Differential Calculus
Differential Calculus
star_batch_jee_advanced_2025
Grade 12
Question:
MATCH THE FOLLOWING:
Column 1:
(A) $f(x) = \begin{cases} e^{-1/x^2}, & x \neq 0 \\ 0, & x = 0 \end{cases}$
(B) $f(x) = \begin{cases} e^{-1/x}, & x > 0 \\ 0, & x \leq 0 \end{cases}$
(C) $f(x) = \begin{cases} e^{-x} + \frac{1}{x}, & x \in (e, \pi) \\ 0, & x \notin (e, \pi) \end{cases}$
(D) $f(x) = \begin{cases} x^2 \sin\frac{1}{x}, & x \neq 0 \\ 0, & x = 0 \end{cases}$
Column 2:
(p) First derivative exists
(q) First derivative is continuous
(r) Second derivative exists
(s) Second derivative is continuous
Step-by-Step Solution
Key Concept: The Gaussian-like function $e^{-1/x^2}$ decays faster than any polynomial, making all derivatives vanish at the origin when divided by appropriate powers of $x$.
For part (A), $f(x) = \begin{cases} \frac{2}{x^3}e^{-1/x^2}, & x \neq 0 \\ 0, & x = 0 \end{cases}$. Computing $f'(0)$ using L'Hôpital's rule on $\lim_{h \to 0} \frac{e^{-1/h^2}}{h}$ gives $0$. Similarly, $f''(0) = 0$. By repeated application of L'Hôpital's rule, $f^{(k)}(0) = 0$ for all $k$, so $f$ is infinitely differentiable at $x = 0$. Part (B) is similar. For part (C), $f_1(x) = g(x-c)$ and $f_2(x) = g(\pi - x)$ where $g(x) = e^{-1/x}$, so $f(x) = f_1(x)f_2(x)$ is infinitely differentiable since both factors are. Part (D) has only first derivative existing and not continuous.
Correct Answer: [A-p, q, r, s] [B-p, q, r, s] [C-p, q, r, s] [D-r]