Maths

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:

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

At $\left(\frac{\pi }{4},\frac{\pi }{4}\right)$ :

$\begin{array}{rr}& ab\frac{dy}{dx}-ab-2b^2\left(-\frac{1}{2}\frac{dy}{dx}-\frac{1}{2}\right)=0\\ & \Rightarrow \frac{dx}{dy}=\frac{ab+b^2}{ab-b^2}=\frac{a+b}{a-b};a,b>0\end{array}$

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

$\begin{array}{l}Firstnnaturalnumbersare1,2,3,\dots ,n.Here,nisodd.\\ \therefore Mean\overline{x}=\frac{1+2+3+\dots +n}{n}=\frac{\frac{n\left(n+1\right)}{2}}{n}=\frac{n+1}{2}\\ Thedeviationsofnumbersfrommean\left(\frac{n+1}{2}\right)are\\ 1-\frac{n+1}{2},2-\frac{n+1}{2},3-\frac{n+1}{2},\dots ,n-\frac{n+1}{2}\\ i.e.,-\frac{n-1}{2},-\frac{n-3}{2},\dots ,-2,-1,0,1,2,\dots ,\frac{n-1}{2}.\\ Theabsolutevaluesofdeviationfromthemeani.e.,\left|x_i-\overline{x}\right|are\\ \frac{n-1}{2},\frac{n-3}{2},\dots ,2,1,0,1,2,\dots ,\frac{n-1}{2}.\\ Thesumofabsolutevaluesofdeviationfromthemeani.e.,\left|x_i-\overline{x}\right|are\\ =2\left(1+2+3+\dots to\frac{n-1}{2}terms\right)\\ =2.\frac{\frac{n-1}{2}\left(\frac{n-1}{2}+1\right)}{2}=\frac{n-1}{2}.\frac{n+1}{2}=\frac{n^2-1}{4}.\\ \therefore Meandeviationaboutthemean=\frac{{\displaystyle \sum _{}^{}\left|x_i-\overline{x}\right|}}{n}=\frac{\frac{n^2-1}{4}}{n}=\frac{n^2-1}{4n}.\end{array}$

$f\left(x\right)=\stackrel{?}{a}\cdot (\stackrel{?}{b}*\stackrel{?}{c})=\left|\begin{array}{ccc}x& -2& 3\\ -2& x& -1\\ 7& -2& x\end{array}\right|$

$=x^3-27x+26$

$f^{\mathrm{\text{'}}}\left(x\right)=3x^2-27=0\Rightarrow x=\pm 3$ and $f^{\mathrm{\text{'}}\mathrm{\text{'}}}(-3)<0$

$\Rightarrow$ local maxima at $\mathrm{x}={\mathrm{x}}_0=-3$

Thus, $\stackrel{?}{a}=-3\hat{i}-2\hat{j}+3\hat{k},\stackrel{?}{b}=2\hat{i}-3\hat{j}-\hat{k}$ , and $\stackrel{?}{c}=7\hat{i}-2\hat{j}-3\hat{k}$

$\Rightarrow \stackrel{?}{a}\cdot \stackrel{?}{b}+\stackrel{?}{b}\cdot \stackrel{?}{c}+\stackrel{?}{c}\cdot \stackrel{?}{a}=9-5-26=-22$

$\bar{x}=10$

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

$\Rightarrow a+b=17$

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 $b=5$

 $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(ab);\frac{2b^2}{a}=10\Rightarrow b^2=5a$

Now, $?\left(t\right)=\frac{5}{12}+t-t^2=\frac{8}{12}-{\left(t-\frac{1}{2}\right)}^2$
$?(\mathrm{t}{)}_{\mathrm{m}\mathrm{a}\mathrm{x}}=\frac{8}{12}=\frac{2}{3}=\mathrm{e}\Rightarrow {\mathrm{e}}^2=1-\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}=\frac{4}{9}$

$\Rightarrow {\mathrm{a}}^2=81\mathrm{}$ (From (i) and (ii)

So, $a^2+b^2=81+45=126$

 $f\left(2\right)=8,f^{\mathrm{\text{'}}}\left(2\right)=5,f^{\mathrm{\text{'}}}\left(x\right)\ge 1,f^{\mathrm{\text{'}}\mathrm{\text{'}}}\left(x\right)\ge 4,\forall x\in \left(\mathrm{1,6}\right)$

Using LMVT

$f^{\mathrm{\text{'}}\mathrm{\text{'}}}\left(x\right)=\frac{f^{\mathrm{\text{'}}}\left(5\right)-f^{\mathrm{\text{'}}}\left(2\right)}{5-2}\ge 4\Rightarrow f^{\mathrm{\text{'}}}\left(5\right)\ge 17$

$f^{\mathrm{\text{'}}}\left(x\right)=\frac{f\left(5\right)-f\left(2\right)}{5-2}\ge 1\Rightarrow f\left(5\right)\ge 11$

Therefore $f^{\mathrm{\text{'}}}\left(5\right)+f\left(5\right)\ge 28$

${\int }_0^3?g\left(x\right)-f\left(x\right)={\int }_0^3?\left|\right|x-2|-2|dx-{\int }_0^3?|x-2|dx$

$=\left(\frac{1}{2}*2*2+1+\frac{1}{2}*1*1\right)-\left(\frac{1}{2}*2*2+\frac{1}{2}*1*1\right)$

$=\left(2+1+\frac{1}{2}\right)-\left(2+\frac{1}{2}\right)=1$