Maths

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

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

i.e. $PM\cdot PN=16$ PM⋅PN=16

$\Delta APC,AC=AP=12$  

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

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

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

$\left(a-1\right)x+0y+z=\alpha$        ……(1)

$x+\left(b-1\right)y+0z=\beta$                       ……(2)

$0x+y+\left(c-1\right)z=\gamma$                       ……(3)

$\left|\begin{array}{ccc}\hfill \left(a-1\right)\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill \left(b-1\right)\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill \left(c-1\right)\hfill \end{array}\right|=0$           For no unique solution D = 0

$\left(a-1\right)\left(b-1\right)\left(c-1\right)+1=0$            

$\therefore a=2;b=2;c=0$            

Hence, |a + b + c| = 4

  $A\left(a\right)=2{\displaystyle \underset{0}{\overset{\sqrt[]{1-a}}{\int }}\left(\left(1-x^2\right)-a\right)dx=\frac{4}{3}{\left(1-a\right)}^{3/2}}$  

  $\therefore A\left(0\right)=\frac{4}{3}$            

and    ^{$A\left(\frac{1}{2}\right)=\frac{4}{3}{\left(\frac{1}{2}\right)}^{\frac{3}{2}}\Rightarrow \frac{A\left(0\right)}{A\left(\frac{1}{2}\right)}=2\sqrt[]{2}$}

$S_1=1+2+3\dots \dots \dots .n$

$S_2=1+3+5\dots \dots \dots .n$

$S_3=1+4+7\dots \dots \dots .$

$\begin{array}{ll}\left(S_1+S_3\right)=\lambda S_2& S_1=\frac{n}{2}[2a+(n-1\left)1\right]\\ \frac{n}{2}[2a+(n-1\left)1\right]+\frac{n}{2}[2a+(n-1\left)3\right]& S_2=\frac{n}{2}[2a+(n-1\left)2\right]\\ =\lambda \frac{n}{2}[2a+(n-1\left)2\right]& S_3=\frac{n}{2}[2a+(n-1\left)3\right]\end{array}$

$\lambda =2$

$\sim (\sim p\vee \sim q)\to (\sim q\vee r)\Rightarrow p\wedge q\to \sim q\vee r$

$\mathrm{}\mathrm{T}\mathrm{F}\mathrm{T}\Rightarrow \mathrm{F}\to \mathrm{T}\to \mathrm{T}$

$\mathrm{F}\mathrm{T}\mathrm{T}\Rightarrow \mathrm{F}\to \mathrm{T}\to \mathrm{T}$

$\mathrm{T}\mathrm{T}\mathrm{T}\Rightarrow \mathrm{T}\to \mathrm{T}\to \mathrm{T}$

$\mathrm{T}\mathrm{T}\mathrm{F}\Rightarrow \mathrm{T}\to \mathrm{F}\to \mathrm{F}$

$\left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}+\overrightarrow{d}\right|=\sum {\left|\overrightarrow{a}\right|}^2+2\sum \overrightarrow{a}.\overrightarrow{b}$  

$=4+2\overrightarrow{d}.\left(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right)$  (a, b, c are mutually ^ r)

Let   $\overrightarrow{d}=\lambda \overrightarrow{a}+\mu \overrightarrow{b}+v\overrightarrow{c}$

Also   ${\lambda }^2+{\mu }^2+v^2=1=3co{s}^2\theta orcos\theta =\frac{1}{\sqrt[]{3}}$

$\therefore {\left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}+\overrightarrow{d}\right|}^2=4\pm 2.\frac{3}{\sqrt[]{3}}$        

$=4\pm 2\frac{3}{\sqrt[]{3}}$       

${\mathrm{l}\mathrm{i}\mathrm{m}}_{x\to 9^{-}}?\left. [x^2+1\right]+\left. [x^2-1\right]+\left\{x\right\}$

$\Rightarrow \mathrm{}{\mathrm{l}\mathrm{i}\mathrm{m}}_{x\to 9^{-}}?2\left. [x^2\right]+\left\{x\right\}\Rightarrow {\mathrm{l}\mathrm{i}\mathrm{m}}_{h\to 0}?2\left. [(9-x{)}^2\right]+\{9-h\}=160+1=161$

Let $z=x+yi\mathrm{}z-2\bar{z}=1\Rightarrow x=-1\mathrm{}y=0$