Maths

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

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

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

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

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

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

 $x^2-3x+p=0$

$\alpha ,\beta ,\gamma ,\delta$ in G.P.

$\alpha +\alpha r=3$

$x^2-6x+q=0$

$\alpha r^2+\alpha r^3=6$

$\left(2\right)÷\left(1\right)\Rightarrow r^2=2$

So,  $\frac{2q+p}{2q-p}=\frac{2r^5+r}{2r^5-r}=\frac{2r^4+1}{2r^4-1}=\frac{9}{7}$

$y+2x=\sqrt[]{11}+7\sqrt[]{7}$ ……… (i)

$2y+x=2\sqrt[]{11}+6\sqrt[]{7}$ ……… (ii)

$\Rightarrow x+y=\sqrt[]{11}+\frac{13}{3}\sqrt[]{7}$ ……… (iii)

Centre of the circle given by solving (i) & (ii)

$as\left(\frac{8\sqrt[]{7}}{3},\sqrt[]{11}+\frac{5\sqrt[]{7}}{3}\right)$

Again $\sqrt[]{11}y-3x=\frac{5\sqrt[]{77}}{3}$ is tangent to the circle.

$\therefore r=\left|\frac{\sqrt[]{11}\left(\sqrt[]{11}+\frac{5\sqrt[]{7}}{3}\right)-8\sqrt[]{7}-\frac{5\sqrt[]{77}}{3}}{\sqrt[]{11+9}}\right|=\left|\frac{11-8\sqrt[]{7}}{\sqrt[]{20}}\right|$

$\therefore {\left(5h-8k\right)}^2+5r^2=816$

$1<\left|z-3+2i\right|<4$

$z=a+iba,b\in I$              

⇒ 1 < (a 3)² + (b + 2)² < 16

in equivalent to 1 < ² + β² < 16  $\alpha =a-3\in l$               

$\left(\pm 2,0\right)\left(\pm 3,0\right),\left(0,\pm 2\right)\left(0,\pm 3\right)$             $\beta =b+2\in l$

Total 40 such points are possible