Class 12th

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

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

$If{\widehat{n}}_1$ is a vector normal to the plane determined by $\widehat{i}and\widehat{i}+\widehat{j}then{\widehat{n}}_1=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \end{array}\right|=\widehat{k}$

$If{\widehat{n}}_2$ is a vector normal to the pane determined by $\widehat{i}-\widehat{j},$ and $\widehat{i}+\widehat{k}$ then ${\widehat{n}}_2$ = $\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right|=\widehat{i}-\widehat{j}+\widehat{k}$

Vector $\widehat{a}$ is parallel to ${\widehat{n}}_1*{\widehat{n}}_2$ i.e. $\widehat{a}$ is parallel to $\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill -1\hfill & \hfill 1\hfill \end{array}\right|=\widehat{i}-\widehat{j}$

Given $\overrightarrow{b}=\widehat{i}-2\widehat{j}+2\widehat{k}$

consine of acute angle between $\widehat{a}and\widehat{b}$ = $\left|\frac{\widehat{a}.\widehat{b}}{\left|\widehat{a}\right|.\left|\widehat{b}\right|}\right|=\frac{1}{\sqrt[]{2}}$

obtuse angle between $\widehat{a}and\widehat{b}=\frac{3\pi }{4}$

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

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

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

Given hyperbola : $\frac{k{x}^2}{6}-\frac{y^2}{6}=1$ so eccentricity e = $\sqrt[]{1+k}$ and directrices $x=\pm \frac{a}{e}$

$\Rightarrow x=\pm \frac{\sqrt[]{6}}{\sqrt[]{k}\sqrt[]{k+1}}\Rightarrow \frac{\sqrt[]{6}}{\sqrt[]{k}\sqrt[]{k+1}}=1$

k = 2 therefore equation of hyperbola is $\frac{x^2}{3}-\frac{y^2}{6}=1$

hence it passes through the point $\left(\sqrt[]{5},-2\right)$

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

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 $b\in [-6,-2)\cup (2,6]$

Abscissae of PQ are roots of x² – 4x – 6 = 0

Ordinates of PQ are roots of y² + 2y – 7 = 0 and PQ is diameter

$\therefore$ Equation of circle is $x^2+y^2-4x+2y-13=0$   ……………. (i)

But, given  $x^2+y^2+2ax+2by+c=0$ ……………. (ii)

By comparison a = -2, b = 1, c = -13 Þ a + b – c = -2 + 1 + 13 = 12

 $f\left(x\right)=\{\begin{array}{cc}\hfill x+a,\hfill & \hfill x\le 0\hfill \\ \hfill \left|x-4\right|,\hfill & \hfill x>0\hfill \end{array}andg\left(x\right)=\{\begin{array}{cc}\hfill x+1,\hfill & \hfill x<0\hfill \\ \hfill {\left(x-4\right)}^2+b,\hfill & \hfill x\ge 0\hfill \end{array}$

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

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

a =   $\frac{qE}{m}=\frac{40*10^{?6}*10^5}{100*10^{?3}*10^{?3}}$

                      = 40 * 10³ m/sec²

              v² = u² + 2a s

$?0={\left(200\right)}^2?2*40*10^{3}s$

$s=\frac{200*200}{2*40*10^3}=0.5m$                

Equation of tangent of slope m to y = x² is y = mx $\frac{1}{4}{m}^2$ - …………. (i)

Equation of tangent of slope m to y = - (x - 2)² is y = m (x – 2) + $\frac{1}{4}{m}^2$  …………… (ii)

If both equation represent the same line therefore on comparing (i) and (ii) we get m = 0, 4

therefore equation of tangent is y = 4x – 4

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

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