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CAT 2025Slot 1QAQues & Sol

Modern MathMedium

Question

A value of c for which the minimum value of $f(x)=x^{2}-4cx+8c$ is greater than the maximum value of $g(x)=-x^{2}+3cx-2c$, is

Options

Solution

1. Concept Used

  • Topic: Quadratic Functions — Vertex (Minimum/Maximum) and Inequality
  • Formula: For a quadratic $ax^2 + bx + c$, the vertex occurs at $x = -\dfrac{b}{2a}$, giving an extremum value of $c - \dfrac{b^2}{4a}$

2. Calculation

Step 1: Find the minimum value of $f(x) = x^2 - 4cx + 8c$

Since the coefficient of $x^2$ is positive ($a = 1 > 0$), this parabola opens upward, so it has a minimum.

The minimum occurs at $x = -\dfrac{-4c}{2(1)} = 2c$.

Substituting $x = 2c$ into $f(x)$:

$f(2c) = (2c)^2 - 4c(2c) + 8c = 4c^2 - 8c^2 + 8c = -4c^2 + 8c$

So, minimum of $f(x) = -4c^2 + 8c$.

Step 2: Find the maximum value of $g(x) = -x^2 + 3cx - 2c$

Since the coefficient of $x^2$ is negative ($a = -1 < 0$), this parabola opens downward, so it has a maximum.

The maximum occurs at $x = -\dfrac{3c}{2(-1)} = \dfrac{3c}{2}$.

Substituting $x = \dfrac{3c}{2}$ into $g(x)$:

$g\left(\dfrac{3c}{2}\right) = -\left(\dfrac{3c}{2}\right)^2 + 3c \cdot \dfrac{3c}{2} - 2c$

$= -\dfrac{9c^2}{4} + \dfrac{9c^2}{2} - 2c = -\dfrac{9c^2}{4} + \dfrac{18c^2}{4} - 2c = \dfrac{9c^2}{4} - 2c$

So, maximum of $g(x) = \dfrac{9c^2}{4} - 2c$.

Step 3: Apply the condition — minimum of $f$ > maximum of $g$

$-4c^2 + 8c > \dfrac{9c^2}{4} - 2c$

Multiply through by 4:

$-16c^2 + 32c > 9c^2 - 8c$

$0 > 25c^2 - 40c$

$25c^2 - 40c < 0$

$5c(5c - 8) < 0$

This inequality holds when: $0 < c < \dfrac{8}{5}$

Step 4: Check which option lies in $\left(0, \dfrac{8}{5}\right)$

  • op1: $c = 2$ → $2 > \dfrac{8}{5}$ ❌
  • op2: $c = \dfrac{1}{2}$ → $0 < \dfrac{1}{2} < \dfrac{8}{5}$ ✅
  • op3: $c = -\dfrac{1}{2}$ → Negative, outside range ❌
  • op4: $c = -2$ → Negative, outside range ❌

3. Solution

Answer = Option B ✅

The value $c = \dfrac{1}{2}$ lies in the valid range $\left(0, \dfrac{8}{5}\right)$, satisfying the condition that the minimum of $f(x)$ is greater than the maximum of $g(x)$.