Question

If \( ax^2 + bx + e>0 \) for all \( x \in \mathbb{R} \) and the expressions \( cx^2 + ax + b \) and \( ax^2 + bx + c \) have their extreme values at the same point \( x \), then for the expression \( cx^2 + ax + b \), find the correct statement regarding its extreme value.

• A. Minimum value = \(\frac{4b}{3}\)
• B. Minimum value = \(\frac{4b}{3}\)
• C. Maximum value = \(\frac{4a}{3}\), Minimum value = \(\frac{3a}{4}\)
• D. Maximum value = \(\frac{3b}{4}\)

Hint:

Equate derivatives of quadratic functions to find common extrema and use relations among coefficients to evaluate maximum or minimum values.

Answer 1

The Correct Option is D

Given \( ax^2 + bx + e>0 \) for all \( x \), the quadratic has no real roots and opens upwards. Hence, \( a>0 \).
The derivatives of the two expressions are:
\[ \frac{d}{dx}(cx^2 + ax + b) = 2cx + a, \quad \frac{d}{dx}(ax^2 + bx + c) = 2ax + b. \]
Since the extreme values occur at the same point \( x_0 \), set derivatives zero and equalize \( x_0 \):
\[ 2cx_0 + a = 0 \implies x_0 = -\frac{a}{2c}, \quad 2ax_0 + b = 0 \implies x_0 = -\frac{b}{2a}. \]
Equate the two expressions for \( x_0 \): \[ -\frac{a}{2c} = -\frac{b}{2a} \implies \frac{a}{c} = \frac{b}{a} \implies a^2 = bc. \]
Calculate the extreme value of \( cx^2 + ax + b \) at \( x_0 \): \[ Q_1(x_0) = c \left(-\frac{a}{2c}\right)^2 + a \left(-\frac{a}{2c}\right) + b = \frac{a^2}{4c} - \frac{a^2}{2c} + b = b - \frac{a^2}{4c}. \]
Substitute \( a^2 = bc \): \[ Q_1(x_0) = b - \frac{bc}{4c} = b - \frac{b}{4} = \frac{3b}{4}. \]