Let $f\left(x\right)=1+e^{\mathcal{l}n\left(\mathcal{l}nx\right)}.\mathcal{l}n(k^2+25)$  and  $g\left(x\right)=\frac{1}{\left|x\right|-1}$. If  $\underset{x\to 1}{\lim }{\left(f\left(x\right)\right)}^{g\left(x\right)}=k(2{\sin }^2\alpha +3\cos \beta +5),k>0$ and $\alpha ,\beta \in R$ , then which of the following is(are) CORRECT? 

As $\underset{x\to 1}{\lim }{(1+\ln x.\ln (k^2+25))}^{\frac{1}{\left|x\right|-1}}=\underset{x\to 1}{\lim }{(1+\ln x.\ln (k^2+25))}^{\frac{1}{x-1}}$

$=e^{\underset{x\to 1}{\lim }\frac{\ln x\ln (k^2+25)}{x-1}}=e^{\ln (k^2+25)}=k^2+25$

$\therefore k^2+25=k(2{\sin }^2\alpha +3\cos \beta +5)$

$\Rightarrow 2{\sin }^2\alpha +3\cos \beta +5=k+\frac{25}{k}$

$\because \frac{k+\frac{25}{k}}{2}\ge \sqrt{k.\frac{25}{k}}\Rightarrow k+\frac{25}{k}\ge 10$

But L.H. S.  $\le 10$
So. both L.H.S. and R.H.S. should be equal to 10.
$\Rightarrow {\sin }^2\alpha =1$ and $\cos \beta =1$

Hence,  $\frac{{\sin }^{10}\alpha +{\cos }^5\beta }{{\sin }^2\alpha +\cos \beta }=\frac{1+1}{1+1}=1$