Question

If \( \alpha, \beta, \gamma \) are the roots of the equation \( 8x^3 - 42x^2 + 63x - 27 = 0 \), If \( \beta<\gamma<\alpha \) and \( \beta, \gamma, \alpha \) are in geometric progression, then the extreme value of the expression \( \gamma x^2 + 4\beta x + \alpha \) is:

• A. \( \frac{3}{4} \)
• B. \( 3 \)
• C. \( \frac{3}{2} \)
• D. \( \frac{21}{4} \) \bigskip

Hint:

In problems involving geometric progressions, use the relationships between the terms to express the roots in terms of a single variable, simplifying calculations.

Answer 1

The Correct Option is C

We are given the cubic equation: \[ 8x^3 - 42x^2 + 63x - 27 = 0 \] with roots \( \alpha, \beta, \gamma \) such that \( \beta<\gamma<\alpha \) and they form a geometric progression. 

Step 1: Relationship Between Roots Since \( \beta, \gamma, \alpha \) are in geometric progression, we let: \[ \beta = \frac{a}{r}, \quad \gamma = a, \quad \alpha = ar \] Using Vieta’s formulas: 1. Sum of roots: \[ \alpha + \beta + \gamma = \frac{42}{8} = \frac{21}{4} \] Substituting values: \[ ar + \frac{a}{r} + a = \frac{21}{4} \] Multiplying both sides by \( r \): \[ a r^2 + a + a r = \frac{21}{4}r \] Factoring: \[ a(r^2 + r + 1) = \frac{21}{4}r \] 2. Product of roots: \[ \alpha \beta \gamma = \frac{27}{8} \] Substituting values: \[ (ar) \cdot \left(\frac{a}{r}\right) \cdot a = \frac{27}{8} \] \[ a^3 = \frac{27}{8} \] \[ a = \frac{3}{2} \] 

Step 2: Finding the Extreme Value of \( \gamma x^2 + 4\beta x + \alpha \) The given quadratic expression: \[ f(x) = \gamma x^2 + 4\beta x + \alpha \] Since the coefficient of \( x^2 \) is \( \gamma \), the extreme value occurs at: \[ x = -\frac{4\beta}{2\gamma} = -\frac{2\beta}{\gamma} \] Substituting \( \beta = \frac{a}{r} \) and \( \gamma = a \): \[ x = -\frac{2(a/r)}{a} = -\frac{2}{r} \] The extreme value is: \[ f\left(-\frac{2}{r}\right) = \gamma \left(-\frac{2}{r}\right)^2 + 4\beta \left(-\frac{2}{r}\right) + \alpha \] \[ = a \left(\frac{4}{r^2}\right) - 8 \frac{a}{r} + ar \] \[ = \frac{4a}{r^2} - \frac{8a}{r} + ar \] Substituting \( a = \frac{3}{2} \) and solving, we get: \[ \frac{3}{2} \] 

Final Answer: \(\boxed{\frac{3}{2}}\)