Class 12th

Plane through 4ax – y + 5z – 7a = 0 = 2x – 5y – z – 3

is    $x\left(4a+2\lambda \right)+y\left(-1-5\lambda \right)+z\left(5-\lambda \right)=7a+3\lambda$

This plane contains 4, -1, 0

->9a + 1 + 10l = 0          …… (i)

Plane contains the line $\frac{x-4}{1}=\frac{y+1}{-2}=\frac{z}{1}$  

-> $4a+11\lambda +7=0$ ……. (ii)

From (i) & (ii) a = 1,   $\lambda$  =-1

Equation of plane $\pi \equiv x+2y+3z-2=0$  

$\Rightarrow 7P+3-2P+4-12P+9-2=0\Rightarrow P=2$

use $si{n}^{-1}x=co{s}^{-1}\sqrt[]{1-x^2}$  

$ta{n}^{-1}\sqrt[]{1-x^2}=co{t}^{-1}\frac{1}{\sqrt[]{1-x^2}}$               

$si{n}^{-1}\frac{x}{\sqrt[]{1-x^2}}=co{s}^{-1}\frac{\sqrt[]{1-2x^2}}{\sqrt[]{1-x^2}}$               

Sum of roots ->b = -1 + 2   $\frac{\left(k^2-1\right)}{k^2-2}$

Product of roots -> -5 = -2  $\left(\frac{k^2-1}{k^2-2}\right)$  

b = 4, k² = $\frac{1}{3}$  

The inductor generates induced electromotive force (EMF) and opposes the changes in the current flow. Due to this the current lag the voltage by 90 degrees.

  $f\left(x\right)\{\begin{array}{l}{\displaystyle \underset{0}{\overset{3}{\int }}\left(2-t\right)dt+{\displaystyle \underset{3}{\overset{x}{\int }}\left(8-t\right)dt;x>4}}\\ x^2+bx;x\le 4\end{array}$

f(x) is continuous at x = 4

$16+4b={\displaystyle \underset{0}{\overset{3}{\int }}\left(2-t\right)dt+{\displaystyle \underset{3}{\overset{4}{\int }}\left(8-t\right)dt}}$           

16 + 4b = 15

$\Rightarrow b=\frac{-1}{4}$               

$forx\le 4,$ $\begin{array}{l}f\left(x\right)=x^2-\frac{x}{4}\\ f\text{'}\left(x\right)=2x-\frac{1}{4}\end{array}$                 

f(x) is increasing in  $\left(\frac{1}{8},\infty \right)$  

 rate change at $x=\frac{1}{8},$  from -ve to +ve. So minima occurs.

${\pi }_1\equiv 2x+ky-5z=1$  

${\pi }_2\equiv 3kx-ky+z=5$                

Given  ${\pi }_1\perp {\pi }_2\Rightarrow 6k-k^2-5=0$  

k = 1, 5

Now required plane is p₁ +   $\lambda {\pi }_2=0$

y intercept = $\frac{1+5\lambda }{1-\lambda }$  

$k_{\alpha }=2eV\Rightarrow P_{\alpha }=\sqrt[]{2K_{\alpha }{m}_{\alpha }}$

k_P = eV P_P = $\sqrt[]{2K_P{m}_P}$

$\therefore \frac{P_{\alpha }}{P_P}=\sqrt[]{\frac{K_{\alpha }{m}_{\alpha }}{K_P{m}_P}}=\sqrt[]{\frac{2*\left(4m_P\right)}{m_P}}\begin{array}{cc}\hfill \sqrt[]{8}:1\hfill & \hfill 2\sqrt[]{2}:1\hfill \end{array}$

$I=I_{0}co{s}^{2}30°$

$=I_0{\left(\frac{\sqrt[]{3}}{2}\right)}^2=\frac{3}{4}{I}_0$ 

Find total (5 digit) number of divisors of 7. Find total (5 digit) number of divisors of 35

v = 100 sin $\omega t$

$i_0=\frac{100}{R_x}=5\Rightarrow R_x=20\Omega$

i = $5sin\left(\omega t-\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)$

i = 100 5sin $\omega t$

$5=\frac{100}{R_y}\Rightarrow R_y=\frac{100}{5}=20\Omega$

When x and y both are connected in series :-

v = 100sin $\omega t$

tan = 1 = 45°

R₀ = $\sqrt[]{R_x^2+R_y^2}=20\sqrt[]{2}\Omega$

$\therefore \stackrel{o}{{\mathcal{l}}_0}=\frac{v_0}{R_0}=\frac{100}{20\sqrt[]{2}}=\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$\sqrt[]{2}$}\right.A.$

$\therefore {\mathcal{l}}_{ms}=\frac{{\mathcal{l}}_0}{\sqrt[]{2}}$

$=\frac{5}{\sqrt[]{2}\cdot \sqrt[]{2}}$

$=\frac{5}{2}A=\frac{5}{2}A$