Maths

$3*7^{22}+2*10^{22}-44=3*{\left(1+6\right)}^{22}+2{\left(1+9\right)}^{22}-44$

$=3\left[1{+}^{22}{C}_1*6{+}^{22}{C}_2*6^2{+}^{22}{C}_3*6^3+....{.}^{22}{C}_{22}{6}^{22}\right]$                

= -39 on division by 18

= (-54 + 15) on division by 18 = 15

$A=\left[\begin{array}{cc}\hfill 0\hfill & \hfill 2\hfill \\ \hfill k\hfill & \hfill -1\hfill \end{array}\right]then{A}^2=\left[\begin{array}{cc}\hfill 0\hfill & \hfill 2\hfill \\ \hfill k\hfill & \hfill -1\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 0\hfill & \hfill 2\hfill \\ \hfill k\hfill & \hfill -1\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 2k\hfill & \hfill -2\hfill \\ \hfill -k\hfill & \hfill 2k+1\hfill \end{array}\right]$

$A^3=\left[\begin{array}{cc}\hfill 2k\hfill & \hfill -2\hfill \\ \hfill -k\hfill & \hfill 2k+1\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 0\hfill & \hfill 2\hfill \\ \hfill k\hfill & \hfill -1\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill -2k\hfill & \hfill 4k+2\hfill \\ \hfill 2k^2+k\hfill & \hfill -4k-1\hfill \end{array}\right]$              

Now, A(A³ + 3l) = 2l gives A³ + 3l = 2A^-1

$\Rightarrow -2k+3=\frac{1}{k}\Rightarrow 2k^2-3k+1=0$    

  $k=\frac{1}{2},1$            

& $4k+2=\frac{2}{k}or2k+1-\frac{1}{k}soor2k^2+k-1=0$

=> $k=\frac{1}{2}$

 Slope of C₁C₂ = $\frac{3}{4}=tan\theta$  

By parametric form C₁ (1 + 5 cosθ, 2 + 5sinθ)

& C₂ (1 – 5 cos θ, 2 – 5 sinθ)

C_{1 $\left(1+5*\frac{4}{5}.2+5*\frac{3}{5}\right)\&C_2\left(1-5*\frac{4}{5}.2-5*\frac{3}{5}\right)$}

So, $\left|\left(\alpha +\beta \right)\left(r+\delta \right)\right|=40$  

${\left(si{n}^{-1}x\right)}^2-{\left(co{s}^{-1}x\right)}^2=a,0<x<1$           

$\left(si{n}^{-1}x+co{s}^{-1}x\right)\left(si{n}^{-1}x-co{s}^{-1}x\right)=a$

$2co{s}^{-1}x=\frac{\pi }{2}-\frac{2a}{\pi }..........\left(i\right)$    

Let $co{s}^{-1}x=\theta thenx=cos\theta$

So, $2x^2-1=2co{s}^2\theta -1=cos2\theta ..........\left(ii\right)$

Now, $2\theta =2co{s}^{-1}x=\frac{\pi }{2}-\frac{2a}{\pi }from\left(i\right)$

So, cos 2θ = cos $\left(\frac{\pi }{2}-\frac{2a}{\pi }\right)=sin\frac{2a}{\pi }$

S = {1, 2, 3, 4, 5, 6, 9}

Elements of type 3n -> 3, 6, 9

Type 3n + 1 ->1, 4

3n + 2 -> 2, 5

Number of subset of S containing one element which are not divisible by  $3{=}^2{C}_1{+}^2{C}_1=4$ number of subset of S containing two numbers whose sum is not divisible $3{=}^3{C}_1{*}^2{C}_1{+}^3{C}_1{*}^2{C}_1{+}^2{C}_2{+}^2{C}_2=14$ by

Number of subset of S containing 3 elements whose sum is not divisible by

Number of subset containing 4 elements whose sum is not divisible by 3

Number of subset of S containing 6 elements = 4

Hence total subset = 80

$\frac{dy}{dx}=2\left(y+2sinx-5\right)x-2cosx$

where P = -2x, Q = 4x sin x – 10x – 2 cosx

Solution be $y.e^{-x^2}={\displaystyle \int \left(4xsinx-10x-2cosx\right).e^{-x^2}dx+c}$

$y.e^{-x^2}{\displaystyle \int \left(4xsinx-10x-2cosx\right).e^{-x^2}dx+c}$     

$y.e^{-x^2}=e^{-x^2}\left(5-2sinx\right)+c........\left(i\right)$  

Put x = 0, y = 7 then 7 = 5 + c i.e. c = 2

Put x = p then $y.e^{-{\pi }^2}=5.e^{-{\pi }^2}+2$

y = 5 + $2e^{x^2}$

sin^4_θ + cos⁴θ - sinθ cosθ = 0

${\left(si{n}^2\theta +co{s}^2\theta \right)}^2-2si{n}^2\theta co{s}^2\theta -sin\theta cos\theta =0$

$\frac{{\left(2sin\theta cos\theta \right)}^2}{2}-\frac{2sin\theta cos\theta }{2}+1=0$               

$si{n}^{2}2\theta +sin2\theta -2=0$              

sin 2θ = 1 .(i)

$as\theta \in \left[0,4\pi \right]so2\theta \in \left[0,8\pi \right]$              

$\left(i\right)2\theta =\frac{\pi }{2},\frac{5\pi }{2},\frac{9\pi }{2},\frac{18\pi }{2}$

Hence $\frac{8s}{\pi }=56$

$y=lo{g}_{10}x+lo{g}_{10}{x}^{1/3}+lo{g}_{10}{x}^{1/9}+........upto\infty terms$

$=lo{g}_{10}x\left(1+\frac{1}{3}+\frac{1}{9}+......\right)$   

y = log₁₀ x * $\frac{1}{1-\frac{1}{3}}=\frac{3}{2}lo{g}_{10}x$  

Now, $\frac{2+4+6+....+2y}{3+6+9+....+3y}=\frac{4}{lo{g}_{10}x}$

$\Rightarrow \frac{2*y\frac{\left(y+1\right)}{2}}{3y\frac{\left(y+1\right)}{2}}=\frac{4}{lo{g}_{10}x}solo{g}_{10}x=6$

$\Rightarrow y=\frac{3}{2}*6=9$

So, (x, y) = (10⁶, 9)

$y=\frac{x^2}{2}+\frac{2}{3}{x}^3+\frac{3}{4}{x}^4+.....$

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

$y=\frac{x^2}{1-x}+log\left(1-x\right)+x=\frac{x}{1-x}+log\left(1-x\right)$

$atx=\frac{1}{2},y=1+ln\frac{1}{2}=1-ln2$

$e^{y+1}=e^{1-ln2+1}=e^{2-ln2}=\frac{e^2}{2}$

Given let α, β be the roots of the equation x² + bx + c = 0

So, ${\alpha }^2+b\alpha +c=0\&{\beta }^2+b\beta +c=0$

Also x² + bx + c = (x - α) (x - β)

$NowL=\underset{x\to \beta }{lim}\frac{e^{2\left(x^2+bx+c\right)}-1-2\left(x^2+bx+c\right)}{{\left(x-\beta \right)}^2}$           

$\underset{x\to \beta }{lim}\frac{2{\left(x-\alpha \right)}^2{\left(x-\beta \right)}^2+\frac{8}{6}{\left(x-\alpha \right)}^3{\left(x-\beta \right)}^3+........}{{\left(x-\beta \right)}^2}$

=  $2{\left(\beta -\alpha \right)}^2=2\left[{\left(\beta +\alpha \right)}^2-4\alpha \beta \right]=2\left[b^2-4c\right]$