Differential Calculus
Differential Calculus
star_batch_jee_advanced_2025
Grade 12
Question:
Which of the following statements are true?
(A) If $f$ is differentiable at $x = c$, then $\lim_{h \to 0} \frac{f(c+h) - f(c-h)}{2h}$ exists and equals $f'(c)$.
(B) Given a function $f$ and a point $c$ in the domain of $f$, if the $\lim_{h \to 0} \frac{f(c+h) - f(c-h)}{h}$ exists, then the function is differentiable at $x = c$
(C) Let $g(x) = \begin{cases} x^2 \sin \frac{1}{x^2}, & x \neq 0 \\ 0, & x = 0 \end{cases}$, then $g'$ exists
(D) Let $g(x) = \begin{cases} x^2 \sin \frac{1}{x^2}, & x \neq 0 \\ 0, & x = 0 \end{cases}$, then $g'$ exists and is continuous.
If $f$ is differentiable at $x = c$, then $lim_{h o 0} rac{f(c+h) - f(c-h)}{2h}$ exists and equals $f'(c)$.
Given a function $f$ and a point $c$ in the domain of $f$, if the $lim_{h o 0} rac{f(c+h) - f(c-h)}{h}$ exists, then the function is differentiable at $x = c$
Let $g(x) = egin{cases} x^2 sin rac{1}{x^2}, & x
eq 0 \ 0, & x = 0 end{cases}$, then $g'$ exists
Let $g(x) = egin{cases} x^2 sin rac{1}{x^2}, & x
eq 0 \ 0, & x = 0 end{cases}$, then $g'$ exists and is continuous.
Step-by-Step Solution
Key Concept: Recognize that the given limit equals the sum of right and left derivatives, which are equal for differentiable functions.
(A) is true: $\lim_{h\to 0}\frac{f(c+h) - f(c) + f(c) - f(c-h)}{h} = f'(c) + f'(c) = 2f'(c)$ for differentiable $f$. (B) is false: a limit existing does not guarantee differentiability. (C) is true. (D) is false.
Correct Answer: 1,3