Question

If \(f(x)\) and \(g(x)\) are differentiable functions for \(0 \leq x \leq 1\) such that, \(f(1)-f(0) = k(g(1)-g(0))\), \(k \neq 0\), and there exists a 'c' satisfying \(0<c<1\). Then, the value of \(\frac{f'(c)}{g'(c)}\) is equal to

• A. \(2k\)
• B. \(k\)
• C. \(-k\)
• D. \(\frac{1}{k}\)

Hint:

When you see an expression involving the ratio of derivatives like \(\frac{f'(c)}{g'(c)}\) and conditions on the function values at the endpoints of an interval (e.g., \(f(b)-f(a)\)), immediately think of Cauchy's Mean Value Theorem.

Answer 1

The Correct Option is B

Step 1: Concept Overview:
This problem utilizes Cauchy's Mean Value Theorem (Extended Mean Value Theorem). This theorem connects the ratio of two functions' derivatives at a point \(c\) to the ratio of their changes over an interval \([a, b]\).

Step 2: Core Formula:
Cauchy's Mean Value Theorem: If \(f(x)\) and \(g(x)\) are continuous on \([a, b]\) and differentiable on \((a, b)\), there exists a \(c \in (a, b)\) such that: \[ \frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)} \] (assuming \(g'(c) eq 0\) and \(g(b) eq g(a)\)).

Step 3: Detailed Solution:
Since \(f(x)\) and \(g(x)\) are differentiable on \([0, 1]\), they are continuous on \([0, 1]\). Apply Cauchy's Mean Value Theorem with \(a=0\) and \(b=1\).
There exists a \(c \in (0, 1)\) such that: \[ \frac{f'(c)}{g'(c)} = \frac{f(1) - f(0)}{g(1) - g(0)} \] Given: \[ f(1) - f(0) = k(g(1) - g(0)) \] Since \(k eq 0\), then \(f(1) - f(0) eq 0\). Also, \(g(1) - g(0) eq 0\). Rearranging the given condition: \[ \frac{f(1) - f(0)}{g(1) - g(0)} = k \] Substituting into Cauchy's Mean Value Theorem: \[ \frac{f'(c)}{g'(c)} = k \]
Step 4: Final Result:
The value of \(\frac{f'(c)}{g'(c)}\) is \(k\).