Maths

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

$\begin{array}{l}Let{x}_i,i=1,2,3,4,\dots n_1\\ and{y}_j,j=1,2,3,4,\dots n_2\\ \therefore \overline{x_1}=\frac{1}{n_1}{\displaystyle \sum _{i=1}^n{x}_i}and\overline{x_2}=\frac{1}{n_2}{\displaystyle \sum _{i=1}^n{y}_j}\\ \Rightarrow {\sigma }_1^2=\frac{1}{n_1}{\displaystyle \sum _{i=1}^{n_1}{\left(x_i-\overline{x_1}\right)}^2}and{\sigma }_2^2=\frac{1}{n_2}{\displaystyle \sum _{j=1}^{n_2}{\left(y_j-\overline{x_2}\right)}^2}\\ Nowmeanofthecombinedseriesisgivenby\\ \overline{x}=\frac{1}{n_1+n_2}\left[{\displaystyle \sum _{i=1}^n{x}_i}+{\displaystyle \sum _{i=1}^n{y}_j}\right]=\frac{n_1\overline{x_1}+n_2\overline{x_2}}{n_1+n_2}\\ Therefore,{\sigma }^{2}ofthecombinedseriesis\\ {\sigma }^2=\frac{1}{n_1+n_2}\left[{\displaystyle \sum _{i=1}^{n_1}{\left(x_i-\overline{x}\right)}^2}+{\displaystyle \sum _{j=1}^{n_2}{\left(y_j-\overline{x}\right)}^2}\right]\\ Now,{\displaystyle \sum _{i=1}^{n_1}{\left(x_i-\overline{x}\right)}^2}={\displaystyle \sum _{i=1}^{n_1}{\left(x_i-\overline{x_j}+\overline{x_j}-\overline{x}\right)}^2}\\ ={\displaystyle \sum _{i=1}^{n_1}{\left(x_i-\overline{x_j}\right)}^2}+n_1{\left(\overline{x_j}-\overline{x}\right)}^2+2\left(\overline{x_j}-\overline{x}\right){\displaystyle \sum _{i=1}^{n_1}{\left(x_i-\overline{x_j}\right)}^2}\\ But{\displaystyle \sum _{i=1}^n\left(x_i-\overline{x_i}\right)=0}\left[\begin{array}{l}?Thealgebraicsumofthedeviationofvaluesoffirst\\ seriesfromtheirmeaniszero.\end{array}\right]\\ Also{\displaystyle \sum _{i=1}^{n_1}{\left(x_i-\overline{x}\right)}^2}=n_1{s}_1^2+n_1{\left(\overline{x_1}-\overline{x}\right)}^2\\ =n_1{s}_1^2+n_1{d}_1^2\\ where{d}_1=\left(\overline{x_1}-\overline{x}\right)\\ Similarly,wehave\\ {\displaystyle \sum _{j=1}^{n_2}{\left(y_j-\overline{x}\right)}^2}={\displaystyle \sum _{j=1}^{n_2}{\left(y_j-\overline{x_i}+\overline{x_i}-\overline{x}\right)}^2}=n_2{s}_2^2+n_2{d}_2^{2}where{d}_2=\left(\overline{x_2}-\overline{x}\right)\\ NowcombinedStandardDeviation\left(SD\right)\\ \sigma =\sqrt[]{\frac{n_1\left(s_1^2+d_1^2\right)+n_2\left(s_2^2+d_2^2\right)}{n_1+n_2}}\\ where{d}_1=\overline{x_1}-\overline{x}=\overline{x_1}-\left(\frac{n_1\overline{x_1}+n_2\overline{x_2}}{n_1+n_2}\right)=\frac{n_2\left(\overline{x_1}-\overline{x_2}\right)}{n_1+n_2}\\ d_2=\overline{x_2}-\overline{x}=\overline{x_2}-(\frac{n_1\overline{x_1}+}{}\end{array}$

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

$\begin{array}{l}Giventhat{n}_1=25,\overline{x_1}=18.2and{\sigma }_1=3.25\\ and{n}_2=15,{\displaystyle \sum }_{i=1}{x}_i=279and{\displaystyle \sum }_{i=1}{x}_i^2=5524\\ Forthefirstset,wehave\\ {\displaystyle \sum _{}^{}{x}_1}=25*18.2=455\\ \therefore {\sigma }_1^2=\frac{{\displaystyle \sum _{}^{}{x}_i^2}}{25}-{\left(18.2\right)}^2\\ \Rightarrow {\left(3.25\right)}^2=\frac{{\displaystyle \sum _{}^{}{x}_i^2}}{25}-331.24\Rightarrow 10.5625+331.24=\frac{{\displaystyle \sum _{}^{}{x}_i^2}}{25}\\ \Rightarrow {\displaystyle \sum _{}^{}{x}_i^2}=25*\left(10.5625+331.24\right)\\ =25*341.8025=8545.06\\ Forthecombinedstandarddeviationofthe40observation,n=40\\ and{\displaystyle \sum _{}^{}{x}_i^2}=5524+8545.06=14069.06\\ \Rightarrow {\displaystyle \sum _{}^{}{x}_i}=455+279=734\\ \therefore SD=\sqrt[]{\frac{14069.06}{40}-{\left(\frac{734}{40}\right)}^2}=\sqrt[]{351.7265-{\left(18.35\right)}^2}\\ =\sqrt[]{351.7265-336.7225}=\sqrt[]{15.004}=3.87\\ Hence,therequiredSD=3.87\end{array}$

