JEE Mathematics 2024 Question with Solution

Given: $f(x) = \frac{x}{2} + \frac{2}{x}$ and

From the solution, for $0$

Hence on $(0,x]$ the minimum is attained at the endpoint, and as used in the source solution,

So for the left side near $x=1$, the value approaches $\frac{5}{2}$.

For $1$

Now compare the one-sided limits at $x=1$:

These are equal and match the value at $x=1$ as concluded in the source solution.

So, $g$ is continuous at $x=1$.

For differentiability, the source solution states that the expressions on the two sides of $x=1$ are different, so the left-hand derivative and right-hand derivative do not match. Therefore, $g$ is not differentiable at $x=1$.

Therefore, the correct option is A: $g$ is continuous but not differentiable at $x=1$.

Note: The question statement gives $f(x)=x^2+\frac{2}{x}$, but the solution works with $f(x)=\frac{x}{2}+\frac{2}{x}$. The extracted answer follows the solution, which explicitly concludes that the correct option is A.