Sets, Relations & Functions
Functions
star_batch_jee_advanced_2025
Grade 11
Question:
Least value of the expression $\frac{1}{2bx - (x^2 + b^2 + \sin^2 x)}$, $x \in [-1, 0]$, $b \in [2, 3]$ is:
\frac{1}{4}
-\frac{1}{4}
\frac{-1}{8 + \sin^2 1}
None of these
Step-by-Step Solution
Key Concept: Rewrite $f(x)$ as a sum of squared terms to identify its minimum and maximum over the given domain constraints.
Given $f(x) = x^2 + b^2 + \sin^2 x - 2bx = (x-b)^2 + \sin^2 x$, with $x \in [-1, 0]$ and $b \in [2, 3]$. As $x$ increases from $-1$ to $0$, $(x-b)^2$ decreases from $(1+b)^2$ to $b^2$ and $\sin^2 x$ decreases from $\sin^2 1$ to $0$. Therefore $f(x) \in [4 + 16 + \sin^2 1]$. The expression $\frac{1}{-f(x)}$ achieves its least (most negative) value when $f(x)$ is maximum, giving $\frac{-1}{4}$.
Correct Answer: 2