Question

If both roots of the equation $x^2 - 5ax + 6a = 0$ exceed 1, then the range of 'a' is

• A. $[-1, 0) \cup [\frac{24}{25}, \infty)$
• B. $[\frac{24}{25}, \infty)$
• C. $[-1, 0)$
• D. $\mathbb{R}$

Hint:

Problems on the location of roots always involve a combination of three conditions: the discriminant ($D$), the value of the function at the given point ($f(k)$), and the position of the vertex ($-B/2A$). It's helpful to visualize the parabola to understand why each condition is necessary. Drawing a number line to find the intersection of the intervals obtained from each condition can prevent errors.

Answer 1

The Correct Option is B

Step 1: Conditions for Location of Roots Let roots be \( \alpha, \beta \). We require \( \alpha>1 \) and \( \beta>1 \). Let \( f(x) = x^2 - 5ax + 6a \). Conditions: 1. Discriminant \( D \ge 0 \) (for real roots). 2. \( f(1)>0 \) (since coefficient of \( x^2>0 \)). 3. Vertex \( x_v>1 \).
Step 2: Solve Inequalities 1. \( D \ge 0 \): \( (-5a)^2 - 4(1)(6a) \ge 0 \) \( 25a^2 - 24a \ge 0 \implies a(25a - 24) \ge 0 \) \( a \in (-\infty, 0] \cup [24/25, \infty) \). 2. \( f(1)>0 \): \( 1^2 - 5a(1) + 6a>0 \) \( 1 + a>0 \implies a>-1 \). 3. \( -\frac{B}{2A}>1 \): \( \frac{5a}{2}>1 \implies 5a>2 \implies a>\frac{2}{5} = 0.4 \).
Step 3: Find Intersection We need the intersection of: - \( a \in (-\infty, 0] \cup [0.96, \infty) \) - \( a \in (-1, \infty) \) - \( a \in (0.4, \infty) \) Comparing \( a>0.4 \) with the discriminant sets: The set \( (-\infty, 0] \) is disjoint from \( (0.4, \infty) \). The set \( [0.96, \infty) \) is fully contained in \( (0.4, \infty) \). Thus, the valid range is \( [24/25, \infty) \).