Maths

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

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

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 $TV\left(q\right)=F$

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

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

 $\begin{array}{l}Giventhat{\sigma }_C=5\\ CV=\frac{SD}{Mean}*100\\ Hence,thevalueofthefillerisSD.\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}MD=\frac{1}{n}{\displaystyle \sum _{i=1}^n\left|x_i-\overline{x}\right|}\\ Hence,thecorrectoptionis\left(b\right).\end{array}$

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l}Giventhatn=10,\overline{x}=45,{\sigma }^2=16\\ \therefore \overline{x}=\frac{{\displaystyle \sum _{}^{}{x}_i}}{n}\Rightarrow 45=\frac{{\displaystyle \sum _{}^{}{x}_i}}{10}\Rightarrow {\displaystyle \sum _{}^{}{x}_i}=450\\ Corrected{\displaystyle \sum _{}^{}{x}_i}=450-52+25=423\\ andCorrectedmean=\frac{423}{10}=42.3\\ Now{\sigma }^2=\frac{{\displaystyle \sum _{}^{}{x}_i^2}}{n}-{\left(\frac{{\displaystyle \sum _{}^{}{x}_i}}{n}\right)}^2\Rightarrow 16=\frac{{\displaystyle \sum _{}^{}{x}_i^2}}{10}-{\left(45\right)}^2\\ \Rightarrow 16=\frac{{\displaystyle \sum _{}^{}{x}_i^2}}{10}-2025\Rightarrow \frac{{\displaystyle \sum _{}^{}{x}_i^2}}{10}=2041\\ \Rightarrow {\displaystyle \sum _{}^{}{x}_i^2}=2041*10\Rightarrow {\displaystyle \sum _{}^{}{x}_i^2}=20410\\ \therefore Corrected{\displaystyle \sum _{}^{}{x}_i^2}=20410-{\left(52\right)}^2+{\left(25\right)}^2\\ =20410-2704+625=18331\\ andCorrectedvariance\\ {\sigma }^2=\frac{18331}{10}-{\left(42.3\right)}^2=1833.1-1789.3=43.8\\ Hence,therequiredmean=42.3andvariance=43.8\end{array}$

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l}Giventhatn=100,\overline{x}=40,\sigma =10\\ \therefore \overline{x}=\frac{{\displaystyle \sum _{}^{}{x}_i}}{n}\Rightarrow 40=\frac{{\displaystyle \sum _{}^{}{x}_i}}{100}\Rightarrow {\displaystyle \sum _{}^{}{x}_i}=4000\\ Corrected{\displaystyle \sum _{}^{}{x}_i}=4000-30-70+3+27=3930\\ andCorrectedmean=\frac{3930}{100}=39.3\\ Now{\sigma }^2=\frac{{\displaystyle \sum _{}^{}{x}_i^2}}{n}-{\left(40\right)}^2\Rightarrow 100=\frac{{\displaystyle \sum _{}^{}{x}_i^2}}{100}-1600\\ \Rightarrow {\displaystyle \sum _{}^{}{x}_i^2}=1700*100\Rightarrow {\displaystyle \sum _{}^{}{x}_i^2}=170000\\ \therefore Corrected{\displaystyle \sum _{}^{}{x}_i^2}=170000-{\left(30\right)}^2-{\left(70\right)}^2+{\left(3\right)}^2+{\left(27\right)}^2\\ =170000-900-4900+9+729=164938\\ \therefore CorrectSD=\sqrt[]{\frac{164938}{100}-{\left(39.3\right)}^2}=\sqrt[]{1649.38-1544.49}=\sqrt[]{104.89}=10.24\\ Hence,therequiredSD=10.24\end{array}$