Maths

$\overline{x}=\frac{{\displaystyle \sum _{i=1}^{10}{x}_i}}{10}=15\frac{{\displaystyle \sum _{i=1}^{10}{x_i}^2}}{10}-{\left(\overline{x}\right)}^2=15$

$\Rightarrow \Sigma x_i=150\Sigma {x_i}^2=2400$

Actual mean $\overline{x}=\frac{\Sigma x_i+15-25}{10}=\frac{140}{10}=14$

Actual variance = $\frac{\Sigma {x_i}^2+15^2-25^2}{10}-{\left(14\right)}^2$

$=\frac{2400-400}{10}-196$

$\begin{array}{l}{\sigma }^2=4\\ \sigma =2\end{array}$

use $si{n}^{-1}x=co{s}^{-1}\sqrt[]{1-x^2}$

$ta{n}^{-1}\sqrt[]{1-x^2}=co{t}^{-1}\frac{1}{\sqrt[]{1-x^2}}$

$si{n}^{-1}\frac{x}{\sqrt[]{1-x^2}}=co{s}^{-1}\frac{\sqrt[]{1-2x^2}}{\sqrt[]{1-x^2}}$

Sum of roots b = 1 + 2 $\frac{\left(k^2-1\right)}{k^2-2}$

Product of roots 5 = 2 $\left(\frac{k^2-1}{k^2-2}\right)$

b = 4, k² = $\frac{1}{3}$

$?x=sin\left(2ta{n}^{-1}\alpha \right)=\frac{2\alpha }{1+{\alpha }^2}$ ……. (i)

and

$y=sin\left(\frac{1}{2}ta{n}^{-1}\frac{4}{3}\right)=sin\left(si{n}^{-1}\frac{1}{\sqrt[]{5}}\right)=\frac{1}{\sqrt[]{5}}$

Now,

$y^2=1-x$

$\therefore \alpha =2,\frac{1}{2}\therefore {\displaystyle \sum _{a\in S}^{}16{\alpha }^3=16*2^3+16*\frac{1}{2^3}=130}$

$4x^3-3x{y}^2+6x^2-5xy-8y^2+9x+14=0$ differentiating both sides we get

$12x^2-3y^2-6xyy\text{'}+12x-5y-5xy=16yy\text{'}+9=0$

At the point (2, 3)

48 – 27 + 36y' – 24 – 15 + 10y' – 48y' + 9 = 0

$\therefore Area=\frac{1}{2}*Base*Height$

$A=\frac{1}{2}*\left(\frac{-4}{3}+\frac{31}{2}\right)\left(3\right)=\frac{1}{2}\left(\frac{85}{0}\right).3=\frac{85}{4}=8A=170$

The circle $x^2+y^2+6x+8y+16=0$ has centre (3, 4) and radius 3 units

The circle

$x^2+y^2+2\left(3-\sqrt[]{3}\right)x+2\left(4-\sqrt[]{6}\right)y=k+6\sqrt[]{3}+8\sqrt[]{6},k>0$ has centre $\left(\sqrt[]{3}-3,\sqrt[]{6}-4\right)$ and radius $\sqrt[]{k+34}$

$?$ These two circles touch internally hence

$\sqrt[]{3+6}=\left|\sqrt[]{k+34}-3\right|$

here, k = 2 is only possible $\left(?k>0\right)$

Equation of common tangent to two circle is

$2\sqrt[]{3}x+2\sqrt[]{6}y+16+6\sqrt[]{3}+8\sqrt[]{6}+k=0$

$?k=2$ then equation is

$x+\sqrt[]{2}y+3+4\sqrt[]{2}+3\sqrt[]{3}=0$ ….(i)

$?\left(\alpha ,\beta \right)$ are foot of perpendicular from (3, 4) to line (i) then

$\frac{\alpha +3}{1}=\frac{\beta +4}{\sqrt[]{2}}=\frac{-3-4\sqrt[]{2}+3+4\sqrt[]{2}+3+\sqrt[]{3}}{1+2}$

