Limits, Continuity & Differentiability
Differential Calculus-1
star_batch_jee_advanced_2025
Grade None
Question:
Let $f: R \to R$ be defined by $f(x) = \begin{cases} x + 2x^2 \sin \frac{1}{x} & \text{for } x \neq 0 \\ 0 & \text{for } x = 0 \end{cases}$ then
$f(x)$ is differentiable for all 'x' but $f'(x)$ is not continuous at $x = 0$
$f'(0) = 1$
$f(x)$ is increasing at $x = 0$
Both $f(x)$ and $f'(x)$ are differentiable for all 'x'
Step-by-Step Solution
Key Concept: Continuity of composite functions at integer points requires checking left and right limits separately when the floor function is involved.
Using the first principle, $f'(0) = \lim_{x \to 0} \frac{x + 2x^2\sin\frac{1}{x} - 0}{x - 0} = \lim_{x \to 0}\left(1 + 2x\sin\frac{1}{x}\right) = 1$. We then find $f'(x) = 1 - 2\cos\frac{1}{x} + 4x\sin\frac{1}{x}$. Although $f$ is differentiable everywhere, $f'(x)$ is discontinuous at $x = 0$ since the term $\cos\frac{1}{x}$ oscillates wildly. Checking continuity: $f'\left(\frac{1}{2k\pi}\right) = -1 1$, we define $g(x) = [x]^2 + \sqrt{[x]^2}$ where $[x]$ is the greatest integer function. Since $f$ is continuous everywhere, $g$ is continuous except possibly at integer points. For $n \in \mathbb{Z}$: $g(n) = n^2$ but $\lim_{h \to 0^-} g(n+h) = n^2$ and $\lim_{h \to 0^+} g(n+h) = (n-1)^2 + |n-1| = (n-1)^2 + 1$. Setting these equal: $n^2 = (n-1)^2 + 1$ gives $n = 1$. Therefore $g$ is continuous only at $x = 1$.
Correct Answer: 1,2