All the values of k for which the quadratic polynomial $\mathrm{f}\left(\mathrm{x}\right)=-2{\mathrm{x}}^2+\mathrm{kx}+{\mathrm{k}}^2+5$ has two distinct zeros and only one of them satisfying 0 < x < 2, lie in the interval (a, b). The value of (a + 10b) is

For two distinct roots, D > 0 i.e., k² + 8(k² + 5) > 0 which is always true

Also let $\mathrm{f}\left(\mathrm{x}\right)=-2{\mathrm{x}}^2+\mathrm{kx}+{\mathrm{k}}^2+5=0$

But f(0) > 0. So, f(2) < 0

$\begin{array}{l}-8+2\mathrm{k}+{\mathrm{k}}^2+5<0\\ \Rightarrow {\mathrm{k}}^2+2\mathrm{k}-3<0\\ \Rightarrow (\mathrm{k}+3)(\mathrm{k}-1)<0\\ \mathrm{k}\in (-3,1)\\ \Rightarrow \mathrm{a}=-3;\mathrm{b}=1\\ \Rightarrow \mathrm{a}+10\mathrm{b}=-3+10=7\end{array}$