Application of Derivatives
Maxima and Minima
IIT-JEE 2009
Grade 12
Question:
Let $f(x) = (1 + b^2)x^2 + 2bx + 1$ and let $m(b)$ denote the minimum value of $f(x)$ for a given $b$. As $b$ varies over $\mathbb{R}$, the range of $m(b)$ is:
(1) $[0, 1]$
(2) $\left(0, \frac{1}{2}\right]$
(3) $\left[\frac{1}{2}, 1\right]$
(4) $(0, 1]$
Step-by-Step Solution
Key Concept: Since $f$ is a quadratic in $x$ with positive leading coefficient, its minimum value occurs at the vertex $x = -\frac{b}{1 + b^2}$ — substitute this back to express $m(b)$ purely in terms of $b$.
The detailed step-by-step mathematical proof is available inside the Mathbee app workspace.
Correct Answer: (1)