Class 12th

$i_{battery}=\frac{\left(20-5\right)}{200}=\frac{15}{200}A$

$i_{300\Omega }=\frac{5}{300}A$

$i_{zener}=\frac{15}{200}-\frac{5}{300}$  

= 58.33 mA

$i=\frac{7}{3.5k+0.9k\Omega }=\frac{7}{4.4k}$

$V_0=i*700\Omega =\frac{7}{4.4k}*7k=\frac{9.9}{4.4}=1.1V$

Kinetic energy: Potential energy = 1 : –2

R₁ = { (1, 1) (1, 2), (1, 3)., (1, 20), (2, 2), (2, 4). (2, 20), (3, 3), (3, 6), . (3, 18),
(4, 4), (4, 8), . (4, 20), (5, 5), (5, 10), (5, 15), (5, 20), (6, 6), (6, 12), (6, 18), (7. 7),
(7, 14), (8, 8), (8, 16), (9, 9), (9, 18), (10, 10), (10, 20), (11, 11), (12, 12), . (20, 20)}

n (R₁) = 66

R₂ = {a is integral multiple of b}

So n (R₁ – R₂) = 66 – 20 = 46

as R₁ Ç R₂ = { (a, a) : a Î s} = { (1, 1), (2, 2), ., (20, 20)}

RHL $\underset{x\to 0^{+}}{lim}\frac{co{s}^{-1}\left(1-x^2\right)si{n}^{-1}\left(1-x\right)}{x-x^3}$

$\underset{x\to 0^{+}}{lim}\frac{\pi }{2}\cdot \frac{co{s}^{-1}\left(1-x^2\right)}{x}$

$\frac{\pi }{2}\underset{x\to 0^{+}}{lim}\frac{-1}{\sqrt[]{{\left(1-\left(1-x^2\right)\right)}^2}}\left(-2x\right)$

$=\frac{\pi }{2}\underset{x\to 2^{+}}{lim}\frac{2x}{\sqrt[]{2x^2-x^4}}=\pi \underset{x\to 0^{+}}{lim}\frac{x}{x\sqrt[]{2-x^2}}$

$=\frac{\pi }{\sqrt[]{2}}$

LHL $\underset{x\to 0^{+}}{lim}\frac{co{s}^{-1}\left(1-{\left(1+x\right)}^2\right)si{n}^{-1}\left(1-\left(1+x\right)\right)}{1\cdot \left(1-{\left(1+x\right)}^2\right)}$

$=\underset{x\to 0^{-}}{lim}\frac{co{s}^{-1}\left(-x^2-2x\right),si{n}^{-1}\left(-x\right)}{-x^2-2x}$            

$=\frac{\pi }{2}\underset{x\to 0^{-}}{lim}\frac{-si{n}^{-1}x}{-x\left(x+2\right)}=\frac{\pi }{2}*\frac{1}{2}=\frac{\pi }{2}$            

$gof\left(x\right)=\{\begin{array}{cc}\hfill f\left(x\right),\hfill & \hfill f\left(x\right)<0\hfill \\ \hfill f\left(x\right),\hfill & \hfill f\left(x\right)>0\hfill \end{array}$

$=\{\begin{array}{lll}{e}^{lnx}=x\hfill & \left(0,1\right)\hfill & e^{-x}\hfill \\ \left(-\infty ,0\right)\hfill & lnx\hfill & \left(1,\infty \right)\hfill \end{array}$

Therefore, gof (x) is many one and into

P (2W and 2B) = P (2B, 6W) * P (2W and 2B)

+ P (3B, 5W) * P (2W and 2B)

+ P (4B, 4W) * P (2W and 2B)

+ P (5B, 3W) * P (2W and 2B)

+ P (6B, 2W) * P (2W and 2B)

(15 + 30 + 36 + 30 + 15)

$=\frac{36}{126}$

$=\frac{18}{63}$

$=\frac{6}{21}$

$=\frac{2}{7}$

5f(x) + 4f $\left(\frac{1}{x}\right)$  = x² – 4           .(1)

Replace x by  $\frac{1}{x}$

5f  $\left(\frac{1}{x}\right)$  + 4f(x) = $\frac{1}{x^2}$  – 4   .(2)

5 * equation (1) – 4 * equation (2)

$9f\left(x\right)=5x^{2-}\frac{4}{x^2}-4$            

$y=9f\left(x\right)\cdot x^2=\frac{5x^4-4-4x^2}{x^2}{x}^2$            

y = 5x⁴ – 4 – 4x²

y = 20x³ – 8x > 0

4x(5x² – 2) > 0

    $x\in \left(-\sqrt[]{\frac{2}{5}},0\right)\cup \left(\sqrt[]{\frac{2}{5}},\infty \right)$

(t + 1)dx = (2x + (t + 1)³)dt

$\frac{dx}{dt}-\frac{2x}{t+1}={\left(t+1\right)}^2$

I.F. $=e^{{\displaystyle \int -\frac{2}{t+1}dt}}=\frac{1}{{\left(t+1\right)}^2}$  

Solution is

$\frac{x}{{\left(t+1\right)}^2}={\displaystyle \int 1dt}$  

x = (t + c) (t + 1)²

$?$ x (0) = 2 then c = 2

x = (t + 2) (t + 1)²

 x (1) = 12