Maths

$A^2=\left(\begin{array}{ll}\mathrm{c}\mathrm{o}\mathrm{s}?2\theta & i\mathrm{s}\mathrm{i}\mathrm{n}?2\theta \\ i\mathrm{s}\mathrm{i}\mathrm{n}?2\theta & \mathrm{c}\mathrm{o}\mathrm{s}?2\theta \end{array}\right)$

Similarly, $A^5=\left(\begin{array}{cc}\mathrm{c}\mathrm{o}\mathrm{s}?5\theta & i\mathrm{s}\mathrm{i}\mathrm{n}?5\theta \\ i\mathrm{s}\mathrm{i}\mathrm{n}?5\theta & \mathrm{c}\mathrm{o}\mathrm{s}?5\theta \end{array}\right)=\left(\begin{array}{ll}a& b\\ c& d\end{array}\right)$

(1) $a^2+b^2={\mathrm{c}\mathrm{o}\mathrm{s}}^2?5\theta -{\mathrm{s}\mathrm{i}\mathrm{n}}^2?5\theta =\mathrm{c}\mathrm{o}\mathrm{s}?10\theta =\mathrm{c}\mathrm{o}\mathrm{s}?{75}^{?}$

(2) $a^2-d^2={\mathrm{c}\mathrm{o}\mathrm{s}}^2?5\theta -{\mathrm{c}\mathrm{o}\mathrm{s}}^2?5\theta =0$

(3) $a^2-b^2={\mathrm{c}\mathrm{o}\mathrm{s}}^2?5\theta +{\mathrm{s}\mathrm{i}\mathrm{n}}^2?5\theta =1$

(4) $a^2-c^2={\mathrm{c}\mathrm{o}\mathrm{s}}^2?5\theta +{\mathrm{s}\mathrm{i}\mathrm{n}}^2?5\theta =1$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Givenobservationarea,b,c,dande\\ \therefore Mean=m=\frac{a+b+c+d+e}{5}\\ \therefore {\displaystyle \sum _{}^{}{x}_i}=5m\\ Nowmeanofa+K,b+K,c+K,d+Kande+Kis\\ =\frac{a+K+b+K+c+K+d+K+e+K}{5}\\ =\frac{\left(a+b+c+d+e\right)+5K}{5}=\frac{5m+5K}{5}=m+K\\ \therefore S.D.=\sqrt[]{\frac{{\displaystyle \sum _{}^{}{\left(x_i+K\right)}^2}}{N}-{\left(\frac{{\displaystyle \sum _{}^{}{x}_i+K}}{N}\right)}^2}\\ =\sqrt[]{\frac{{\displaystyle \sum _{}^{}\left(x_i^2+K^2+2x_{i}K\right)}}{N}-{\left(m+K\right)}^2}\\ =\sqrt[]{\frac{{\displaystyle \sum _{}^{}{x}_i^2}}{N}+\frac{{\displaystyle \sum _{}^{}{K}^2}}{N}+\frac{2K{\displaystyle \sum _{}^{}{x}_i}}{N}-m^2-K^2-2mK}\\ =\sqrt[]{\frac{{\displaystyle \sum _{}^{}{x}_i^2}}{N}+K^2+2Km-m^2-K^2-2mK}=\sqrt[]{\frac{{\displaystyle \sum _{}^{}{x}_i^2}}{N}-m^2}\left[?\frac{{\displaystyle \sum _{}^{}{x}_i}}{N}=m\right]\\ =S\\ Hence,thecorrectoptionis\left(a\right).\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Here\overline{x}=\frac{{\displaystyle \sum _{}^{}{x}_i}}{n}\\ 50=\frac{{\displaystyle \sum _{}^{}{x}_i}}{100}\Rightarrow {\displaystyle \sum _{}^{}{x}_i}=5000\\ \therefore S.D.=\sqrt[]{\frac{{\displaystyle \sum _{}^{}{x}_i^2}}{n}-{\left(\frac{{\displaystyle \sum _{}^{}{x}_i}}{n}\right)}^2}\\ 5=\sqrt[]{\frac{{\displaystyle \sum _{}^{}{x}_i^2}}{100}-{\left(\frac{5000}{100}\right)}^2}\Rightarrow 25=\frac{{\displaystyle \sum _{}^{}{x}_i^2}}{100}-2500\\ \Rightarrow \frac{{\displaystyle \sum _{}^{}{x}_i^2}}{100}=2500+25\Rightarrow \frac{{\displaystyle \sum _{}^{}{x}_i^2}}{100}=2525\\ \therefore {\displaystyle \sum _{}^{}{x}_i^2}=2525*100=252500\\ Hence,thecorrectoptionis\left(c\right).\end{array}$

 $\int {\left(\frac{x}{x\mathrm{s}\mathrm{i}\mathrm{n}?x+\mathrm{c}\mathrm{o}\mathrm{s}?x}\right)}^{2}dx=\int \left(\frac{x}{\mathrm{c}\mathrm{o}\mathrm{s}?x}\right)\cdot \frac{x\mathrm{c}\mathrm{o}\mathrm{s}?xdx}{(x\mathrm{s}\mathrm{i}\mathrm{n}?x+\mathrm{c}\mathrm{o}\mathrm{s}?x{)}^2}$

$\begin{array}{rr}& =\frac{x}{\mathrm{c}\mathrm{o}\mathrm{s}?x}\left(-\frac{1}{x\mathrm{s}\mathrm{i}\mathrm{n}?x+\mathrm{c}\mathrm{o}\mathrm{s}?x}\right)+\int \left(\frac{\mathrm{c}\mathrm{o}\mathrm{s}?x+x\mathrm{s}\mathrm{i}\mathrm{n}?x}{{\mathrm{c}\mathrm{o}\mathrm{s}}^2?x}\right)\left(\frac{1}{x\mathrm{s}\mathrm{i}\mathrm{n}?x+\mathrm{c}\mathrm{o}\mathrm{s}?x}\right)dx\\ & =-\frac{x\mathrm{s}\mathrm{e}\mathrm{c}?x}{x\mathrm{s}\mathrm{i}\mathrm{n}?x+\mathrm{c}\mathrm{o}\mathrm{s}?x}+\int {\mathrm{s}\mathrm{e}\mathrm{c}}^2?xdx\\ & =-\frac{x\mathrm{s}\mathrm{e}\mathrm{c}?x}{x\mathrm{s}\mathrm{i}\mathrm{n}?x+\mathrm{c}\mathrm{o}\mathrm{s}?x}+\mathrm{t}\mathrm{a}\mathrm{n}?x+C\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}TheformulaforS.D.=\sigma =\sqrt[]{\frac{{\displaystyle \sum _{}^{}{\left(x_i-\overline{x}\right)}^2}}{n}}\\ Hence,thecorrectoptionis\left(c\right).\end{array}$

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}$