Differential Calculus
Differential Calculus
star_batch_jee_advanced_2025
Grade 12
Question:
Let $f$ be a twice differentiable function defined in $[-3,3]$ such that $f(0) = -4, f'(3) = 0, f'(-3) = 12$ and $f''(x) \geq -2\forall x \in [-3,3]$. If $g(x) = \int_0^x f(t)dt$ then find maximum value of $g(x), x \in [-3,3]$
Step-by-Step Solution
Key Concept: Use mean value theorem and integration to find explicit form of $f(x)$, then optimize the resulting integral function $g(x)$ on the closed interval.
Given $f'(x) ≥ -2√x$ on $[-3,3]$, integrating from $-3$ to $3$ yields $∫_{-3}^{3} f'(x)dx ≥ -2∫_{-3}^{3} √x dx$, which gives $f(3) - f(-3) ≥ -12$. By LMVT, $f'(x) = -2$ at some point. Integrating $f'(x) = -2x + C$ with $f'(3) = 6 - C = 0$ gives $C = 6$, so $f(x) = 6x - x^2 - 4$. Computing $g(x) = ∫_0^x f(t)dt = -\frac{x^3}{3} + 3x^2 - 4x$ and finding critical points: $g'(x) = -x^2 + 6x - 4 = 0$ yields $x = 3 ± √5$. Evaluating at $x = -3$ gives maximum $g(-3) = 48$.
Correct Answer: 48