Class 12th

The line x + y – z = 0 = x – 2y + 3z – 5 is parallel to the vector

  $\overrightarrow{b}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill -2\hfill & \hfill 3\hfill \end{array}\right|=\left(1,4,-3\right)$ Equation of line through P(1, 2, 4) and parallel to $\overrightarrow{b}\frac{x-1}{1}=\frac{y-2}{-4}=\frac{z-4}{-3}$  

Let  $N\equiv \left(\lambda +1,-4\lambda +2,-3\lambda +4\right)\overline{QN}=\left(\lambda ,-4\lambda +4,-3\lambda -1\right)$  

$\overline{QN}$ is perpendicular to $\overrightarrow{b}\Rightarrow \left(\lambda ,-4\lambda +4,-3\lambda -1\right).\left(1,4,-3\right)=0\Rightarrow \lambda =\frac{1}{2}.$  

Hence  $\overline{QN}\left(\frac{1}{2},2,\frac{-5}{2}\right)and\overline{\left|QN\right|}=\sqrt[]{\frac{21}{2}}$  

$\frac{dy}{dx}+2ytanx=sinx,\therefore I.F.e^{{\displaystyle \int 2tanxdx}}=se{c}^{2}x$  

= cos x – 2 cos² x = $-2{\left(cosx-\frac{1}{4}\right)}^2+\frac{1}{8}\therefore y_{max}=\frac{1}{8}$

$a=\underset{n\to \infty }{lim}{\displaystyle \sum _{k=1}^n\frac{2n}{n^2+k^2}=\underset{n\to \infty }{lim}\frac{1}{n}}{\displaystyle \sum _{k=1}^n\frac{2}{1+{\left(\frac{k}{n}\right)}^2}}$

 $\therefore a={\displaystyle {\int }_0^1\frac{2}{1+x^2}dx=2ta{n}^{-1}x}]{}_0^1=\frac{\pi }{2}$

⇒ $f\text{'}\left(\frac{a}{2}\right)=\sqrt[]{2}f\left(\frac{a}{2}\right)$

$f\left(x\right)=\{\begin{array}{cc}\hfill x^3-x^2+10x-7,\hfill & \hfill x\le 1\hfill \\ \hfill -2x+lo{g}_2\left(b^2-4\right),\hfill & \hfill x>1\hfill \end{array}$  

If f(x) has maximum value at x = 1 then

$f\left(1\right)\le f\left(1\right)\Rightarrow -2+lo{g}_2\left(b^2-4\right)\le 1-1+10-7$

$\Rightarrow lo{g}_2\left(b^2-4\right)\le 5\Rightarrow 0<b^2-4\le 32$

$\Rightarrow b^2-4>0\Rightarrow b\in \left(-\infty ,-2\right)\cup \left(2,\infty \right)$         …….(i)

$And{b}^2-4\le 32\Rightarrow b\in \left[-6,6\right]$                      …….(ii)

From (i) and (ii) we get  $b\in [-6,-2)\cup (2,6]$  

$Q=\Delta \upsilon +w\Rightarrow nC\Delta T=n{C}_v\Delta T=\frac{nC\Delta T}{4}=\frac{3}{4}nC\Delta T=n{C}_v\Delta T$

$C=\frac{4}{3}{C}_v=\frac{4}{3}*\frac{3}{2}R=2R$

$f\left(x\right)=\{\begin{array}{cc}\hfill x+a,\hfill & \hfill x?0\hfill \\ \hfill \left|x?4\right|,\hfill & \hfill x>0\hfill \end{array}\text{?}\text{?}\text{?}\text{?}and\text{?}\text{?}\text{?}g\left(x\right)=\{\begin{array}{cc}\hfill x+1,\hfill & \hfill x<0\hfill \\ \hfill {\left(x?4\right)}^2+b,\hfill & \hfill x?0\hfill \end{array}$

$?$    f (x) and g (x) are continuous on R $?$  a = 4 and b = 1 – 16 = 15

then (gof) (2) + (fog) (-2) = g (2) + f (-1) = -11 + 3 = -8

given N = 1000 turns

A = 1m²

$\omega =1rev/sec$

$=1*2\pi rad/sec$

$V=\frac{-d{?}_B}{dt}=\frac{-d}{dt}\left(NBAcos\omega t\right)$  

=  $140*\frac{22}{7}$  

= 20 * 22

= 440 volts

$f\left(x\right)=\{\begin{array}{l}\frac{lo{g}_e\left(1-x+x^2\right)+lo{g}_e\left(1+x+x^2\right)}{secx-cosx},x\in \left(\frac{-\pi }{2},\frac{\pi }{2}\right)-\left\{0\right\}\\ k,x=0\end{array}$ for continuity at x = 0

      $\underset{x\to 0}{lim}f\left(x\right)=k\therefore k=\underset{x\to 0}{lim}\frac{lo{g}_e\left(1+x^2+x^4\right)}{secx-cosx}\left(\frac{0}{0}form\right)=\underset{x\to 0}{lim}\frac{cosxlo{g}_e\left(1+x^2+x^4\right)}{si{n}^{2}x}=1$  

$d=4\sqrt[]{3}$ cm (Lateral shift)

By Snell's law

${\mu }_{air}sin60°={\mu }_{g}sin\theta$

θ = 30

$\Rightarrow 1*\frac{\sqrt[]{3}}{2}=\sqrt[]{3}sin\theta$  

$sin\theta =\frac{1}{2}$  

t = 12 cm

Since a is a odd natural number then $\left|{\displaystyle \underset{1}{\overset{3}{\int }}{y}^{a}dy}\right|=\frac{364}{3}\Rightarrow \left|{\left(\frac{y^{a+1}}{a+1}\right)}_1^3\right|=\frac{364}{3}\Rightarrow \frac{3^{a+1}}{a+1}=\frac{364}{3}$  

 Þ a = 5