Class 12th

$\frac{3x^2-9x+17}{x^2+3x+10}=\frac{5x^2-7x+19}{3x^2+5x+12}$

$1+\frac{2x^2-12x+7}{x^2+3x+10}=1+\frac{2x^2-12x+7}{3x^2+5x+12}$

$\left(2x^2-12x+7\right)\left(\frac{1}{x^2+3x+10}-\frac{1}{3x^2+5x+12}\right)=0$

$\underset{x\to 10}{lim}{\left(\frac{{\left(x+2cosx\right)}^3+2{\left(x+2cosx\right)}^2+3sin\left(x+2cosx\right)}{{\left(x+2\right)}^3\pm 2{\left(x+2\right)}^2+3sin\left(x+2\right)}\right)}^x$

$e^{\frac{100}{16+3sin2}\left[\frac{12-3\left(4\right)+8*1-8+3cos2-3cos2}{1}\right]}$ , using L.H.L Rule, e0 = 1

$2x^2-12x+7=0or3x^2+5x+12=x^2+3x+10$

$x^2+x+1=0$

Sum of roots = 6, no solution.

$f\left(x\right)=0\Rightarrow {\left(x-p\right)}^2-q=0$  

Roots are  $p+\sqrt[]{q},p-\sqrt[]{q}$

Now, $\left|f\left(a_i\right)\right|=500$  

$Let{a}_1,a_2,a_3...ara,a+d,a+2d,a+3d$               

=>  $\frac{9}{4}{d}^2-q=500$           …….(i)

and ${\left|f\left(a_1\right)\right|}^2={\left|f\left(a_2\right)\right|}^2$

From equation (i)

$\frac{9}{4}.\frac{4q}{5}-q=500$     

$\frac{4q}{5}=500$           

and  $2\sqrt[]{q}=2*\frac{50}{2}=50$

$A+B=\left[\begin{array}{cc}\hfill \beta +1\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill \alpha \hfill \end{array}\right]$

${\left(A+B\right)}^2=\left[\begin{array}{cc}\hfill {\left(\beta +1\right)}^2\hfill & \hfill 0\hfill \\ \hfill 3\left(\beta +1\right)+3\alpha \hfill & \hfill {\alpha }^2\hfill \end{array}\right]$

$\therefore \left[\begin{array}{cc}\hfill 1\hfill & \hfill -\alpha +1\hfill \\ \hfill 2\alpha +4\hfill & \hfill {\alpha }^2\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill {\left(\beta +1\right)}^2\hfill & \hfill 0\hfill \\ \hfill 3\left(\alpha +\beta +1\right)\hfill & \hfill {\alpha }^2\hfill \end{array}\right]$

= 1 = 1

$B^2=\left[\begin{array}{cc}\hfill \beta \hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right]\left[\begin{array}{cc}\hfill \beta \hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right]$

$\beta =0,\alpha =-1={\alpha }_2$

$\left|{\alpha }_1-{\alpha }_2\right|=\left|1-\left(-1\right)\right|=2$

$f\text{'}\left(a\right)=f\text{'}\left(b\right)=f\text{'}\left(c\right)=2$

f'' (x) is zero for atleast $x_1\in \left(a,b\right)\&x_2\in \left(b,c\right)$ ${}_{}{}_{}$

$RS\equiv \left(\alpha ,-,1\beta \right)$

$DRofPQ\equiv \left(\frac{56}{17}+2,\frac{43}{17}+1,\frac{111}{17}-1\right)$

$\equiv \left(\frac{90}{17},\frac{60}{17},\frac{94}{17}\right)$

$90\alpha +94\beta =60$

$\begin{array}{l}\beta =-15,\alpha =-15\\ {\alpha }^2+{\beta }^2=450\end{array}$

$\left[Cu{\left(en\right)}_2{\left(SCN\right)}_2\right]$

More stable isomers = 3 (trans isomers)

  $Mn{O}_4^{-2}\to A+B$

Oxidation state of Mn in B < A

  $Mn{O}_4^{-2}+H^{+}\to Mn{O}_4^{-}+Mn{O}_2$             

B is MnO₂

$\therefore$ Oxidation state of Mn = +4

    ${\therefore }_{25}M{n}^{+4}=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{0}3d^3$          

$\therefore$ unpaired electron = 3

Spin only magnetic moment  $\left(\mu \right)$

$=\sqrt[]{n\left(n+2\right)}=\sqrt[]{3\left(3+2\right)}=\sqrt[]{15}$

$=3.87\approx 4$

  $C\mathcal{l}{F}_3$ -> T-shaped (sp³d)

 IF₇ -> Pentagonal bipyramidal (sp³d³)

BrF₅ -> Square pyramidal (sp³d²)

BrF₃ -> T-Shaped (sp³d)

I₂Cl₆ -> Triangular bipyramidal (sp³d)

$I{F}_5\to$ Square Pyramidal (sp³d²)

ClF -> (sp³)

ClF₅ -> Square Pyramidal (sp³d²)

Br₅, IF₅ & ClF₅ -> Square Pyramidal

t1/2 = 0.301 min

t = 2 min

$K=\frac{2.303}{t}\mathcal{l}og\left(\frac{Co}{Ct}\right)$

$\frac{0.693}{0.301}=\frac{2.303}{2}\mathcal{l}og\left(\frac{Co}{Ct}\right)$

$\frac{2.303*0.301}{0.301}=\frac{2.303}{2}\mathcal{l}og\left(\frac{Co}{Ct}\right)$

$\therefore 2=\mathcal{l}og\left(\frac{Co}{Ct}\right)$

$\frac{Co}{Ct}=10^2=100$

Ans. 100