Let $f\left(x\right)=\{\begin{array}{cc}\hfill 2+2x,\hfill & \hfill x\in \left(-1,0\right)\hfill \\ \hfill 1-\frac{x}{3},\hfill & \hfill x\in [0,3)\hfill \end{array}$

$g\left(x\right)=\{\begin{array}{cc}\hfill x,\hfill & \hfill x\in [0,1)\hfill \\ \hfill -x,\hfill & \hfill x\in \left(-3,0\right)\hfill \end{array}$           

The range of fog(x) is

  $\left|\frac{1}{2}-2i+1\right|=\alpha \left(\frac{1}{2}-2i\right)+\beta \left(1+i\right)$  

$\sqrt[]{\frac{9}{4}+4}=\alpha \left(\frac{1}{2}-2i\right)+\beta \left(1+i\right)$

$\frac{5}{2}=\alpha \left(\frac{1}{2}\right)+\beta +i\left(-2\alpha +\beta \right)$             

$\frac{\alpha }{2}+\beta =\frac{5}{2}$      ...(1)

 –2α + β = 0                    …(2)

Solving (1) and (2)

$\frac{\alpha }{2}+2\alpha =\frac{5}{2}$

$\frac{5}{2}\alpha =\frac{5}{2}$            

a =

Variance = $\frac{\sum x^2}{n}-{\left(\overline{x}\right)}^2$  

$\frac{60^2+60^2+44^2+58^2+68^2+{\alpha }^2+{\beta }^2+56^2}{8}={\left(58\right)}^2=66.2$            

$\frac{7200+1936+3364+4624+3136+{\alpha }^2+{\beta }^2}{8}=3364=66.2$             

$2532.5+\frac{{\alpha }^2+{\beta }^2}{8}-3364=66.2$            

α² + β² = 897.7 × 8

= 7181.6

Start with

(1) $\overline{E}:\frac{6!}{2!}=360$  

(2)    $\overline{GE}:\frac{5!}{2!},$ $\overline{GN}:\frac{5!}{2!}$  

(3) GTE : 4!, GTN: 4!, GTT : 4!

(4) GTWENTY = 1

⇒ 360 + 60 + 60 + 24 + 24 + 24 + 1 = 553

${\left(1+x\right)}^{11}={}^{11}{C}_0+{}^{11}{C}_{1}x+{}^{11}{C}_2{x}^2+.....{+}^{11}{C}_{11}{x}^{11}$

$=\frac{2^{12}-2-24}{12}$

$=\frac{2^{12}-26}{12}=\frac{4070}{12}=\frac{2035}{6}=\frac{m}{n}$

m + n = 2035 + 6 = 2041

Reflexive :for (a, b) R (a, b)

-> ab– ab = 0 is divisible by 5.

So (a, b) R (a, b) " a, b Î Z

R is reflexive

Symmetric:

For (a, b) R (c, d)

If ad – bc is divisible by 5.

Then bc – ad is also divisible by 5.

-> (c, d) R (a, b) "a, b, c, dÎZ

R is symmetric

Transitive:

If (a, b) R (c, d) ->ad –bc divisi