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 $\left|\begin{array}{ccc}\hfill \left(a+1\right)\left(a+2\right)\hfill & \hfill a+2\hfill & \hfill 1\hfill \\ \hfill \left(a+2\right)\left(a+3\right)\hfill & \hfill a+3\hfill & \hfill 1\hfill \\ \hfill \left(a+3\right)\left(a+4\right)\hfill & \hfill a+4\hfill & \hfill 1\hfill \end{array}\right|=\left|\begin{array}{ccc}\hfill \left(a+1\right)\left(a+2\right)\hfill & \hfill a\hfill & \hfill 1\hfill \\ \hfill \left(a+2\right)\left(a+3\right)\hfill & \hfill a\hfill & \hfill 1\hfill \\ \hfill \left(a+3\right)\left(a+4\right)\hfill & \hfill a\hfill & \hfill 1\hfill \end{array}\right|+\left|\begin{array}{ccc}\hfill \left(a+1\right)\left(a+2\right)\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill \left(a+2\right)\left(a+3\right)\hfill & \hfill 3\hfill & \hfill 1\hfill \\ \hfill \left(a+3\right)\left(a+4\right)\hfill & \hfill 4\hfill & \hfill 1\hfill \end{array}\right|,$ [using properties]

$=0+\left(a+1\right)\left(a+2\right)\left(-1\right)+2\left(a+3\right)\left(a+4\right)0\left(a+2\right)\left(a+3\right)+4\left(\left(a+2\right)\left(a+3\right)-3\left(a+3\right)\left(a+4\right)\right)$

$=2\left(a+2\right)-2\left(a+3\right)=2\left(a+2-a-3\right)=-2$

The grading system differs per country in Europe. In some European countries, grades are based on percentages from 0 to 100 and in others, grades range from A to F or from 1 to 10. Each university has the same grading system more or less, there may be slight variations based on the geographic location of the European university. Students can still research and check out the university's grading system in which they wish to seek admission in future. 

$V=k{T}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

$T{V}^{-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=K$

$\Rightarrow \gamma -1=-3/2$

$\Rightarrow \gamma =-1/2$

Work done = $\frac{nR\Delta T}{-y+1}=\frac{1\times R\times 90}{3/2}=6R$

n = 60

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Total oscillation = $\frac{1}{2}+\frac{1}{8}=\frac{5}{8}$ $\frac{}{}\frac{}{}\frac{}{}$

$\Rightarrow timetaken=\frac{T}{2}+\frac{T}{12}=\frac{7T}{12}$

= 7

$\frac{si{n}^{-1}x}{a}=\frac{co{s}^{-1}x}{b}=\frac{ta{n}^{-1}y}{c}=\lambda$

$\therefore \frac{si{n}^{-1}x}{a}=\lambda$

$\therefore si{n}^{-1}x=a\lambda$

$x=sina\lambda .........\left(i\right)$

$x=cosb\lambda ..........\left(ii\right)$

$y=tanc\lambda ..........\left(iii\right)$

From (i) & (ii), we get, $sina\lambda =cosb\lambda =sin\left(\frac{\pi }{2}-b\lambda \right)$

$\therefore a\lambda =\frac{\pi }{2}-b\lambda \Rightarrow \left(a+b\right)\lambda =\frac{\pi }{2}\Rightarrow \lambda =\frac{\pi }{2\left(a+b\right)}...........\left(iv\right)$

From (iii) y = tan c $\lambda$

$\Rightarrow cos\left(\frac{2c\pi }{2\left(a+b\right)}\right)=\frac{1-y^2}{1+y^2}\Rightarrow cos\left(\frac{\pi c}{a+b}\right)=\frac{1-y^2}{1+y^2}$

= A + B (say)

(1 + x)¹⁵ =

differentiating 15 (1 + x)¹⁴ =   $\sum r{.}^{15}{C}_r{x}^{r-1}$

put  = x = -1

$\Rightarrow 0={-}^{15}{C}_1+2{.}^{15}{C}_2-.......=A$      

$B{=}^{14}{C}_{13}=2^{13}$    

$\therefore A+B=2^{13}-14$     

$y=\overline{\overline{A}+B}=A.\overline{B}$

Quantum University is basically known for its placement records in north India. It holds a record of more than 85% batch getting placement through campus placement. and other students go for higher studies as well.

