Class 11th

Only tritium is the isotope of hydrogen which is radioactive in nature that emits low energy b^- particles since its $\frac{n}{p}$ ratio is above stability belt.

$?f\left(x\right)=ln\left(x+\sqrt[]{1+x^2}\right)\therefore f\left(-x\right)=-f\left(x\right)$ .

Hence f(x) is an odd function.

$Nowg\left(t\right)={\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}cos\left(\frac{\pi }{4}t+f\left(x\right)\right)dx}$

Put t = 0, g(0) = ${\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}cos\left(f\left(x\right)\right)dx=2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}cos\left(f\left(x\right)\right)dx}...............\left(i\right)$

where cos(f(x)) is an even function.

Now again put t = 1,

$\therefore g\left(1\right)=\frac{1}{\sqrt[]{2}}*g\left(0\right),from\left(i\right)$

$\therefore g\left(0\right)=\sqrt[]{2}g\left(1\right)$

$f\left(x\right)=\frac{5x+3}{6x-\alpha }=y.....................(i)$

$\Rightarrow x=\frac{\alpha y+3}{6y-5}\Rightarrow f^{-1}\left(x\right)=\frac{\alpha x+3}{6x-5}..............(ii)$

According to question, $f\left(x\right)=f^{-1}\left(x\right)\therefore$  from (i) and (ii) we get $\propto =5$

$\frac{x+1}{a}=\frac{y}{1}=\frac{z-1}{a}...........\left(i\right)$

$\frac{x+2}{a}=\frac{y}{1}=\frac{z}{\frac{3}{b}}..........\left(ii\right)$

Let A (-1, 0, 1) and B (-2, 0, 0) $$ $\therefore$ direction ratios of AB = -1, 0, -1

$$ $\therefore$  lines are coplanar

$\therefore \left|\begin{array}{ccc}\hfill a\hfill & \hfill 1\hfill & \hfill a\hfill \\ \hfill 3\hfill & \hfill 1\hfill & \hfill \frac{3}{b}\hfill \\ \hfill -1\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right|=0\Rightarrow b=1anda\in R-\left\{0\right\}$

10 =   $\frac{7+10+11+15+a+b}{6}\Rightarrow a+b=17$  . (i)

$\Rightarrow \frac{20}{3}=\frac{7^2+10^2+11^2+15^2+a^2+b^2}{6}-10^2$

a² + b² = 145       . (ii)

solving (i) and (ii) we get a = 8, b = 9 or a = 9, b = 8

|a – b|= 1

Centre of the smallest circle is A

Centre of the largest circle is B

$r_1=\left|CP+CB\right|=3\sqrt[]{2}+3$ and

$r_2=\left|CP-CB\right|=3\sqrt[]{2}-3$    

$\frac{r_1}{r_2}=\frac{3\sqrt[]{2}+3}{3\sqrt[]{2}-3}=3+2\sqrt[]{2}$

Electronic configuration of Ga⁺ ion = [Ar] 3d¹⁰ 4s² 4p⁰

Last electron goes into S-orbital, hence

Azimuthal quantum number $\left(\mathcal{l}\right)$ for last electron = 0

Let mid point of PQ is R (h, k)

$\therefore h=\frac{\alpha +\frac{4{\alpha }^2+\alpha +1}{2}}{2}and$

$k=\frac{4{\alpha }^2+1+\frac{4{\alpha }^2+\alpha +1}{2}}{2}$   

Eliminate a from above these two, we get

2 (3x – y)² + (x – 3y) + 2 = 0

$ta{n}^{-1}\left(\frac{2*\frac{3}{5}}{1-\frac{9}{25}}\right)+si{n}^{-1}\frac{5}{13}\Rightarrow ta{n}^{-1}\frac{15}{8}+ta{n}^{-1}\frac{5}{12}=ta{n}^{-1}\frac{\frac{15}{8}+\frac{5}{12}}{1-\frac{15}{8}\frac{5}{12}}=ta{n}^{-1}\frac{220}{21}$

$\therefore tan\left(ta{n}^{-1}\frac{220}{21}\right)=\frac{220}{21}$