si{n}^{-1}\frac{1}{\sqrt[]{5}}\right)=\frac{1}{\sqrt[]{5}}$

Now,

$y^2=1-x$

$\therefore \alpha =2,\frac{1}{2}\therefore {\displaystyle \sum _{a\in S}^{}16{\alpha }^3=16*2^3+16*\frac{1}{2^3}=130}$

$4x^3-3x{y}^2+6x^2-5xy-8y^2+9x+14=0$ differentiating both sides we get

$12x^2-3y^2-6xyy\text{'}+12x-5y-5xy=16yy\text{'}+9=0$

At the point (2, 3)

48 – 27 + 36y' – 24 – 15 + 10y' – 48y' + 9 = 0

$\therefore Area=\frac{1}{2}*Base*Height$

$A=\frac{1}{2}*\left(\frac{-4}{3}+\frac{31}{2}\right)\left(3\right)=\frac{1}{2}\left(\frac{85}{0}\right).3=\frac{85}{4}=8A=170$

$f\left(x\right)+{\displaystyle {\int }_0^x\left(x-t\right)f\text{'}\left(t\right)dt=\left(e^{2x}+e^{-2x}\right)cos2x+\frac{2x}{a}.......(i)}$

Here f (0) = 2 ………. (ii)

On differentiating equation (i) w.r.t. x we get :

$f\text{'}\left(x\right)+f{\displaystyle {\int }_0^{x}f\text{'}\left(t\right)dt+xf\text{'}\left(x\right)-xf\text{'}\left(x\right)}$

$4=\frac{2}{a}\Rightarrow a=\frac{1}{2}$

$\therefore {\left(2a+1\right)}^5.a^2=2^5.\frac{1}{2^2}=2^3=8$

$\frac{dy}{dx}=\frac{y}{x}\frac{\left(4y^2+2x^2\right)}{\left(3y^2+x^2\right)}$

Put y = vx

$\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$

$\Rightarrow v+x\frac{dv}{dx}=\frac{v\left(4v^2+2\right)}{\left(3v^2+1\right)}$

$\mathcal{l}n\left(\frac{y^2}{8}+\frac{y}{2}\right)=2\mathcal{l}n2\Rightarrow \frac{y^3}{8}+\frac{y}{2}=4$

$\Rightarrow \left[y\left(2\right)\right]=2$

$\Rightarrow n=3$

$f\left(x\right)=\left|5x-7\right|+\left[x^2+2x\right]$

$=\left|5x-7\right|+\left[{\left(x+1\right)}^2\right]-1$

f $\left(\frac{7}{4}\right)=0+4=4$

as both |5x – 7| and x² + 2x are increasing in nature after x = 7/5

f (2) = 3 + 8 = 11

$f{\left(\frac{7}{5}\right)}_{min}-4andf{\left(2\right)}_{max}=11$

Sum is 4 + 11 = 15

$A=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]+\left[\begin{array}{ccc}\hfill 0\hfill & \hfill a\hfill & \hfill a\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]=l+B$

$B^2=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill a\hfill & \hfill a\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]*\left[\begin{array}{ccc}\hfill 0\hfill & \hfill a\hfill & \hfill a\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill ab\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$

B³ = 0

$\therefore A^n={\left(1+B\right)}^n{=}^n{C}_{0}l{+}^n{C}_{1}B{+}^n{C}_2{B}^2{+}^n{C}_3{B}^3+....$

On comparing we get na = 48, nb = 96 and na + $\frac{n\left(n-1\right)}{2}ab=2160$

$\Rightarrow a=4,n=12andb=8$

$\Rightarrow n+a+b=24$