Given, ${\int }_0^1\frac{{\tan }^{-1}x}{1+x^2}dx$.

Let $t={\tan }^{-1}x$, $\dots \dots (i)$

Differentiating with respect to $x$,

Then $\frac{dt}{dx}=\frac{1}{1+x^2}$

$dt=\frac{1}{1+x^2}dx$

The new limits to $e{q}^n(i)$,

When $x=0$, $t=0$

And when $x=1,t=\frac{\pi }{4}$

Therefore,

${\int }_0^1\frac{{\tan }^{-1}x}{1+x^2}dx$ $={\int }_0^{\frac{\pi }{4}}tdt$

$={\left[\frac{t^2}{2}\right]}_0^{\frac{\pi }{4}}$

Put the value of limit,

$=\frac{1}{2}[\frac{{\pi }^2}{16}-0]$

$=\frac{{\pi }^2}{32}$

Given:

$A=\{x\in R:x^2=2\}$

And, $B=\{y\in R:y^2-5y+6=0\}$

Then,

$x^2-2=0$ (Given)

$\Rightarrow x=\pm \sqrt{2}$

$\Rightarrow A=\{\sqrt{2},-\sqrt{2}\}$

$\therefore n(A)=2$.

Similarly,

$y^2-5y+6=0$ (Given)

$\Rightarrow y^2-3y-2y+6=0$

$\Rightarrow (y-3)(y-2)=0$

$\Rightarrow y=2,3$

$\Rightarrow B=\{2,3\}$

$\therefore n(B)=2$.

As we know that, if $n(A)=p,n(B)=q$, then:

$n(A\times B)=n(A)\times n(B)=p\times q$.

$\Rightarrow n(A\times B)=n(A)\times n(B)$

$=2\times 2$

$=4$

$A=\left[\begin{array}{cc}0& -2\\ 2& 0\end{array}\right]$

$A^2=\left[\begin{array}{cc}0& -2\\ 2& 0\end{array}\right]\left[\begin{array}{cc}0& -2\\ 2& 0\end{array}\right]=\left[\begin{array}{cc}-4& 0\\ 0& -4\end{array}\right]=-4I$

$\mathrm{M}={\mathrm{A}}^2+{\mathrm{A}}^4+{\mathrm{A}}^6+\dots .+{\mathrm{A}}^{20}$

$=-4\mathrm{I}+16\mathrm{I}-64\mathrm{I}+\dots .\text{ upto }10\text{ terms }$

$=-\mathrm{I}[4-16+64\dots +\text{ upto }10\text{ terms }]$

$=-\mathrm{I}.4\left[\frac{(-4{)}^{10}-1}{-4-1}\right]=\frac{4}{5}(2^{20}-1)\mathrm{I}$

$\mathrm{N}={\mathrm{A}}^1+{\mathrm{A}}^3+{\mathrm{A}}^5+\dots .+{\mathrm{A}}^{19}$

$=\mathrm{A}-4\mathrm{A}+16\mathrm{A}+\dots .\text{ upto }10\text{ terms }$

$=\mathrm{A}\left(\frac{(-4{)}^{10}-1}{-4-1}\right)=-\left(\frac{2^{20}-1}{5}\right)\mathrm{A}$

${\mathrm{N}}^2=\frac{{(2^{20}-1)}^2}{2^5}{\mathrm{A}}^2=\frac{-4}{24}{(2^{20}-1)}^2\mathrm{I}$

${\mathrm{MN}}^2=\frac{-16}{125}{(2^{20}-1)}^3\mathrm{I}=\mathrm{KI}(\mathrm{K}\ne \pm 1)$

${\left({\mathrm{MN}}^2\right)}^{\mathrm{T}}=(\mathrm{KI}{)}^{\mathrm{T}}=\mathrm{KI}$

$\therefore \mathrm{A}\text{ is correct }$

Refer diagram, the normal vector be $\mathrm{n}$ and it is perpendicular to both

$\mathrm{AB}$ and $\mathrm{AC}$

$\mathrm{AB}\times \mathrm{AC}=\mathrm{n}$

$\begin{array}{rl}& \text{ Now, }A(1,2,3),B(2,3,1)\text{ and }C(2,4,2)\\ & \text{ Then, }AB=(2-1)\hat{i}+(3-2)\hat{j}+(1-3)\hat{k}\\ & =\hat{i}+\hat{j}-2\hat{k}\\ & -1)\hat{i}+(4-2)\hat{j}+(2-3)\hat{k}\\ & \begin{array}{c}AC=(2-1)\hat{i}+(4\\ =\hat{i}+2\hat{j}-\hat{k}\end{array}\end{array}$

