Class 11th

Required number =  - number of solution of x₁ + x₂ + x₃ = 15, where x₁ < x₂ < x₃

For x₂ = x₁ + a, a   $\ge 1$

->3x₁ + 2a + b = 15

Coefficient of x¹⁵ in

$\left(x^3+x^6+x^9+x^{12}+x^{15}\right)$

$\left(x^1+x^2+x^3+.....+x^{10}\right)=12$

Required number = 455 – 12 = 443

$x+1-2lo{g}_2\left(3+2^x\right)+2lo{g}_4\left(10-2^{-x}\right)=0$

$x+1+lo{g}_2\left[\frac{10.2^x-1}{{\left(3+2^x\right)}^2}\right]-x=0$

${\left(2^x\right)}^2-14.2^x+11=0$

2x + y

$x_1+x_2=lo{g}_2\left(49-\frac{152}{4}\right)$

$x_1+x_2=lo{g}_{2}11$

Minimum distance Andhra Pradesh = OP – OA

$\begin{array}{l}=3\sqrt[]{2}-\sqrt[]{2}\\ =2\sqrt[]{2}\end{array}$

As $\sim \left(A\Rightarrow B\right)=A\wedge \sim B$

$\sim \left(\left(pvr\right)\Rightarrow \left(qvr\right)\right)$

$=\left(pvr\right)\wedge \left(\sim r\wedge \sim q\right)$

$=\left(p\wedge \sim r\right)\vee \left(r\wedge \sim r\right)\wedge \left(\sim q\right)$

= $\left(p\wedge \sim r\right)$

$b^2{x}^2+a^2{y}^2=a^2{b}^2$

$\left(b^2{x}^2+a^2\left(ab-x^2\right)\right)=a^2{b}^2$

$x^2=\frac{b{a}^2\left(b-a\right)}{b^2-a^2},y^2=\frac{a{b}^2}{a+b}$

points of intersection

$\left(a\sqrt[]{\frac{b}{a+b}},b\sqrt[]{\frac{a}{a+b}}\right)$

$tan\theta =\left|\frac{m_1-m_2}{1+m_1.m_2}\right|$

$=\left|\frac{a-b}{\sqrt[]{ab}}\right|$

$f\left(x\right)=lo{g}_{\sqrt[]{5}}\left(3+\sqrt[]{2}\left(cosx-sinx\right)\right)$

$-\sqrt[]{2}\le cosx-sinx\le \sqrt[]{2}$

$\Rightarrow 0\le f\left(x\right)\le 2$

$\left|\begin{array}{ccc}\hfill \alpha \hfill & \hfill \beta \hfill & \hfill 1\hfill \\ \hfill 5\hfill & \hfill 6\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill 2\hfill & \hfill 1\hfill \end{array}\right|=24$

$4\alpha -2\beta -8=\pm 24$

$4\alpha -2\beta =24+8,4\alpha -2\beta =-24+8$

$2\left(2\alpha -\beta \right)=322\alpha -\beta =-8$

Distance of origin

$D=\sqrt[]{{\alpha }^2+{\left(2\alpha +8\right)}^2}$

$\alpha =-\frac{16}{5}$

$D=\sqrt[]{{\left(\frac{-16}{5}\right)}^2+{\left(\frac{8}{5}\right)}^2}$

if 2 - = 16

$D=\frac{16}{\sqrt[]{5}}$

$S_{20}={\displaystyle \underset{n=1}{\overset{20}{?}}\frac{1}{d}\left(\frac{1}{a_n}?\frac{1}{a_{n+1}}\right)}$

$=\frac{1}{d}\left(\frac{1}{a_1}?\frac{1}{a_{21}}\right)$

$?a\left(a+20d\right)=45.......\left(i\right)$

$?a+10d=9.........\left(ii\right)$

$\left(i\right)\&\left(ii\right)?d^2=\frac{36}{100}$

$DE:\frac{dy}{dx}+\frac{y}{x^2}=\frac{1}{x^3}$  

IF =    $e^{-\frac{1}{x}}$

Solution : $y{e}^{-\frac{1}{x}}={\displaystyle \int e^{-\frac{1}{x}}.\frac{1}{x^3}dx=\frac{1}{x}{e}^{-\frac{1}{x}}+e^{-\frac{1}{x}}+C}$  

Point (1, 1) -> C = $-\frac{1}{e}$  

$x=\frac{1}{2}\Rightarrow y=3-e$            

$\alpha =\underset{x\to \pi /4}{lim}\frac{ta{n}^{3}x-tanx}{cos\left(x+\pi /4\right)}$

a = -4

$\beta =\underset{x\to 0}{lim}{\left(cosx\right)}^{cotx}$      

$\beta =e^0=1$

Equation whose roots are a and b

x² + 3x – 4 = 0

a = 1, b = 3