Differential Calculus
Differential Calculus
star_batch_jee_advanced_2025
Grade None
Question:
Given $f(x) = \frac{e^x - \cos 2x - x}{x^2}$ for $x \in \mathbb{R} - \{0\}$, $\{x\}$ is fractional part function
$$g(x) = \begin{cases} f\{x\} & n 1)$ is equal to :
$1$
$0$
$\frac{\pi}{2}$
Does not exist
Step-by-Step Solution
Key Concept: Continuity requires matching left and right limits with the function value using L'Hôpital's rule or Taylor expansions.
We verify continuity of $g(x)$ at $x = b$ using left and right limits. $\lim_{h \to 0^+} g(b+h) = \lim_{h \to 0} \frac{e^h - \cos 2h - h}{h^2} = \lim_{h \to 0} \frac{e^h - h - 1}{h^2} + \lim_{h \to 0} \frac{1 - \cos 2h}{4h^2} = \frac{1}{2} + 2 = \frac{5}{2}$. Similarly, $\lim_{h \to 0^-} g(b-h) = g(b) = \frac{5}{2}$. Since all limits equal $\frac{5}{2}$, $g(x)$ is continuous at $b$ and hence on all of $\mathbb{R}$.
Correct Answer: 4