Maths

$sin\left(2x^2\right).109_e\left(tan{x}^2\right)dy+4xydx=4\sqrt[]{2}.x.\left(sin{x}^{2}cos\frac{\pi }{4}-cos{x}^{2}sin\frac{\pi }{4}\right)dx$

$\Rightarrow \mathcal{l}n\left(tan{x}^2\right)dy+\frac{4x}{sin\left(2x^2\right)}ydx=\frac{4x\left(sin{x}^2-cos{x}^2\right)}{sin\left(2x^2\right)}dx$

Integrate

$\Rightarrow y.\mathcal{l}n\left(tan{x}^2\right)=2.\mathcal{l}n\left(\frac{sin{x}^2+cos{x}^2-1}{sin{x}^2+cos{x}^2+1}\right)+C$

$x=\sqrt[]{\frac{\pi }{6}},y=1$ calculate C.

$e_H=\sqrt[]{1+\frac{64}{49}}=\frac{\sqrt[]{113}}{7}$

$\Rightarrow e_H.e_E=\frac{1}{2}$

$\Rightarrow \frac{113}{49}.\frac{\left(64-a^2\right)}{64}=\frac{1}{4}\Rightarrow a^2-64=\frac{32^2}{113}$

$\mathcal{l}=\frac{2a^2}{b}=2\left(64+\frac{32^2}{113}\right).\frac{1}{8}$

$113\mathcal{l}=1552$

Plane through 4ax – y + 5z – 7a = 0 = 2x – 5y – z – 3

is $x\left(4a+2\lambda \right)+y\left(-1-5\lambda \right)+z\left(5-\lambda \right)=7a+3\lambda$

This plane contains 4, -1, 0

9a + 1 + 10 = 0…… (i)

Plane contains the line $\frac{x-4}{1}=\frac{y+1}{-2}=\frac{z}{1}$

$4a+11\lambda +7=0$ ……. (ii)

From (i) & (ii) a = 1,  $\lambda$ =1

Equation of plane $\pi \equiv x+2y+3z-2=0$

$\Rightarrow$$7$$P$$+$$3$$-$$2$$P$$+$$4$$-$$1$$2$$P$$+$$9$$-$$2$$=$$0$$\Rightarrow$$P$$=2$$\mathrm{}$

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

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

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 :

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

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

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