Maths

$32^{ta{n}^{2}x}+32^{se{c}^{2}x}=81$

$32^{ta{n}^{2}x}+32^{1+ta{n}^{2}x}=81$

$33.32^{ta{n}^{2}x}=81$

$32^{ta{n}^{2}x}=\frac{81}{33}$   

$forx\in \left[0,\pi /4\right]ta{n}^{2}x\in \left[0,1\right]$

One solution

Given expression

= 2π - 5 + 6 - 2π - (12 - 4π) = 4π - 11

f (m + n) = f (m) + f (n)

put m = n = 1, f (2) = f (1) + f (1)

again put m = 2, n = 1, f (3) = f (2) + f (1)

and put m = 3, n = 3, f (3 + 3) = f (3) + f (3), 2f (3) = f (6) = 18 Þ f (3) = 9

f (3) = 3f (1)

f (1) = 3, f (2) = 6

f (2).f (3) = 54

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

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

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

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

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

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

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