Maths

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

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

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

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

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

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

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

$x+\sqrt[]{3}y=2\sqrt[]{3},andm=-\frac{1}{\sqrt[]{3}},C=2$          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 $x_1=y_1^4\&x_1{y}_1=k$

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

$m_1{m}_2=-1$

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

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

Area of shaded region

$2{\displaystyle \underset{0}{\overset{\sqrt[]{3}}{\int }}\left(2x^2+9-5x^2\right)dx}$

$=2{\displaystyle \underset{0}{\overset{\sqrt[]{3}}{\int }}\left(9-3x^2\right)dx}$

${\displaystyle \underset{0}{\overset{\sqrt[]{3}}{\int }}\left(3-x^2\right)dx=6{\left[3x-\frac{x^3}{3}\right]}_0^{\sqrt[]{3}}=6\left[\frac{9\sqrt[]{3}-3\sqrt[]{3}}{3}\right]}=12\sqrt[]{3}$

 $f\left(x\right)=\left\{\begin{array}{cc}\hfill min\left\{\left|x\right|,2-x^2\right\},\hfill & \hfill -2\le x\le 2\hfill \\ \hfill \left[\left|x\right|\right],\hfill & \hfill 2\le \left|x\right|\le 3\hfill \end{array}\right\}$

Number of points where f is not differentiable = 5

$I=3{\displaystyle \underset{-2}{\overset{2}{\int }}\left|x^2-x-2\right|dx}$

$=3\left({\displaystyle \underset{-2}{\overset{-1}{\int }}\left(x^2-x-2\right)dx-{\displaystyle \underset{-1}{\overset{2}{\int }}\left(x^2-x-2\right)dx}}\right)$

$=3\left[\left(\frac{7}{6}+\frac{2}{3}\right)-\left\{-\frac{10}{3}-\frac{7}{6}\right\}\right]$

$=3\left(\frac{11}{6}+\frac{27}{6}\right)=\frac{38}{2}=19$

Total possibilities = 2⁵ * 2⁵

Farounable case = ⁵C₂ * 3³ = 10 * 3³

$\therefore requiredprobability=\frac{10*3^3}{2^5*2^5}=\frac{5*27}{2^9}=\frac{135}{2^9}$