Let direction ratio of the normal to the required plane are l, m, n

$\begin{array}{l}\therefore 3l+m-2n=0\\ \therefore 2l-5m-n=0\\ \therefore \frac{l}{-11}=\frac{m}{-1}=\frac{n}{-17}\end{array}$             

Equation of required plane

11 (x – 1) + 1 (y – 2) + 17 (z + 3) = 0

All the charge given to a conducting sphere resides on outer surface.

  $e^{\left(co{s}^{2}x+co{s}^{4}x+co{s}^{6}x+.....\infty \right)}lo{g}_{e}2$

$=e^{\frac{co{s}^{2}x}{1-co{s}^{2}x}lo{g}_{e}2}=e^{co{t}^{2}xlo{g}_{e}2}=2^{co{t}^{2}x}$            

t² – 9t + 8 = 0

(t – 8) (t – 1) = 0

$t=2^{co{t}^{2}x}=8=2^3$        

$\Rightarrow co{t}^{2}x=3=co{t}^2\frac{\pi }{6}$       

$\frac{2sinx}{sinx+\sqrt[]{3}cosx}=\frac{2\times \frac{1}{2}}{\frac{1}{2}+\sqrt[]{3}\times \frac{\sqrt[]{3}}{2}}=\frac{1}{2}$

$\Delta E=13.6\left(\frac{1}{1^2}-\frac{1}{5^2}\right)=13.6\times \frac{24}{25}eV$

$\Rightarrow \frac{hc}{\lambda }=13.6\times \frac{24}{25}eV..........\left(1\right)$

With the help of conservation of linear momentum, we can write

$\frac{h}{\lambda }=m_H{v}_H\Rightarrow \frac{hc}{\lambda }=c{m}_H{v}_H\Rightarrow v_H=\frac{\frac{hc}{\lambda }}{c{m}_H}=\frac{13.6\times \frac{24}{25}\times 1.6\times 10^{-19}}{3\times 10^8\times 1.67\times 10^{-27}}=4.17m/s$

Any point on line  $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}=r$ is P (r + 3, 2r + 4, 2r + 5) lies on x + y + z = 17,

5r + 12 = 17

r = 1

$P\left(4,6,7\right)A\left(1,1,9\right)distAP=\sqrt[]{9+25+4}=\sqrt[]{38}$  

$R_i=\frac{\rho \mathcal{l}}{A}$

$R_f=\frac{\rho \left(1.25\mathcal{l}\right)}{\left(A/1.25\right)}={\left(1.25\right)}^2\times \frac{\rho \mathcal{l}}{A}$

$\Rightarrow R_f=1.5625\times R_i$

$\Rightarrow \frac{R_f-R_l}{R_l}\times 100=56.25\mathrm{\%}$

x² + y² = 36         ……(i)

y² = 9x      .(ii)

Solving (i) & (ii)

x² + 9x – 36 = 0

(x + 12) (x – 3) = 0

x = 3

Let $A={\displaystyle \underset{0}{\overset{3}{\int }}\left(\sqrt[]{36-x^2}-3\sqrt[]{x}\right)dx}$  

$={\left[\frac{x\sqrt[]{36-x^2}}{2}+18si{n}^{-1}\frac{x}{6}\right]}_0^3-3.{\frac{x^{3/2}}{3/2}]}_0^3$       

$=3\pi -\frac{3\sqrt[]{3}}{2}$        

Required area = $=\frac{1}{2}\pi {\left(6\right)}^2+2\left(3\pi -\frac{3\sqrt[]{3}}{2}\right)=\left(24\pi -3\sqrt[]{3}\right)sq.unit$

$v=\frac{\omega }{k}=\frac{30}{1}=30m/s$

$v=\sqrt[]{\frac{T}{\mu }}\Rightarrow T=\mu v^2=\frac{0.135\times 10^{-3}}{10^{-2}}\times {\left(30\right)}^2=12.15N=1215\times 10^{-2}N$