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

$\begin{array}{l}{\displaystyle \sum _{}^{}{x}_i}=1+2+3+4+\dots +n=\frac{n\left(n+1\right)}{2}\\ {\displaystyle \sum _{}^{}{x}_i^2}=1^2+2^2+3^2+4^2+\dots +n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\\ \therefore S.D.\left(\sigma \right)=\sqrt[]{\frac{{\displaystyle \sum _{}^{}{x}_i^2}}{n}-{\left(\frac{{\displaystyle \sum _{}^{}{x}_i}}{n}\right)}^2}\\ =\sqrt[]{\frac{n\left(n+1\right)\left(2n+1\right)}{6n}-\frac{n^2{\left(n+1\right)}^2}{4n^2}}\\ =\sqrt[]{\frac{\left(n+1\right)\left(2n+1\right)}{6}-\frac{{\left(n+1\right)}^2}{4}}\\ =\sqrt[]{\frac{2n^2+3n+1}{6}-\frac{n^2+2n+1}{4}}\\ =\sqrt[]{\frac{4n^2+6n+2-3n^2-6n-3}{12}}=\sqrt[]{\frac{n^2-1}{12}}\\ Hence,therequiredSD=\sqrt[]{\frac{n^2-1}{12}}.\end{array}$

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

$\begin{array}{l}Firstnnaturalnumbersare1,2,3,4,5,6,\dots ,n.Here,niseven.\\ \therefore Mean\overline{x}=\frac{1+2+3+4+\dots +n}{n}=\frac{n\left(n+1\right)}{2n}=\frac{n+1}{2}\\ \therefore MD=\frac{1}{n}\left[\begin{array}{l}\left|1-\frac{n+1}{2}\right|+\left|2-\frac{n+1}{2}\right|+\left|3-\frac{n+1}{2}\right|+\dots +\left|\frac{n-2}{2}-\frac{n+1}{2}\right|+\left|\frac{n}{2}-\frac{n+1}{2}\right|+\\ \left|\frac{n+2}{2}-\frac{n+1}{2}\right|+\dots +\left|n-\frac{n+1}{2}\right|\end{array}\right]\\ =\frac{1}{n}\left[\left|\frac{1-n}{2}\right|+\left|\frac{3-n}{2}\right|+\left|\frac{5-n}{2}\right|+\dots +\left|\frac{-3}{2}\right|+\left|-\frac{1}{2}\right|+\left|\frac{1}{2}\right|+\dots +\left|\frac{n-1}{2}\right|\right]\\ =\frac{1}{n}\left[\frac{1}{2}+\frac{3}{2}+\dots +\frac{n-1}{2}\right]\left(\frac{n}{2}\right)terms\\ =\frac{1}{n}{\left(\frac{n}{2}\right)}^2=\frac{1}{n}.\frac{n^2}{4}=\frac{n}{4}\left[?Sumoffirstoddnnaturalnumbers=n^2\right]\\ Hence,therequiredMD=\frac{n}{4}.\end{array}$

$\mathrm{t}\mathrm{a}\mathrm{n}?\theta =\frac{10}{\mathrm{x}}=\frac{\mathrm{h}}{{\mathrm{x}}_2}\Rightarrow {\mathrm{x}}_2=\frac{\mathrm{h}\mathrm{x}}{10}$

$\mathrm{t}\mathrm{a}\mathrm{n}??=\frac{15}{\mathrm{x}}=\frac{\mathrm{h}}{\mathrm{x}}\Rightarrow {\mathrm{x}}_1=\frac{\mathrm{h}\mathrm{x}}{15}$

Now,  $x_1+x_2=x=\frac{hx}{15}+\frac{hx}{10}$

$\Rightarrow 1=\frac{\mathrm{h}}{10}+\frac{\mathrm{h}}{15}\Rightarrow \mathrm{h}=6$

$(a+\sqrt[]{2}b\mathrm{c}\mathrm{o}\mathrm{s}?x)(a-\sqrt[]{2}b\mathrm{c}\mathrm{o}\mathrm{s}?y)=a^2-b^2$

$\Rightarrow a^2-\sqrt[]{2}ab\mathrm{c}\mathrm{o}\mathrm{s}?y+\sqrt[]{2}ab\mathrm{c}\mathrm{o}\mathrm{s}?x-2b^2\mathrm{c}\mathrm{o}\mathrm{s}?x\mathrm{c}\mathrm{o}\mathrm{s}?y=a^2-b^2$

Differentiating both sides:

$0-\sqrt[]{2}ab\left(-\mathrm{s}\mathrm{i}\mathrm{n}?y\frac{dy}{dx}\right)+\sqrt[]{2}ab(-\mathrm{s}\mathrm{i}\mathrm{n}?x)-2b^2\left[\mathrm{c}\mathrm{o}\mathrm{s}?x\left(-\mathrm{s}\mathrm{i}\mathrm{n}?y\frac{dy}{dx}\right)+\mathrm{c}\mathrm{o}\mathrm{s}?y(-\mathrm{s}\mathrm{i}\mathrm{n}?x)\right]=0$