Limits, Continuity & Differentiability
Differential Calculus-1
star_batch_jee_advanced_2025
Grade 12
Question:
Let $f(x) = \min(x^3, x^2)$ and $g(x) = |x|^2 + \sqrt{[x]^2}$, where $[x]$ denotes the greatest integer and $\{x\}$ denotes the fractional part function. Then which of the following holds?
$f$ is continuous for all $x$
$g$ is discontinuous for all $x \in I$
$f$ is differentiable for all $x \in (1, \infty)$
$g$ is not differentiable for all $x \in I$
Step-by-Step Solution
Key Concept: Identify the transition points of piecewise functions ($x=0,1$ for $f$) and verify continuity/differentiability by checking left and right derivatives; for $g$, the floor function is discontinuous at integers.
For $f(x) = \min(x^3, x^2)$, we need to find where $x^3 = x^2$, giving $x = 0$ or $x = 1$. For $x \in (-\infty, 0)$: $x^3 < x^2$, so $f(x) = x^3$. For $x \in [0,1]$: $x^2 \leq x^3$, so $f(x) = x^2$. For $x \in (1, \infty)$: $x^3 > x^2$, so $f(x) = x^2$. Thus $f$ is continuous everywhere (continuous at $x=0$ and $x=1$), making option 1 correct. For $x > 1$, $f(x) = x^2$ which is differentiable, confirming option 3. For $g(x) = |x|^2 + \sqrt{[x]^2} = x^2 + |[x]|$, at integer points $g$ has jump discontinuities in the floor function's derivative, making options 2 and 4 incorrect.
Correct Answer: 1,3