Let the mean and the variance of 20 observations x1, x2,…….x20 be 15 and 9, respectively. For $\in R$ , if the mean of ${\left(x_1+\alpha \right)}^2,{\left(x_2+\alpha \right)}^2,....,{\left(x_{20}+\alpha \right)}^2$ is 178, then the square of the maximum value of is equal to………….

  $\frac{x_1+x_2+x_3+x_4}{4}=\frac{7}{2}$

$x_1+x_2+x_3+x_4=14$

and $\frac{x_1+x_2+x_3+x_4+x_5}{5}=\frac{24}{5}$

x₅ = 10

Variance $\frac{{\displaystyle \sum _{i=1}^4{x}_i^2}}{4}-{\left(\frac{\Sigma xi}{4}\right)}^2=a$

$\frac{x_1^2+x_2^2+x_3^2+x_4^2}{4}-\frac{49}{4}=a$

$x_1^2+x_2^2+x_3^2+x_4^2=4a+49$

and $\frac{x_1^2+x_2^2+x_5^2+x_4^2+x_5^2}{5}-{\left(\frac{24}{5}\right)}^2=\frac{194}{25}$

$\frac{4a+49+x_5^2}{5}=\frac{576+194}{25}$

49 + 49 + $x_5^2=\frac{770}{5}$

49 $+x_5^2+49=154$

4a + 149 = 154

4a = 5

now 4a + x₅ = 15

 $\overline{x}=\frac{\sum xi}{40}=30\Rightarrow \sum xi=1200$ ………. (i)

${\alpha }^2=\frac{1}{40}\sum x{i}_i^2-{\left(30\right)}^2=25$

$\Rightarrow \sum x_i^2=37000$

after omitting two wrong observations

$\sum y_i^2=37000-144-100=36756$

$\therefore 38a^2=36756-36158=238$

$\bar{x}=10$

$\Rightarrow \stackrel{?}{\mathrm{x}}=\frac{63+\mathrm{a}+\mathrm{b}}{8}=10$

$\Rightarrow a+b=17$

Since, variance is independent of origin.

So, we subtract 10 from each observation.

So,  ${\sigma }^2=13.5=\frac{79+(a-10{)}^2+(b-10{)}^2}{8}$

$\Rightarrow {\mathrm{a}}^2+{\mathrm{b}}^2-20(\mathrm{a}+\mathrm{b})=-171$

$\Rightarrow a^2+b^2=169$

From (1) and (2) ; $a=12$ and $b=5$

 Since,  ${\mathrm{l}\mathrm{i}\mathrm{m}}_{x\to 0}?\frac{f\left(x\right)}{x}$ exist $\Rightarrow f\left(0\right)=0$

Now,  $f^{\mathrm{\text{'}}}\left(x\right)={\mathrm{l}\mathrm{i}\mathrm{m}}_{h\to 0}?\frac{f(x+h)-f\left(x\right)}{h}={\mathrm{l}\mathrm{i}\mathrm{m}}_{h\to 0}?\frac{f\left(h\right)+x{h}^2+x^{2}h}{h}($ take $y=h)$

$={\mathrm{l}\mathrm{i}\mathrm{m}}_{h\to 0}?\frac{f\left(h\right)}{h}+{\mathrm{l}\mathrm{i}\mathrm{m}}_{h\to .0}?\left(xh\right)+x^2$

$\Rightarrow {\mathrm{f}}^{\mathrm{\text{'}}}\left(\mathrm{x}\right)=1+0+{\mathrm{x}}^2$

$\Rightarrow f^{\mathrm{\text{'}}}\left(3\right)=10$

 $=\frac{4+5+6+6+7+8+x+y}{8}=6$

$\therefore x+y=12$ …. (i)

And variance

$=\frac{2^2+1^2+0^2+0^2+1^2+2^2+{\left(x-6\right)}^2+{\left(y-6\right)}^2}{8}$

$=\frac{9}{4}$

$\therefore {\left(x-6\right)}^2+{\left(y-6\right)}^2=8$ ……. (ii)

From (i) and (ii)

x = 4 and y = 8

$\therefore x^4+y^2=320$