Differential Calculus
Differential Calculus
star_batch_jee_advanced_2025
Grade None
Question:
Let $f$ be a function with two continuous derivatives and $f(0) = 0$, $f'(0) = 0$, $f''(0) = 0$. Function $g$ is defined by $g(x) = \begin{cases} \frac{f(x)}{x}, & x \neq 0 \\ 0, & x = 0 \end{cases}$
Then which of the following statements are correct?
g has a continuous first derivative
$g'(x)$ exist at $x = 0$
$g(x)$ is continuous but $g'(x)$ do not exist
$g(x)$ is continuous, but the first derivative of $g$ is not continuous
Step-by-Step Solution
Key Concept: Use the quotient rule and the definition of derivative at a point to verify continuity of the derived function.
For $g(x) = \frac{f(x)}{x}$ when $x \neq 0$ and $g(0) = 0$, compute $h(x) = g'(x) = \frac{xf'(x) - f(x)}{x^2}$ for $x \neq 0$. At $x = 0$, use the definition: $L.H.S. = \lim_{h\to 0}\frac{-hf'(-h) - f'(-h)}{2h} = \frac{f''(0)}{2} = 0$ and $R.H.L. = 0$, confirming continuity of $h(x)$ at $x=0$.
Correct Answer: 1,2