Maths

^{$y.\frac{dy}{dx}=x\left[\frac{y^2}{x^2}+\frac{?\left(y^2/x^2\right)}{?\text{'}\left(y^2/x^2\right)}\right],x>0,?>0$}

$Let\frac{y}{x}=t$

$\frac{dy}{dx}=t+x.\frac{dt}{dx}$

$ln\left(?\left(\frac{y^2}{4}\right)\right)=ln4+ln\left(?\left(1\right)\right)$

$=ln4\left(?\left(1\right)\right)$

$?\left(\frac{y^2}{4}\right)=4.?\left(1\right)$

line: 3y – 2z – 1 = 0, 3x – z + 4 = 0

$\Rightarrow apointP\left(\frac{t-4}{3},\frac{2t+1}{3},t\right)$  on the line, Q (2, -1, 6)

$P{Q}^2=\frac{2}{9}\left(7t^2-56t+220\right)$

${\left(PQ\right)}_{min}=2\sqrt[]{6}$  

  $f\left(x\right)=si{n}^{-1}\left(\frac{3x^2+x-1}{{\left(x-1\right)}^2}\right)+co{s}^{-1}\left(\frac{x-1}{x+1}\right)$ ${}^{}\frac{{}^{}}{{}^{}}{}^{}\frac{}{}$

Domain of sin1 $\left(\frac{3x^2+x-1}{{\left(x-1\right)}^2}\right)$ $\frac{{}^{}}{{}^{}}$ is

$-1\le \frac{3s^2+x-1}{{\left(x-1\right)}^2}\le 1$

${}^{}{}^{}$  $-x^2-1+2x\le 3x^2+x-1$ and $3x^2+x-1\le x^2+1-2x$ ${}^{}{}^{}$

Domain of the function is $\left[\frac{1}{4},\frac{1}{2}\right]\cup \left\{0\right\}$ .

$2cosx\left(4sin\left(\frac{\pi }{4}+x\right)sin\left(\frac{\pi }{4}-x\right)-1\right)=1$  

->2cos x (2 cos 2x – 1) = 1

$\Rightarrow cos3x=\frac{1}{2},For0\le x\le \pi ,x=\frac{\pi }{9},\frac{5\pi }{9},\frac{7\pi }{9}$                

$dist.=\frac{{\overrightarrow{A}}_1{A}_2.\left({\overrightarrow{b}}_1*{\overrightarrow{b}}_2\right)}{\left|{\overrightarrow{b}}_1*{\overrightarrow{b}}_2\right|}$

$=\frac{\left(-4-\alpha ,-2,-3\right).\left(2,2,1\right)}{3}$

$\left|-5-\frac{2\alpha }{3}\right|=9\Rightarrow -5-\frac{2\alpha }{3}=\pm 9$

$\Rightarrow \frac{2\alpha }{3}=4,-14,$

but a > 0

a = 6

$\underset{x\to 0}{lim}{\left(2-cosx.\sqrt[]{cos2x}\right)}^{\frac{x+2}{x^2}}$ $\left(1^{\infty }\right)$

= $\underset{x\to 0}{lim}{e}^{2\left(\frac{sinx\sqrt[]{cos2x}-cosx.\frac{1\left(-2sin2x\right)}{2\sqrt[]{cos2x}}}{2x}\right)}$

$=\underset{x\to 0}{lim}{e}^{2\left(\frac{1}{2}+1\right)}=e^3=e^a$

a = 3

For the plane P,

$\overrightarrow{n}=\frac{1}{2}\left(-2\widehat{j}\right)*\left(-\widehat{i}+\widehat{j}-3\widehat{k}\right)=\widehat{j}*\widehat{i}+3\widehat{j}*\widehat{k}=-\widehat{k}+3\widehat{i}$

$\overrightarrow{a}.\overrightarrow{n}=0\Rightarrow 3\alpha -\gamma =0.......\left(i\right)$

$\overrightarrow{a}.\left(1,2,3\right)=0\Rightarrow \alpha +2\beta +3\gamma =0..........\left(ii\right)$

$\overrightarrow{a}.\left(1,1,2\right)=2\Rightarrow \alpha +\beta +2\gamma =2...........\left(iii\right)$

(i), (ii) & (iii) Þ a = 1, β = -5, $\gamma =$ 3

$T_{r+1}{=}^{120}{C}_r{4}^{\frac{120-r}{4}}.5^{r/6}$

For to be rational

r should be multiple of 6

Number of such terms = 21

$Area=\frac{1}{2}\left(\sqrt[]{5}-1\right)\frac{9-\sqrt[]{5}}{4}-{\displaystyle \underset{1}{\overset{\sqrt[]{5}}{\int }}\frac{1}{2}\sqrt[]{5-x^2}dx}$

$=\frac{1}{8}\left(-14+10\sqrt[]{5}\right)-\frac{1}{2}{\displaystyle \underset{co{s}^{-1}\frac{1}{\sqrt[]{5}}}{\overset{0}{\int }}\left(-si{n}^2\theta \right)d\theta }$

$A=\frac{-14}{8}+\frac{5}{4}\sqrt[]{5}-\left(\frac{5}{4}co{s}^{-1}\frac{1}{\sqrt[]{5}}-\frac{1}{2}\right)$

$=\frac{5}{4}\sqrt[]{5}-\frac{5}{4}-\frac{5}{4}co{s}^{-1}\frac{1}{\sqrt[]{5}}$

$\alpha =\frac{5}{4},\beta =\frac{-5}{4},\gamma =\frac{-5}{4}$

$\left|\alpha +\beta +\gamma \right|=\frac{5}{4}$