Question

\( f(x) \) and \( g(x) \) are differentiable functions such that \( \frac{f(x)}{g(x)} = \) a non-zero constant. If \( \frac{f'(x)}{g'(x)} = \alpha(x) \) and \( \left( \frac{f(x)}{g(x)} \right)' = \beta(x) \), then \( \frac{\alpha(x) - \beta(x)}{\alpha(x) + \beta(x)} = \):

• A. \( 0 \)
• B. \( f(x) + g(x) \)
• C. \( 1 \)
• D. \( f'(x) + g'(x) \)

Hint:

The derivative of a constant is zero, which is a key step in solving this problem.

Answer 1

The Correct Option is C

Step 1: Use the given information that \( \frac{f(x){g(x)} \) is a non-zero constant.} Let \( \frac{f(x)}{g(x)} = c \), where \( c \) is a non-zero constant.
Step 2: Find the derivative of \( \frac{f(x){g(x)} \) with respect to \( x \).} Using the quotient rule, \( \left( \frac{f(x)}{g(x)} \right)' = \frac{g(x) f'(x) - f(x) g'(x)}{(g(x))^2} \). We are given \( \left( \frac{f(x)}{g(x)} \right)' = \beta(x) \). Since \( c \) is a constant, \( \beta(x) = 0 \).
Step 3: Use the given information that \( \frac{f'(x){g'(x)} = \alpha(x) \).} This implies \( f'(x) = \alpha(x) g'(x) \).
Step 4: Substitute the expressions into the derivative of the quotient. \[ 0 = \frac{g(x) (\alpha(x) g'(x)) - f(x) g'(x)}{(g(x))^2} \] \[ g'(x) (g(x) \alpha(x) - f(x)) = 0 \] Assuming \( g'(x) \neq 0 \) (otherwise \( f \) and \( g \) are constants, leading to issues with \( \alpha(x) \)), we have \( g(x) \alpha(x) - f(x) = 0 \), so \( g(x) \alpha(x) = f(x) \).
Step 5: Find the value of \( \frac{\alpha(x) - \beta(x){\alpha(x) + \beta(x)} \).} Substitute \( \beta(x) = 0 \): \[ \frac{\alpha(x) - 0}{\alpha(x) + 0} = \frac{\alpha(x)}{\alpha(x)} = 1 \] (Assuming \( \alpha(x) \neq 0 \)).