CAT 2025 Slot1 Quant: Finding the value of c for which the minimum of a function f(x) = x^2 – 4cx + 8c is greater than the maximum of g(x) = -x^2 + 3cx – 2c

Step 1: Find the Minimum of $f(x)$
For $f(x) = x^2 - 4cx + 8c$, the minimum occurs at $x = -b/2a = 4c/2 = 2c$.
$f(2c) = (2c)^2 - 4c(2c) + 8c = 4c^2 - 8c^2 + 8c = \mathbf{8c - 4c^2}$.

Step 2: Find the Maximum of $g(x)$
For $g(x) = -x^2 + 3cx - 2c$, the maximum occurs at $x = -b/2a = -3c/-2 = 1.5c$.
$g(1.5c) = -(1.5c)^2 + 3c(1.5c) - 2c = -2.25c^2 + 4.5c^2 - 2c = \mathbf{2.25c^2 - 2c}$.

Step 3: Set up the Inequality
$8c - 4c^2 > 2.25c^2 - 2c \implies 10c > 6.25c^2$.
Dividing by $c$ (checking $c > 0$ from options): $10 > 6.25c \implies c < 10/6.25 = 1.6$.
Also, if $c$ is negative, the inequality reverses. Testing $c = 1/2$:
$f_{min} = 8(0.5) - 4(0.25) = 4 - 1 = 3$.
$g_{max} = 2.25(0.25) - 2(0.5) = 0.5625 - 1 = -0.4375$.
Since $3 > -0.4375$, option **D** is correct.