The range of values of ‘k’ for which the equation ${\mathrm{x}}^4-14{\mathrm{x}}^2+24\mathrm{x}-\mathrm{k}=0$ has four roots, is

Let $$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^4-14{\mathrm{x}}^2+24\mathrm{x}-\mathrm{k} \\ \Rightarrow {\mathrm{f}}^1\left(\mathrm{x}\right)=4{\mathrm{x}}^3-28\mathrm{x}+24 \\ \Rightarrow {\mathrm{f}}^1\left(\mathrm{x}\right)=4(x-1)(x-2)(x+3) \end{gathered}$$$\Rightarrow \mathrm{Roots} \mathrm{of} {\mathrm{f}}^1\left(\mathrm{x}\right)=0 \mathrm{are} -3,2,1$

$\text{ Now, }f(-3)=-117-k,f\left(1\right)=11-k\text{ and }f\left(2\right)=8-k$

Clearly,  $\mathrm{f}(-3)<0 \mathrm{and} \mathrm{f}\left(1\right)>0 \mathrm{and} \mathrm{f}\left(2\right)<0$

$\begin{array}{l}\Rightarrow (8-k)<0,11-\mathrm{k}>0\mathrm{\&}-117-\mathrm{k}<0\\ \Rightarrow \mathrm{k}>8,\mathrm{k}<11\text{ and }\mathrm{k}>-117\\ \Rightarrow 8<\mathrm{k}<11\end{array}$