Maths

$f\left(x\right)=\frac{4x^3-3x^2}{6}-2sinx+\left(2x-1\right)cosx$      

$f\text{'}\left(x\right)=2x^2-x-2cosx+2cosx-\left(2x-1\right)sinx$     

= $\left(2x-1\right)\left(x-sinx\right)\ge 0forx\ge 0$  

  $f\text{'}\left(x\right)\ge 0\forall x\ge 1/2$          

$\therefore f\left(x\right)$ is increasing in $\left[\frac{1}{2},\infty \right)$

$h=\frac{a\left(1+t^2\right)}{2}.......(i)$

k = at        ……… (ii)

From (i) & (ii) $\frac{2h}{a}=1+\frac{k^2}{a^2}$  

$\therefore$ required locus of Q is y² = 2a (x – a/2)

Equation of directrix x – a/2 = -a/2-> x = 0

Let total number of throws = n

Probability of getting 2 times = Probability of getting an even number 3 times.

[as probability of getting odd number = probability of getting even number = $\frac{1}{2}$ ]

Probability of getting an odd number for odd number of times =

I (6)    F (8)

Case I       2            4

Case II      3            6

Case III     4            8

Total =   

  ${\displaystyle \underset{-a}{\overset{a}{\int }}\left(\left|x\right|+\left|x-2\right|\right)dx=22,a>2}$

${\displaystyle \underset{-a}{\overset{0}{\int }}\left(-2x+2\right)dx+{\displaystyle \underset{0}{\overset{2}{\int }}\left(x-x+2\right)dx+{\displaystyle \underset{2}{\overset{a}{\int }}\left(2x-2\right)dx=22}}}$

$\Rightarrow 2a^2+2=20\Rightarrow a^2=9\Rightarrow a=3$

$\therefore {\displaystyle \underset{3}{\overset{-3}{\int }}\left(x+\left[x\right]\right)dx=-{\displaystyle \underset{-3}{\overset{3}{\int }}\left(2x-\left\{x\right\}\right)dx=}}-{\displaystyle \underset{-3}{\overset{3}{\int }}2xdx+6}{{\displaystyle \underset{0}{\overset{1}{\int }}xdx=6.\frac{x^2}{2}}|}_0^1=3$           

$P=\left[\begin{array}{ccc}\hfill 3\hfill & \hfill -1\hfill & \hfill -2\hfill \\ \hfill 2\hfill & \hfill 0\hfill & \hfill \alpha \hfill \\ \hfill 3\hfill & \hfill -5\hfill & \hfill 0\hfill \end{array}\right]andQ=\left[q_{ij}\right]\Rightarrow PQ=k{l}_3$           

$q_{23}=-\frac{k}{8}and\left|Q\right|=\frac{k^2}{2}$            

  $PQ=k{l}_3\Rightarrow P^{-1}=\frac{Q}{k}={\left(\begin{array}{ccc}\hfill 3\hfill & \hfill -1\hfill & \hfill -2\hfill \\ \hfill 2\hfill & \hfill 0\hfill & \hfill \alpha \hfill \\ \hfill 3\hfill & \hfill -5\hfill & \hfill 0\hfill \end{array}\right)}^{-1}$

$\left|p\right|\left|Q\right|=\left(k{l}_3\right)\Rightarrow 8.\frac{k^2}{2}=k^3$

$k\ne 0\Rightarrow k=4$

$\therefore {\alpha }^2+k^2=1+16=17.$         

$\overrightarrow{c}=\alpha \overrightarrow{a}+\beta \overrightarrow{b}.....\left(i\right)$

$\overrightarrow{a}.\overrightarrow{c}=7\overrightarrow{b}.\overrightarrow{c}=0$           

$\overrightarrow{a}=-\widehat{i}+\widehat{j}+\widehat{k}\Rightarrow \left|\overrightarrow{a}\right|=\sqrt[]{3}$          

$\overrightarrow{b}=2\widehat{i}+\widehat{k}\Rightarrow \left|\overrightarrow{b}\right|=\sqrt[]{5}\overrightarrow{a}.\overrightarrow{b}=-2+1=-1$           

From   $\left(i\right)\overrightarrow{a}.\overrightarrow{c}=\alpha {\left|\overrightarrow{a}\right|}^2-\beta$

$3\alpha -\beta =7..........\left(ii\right)$        

$\overline{b}.\overline{c}=\alpha \overline{b}.\overline{a}+\beta {\left|\overrightarrow{b}\right|}^2\Rightarrow -\alpha +5\beta =0.......\left(iii\right)$     

Solving $\alpha =\frac{5}{2}and\beta =\frac{1}{2}$  

$\Rightarrow 2{\left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right|}^2=75$       

Let P (B₁) = a      P (B₂) = b            P (B₃) = c

Given a (1 – b) (1 – c) = a . (i)

b (1 – a) (1 – c) = b              . (ii)

c (1 – b) (1 – a) = $\gamma$             . (iii)

(1 – a) (1 – b) (1 – c) = p     . (iv)

$\left(\alpha -2\beta \right)p=\alpha \beta$  

->a – ab – 2b + 2ab = ab Þ a = 2b . (v)

Again $\left(\beta -3\gamma \right)p=2\beta \gamma$  

->b – bc – 3c + 3bc = 2bc Þ b = 3c   . (vi)

$\Rightarrow \frac{P\left(B_1\right)}{P\left(B_3\right)}=\frac{a}{c}=\frac{2b}{b/3}=6$      

$x^2+y^2-2x-6y+6=0$  centre (1, 3)

$\begin{array}{l}r=\sqrt[]{1+9-6}=2\\ CM=\sqrt[]{1+4}=\sqrt[]{5}\end{array}$

$r=\sqrt[]{5+4}=3$

Sum of elements $A\cap \left(B\cup C\right)=274*400$  

In set B numbers of the form 9k + 2 are {101, 109, .992}

$\therefore sum=\frac{100}{2}\left(101+992\right)=\frac{100*1093}{2}......\left(i\right)$          

Another possible number is 9k + 5 forms are {104, .995}

$\therefore sum=\frac{100}{2}\left(104+995\right)=\frac{100}{2}*1099.........\left(ii\right)$  

$\therefore Total=\frac{100}{2}*\left[1093+1099\right]=100*1096=274*4*100=274*400$           

$\therefore$  possible value of $\mathcal{l}$ = 5