Let f(x) and g(x) be two differentiable functions in R such that f(2)=8,g(2)=0,f(4)=10,g(4)=8 and for atleast one x∈(2,4),g' (x)=kf'(x), then k is

Given: f(x) and g(x) be two differentiable functions in R and f(2)=8,g(2)=0,f(4)=10 and g(4)=8
Now, According to cauchy mean value theorem there exists one c∈(2,4) st.
$\frac{{\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{c}\right)}{{\mathrm{g}}^{\mathrm{\prime }}\left(\mathrm{c}\right)}=\frac{\mathrm{f}\left(4\right)-\mathrm{f}\left(2\right)}{\mathrm{g}\left(4\right)-\mathrm{g}\left(2\right)}-----\left(1\right)$
Putting the given values in (1)
$\begin{array}{l}\Rightarrow \frac{{\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{c}\right)}{{\mathrm{g}}^{\mathrm{\prime }}\left(\mathrm{t}\right)}=\frac{10-8}{8-0}\\ \Rightarrow \frac{{\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{c}\right)}{{\mathrm{g}}^{\mathrm{\prime }}\left(\mathrm{c}\right)}=\frac{2}{8}\\ \Rightarrow \frac{{\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)}{{\mathrm{g}}^{\mathrm{\prime }}\left(\mathrm{x}\right)}=\frac{1}{4}\text{ (replacing }\mathrm{c}\text{ by }\mathrm{x}\text{ ) }\\ \Rightarrow 4{\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)={\mathrm{g}}^{\mathrm{\prime }}\left(\mathrm{x}\right)\text{ for atleast one }\\ \mathrm{x}\in (2,4).\end{array}$
∴ option D. g^' (x)=4f'(x) for at least one x∈(2,4) is correct Answer.