A value of c for which the minimum value of f(x) = x2 – 4cx + 8c is greater than the maximum value of g(x) = – x2 + 3cx – 2c, is
\(\dfrac{1}{2}\)
Minimum of \(f\): \(f(x)=(x-2c)^{2}-4c^{2}+8c \Rightarrow f_{\min}=8c-4c^{2}\).
Maximum of \(g\): \(g(x)=-\left(x-\tfrac{3c}{2}\right)^{2}+\tfrac{9c^{2}}{4}-2c \Rightarrow g_{\max}=\tfrac{9c^{2}}{4}-2c\).
\(8c-4c^{2}>\tfrac{9c^{2}}{4}-2c \Rightarrow 10c>\tfrac{25c^{2}}{4} \Rightarrow c(25c-40)<0 \Rightarrow 0<c<1.6\).
Among the given values only \(c=\tfrac{1}{2}\) lies in this range.
Hence, \(\dfrac{1}{2}\).
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