Question

Consider the formula, \( S = \dfrac{\alpha \times \omega}{\tau + \rho \times \omega} \) with positive integers. If \(\omega\) is increased and \(\alpha, \tau, \rho\) are kept constant, then \(S\):

• A. increases
• B. decreases
• C. increases and then decreases
• D. decreases and then increases
• E. cannot be determined

Hint:

When analyzing rational functions like \(\frac{a\omega}{b+c\omega}\), differentiate to test monotonicity. The sign of the derivative directly shows whether the function is increasing or decreasing.

Answer 1

The Correct Option is A

Step 1: Expression for \(S\).
\[ S(\omeg(A) = \frac{\alpha \omega}{\tau + \rho \omega}. \] Here, \(\alpha, \tau, \rho>0\) and \(\omega>0\). Step 2: Differentiate to check monotonicity.
Treating \(\omega\) as a continuous variable: \[ \frac{dS}{d\omega} = \frac{(\alph(A)(\tau + \rho \omeg(A) - (\alpha \omeg(A)(\rho)}{(\tau + \rho \omeg(A)^2}. \] Simplify numerator: \[ \frac{dS}{d\omega} = \frac{\alpha \tau + \alpha \rho \omega - \alpha \rho \omega}{(\tau + \rho \omeg(A)^2} = \frac{\alpha \tau}{(\tau + \rho \omeg(A)^2}. \] Step 3: Sign of derivative.
Since \(\alpha, \tau, \rho, \omega>0\), both numerator and denominator are positive. Hence: \[ \frac{dS}{d\omega}>0. \] Therefore, \(S\) is a strictly increasing function of \(\omega\). Step 4: Behavior as \(\omega \to \infty\).
As \(\omega \to \infty\), \[ S(\omeg(A) \to \frac{\alpha}{\rho}, \] a finite asymptote, but the increase is monotonic from \(0\) upwards. \[ \boxed{S \text{ increases as } \omega \text{ increases.}} \]