CAT 2025

Slot 1

QA

Question & 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 $$c - \dfrac{b^2}{4a}$$

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

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

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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: $$0 < c < \dfrac{8}{5}$$

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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 ❌

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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 $$g(x)$$.