Maths

In the integral J, substitute x + 1 = t

$\Rightarrow dx=dtand{x}^2+2x=\left(t^2-1\right)$            

Now   $J={\displaystyle \underset{1}{\overset{e}{\int }}\frac{e^{\frac{t^2-2}{2}}}{t}}dtandK={\displaystyle \underset{1}{\overset{e}{\int }}t}lnt{e}^{\frac{t^2-2}{2}}dt$

Hence   $\left(J+K\right)={\displaystyle \underset{1}{\overset{e}{\int }}{e}^{\frac{t^2-2}{2}}}\left(\frac{1}{t}+tlnt\right)dt$

$={\left(e^{\frac{t^2-2}{2}}lnt\right)}_{t=1}^{t=e}=e^{\frac{e^2-2}{2}}={\left(\sqrt[]{e}\right)}^{e^2-2}$

TSA  of cube $=6a^2$

$\frac{d}{dt}\left(6a^2\right)=4.8;12a\frac{da}{dt}=4.8;a\frac{da}{dt}=0.4;\frac{dv}{dt}=\frac{d}{dt}\left(a^3\right)\Rightarrow 3a\left(a\frac{da}{dt}\right)$

$3*15*0.4=3*15*\frac{04}{10}\Rightarrow 18$

$?arg\left(z\right)=\pi ifz=x+iy$

$\therefore y=0andx<0$

$\therefore z=-3+i.0\Rightarrow \left|z\right|=3$

P(W) = $\frac{1}{3};$  P(B) =   $\frac{2}{3}$

$\Rightarrow p=\frac{1}{3};q=\frac{2}{3}$           and r = 4 or 5 and n = 5

Use   $P\left(r\right){=}^n{C}_r{p}^r{q}^{n-r}$

 P(4) + P(5)

${=}^5{C}_4{\left(\frac{1}{3}\right)}^4{\left(\frac{2}{3}\right)}^1{+}^5{C}_5{\left(\frac{1}{3}\right)}^5$            

$=\frac{10}{3^5}+\frac{1}{3^5}=\frac{11}{3^5}=\frac{11}{243}$  

${\mathrm{l}\mathrm{o}\mathrm{g}}_{(2x+3)}?x^2<{\mathrm{l}\mathrm{o}\mathrm{g}}_{(2x+3)}?(2x+3)$

Case-I: $0<2x+3<1$  

$x^2>2x+3$

Case-II: $2x+3>1$

$0<x^2<2x+3$  

By solving (1) & (2) $\begin{array}{rr}& \frac{-3}{2}<x<-1\&(x-3)(x+1)>0.\\ & \frac{-3}{2}<x<-1,x>3\end{array}$

Equation (4) has no solution $\frac{-3}{2}<x<-1,x<-1$

.(5) By equation (5) . $x\in \left(\frac{-3}{2},-1\right)$

We obtain solving equation (2) $x>-1\mathrm{}x>-1\mathrm{}\Rightarrow x\in (-\mathrm{1,3})$

or $(x-3)(x+1)<0\mathrm{}-1<x<3,x^2>0\Rightarrow x\ne 0$

$x\in \left(\frac{-3}{2},-1\right)\cup (-\mathrm{1,3})-\left\{0\right\}$

Note that D^r < 0, hence given inequality (1) is true only if $N^r\ge 0$  

  i.e.   $\left(x-8\right)\left(x-2\right)\le 0and{2}^{x-3}>31$

i.e.    $2\le x\le 8and{2}^{x-3}\ge 31$

only x = 8 satisfies both the inequality.

$\boxed{\begin{array}{lll}\text{12.(D)}& {\int }_0^{\frac{\pi }{6}}??\frac{4{\mathrm{s}\mathrm{i}\mathrm{n}}^2?\theta \frac{2}{5}*2\mathrm{c}\mathrm{o}\mathrm{s}?\theta d\theta }{\left(4-4{\mathrm{s}\mathrm{i}\mathrm{n}}^2?\theta \right)}=\frac{16}{5*8}\int \frac{{\mathrm{s}\mathrm{i}\mathrm{n}}^2?\theta \mathrm{c}\mathrm{o}\mathrm{s}?d\theta }{{\mathrm{c}\mathrm{o}\mathrm{s}}^3?\theta }& x^{5/2}=2\mathrm{s}\mathrm{i}\mathrm{n}?\theta \\ & & \frac{5}{2}{x}^{3/2}dx=2\mathrm{c}\mathrm{o}\mathrm{s}?\theta d\theta \\ & & =\frac{2}{5}{\int }_0^{\frac{\pi }{6}}??{\mathrm{t}\mathrm{a}\mathrm{n}}^2?\theta d\theta =\frac{2}{5}{\int }_0^{\frac{\pi }{6}}??\left({\mathrm{s}\mathrm{e}\mathrm{c}}^2?\theta -1\right)d\theta =\frac{2}{5}[\mathrm{t}\mathrm{a}\mathrm{n}?\theta -\theta {]}_0^{\pi /6}=\frac{2}{5}\left(\frac{1}{\sqrt[]{3}}-\frac{\pi }{6}\right)\end{array}}$

Clearly PM. PN $={\mathrm{O}\mathrm{P}}^2$ =OP2

i.e. $PM\cdot PN=16$ PM⋅PN=16

$\Delta APC,AC=AP=12$  

$\Delta ABP,tan60°=\frac{12}{AB}$           

$AB=\frac{12}{\sqrt[]{3}}=4\sqrt[]{3}$            

$Area=4\sqrt[]{3}.4\sqrt[]{6}=48\sqrt[]{2}$