$\therefore {\left(\alpha +\sqrt[]{3}\right)}^2+{\left(\beta +\sqrt[]{6}\right)}^2=25$

 $?a_n={\displaystyle {\int }_{-1}^0\left(1+\frac{x}{2}+\frac{x^2}{2}+....+\frac{x^{n-1}}{n}\right)dx}$

$={\left[x+\frac{x^2}{2^2}+\frac{x^3}{3^2}+.....+\frac{x^n}{n^2}\right]}_{-1}^n$

$a_n=\frac{n+1}{1^2}+\frac{n^2-1}{2^2}=\frac{n^3+1}{3^2}+\frac{n^4-1}{4^2}+.....+\frac{n^n+{\left(-1\right)}^{n+1}}{n^2}$

Here a₁ = 2, $a_2=\frac{2+1}{1}+\frac{2^2-1}{2}=3+\frac{3}{2}=\frac{9}{2}$

$a_4=5+\frac{15}{4}+\frac{65}{9}+\frac{255}{16}>31$

$\therefore$ The required set is {2, 3}

$?a_n\in \left(2,30\right)$

$\therefore$ Sum of elements = 5.

$f\left(x\right)+{\displaystyle {\int }_0^x\left(x-t\right)f\text{'}\left(t\right)dt=\left(e^{2x}+e^{-2x}\right)cos2x+\frac{2x}{a}.......(i)}$

Here f (0) = 2 ………. (ii)

On differentiating equation (i) w.r.t. x we get :

$f\text{'}\left(x\right)+f{\displaystyle {\int }_0^{x}f\text{'}\left(t\right)dt+xf\text{'}\left(x\right)-xf\text{'}\left(x\right)}$

$4=\frac{2}{a}\Rightarrow a=\frac{1}{2}$

$\therefore {\left(2a+1\right)}^5.a^2=2^5.\frac{1}{2^2}=2^3=8$

$\frac{dy}{dx}=\frac{y}{x}\frac{\left(4y^2+2x^2\right)}{\left(3y^2+x^2\right)}$

Put y = vx

$\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$

$\Rightarrow v+x\frac{dv}{dx}=\frac{v\left(4v^2+2\right)}{\left(3v^2+1\right)}$

$\mathcal{l}n\left(\frac{y^2}{8}+\frac{y}{2}\right)=2\mathcal{l}n2\Rightarrow \frac{y^3}{8}+\frac{y}{2}=4$

$\Rightarrow \left[y\left(2\right)\right]=2$

$\Rightarrow n=3$

$f\left(x\right)=\left|5x-7\right|+\left[x^2+2x\right]$

$=\left|5x-7\right|+\left[{\left(x+1\right)}^2\right]-1$

f $\left(\frac{7}{4}\right)=0+4=4$

as both |5x – 7| and x² + 2x are increasing in nature after x = 7/5

f (2) = 3 + 8 = 11

$f{\left(\frac{7}{5}\right)}_{min}-4andf{\left(2\right)}_{max}=11$

Sum is 4 + 11 = 15

$A=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]+\left[\begin{array}{ccc}\hfill 0\hfill & \hfill a\hfill & \hfill a\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]=l+B$

$B^2=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill a\hfill & \hfill a\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]*\left[\begin{array}{ccc}\hfill 0\hfill & \hfill a\hfill & \hfill a\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill ab\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$

B³ = 0

$\therefore A^n={\left(1+B\right)}^n{=}^n{C}_{0}l{+}^n{C}_{1}B{+}^n{C}_2{B}^2{+}^n{C}_3{B}^3+....$

On comparing we get na = 48, nb = 96 and na + $\frac{n\left(n-1\right)}{2}ab=2160$

$\Rightarrow a=4,n=12andb=8$

$\Rightarrow n+a+b=24$