Applications of Derivatives
Differential Calculus-2
star_batch_jee_advanced_2025
Grade 12
Question:
Let $f(x) = ax^2 - b|x|$, where $a$ and $b$ are constants. Then at $x = 0$, $f(x)$ has:
a maxima whenever $a > 0, b > 0$
a maxima whenever $a > 0, b 0, b > 0$
neither a maxima nor minima whenever $a > 0, b < 0$
Step-by-Step Solution
Key Concept: The absolute value term $-b|x|$ creates a 'V-shaped' downward dip that dominates the parabola $ax^2$ near the origin when $b > 0$, producing a local maximum at $x = 0$ regardless of the sign of $a$ in that regime.
We analyze $f(x) = ax^2 - b|x|$ near $x = 0$. Since $f(x)$ is even (as $f(-x) = a(-x)^2 - b|-x| = ax^2 - b|x| = f(x)$), we examine the behavior for small positive $x$. For $x > 0$: $f(x) = ax^2 - bx$, so $f'(x) = 2ax - b$ and $f''(x) = 2a$. At $x = 0^+$, we have $f'(0^+) = -b$. For a maxima, we need $f'(0^+) ≥ 0$, which requires $b ≥ 0$. Additionally, the second derivative test around $x=0$ requires $f''(0) = 2a < 0$, meaning $a < 0$. However, examining the actual behavior: when $a > 0, b > 0$, near $x = 0$ the term $-b|x|$ dominates over $ax^2$, making $f$ decrease as $|x|$ increases from 0, confirming a local maxima at $x = 0$.
Correct Answer: 1