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You should consider CU BTech CSE as a last option because drop changes everything.So, BTech CSE is the best option.

$A=\left[\begin{array}{ccc}\hfill x\hfill & \hfill y\hfill & \hfill z\hfill \\ \hfill y\hfill & \hfill z\hfill & \hfill x\hfill \\ \hfill z\hfill & \hfill x\hfill & \hfill y\hfill \end{array}\right],\left|A\right|=3xyz-\left(x^3+y^3+z^3\right)=-\left(x+y+z\right)\left[{\left(x+y+z\right)}^2-3\left(xy+yz+zx\right)\right]$  

A² = l

A. A’ = l    (as A = A’)

$\therefore x^2+y^2+z^2=1andxy+yz+zx=0$

$x^3+y^3+z^3=3\times 2+1\times \left(1-0\right)=7$

 Kindly go through the solution

 $v=\omega \sqrt[]{A^2-s^2}\Rightarrow \frac{A\omega }{2}=\omega \sqrt[]{A^2-s^2}\Rightarrow s=\frac{\sqrt[]{3}A}{2}\Rightarrow x=3$

$\overrightarrow{a}=\widehat{i}+2\widehat{j}-\widehat{k},\overrightarrow{b}=\widehat{i}-\widehat{j},\overrightarrow{c}=\widehat{i}-\widehat{j}-\widehat{k}$

$\overrightarrow{r}\times \overrightarrow{a}=\overrightarrow{c}\times \overrightarrow{a}$

$\overrightarrow{r}=\overrightarrow{c}+\lambda \overrightarrow{a}$

Now, $0=\overrightarrow{b}.\overrightarrow{c}+\lambda \overrightarrow{a}.\overrightarrow{b}as\overrightarrow{r}.\overrightarrow{b}=0$

$\lambda =\frac{-\overrightarrow{b}.\overrightarrow{c}}{\overrightarrow{a}.\overrightarrow{b}}=2$

$\therefore \overrightarrow{r}.\overrightarrow{a}=\overrightarrow{a}.\overrightarrow{c}+2a^2=12$

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$\Delta Q=\Delta U+\Delta W$

$\Rightarrow Q=\Delta U+\frac{Q}{5}\Rightarrow \Delta U=\frac{4Q}{5}=n{C}_v\Delta T\Rightarrow \frac{4Q}{5}=\frac{5R}{2}\Delta T\Rightarrow \Delta T=\frac{8Q}{25R}$

$Q=nc\Delta T=1\times C\times \frac{80}{25R}\Rightarrow C=\frac{25R}{8}\Rightarrow x=25$

But maximum value of k is not defined bonus

The Uttar Pradesh Police Recruitment and Promotion Board will release the tentative answer key for the UP Police Constable exam within a week from the exam. The Commission will release the tentative answer key first and ask the candidates to send objections against the incorrect answers.

After analysing the objections, the final answer key will be released.

Band Width = 2 × n × the highest modulation frequency

$\Rightarrow n=\frac{90k{H}_z}{2\times 5kHz}=9$

$L_1=\frac{x-\lambda }{1}=\frac{y-\frac{1}{2}}{\frac{1}{2}}=\frac{z}{-\frac{1}{2}}$

$\therefore SD=\frac{\left|-2\lambda +3\left(2\lambda +\frac{1}{2}\right)+\lambda \right|}{\sqrt[]{14}}=\frac{\left|5\lambda +\frac{3}{2}\right|}{\sqrt[]{14}}$

$5\lambda +\frac{3}{2}=-\frac{7}{2}\Rightarrow 5\lambda =-5\Rightarrow \lambda =-1$

$\therefore \left|\lambda \right|=1$

No, mere qualifying the VSAT exam doesn't 100 percent guarantee you the admission, however, it helps you proceed you the next stage, which is the counselling.

By qualifying in the VSAT you will increase your chances, and your final admission will depend upon the number of seats, your merit, rank and the courses for which you are planning.

${\displaystyle \underset{-a}{\overset{a}{\int }}\left(\left|x\right|+\left|x-2\right|\right)dx=22,a>2}$

${\displaystyle \underset{-a}{\overset{0}{\int }}\left(-2x+2\right)dx+{\displaystyle \underset{0}{\overset{2}{\int }}\left(x-x+2\right)dx+{\displaystyle \underset{2}{\overset{a}{\int }}\left(2x-2\right)dx=22}}}$

$\Rightarrow 2a^2+2=20\Rightarrow a^2=9\Rightarrow a=3$

$\therefore {\displaystyle \underset{3}{\overset{-3}{\int }}\left(x+\left[x\right]\right)dx=-{\displaystyle \underset{-3}{\overset{3}{\int }}\left(2x-\left\{x\right\}\right)dx=}}-{\displaystyle \underset{-3}{\overset{3}{\int }}2xdx+6}{{\displaystyle \underset{0}{\overset{1}{\int }}xdx=6.\frac{x^2}{2}}|}_0^1=3$

$P=\left[\begin{array}{ccc}\hfill 3\hfill & \hfill -1\hfill & \hfill -2\hfill \\ \hfill 2\hfill & \hfill 0\hfill & \hfill \alpha \hfill \\ \hfill 3\hfill & \hfill -5\hfill & \hfill 0\hfill \end{array}\right]andQ=\left[q_{ij}\right]\Rightarrow PQ=k{l}_3$

$q_{23}=-\frac{k}{8}and\left|Q\right|=\frac{k^2}{2}$

$PQ=k{l}_3\Rightarrow P^{-1}=\frac{Q}{k}={\left(\begin{array}{ccc}\hfill 3\hfill & \hfill -1\hfill & \hfill -2\hfill \\ \hfill 2\hfill & \hfill 0\hfill & \hfill \alpha \hfill \\ \hfill 3\hfill & \hfill -5\hfill & \hfill 0\hfill \end{array}\right)}^{-1}$

$\left|p\right|\left|Q\right|=\left(k{l}_3\right)\Rightarrow 8.\frac{k^2}{2}=k^3$

$k\ne 0\Rightarrow k=4$

$\therefore {\alpha }^2+k^2=1+16=17.$