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: Understanding the Concept:
This problem is a direct application of Cauchy's Mean Value Theorem (also known as the Extended Mean Value Theorem). This theorem relates the ratio of the derivatives of two functions at a point \(c\) to the ratio of the change in the functions over an interval \([a, b]\).

Step 2: Key Formula or Approach:
Cauchy's Mean Value Theorem states that if \(f(x)\) and \(g(x)\) are continuous on \([a, b]\) and differentiable on \((a, b)\), then there exists some \(c \in (a, b)\) such that: \[ \frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)} \] (assuming \(g'(c) \neq 0\) and \(g(b) \neq g(a)\)).

Step 3: Detailed Explanation:
The problem provides that \(f(x)\) and \(g(x)\) are differentiable on \([0, 1]\). This implies they are also continuous on \([0, 1]\). We can apply Cauchy's Mean Value Theorem with \(a=0\) and \(b=1\).
According to the theorem, there exists a \(c \in (0, 1)\) such that: \[ \frac{f'(c)}{g'(c)} = \frac{f(1) - f(0)}{g(1) - g(0)} \] We are given the condition: \[ f(1) - f(0) = k(g(1) - g(0)) \] Since \(k \neq 0\), it implies \(f(1) - f(0) \neq 0\). Also, for the expression to be meaningful, we must have \(g(1) - g(0) \neq 0\). We can rearrange the given condition as: \[ \frac{f(1) - f(0)}{g(1) - g(0)} = k \] By substituting this result into the formula from Cauchy's Mean Value Theorem, we get: \[ \frac{f'(c)}{g'(c)} = k \]
Step 4: Final Answer:
The value of \(\frac{f'(c)}{g'(c)}\) is equal to \(k\).