Now, $AB\times AC=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ \hat{j}& \mathrm{k}& \\ 1& 1& -2\\ 1& 2& -1\end{array}\right|$

$\begin{array}{rl}& =\hat{i}(-1+4)-\hat{j}(-1+2)+\hat{k}(2-1)\\ & =3\hat{i}-\hat{j}+\hat{k}\\ n& =3\hat{i}-\hat{j}+\hat{k}\end{array}$

Let P be any point on normal vector and O be origin. Then refer the diagram, projection of OP on plane have length OM .

$\begin{array}{rl}& \text{ OP }=2\hat{i}-\hat{j}+\hat{k}\text{ and }n=3\hat{i}-\hat{j}+\hat{k}\\ & \text{ OP. }n=|OP||n|\cos \theta \\ & (6+1+1)=\sqrt{4+1+1}(\sqrt{9+1+1})\cos \theta \\ & 8=\sqrt{6}\sqrt{11}\cos \theta \Rightarrow \cos \theta =\frac{8}{\sqrt{66}}\\ & \text{ Again, }\sin \theta =\frac{|\mathrm{OM}|}{|\mathrm{OP}|},\text{ gives }|\mathrm{OM}|=\sin \theta \cdot |\mathrm{OP}|\\ & \Rightarrow |\mathrm{OM}|=\sqrt{1-{\cos }^2\theta |\mathrm{OP}|}\\ & =\sqrt{1-\frac{64}{66}}\sqrt{4+1+1}\text{ (use }\cos \theta =8\sqrt{66})\\ & =\sqrt{\frac{2}{66}}\cdot \sqrt{6}\\ & \therefore |OM|=\sqrt{\frac{2}{11}}\end{array}$

If G is a group, then the subgroup consisting of G itself is the improper subgroup of G. All other subgroups are proper subgroups

In any group, the number of improper subgroups is 2.

If there is no point in the feasible set, there is no feasible solution of the linear programming model.

In linear programming, lack of points for a solution set is said to have no feasible solution

Given curve and lines are:

$y=x-2$ and $x=0$ to $x=4$

In figure $\mathrm{△}\mathrm{ABC}$ and $\mathrm{△}\mathrm{AOD}$ is similar.

So, Area of the region $=2\times ($ Area of $\mathrm{△}\mathrm{ABC})$

For the area of $\mathrm{△}ABC$

We know that:

$={\int }_{{\mathrm{x}}_1}^{{\mathrm{x}}_2}|{\mathrm{y}}_2-{\mathrm{y}}_1|\mathrm{dx}$

$={\int }_2^4(\mathrm{x}-2)\mathrm{dx}$

$={[\frac{{\mathrm{x}}^2}{2}-2\mathrm{x}]}_2^4$

$=\frac{4^2}{2}-2(4)-(\frac{2^2}{2}-2\times 2)$

$=8-8-2+4$

$=2$ sq. unit

So,

Area of the region $=2\times ($ Area of $\mathrm{△}\mathrm{ABC})$

$=2\times 2$

$=4$ sq. unit

Given lines are $\frac{2x-2}{2k}=\frac{4-y}{3}=\frac{z+2}{-1}$ and $\frac{x-5}{1}=\frac{y}{k}=\frac{z+6}{4}$

Write the above equation of a line in the standard form of lines

$\Rightarrow \frac{2(x-1)}{2k}=\frac{-(y-4)}{3}=\frac{z+2}{-1}\iff \frac{(x-1)}{k}=\frac{y-4}{-3}=\frac{z+2}{-1}$

So, the direction ratio of the first line is $(k,-3,-1)$

$\frac{x-5}{1}=\frac{y}{k}=\frac{z+6}{4}$

So, direction ratio of second line is $(1,k,4)$

Lines are perpendicular,

$\therefore (k\times 1)+(-3\times k)+(-1\times 4)=0$

$\Rightarrow k-3k-4=0$

$\Rightarrow -2k-4=0$

$\therefore k=-2$