Class 12th

$\frac{log\left(\left(2x+5\right)\left(x+1\right)\right)}{log\left(x+1\right)}+\frac{log\left(x+1\right)}{log\left(2x+5\right)}-4=0\Rightarrow \frac{log\left(2x+5\right)}{log\left(x+1\right)}+\frac{log\left(x+1\right)}{log\left(2x+5\right)}-3=0$

$\Rightarrow x\in \left(-1,0\right)\cup \left(0,\infty \right)$

$\frac{log\left(2x+5\right)}{log\left(x+1\right)}=1\Rightarrow x=-4$ (not possible) and $\frac{log\left(2x+5\right)}{log\left(x+1\right)}=2\Rightarrow x=2$

Hence only one solution is possible.

${\displaystyle \int dy}={\displaystyle \int \frac{cos\left(\frac{1}{2}co{s}^{-1}\left(e^{-x}\right)\right)}{e^x\sqrt[]{1-{\left(e^{-x}\right)}^2}}}dxputcos2\theta =e^{-x}\Rightarrow -2sin2\theta d\theta =-e^{-x}dx$

$\therefore {\displaystyle \int dy}={\displaystyle \int \frac{2cos\theta sin2\theta d\theta }{\sqrt[]{1-co{s}^{2}2\theta }}\Rightarrow y=2sin\theta +c\Rightarrow y=-1,\theta =0,c=0}$

$\therefore y=2\sqrt[]{\frac{1-e^{-x}}{2}}-1at\left(\alpha ,0\right),e^{\alpha }=2$

$?\left|{\overrightarrow{v}}_1\right|=\left|{\overrightarrow{v}}_2\right|\Rightarrow p^2-p-2=0\Rightarrow p-1,2$ we take only p = 2 (p > 0)

$\therefore cos\theta =\frac{\overrightarrow{v_2}.\overrightarrow{v_2}}{\left|\overrightarrow{v_1}\right|\left|\overrightarrow{v_2}\right|}=\frac{4\sqrt[]{3}+3}{13}\Rightarrow tan\theta =\frac{6\sqrt[]{3}-2}{4\sqrt[]{3}+3}=\frac{\alpha \sqrt[]{3}-2}{4\sqrt[]{3}+3}\therefore \alpha =6$

Compound A is phenol since phenol gives dark green colour with FeCl₃.

$A=\left[\begin{array}{ccc}\hfill 2\hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 2\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 2\hfill \end{array}\right]\Rightarrow \left|A\right|=4\Rightarrow \left|3adj\left(2A^{-1}\right)\right|=\left|3.2^{2}adj\left(A^{-1}\right)\right|$

$\Rightarrow 12^3\left|adj\left(A^{-1}\right)\right|=12^3{\left|A^{-1}\right|}^2=\frac{12^3}{{\left|A\right|}^2}=\frac{12*12*12}{16}=108$

Let equation of normal PQ is

$y=-tx+3t+\frac{3}{2}{t}^3$ and it is passing through

$P\left(3,\frac{3}{2}\right)\therefore t=1$

$\therefore Q\left(\frac{3}{2},3\right)\Rightarrow a=\frac{3}{2}andb=3$     

2 (a + b) = 9

(A) $N{i}^{2+}-4s^{0}3d^8$ - Paramagnetic & coloured

$M{n}^{7+}-4s^{0}3d^0$       - Diamagnetic & colourless

$H{g}^{2+}-6s^{0}4f^{14}5d^{10}$ - Diamagnetic & colourless

(B)  $C{u}^{+}-4s^{0}3d^{10}$    - Diamagnetic & colourless

$Z{n}^{2+}-4s^{0}4d^{10}$          - Diamagnetic & colourless

$M{n}^{4+}-4s^{0}3d^3$    - paramagnetic & coloured

(C)  $S{c}^{3+}-4s^{0}3d^0$

$\begin{array}{l}{V}^{5+}-4s^{0}3d^0\\ T{i}^{4+}-4s^{0}3d^0\end{array}$  

All are diamagnetic & colourless

(D)  $C{u}^{2+}-4s^{0}3d^9$  

$\begin{array}{l}C{r}^{3+}-4s^{0}3d^3\\ S{c}^{+}-4s^{1}3d^1\end{array}$

All are paramagnetic & coloured

All d- & f block paramagnetic cations are coloured also.

$\underset{x\to 0}{lim}\frac{\alpha x\left(1+x+\frac{x^2}{2}+....\right)-\beta \left(x-\frac{x^2}{2}+\frac{x^3}{3}+....\right)+\gamma x^2\left(1-x\right)}{x^3}=10$

For limit to exist a - b = 0 . (i), $\alpha +\frac{\beta }{2}+\gamma =0.........(ii)$

and $\frac{\alpha }{2}-\frac{\beta }{2}-\gamma =10$ . (iii)

Solving (i), (ii) and (iii) we get α = 6, β = 6 and $\lambda =-9\Rightarrow \alpha +\beta +\lambda =3$

  $P\left(\overline{A}\cap B\right)+P\left(A\cap \overline{B}\right)=1-k,$

$P\left(A\cap B\cap C\right)=k^2$             

$P\left(A\right)+P\left(B\right)-2P\left(A\cap B\right)=1-k$ .(i)

$P\left(B\right)+P\left(C\right)-2P\left(B\cap C\right)=1-k$ .(ii)

$P\left(A\right)+P\left(C\right)-2P\left(A\cap C\right)=1-2k$ .(iii)

Adding (i), (ii) and (iii) we get $P\left(A\cup B\cup C\right)=\frac{-4k+3}{2}+k^2$

$\Rightarrow P\left(A\cup B\cup C\right)=\frac{2k^2-4k+2+1}{2}=\frac{2{\left(k-1\right)}^2+1}{2}\Rightarrow P\left(A\cup B\cup C\right)>\frac{1}{2}$

1, 2 and 3 talks about a black hole and its distinctive features. However, 4^th sentence talks about a star and its life.