Maths

${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left[\frac{1+si{n}^{2}x}{1+{\pi }^{sinx}}+\frac{1+si{n}^{2}x}{1+{\pi }^{-sinx}}\right]dx}$

$={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(1+si{n}^{2}x\right)dx=\frac{\pi }{2}+\frac{1}{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(1-cos2x\right)dx}}$

$=\frac{\pi }{2}+\frac{1}{2}{\left[x-\frac{sin2x}{2}\right]}_0^{\frac{\pi }{2}}$
$=\frac{\pi }{2}+\frac{\pi }{4}=\frac{3\pi }{4}$

$A^2=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)$

$A^3=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)$

$A^{2025}-A^{2020}$

$=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 2024\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)-\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 2019\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 5\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)=A^6-A$

Equation of required circle

$C:{\left(x-2\right)}^2+{\left(y-1\right)}^2+\lambda \left(2y-x\right)=0..........\left(i\right)$              

Intersect C₁x² + y² + 2y – 5 = 0 .(ii)

$\therefore$ Equation of radical axis

    $-4x-4y+10+\lambda \left(2y-x\right)=0..........\left(ii\right)$          

Centre of C₁ (0, -1) lies on .(ii)

$\Rightarrow 4+10-2\lambda =0\Rightarrow \lambda =7$        

Equation of circle C is

$x^2+y^2-11x+12y+5=0$       

Diameter = $\sqrt[]{245}=7\sqrt[]{5}$  

Given limit
$={\displaystyle \underset{0}{\overset{2}{\int }}\frac{dx}{1+4x^2}}$

$={\left[\frac{1}{2}ta{n}^{-1}2x\right]}_0^2=\frac{1}{2}ta{n}^{-1}4$

$\frac{a}{1-r}=15,\frac{a^2}{1-r^2}=150$

$\Rightarrow r=\frac{1}{5},a=12\Rightarrow \frac{a{r}^2}{1-r^2}=\frac{1}{2}$

$\overrightarrow{a}.\overrightarrow{b}*\overrightarrow{c}$

$=\overrightarrow{b}.\overrightarrow{c}*\overrightarrow{a}$

$=\overrightarrow{b}.\left(-\overrightarrow{b}\right)$

$=-{\left|\overrightarrow{b}\right|}^2=2$

$\frac{dy}{dx}+\frac{se{c}^{2}x}{tanx}y=-tanx$

IF = tan x

Solution : y tan x = x – tan x + C

$\underset{x\to 0^{+}}{lim}xy=1\Rightarrow C=1$

For $x=\frac{\pi }{4},y=\frac{\pi }{4}$

Line BM : 2x + y = 3 Þ M (0, 3)

Line CD : 7x – 4y = 1 Þ C (3, 5)

Mirror image of A (-3, 1) in the line CD is $\left(\frac{13}{5},\frac{-11}{5}\right)$  and it will lie on BC.

Slope of AC is 2/3

Slope of BC is 18.

$tan\theta =\frac{18-\frac{2}{3}}{1+18\left(\frac{2}{3}\right)}=\frac{4}{3}$   

 where θ = $\angle ACB$  

$\sum X=200$

$\sum x^2=2125$

For new data

$\sum x=200-25+35=210$

$\overline{x}=10.5$

$\sum x^2=2725$

${\sigma }^2=\frac{2725}{20}-{\left(10.5\right)}^2=26$

Kindly consider the following figure