Question

Consider the function \( f : (0, 2) \to \mathbb{R} \) defined by \[ f(x) = \frac{x}{2} + \frac{2}{x} \] and the function \( g(x) \) defined by \[ g(x) = \begin{cases} \min\{f(t)\}, & 0 < t \leq x \text{ and } 0 < x \leq 1 \\ \frac{3}{2} + x, & 1 < x < 2 \end{cases} \] Then:

• A. g is continuous but not differentiable at x = 1
• B. g is not continuous for all x (0,2)
• C. g is neither continuous nor differentiable at x = 1
• D. g is continuous and differentiable for all x (0,2) 

Answer 1

The Correct Option is A

To determine the continuity and differentiability of \(g\) at \(x = 1\), we need to check the left-hand limit (LHL) and right-hand limit (RHL) as well as the derivative behavior at \(x = 1\).

Continuity Check:

For \(0 < x \leq 1\), \(g(x) = \min\{f(t)\}\) where \(f(t) = \frac{t}{2} + \frac{2}{t}\).

At \(x = 1\), \(f(1) = \frac{1}{2} + 2 = \frac{5}{2}\).

For \(1 < x < 2\), \(g(x) = \frac{3}{2} + x\).

- Left-hand limit as \(x\) approaches \(1\) from the left:
\(\lim_{x \to 1^-} g(x) = \min\left\{\frac{5}{2}\right\} = \frac{5}{2}.\)

- Right-hand limit as \(x\) approaches \(1\) from the right:
\(\lim_{x \to 1^+} g(x) = \frac{3}{2} + 1 = \frac{5}{2}.\)

Since the left-hand limit equals the right-hand limit and equals \(g(1)\), \(g(x)\) is continuous at \(x = 1\).

Differentiability Check:

The derivative from the left side, \(\frac{d}{dx} (\min\{f(t)\})\) at \(x = 1\), does not match the derivative of \(\frac{3}{2} + x\) from the right side.

Therefore, \(g(x)\) is not differentiable at \(x = 1\).

Thus, \(g\) is continuous but not differentiable at \(x = 1\).

The Correct answer is: g is continuous but not differentiable at x = 1