Differential Calculus
Differential Calculus
star_batch_jee_advanced_2025
Grade None
Question:
If $f(x) = 4x^3 - x^2 - 2x + 1$ and $g(x) = \begin{cases} \min\{f(t): 0 \leq t \leq x\}, & 0 \leq x \leq 1 \\ 3 - x & , 1 h''(x) h(x)$ for each $x \in J$. Then:
$g$ is increasing on $J$
$g$ is decreasing on $J$
$g$ is concave up on $J$
$g$ is concave down on $J$
Step-by-Step Solution
Key Concept: Decompose telescoping sums by separating into standard summation formulas and use asymptotic analysis to evaluate the limit.
Let $N_r = n(1^2+2^2+\cdots+n^2) - (1^2\cdot 2^2+2^2\cdot 3^2+\cdots+(n-1)^2\cdot n^2)$. Factor this as $N_r = n\sum r^2 - \sum(r^3-r^2) = (n+1)\sum r^2 - \sum r^3$. Taking the limit: $\lim_{n\to\infty}\frac{(n+1)\sum r^2 - \sum r^3}{\sum r^3} = \lim_{n\to\infty}\frac{(n+1)\frac{n(n+1)(2n+1)}{6} - \frac{n^2(n+1)^2}{4}}{\frac{n^2(n+1)^2}{4}} = \frac{4}{3}-1 = \frac{1}{3}$.
Correct Answer: 1