Maths

$x_1=\underset{x\to \infty }{lim}\frac{2^{\frac{x^n}{e^x}}-3^{\frac{x^n}{e^x}}}{\frac{x^n}{e^x}},put\frac{x^n}{e^x}=t$

$x_2=\underset{x\to \infty }{lim}\frac{co{t}^{-1}\left(\sqrt[]{x+1}-\sqrt[]{x}\right)}{se{c}^{-1}\left({\left(\frac{2x+1}{x-1}\right)}^x\right)}$

$x_2=\frac{2}{\pi }\underset{x\to \infty }{lim}ta{n}^{-1}\left(\sqrt[]{x+1}+\sqrt[]{x}\right)$

$x_2=\frac{2}{\pi }.\frac{\pi }{2}=1$

$g\left(f\left(x\right)\right)=x\Rightarrow g\text{'}\left(f\left(x\right)\right).f\text{'}\left(x\right)=1$

$f\left(x\right)=1\Rightarrow x=0$

$g\text{'}\left(1\right).f\text{'}\left(0\right)=1$

$f\text{'}\left(x\right)=2x+e^{-x}$

$\begin{array}{l}f\text{'}\left(0\right)=1\\ g\text{'}\left(1\right)=1\end{array}$

$\left|\begin{array}{ccc}\hfill 1-\alpha \hfill & \hfill 6\hfill & \hfill 6\hfill \\ \hfill 4\hfill & \hfill -1-\alpha \hfill & \hfill 4\hfill \\ \hfill 2\alpha \hfill & \hfill 2\alpha \hfill & \hfill \alpha -5\hfill \end{array}\right|=0$

$\alpha =-5$

$\frac{P\left(A\cap B\right)}{P\left(B\right)}=\frac{1}{4},\frac{P\left(A\cap B\right)}{P\left(A\right)}=\frac{1}{12}$

divide both $\frac{P\left(B\right)}{P\left(A\right)}=\frac{1}{3}$

$P\left(A\right)=\frac{1}{11},P\left(B\right)=\frac{1}{33}$

$\frac{dy}{dx}=3a{x}^2-2bx$

${\frac{dy}{dx}|}_{x=1}=3a-2b=3$

$a=\frac{2b+3}{3}\in \left[1,2\right]$

$2b+3\in \left[3,6\right]$

$2b\in \left[0,3\right]$

$b\in \left[0,\frac{3}{2}\right]$

$\frac{3z_1+i}{4}=\frac{2z_2+2z_3}{4}$

P = point of intersection of AD & BC

$AD=1\Rightarrow DP=\frac{3}{4}$

$BP=\sqrt[]{1-\frac{9}{16}}=\frac{\sqrt[]{7}}{4}\Rightarrow BC=\frac{\sqrt[]{7}}{2}$

area of Quad. ABCD = $\frac{1}{2}.AD*BC$

= $\frac{1}{2}*1*\frac{\sqrt[]{7}}{2}=\frac{\sqrt[]{7}}{4}$

${\displaystyle \int e^{sinx}}\left(\frac{5+co{s}^{2}x}{\left(2-sinx\right)\sqrt[]{3+co{s}^{2}x}}\right)cosxdx$

put sin x = t

${\displaystyle {}^{}}\frac{{}^{}}{\text{exception 25:}}$  ${\displaystyle \int e^t}\left(\frac{6-t^2}{\left(2-t\right)\sqrt[]{4-t^2}}\right)$ dt

${\displaystyle \int e^t}\left(\sqrt[]{\frac{2+t}{2-t}}+\frac{2}{{\left(2-t\right)}^{3/2}{\left(2+t\right)}^{1/2}}\right)dt$

If $g\left(t\right)=\sqrt[]{\frac{2+t}{2-t}},g\text{'}\left(t\right)=\frac{2}{{\left(2-t\right)}^{3/2}{\left(2+t^2\right)}^{1/2}}$ $\text{exception 25:}\frac{}{{}^{}{{}^{}}^{}}$

$e^t\sqrt[]{\frac{2+t}{2-t}}+c$

$f\left(\frac{\pi }{2}\right)=\sqrt[]{3}e$

Line L is   $\frac{x-1}{1}=\frac{y-2}{2\sqrt[]{2}}=\frac{z}{0}$

this line L makes an angle of 45° with the plane  $\sqrt[]{2}x+y-z=1$

$\therefore$ Required distance PQ is perpendicular distance of plane from P is i.e.,

$PQ=\frac{\left|\sqrt[]{2}+7-\sqrt[]{2}-1\right|}{\sqrt[]{2+1+1}}=3$

$\left|A-xI\right|=\left|\begin{array}{cc}\hfill 1/2-x\hfill & \hfill 1/2\hfill \\ \hfill a\hfill & \hfill b-x\hfill \end{array}\right|=x^2-\left(b+\frac{1}{2}\right)x+\frac{b-a}{2}$

from Cayley – Hamilton Theorem

$A^2-\left(b+\frac{1}{2}\right)A+\left(\frac{b-a}{2}\right)I=0$

$\Rightarrow A^3=\left({\left(b+\frac{1}{2}\right)}^2-\left(\frac{b-a}{2}\right)\right)A-\left(b+\frac{1}{2}\right)\left(\frac{b-a}{2}\right)I$

$\therefore {\left(b+\frac{1}{2}\right)}^2-\left(\frac{b-a}{2}\right)=1\&\left(b+\frac{1}{2}\right)\left(\frac{b-a}{2}\right)=0$

$\therefore \left(a,b\right)=\left(\frac{1}{2},\frac{1}{2}\right),\left(-\frac{3}{2},-\frac{3}{2}\right),\left(\frac{3}{2},-\frac{1}{2}\right)$

A = {1, 2, 3, 4}. As (4, 4) $\notin$  R so not reflexive

Also not transitive