Maths

$tan\left(\frac{1}{4}si{n}^{-1}\frac{\sqrt[]{63}}{8}\right)$

$Put\frac{1}{4}si{n}^{-1}\frac{\sqrt[]{63}}{8}=\theta$

$sin4\theta =\frac{\sqrt[]{63}}{8}\therefore cos4\theta =\frac{1}{8}$

$co{s}^2\theta =\frac{7}{8}\Rightarrow se{c}^2\theta =\frac{8}{7}$

$ta{n}^2\theta =\frac{8}{7}-1=\frac{1}{7}$

$tan\theta =\frac{1}{\sqrt[]{7}}$

y = x² + 4

x² = y – 4

y = 4x – 1

$PQ=\frac{\left|4*\frac{1}{2}t-\frac{1}{4}{t}^2\right|-4-1}{\sqrt[]{17}}$           

$p=\frac{\left|t^2-8t-2\right|}{4\sqrt[]{17}}$          

$\frac{dp}{dt}=\frac{1}{4\sqrt[]{17}}$           (2t – 8) for maximum / minimum, $\frac{dp}{dt}=0\Rightarrow t=4$  

$also\frac{d^{2}p}{d{t}^2}>0att=4$           

Hence the closest point becomes at t = 4 is (2, 8)

  $\left|\begin{array}{cc}\hfill f\left(X\right)\hfill & \hfill f\text{'}\left(x\right)\hfill \\ \hfill f\text{'}\left(x\right)\hfill & \hfill f\text{'}\text{'}\left(x\right)\hfill \end{array}\right|=0$         

$f\left(x\right)f\text{'}\text{'}\left(x\right)=f\text{'}\left(x\right)f\text{'}\left(x\right)$

$\frac{f\text{'}\text{'}\left(x\right)}{f\text{'}\left(x\right)}=\frac{f\text{'}\left(x\right)}{f\left(x\right)},$ Integrating on both sides

$\frac{f\text{'}\left(x\right)}{f\left(x\right)}=2,$ again integrating on both side

ln f(x) = 2^X + k

f(x) = e^{2x + k}

f(0) = e^k = e^k = 1-> k = 0

$\therefore f\left(x\right)=e^{2x}\left[\therefore e=2.718\right]$           

$\therefore e^2\in \left(6,9\right)$           

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

$\begin{array}{l}=\left(pv\sim p\right)\wedge \left(pv\sim q\right)\\ =pv\sim q\end{array}$

$\overrightarrow{r}.\left(\widehat{i}+\widehat{j}+\widehat{k}\right)-1+\lambda \left(\overrightarrow{r}.\left(\widehat{i}-2\widehat{j}\right)+2\right)=0$

$\therefore point\left(1,0,2\right)=\widehat{i}+2\widehat{k}$

$\therefore \overrightarrow{r}\left(\frac{\widehat{i}}{3}+\frac{7\widehat{j}}{3}+\widehat{k}\right)-\frac{7}{3}=0$

$\overrightarrow{r}.\left(\widehat{i}+7\widehat{j}+3\widehat{k}\right)=7$

${=}^{n+1}{C}_2+2{\displaystyle \sum _{r=1}^{n-1}\frac{\left(r+1\right)!}{\left(r-1\right)!2!}}{=}^{n+1}{C}_2+{\displaystyle \sum _{r=1}^{n-1}\left(r+1\right)r}$

  =   $\frac{\left(n+1\right)!}{2!\left(n+1\right)!}+\frac{\left(n-1\right)n\left(2\left(n-1\right)+1\right)}{6}+\frac{\left(n-1\right)n}{2}=\frac{n\left(n+1\right)}{2}+\frac{n\left(n-1\right)\left(2n-1\right)}{6}+\frac{n\left(n-1\right)}{2}$

=   $\frac{n\left(6n+2n^2-3n+1\right)}{6}=\frac{\left(2n^2+3n+1\right)}{6}=\frac{n\left(2n+1\right)\left(n+1\right)}{6}$

$x+\sqrt[]{3}y=2\sqrt[]{3},andm=-\frac{1}{\sqrt[]{3}},C=2$          not possible

 (2) , $\left(1\right)\frac{x^2}{\frac{9}{2}}-\frac{y^2}{\frac{1}{2}}=1\Rightarrow c=\sqrt[]{a^2{m}^2-b^2}=\sqrt[]{\frac{9}{2}*\frac{1}{3}-\frac{1}{2}}=1,$ not possible

(3) $c=a\sqrt[]{1+m^2}=\sqrt[]{7}*2=2\sqrt[]{7},$ not possible

(4) $\frac{x^2}{9}+\frac{y^2}{1}=1,C=\sqrt[]{9m^2+1}=\sqrt[]{9*\frac{1}{3}+1}=2$  

$A^T=Aand{B}^T=-B$           

$C=A^2{B}^2-B^2{A}^2$           

$C^T={\left(A^2{B}^2\right)}^T-{\left(B^2{A}^2\right)}^T=B^2{A}^2-A^2{B}^2$

 C^T = -C. Hence C is skew symmetric metrix

$\therefore$   det (C) = 0

Hence system have infinite solution

 $\overrightarrow{a}*\overrightarrow{b}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill \alpha \hfill & \hfill 3\hfill \\ \hfill 3\hfill & \hfill -\alpha \hfill & \hfill 1\hfill \end{array}\right|=\left(4\alpha \widehat{i}+8\widehat{j}-4\alpha \widehat{k}\right)$

$\left|\overrightarrow{a}*\overrightarrow{b}\right|=\sqrt[]{32{\alpha }^2+64}=8\sqrt[]{3}$

322 + 64 = 192

2 = $\frac{128}{32}=4$

$\overrightarrow{a}.\overrightarrow{b}=3-{\alpha }^2+3=6-{\alpha }^2=6-4=2$

 $x=y^4,xy=k$

$\frac{dy}{dx}=\frac{1}{4y^3},\frac{dy}{dx}=\frac{-k}{x^2}$

P (x₁, y₁)

where $x_1=y_1^4\&x_1{y}_1=k$

$\Rightarrow y_1=k^{1/5},x_1{y}_1=k$

$m_1{m}_2=-1$

$\Rightarrow \frac{-1}{4.k^{6/5}}=-1\Rightarrow k^{6/5}=\frac{1}{4}\Rightarrow k^6=\frac{1}{4^5}=\frac{1}{2024}$

${\left(4k\right)}^6=2^{12}.\frac{1}{2^{10}}=2^2=4$