Differential Calculus
Differential Calculus
star_batch_jee_advanced_2025
Grade 12
Question:
Let $a = \min(x^2 + 2x + 3, x \in \mathbb{R})$ and $b = \lim_{x \to 0} \frac{\sin x \cos x}{e^x - e^{-x}}$. Then the value of $\sum_{r=0}^{n} a^r b^{n-r}$ is :
$\frac{2^{n+1} + 1}{3 \cdot 2^n}$
$\frac{2^{n+1} - 1}{3 \cdot 2^n}$
$\frac{2^n - 1}{3 \cdot 2^n}$
$\frac{4^{n+1} - 1}{3 \cdot 2^n}$
Step-by-Step Solution
Key Concept: Find the minimum of a quadratic function, evaluate a limit using standard Taylor approximations, then apply geometric series formula.
From $a = [(x+1)^2 + 2]_{\min}$, setting derivative to zero gives $a = 2$. For $b = \lim_{x \to 0} \frac{\sin 2x}{2(e^x - 1) \cdot 2x}$, using $\sin 2x \sim 2x$ and $e^x - 1 \sim x$ as $x \to 0$: $b = \lim_{x \to 0} \frac{2x}{2 \cdot x \cdot 2x} = \frac{1}{2}$. The geometric series sum gives $\sum_{r=0}^{n} a^r b^{n-r} = \frac{4^{n+1} - 3^{n+1}}{3 \cdot 2^n}$.
Correct Answer: 4