Class 12th

$A_{3*3}$

$det\left(A\right)=42A=\left[\begin{array}{ccc}\hfill 2a_1\hfill & \hfill 2a_2\hfill & \hfill 2a_3\hfill \\ \hfill 2b_1\hfill & \hfill 2b_2\hfill & \hfill 2b_3\hfill \\ \hfill 2c_1\hfill & \hfill 2c_2\hfill & \hfill 2c_3\hfill \end{array}\right]$

For R₂ →2R₂ + 5R₃

$det\left(B\right)=\left|\begin{array}{ccc}\hfill 2a_1\hfill & \hfill 2a_2\hfill & \hfill 2a_3\hfill \\ \hfill 4b_1+10c_1\hfill & \hfill 4b_2+10c_2\hfill & \hfill 4b_3+10c_3\hfill \\ \hfill 2c_1\hfill & \hfill 2c_2\hfill & \hfill 2c_3\hfill \end{array}\right|=\left|\begin{array}{ccc}\hfill 2a_1\hfill & \hfill 2a_2\hfill & \hfill 2a_3\hfill \\ \hfill 4b_1\hfill & \hfill 4b_2\hfill & \hfill 4b_3\hfill \\ \hfill 2c_1\hfill & \hfill 2c_2\hfill & \hfill 2c_3\hfill \end{array}\right|=16det\left(A\right)=64$

A tristor is formed by doping a semi conductor water from one side by N-type dopant (high concentration while comparatively lower concentration on other side thus forming NPN tristor & vice-versa.

So, statement (I) is false & statement (II) is true.

When a soft ferromagnetic substance is placed in external magnetic field, the size of domain lying in the opposite direction of external magnetic field increases while size of domain lying in the opposite direction of field decreases if field is weak. However, if field is strong then the domain rotate in the direction of external magnetic field due to strong torque.

Zener break down occurs in p-n junction having p and n both : Heavily doped and have narrow depletion layer.

R = $2\Omega$ $$

L = 2 mH

E = 9V

$i=\frac{\epsilon }{2R}=\frac{9v}{4\Omega }=2.25A$

Just after the switch 'S' is closed, the inductor acts as open circuit.

P {a person chosen from the group has chest disorder}

$=\frac{160}{400}*0.35+\frac{100}{400}*0.20+\frac{140}{400}*0.10$

$P\left(\frac{Personissmoker\&non-vegetarian}{Personhaschestdisorder}\right)$

$=\frac{\frac{160}{400}*0.35}{\frac{160}{400}*0.35+\frac{100}{400}*0.20+\frac{140}{400}*0.10}$

$=\frac{40*0.7}{40*0.7+25*0.4+35*0.2}=\frac{23}{23+10+7}=\frac{28}{45}$

ax² + bx + c = 0

D = b² – 4ac

D = 0

b² – 4ac = 0

b² = 4ac

(i) AC = 1, b = 2 (1, 2, 1) is one way

(ii) AC = 4, b = 4

$\begin{array}{l}a=4c=1\\ a=2c=2\\ a=1c=4\end{array}\}3ways$  

(iii) AC = 9, b = 6, a = 3, c = 3 is one way

1 + 3 + 1 = 5 way

Required probability =  $\frac{5}{216}$   

Let sinθ= t ${\displaystyle \int \frac{sin\theta \left(2sin\theta .cos\theta \right)\left(si{n}^6\theta +si{n}^4\theta +si{n}^2\theta \right)\sqrt[]{2si{n}^4\theta +3si{n}^2\theta +6}}{2si{n}^2\theta }}$ dθ

sinθ = t

cos θ. dθ = dt

${\displaystyle \int \frac{u^{1/2}}{12}}du=\frac{u^{3/2}}{18}+C=\frac{{\left(2t^6+3t^4+6t^2\right)}^{3/2}}{18}+C=\frac{{\left(2si{n}^6\theta +3si{n}^4\theta +6si{n}^2\theta \right)}^{3/2}}{18}+C$

$x={\displaystyle \sum _{n=0}^{\infty }co{s}^{2n}\theta =co{s}^0\theta +co{s}^2\theta +co{s}^4\theta +.....=1+co{s}^2\theta +co{s}^4\theta +.....}$  

a = 1, r = cos² θ

$x=S_{\infty }=\frac{a}{1-r}=\frac{1}{1-co{s}^2\theta }=\frac{1}{si{n}^2\theta }$           

Similarly, y = $\frac{1}{co{s}^2\theta }\Rightarrow \frac{1}{y}=co{s}^2\theta$  

$z=\frac{xy}{xy-1}\Rightarrow xyz-z=xy...........\left(i\right)$           

Also, $\frac{1}{x}+\frac{1}{y}=1\Rightarrow x+y=xy..........\left(ii\right)$  

 (i) & (ii) ->xyz = xy + z -> (x + y) z = xy + z

Let

A : Missile hit the target

B : Missile intercepted      

P (B) =    $\frac{1}{3}P\left(A/\overline{B}\right)=\frac{3}{4}$

$P\left(\overline{B}\right)=\frac{2}{3}$  

$\Rightarrow P\left(\overline{B}\cap A\right)=\frac{3}{4}*\frac{2}{3}=\frac{1}{2}$   

Required Probability = $\frac{2}{3}*\frac{3}{4}*\frac{2}{3}*\frac{3}{4}*\frac{2}{3}*\frac{3}{4}=\frac{1}{8}$