Maths

Since $\left(\mathrm{3,3}\right)$ lies on $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

$\frac{9}{a^2}-\frac{9}{b^2}=1$

Now, normal at $\left(\mathrm{3,3}\right)$ is $y-3=-\frac{a^2}{b^2}(x-3)$ ,

which passes through $\left(\mathrm{9,0}\right)\Rightarrow b^2=2a^2$

So,  ${\mathrm{e}}^2=1+\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}=3$

Also,  ${\mathrm{a}}^2=\frac{9}{2}$

(From (i) and (ii)

Thus,  $\left({\mathrm{a}}^2,{\mathrm{e}}^2\right)=\left(\frac{9}{2},3\right)$

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Thestandarddeviationsisgreaterthanorequaltothemeandeviationtakenfromthe\\ arithmeticmean.\\ Hence,thevalueofthefillerisgreaterthanorequal.\end{array}$

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Themeandeviationofthedataisleastwhenmeasuredfromthemedian.\\ Hence,thevalueofthefillerisleast.\end{array}$

$\left(\frac{\mathrm{K}-2}{\mathrm{h}-1}\right)\left(\frac{1-2}{3-1}\right)=-1\Rightarrow \mathrm{K}=2\mathrm{h}$

$?[?\mathrm{A}\mathrm{B}\mathrm{C}]=5\sqrt[]{5}$

$\Rightarrow \frac{1}{2}\left(\sqrt[]{5}\right)\sqrt[]{(\mathrm{h}-1{)}^2+(\mathrm{K}-2{)}^2}=5\sqrt[]{5}$

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Thesumofthesquaresofthedeviationsofthevalueofis\text{\hspace{0.17em}minimum}whentaken\\ abouttheirarithmeticmean.\\ Hence,thevalueofthefilleris\text{\hspace{0.17em}minimum}\end{array}$$\begin{array}{l}\end{array}$$\begin{array}{l}.\end{array}$

$\mathrm{u}=\frac{2\mathrm{z}+\mathrm{i}}{\mathrm{z}-\mathrm{k}\mathrm{i}}$

$=\frac{2x^2+(2y+1)(y-k)}{x^2+(y-k{)}^2}+i\frac{\left(x\right(2y+1)-2x(y-k\left)\right)}{x^2+(y-k{)}^2}$

Since $\mathrm{R}\mathrm{e}?\left(u\right)+\mathrm{I}\mathrm{m}?\left(u\right)=1$

$\Rightarrow 2{\mathrm{x}}^2+(2\mathrm{y}+1)(\mathrm{y}-\mathrm{k})+\mathrm{x}(2\mathrm{y}+1)-2\mathrm{x}(\mathrm{y}-\mathrm{k})={\mathrm{x}}^2+(\mathrm{y}-\mathrm{k}{)}^2$

$\begin{array}{c}\mathrm{P}\left(0,{\mathrm{y}}_1\right)\\ \mathrm{Q}\left(0,{\mathrm{y}}_2\right)\end{array}?\Rightarrow {\mathrm{y}}^2+\mathrm{y}-\mathrm{k}-{\mathrm{k}}^2=0?\begin{array}{c}{\mathrm{y}}_1+{\mathrm{y}}_2=-1\\ {\mathrm{y}}_1{\mathrm{y}}_2=-\mathrm{k}-{\mathrm{k}}^2\end{array}$

$?\mathrm{P}\mathrm{Q}=5$

$\Rightarrow \left|{\mathrm{y}}_1-{\mathrm{y}}_2\right|=5\Rightarrow {\mathrm{k}}^2+\mathrm{k}-6=0$

$\Rightarrow \mathrm{k}=-\mathrm{3,2}$

So, $k=2(k>0)$

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l}\text{\hspace{0.17em}Since}the\text{\hspace{0.17em}standard}deviationofanydataisindependentofanychangeinoriginbutis\\ dependentofanychangeofscale.\\ Hence,thevalueofthefillerareindependentanddependent.\end{array}$

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$WeknowthatSD=$=√variance=$\begin{array}{l}\sqrt[]{121}=11\end{array}$

$\begin{array}{l}\\ Hence,thevalueofthefilleris11.\end{array}$

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l}If\overline{x}isthemeanofnobservationsofx,then{\displaystyle \sum _{i=1}^n{x}_i-\overline{x}=0}andif\text{'}a\text{'}hasthevalueother\\ than\overline{x},then{\displaystyle \sum _{i=1}^n{\left(x_i-\overline{x}\right)}^{2}islessthanthan{\displaystyle \sum _{}^{}{\left(x_i-a\right)}^2.}}\\ Hence,thevalueofthefillerare0andless.\end{array}$

Let TV (r) denotes truth value of a statement $r$ .

$\text{Now, if}TV\left(p\right)=TV\left(q\right)=T$

$\Rightarrow \mathrm{T}\mathrm{V}\left({\mathrm{S}}_1\right)=\mathrm{F}$

Also, if $TV\left(p\right)=T$ and $TV\left(q\right)=F$

$\Rightarrow \mathrm{T}\mathrm{V}\left({\mathrm{S}}_2\right)=\mathrm{T}$