Step 1: Find the derivative of \( f(x) \)
The given function is \( f(x) = \frac{x}{2} + \frac{2}{x} \). Differentiate: \[ f'(x) = \frac{1}{2} - \frac{2}{x^2}. \] Step 2: Find critical points
Set \( f'(x) = 0 \): \[ \frac{1}{2} - \frac{2}{x^2} = 0 \implies \frac{2}{x^2} = \frac{1}{2} \implies x^2 = 4 \implies x = 2. \] Step 3: Evaluate \( f(x) \) at critical points and endpoints
At \( x = 1 \): \[ f(1) = \frac{1}{2} + \frac{2}{1} = \frac{1}{2} + 2 = \frac{5}{2}. \] At \( x = 2 \): \[ f(2) = \frac{2}{2} + \frac{2}{2} = 1 + 1 = 2. \]
Step 4: Conclude the result
The absolute maximum value is \( \frac{5}{2} \) at \( x = 1 \), and the absolute minimum value is \( 2 \) at \( x = 2 \).

Comparative Financial Data as on 31st March, 2024 and 2023
| Particulars | 31.03.2024 (₹) | 31.03.2023 (₹) |
|---|---|---|
| Surplus (P&L) | 17,00,000 | 8,00,000 |
| Patents | -- | 50,000 |
| Sundry Debtors | 5,80,000 | 4,20,000 |
| Sundry Creditors | 1,40,000 | 60,000 |
| Cash and Cash Equivalents | 2,00,000 | 90,000 |