Class 12th

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.

Depression in freezing point will be maximum for Al₂ (SO₄)₃ since its Van't Hoff factor is the highest. So its solution will have the lowest freezing point.

 Let the smallest angle is C =?

Therefore angle A = 90° -?

And angle B = 90°

i.e. b > a > c

Here, a = 2 R cos? , b = 2R, c = 2R sin?

b² = a² + c²

according to question,

$\begin{array}{l}\frac{1}{c^2}=\frac{1}{a^2}+\frac{1}{b^2}\\ ?\frac{a^2{b}^2}{c^2}=a^2+b^2\end{array}$     

=> $sin?=?\frac{\sqrt[]{5}?1}{2}$

  $l={\displaystyle {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\left(\left[x\right]+\left[-sinx\right]\right)dx}$ . (i)

$I={\displaystyle {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\left(\left[-x\right]+\left[sinx\right]\right)dx}$ . (ii)

by using property $\left({\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx=}{\displaystyle \underset{a}{\overset{b}{\int }}f\left(a+b-x\right)dx}\right)$

Adding (i) and (ii) we get 2l = ${\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\left(-2\right)dx=-2\pi \Rightarrow l=-\pi }$