Question

Let $f:[a, b] \to \mathbb{R}$ be continuous on $[a, b]$, differentiable on $(a, b)$, and $f(a) = 0 = f(b)$. Then (A) there exists at least one point $c \in (a, b)$ for which $f'(c) = f(c)$ (B) $f'(x) = f(x)$ does not hold at any point of $(a, b)$ (C) at every point of $(a, b)$, $f'(x)>f(x)$ (D) at every point of $(a, b)$, $f'(x)<f(x)$

• A. there exists at least one point c∈(a, b) for which f'(c) = f (c)
• B. f'(x) = f (x) does not hold at any point of (a, b)
• C. at every point of (a, b), f'(x)>f (x)
• D. at every point of (a, b), f'(x)<f (x)

Answer 1

The Correct Option is A

The function \( f \) is continuous on \( [a, b] \), differentiable on \( (a, b) \), and vanishes at both endpoints, i.e., \( f(a) = f(b) = 0 \).

This is a classic situation for applying the Rolle's Theorem, which states:

If a function is continuous on \( [a, b] \), differentiable on \( (a, b) \), and \( f(a) = f(b) \), then there exists at least one point \( c \in (a, b) \) such that: \[ f'(c) = 0 \]

But the question is not asking whether \( f'(c) = 0 \), rather whether \( f'(c) = f(c) \) holds for some \( c \in (a, b) \).

Define a new function: \[ g(x) = f(x) e^{-x} \]

Why define this? Because we want to relate \( f'(x) \) and \( f(x) \). This substitution simplifies comparison.

Now compute \( g'(x) \): \[ g'(x) = \frac{d}{dx}[f(x) e^{-x}] = f'(x)e^{-x} - f(x)e^{-x} = e^{-x}(f'(x) - f(x)) \]

So: \[ g'(x) = 0 \iff f'(x) = f(x) \]

Now, note that:

• \( g(x) \) is continuous on \( [a, b] \) and differentiable on \( (a, b) \) (since both \( f \) and \( e^{-x} \) are).
• \( g(a) = f(a) e^{-a} = 0 \), \( g(b) = f(b) e^{-b} = 0 \)

Therefore, by Rolle’s Theorem, there exists a point \( c \in (a, b) \) such that: \[ g'(c) = 0 \Rightarrow f'(c) - f(c) = 0 \Rightarrow f'(c) = f(c) \]

This confirms option (A) is true.

Answer:

\[ \boxed{\text{There exists at least one point } c \in (a, b) \text{ for which } f'(c) = f(c)} \]