Question

Let \( f \) be any continuous function on \( [0, 2] \) and twice differentiable on \( (0, 2) \). If \( f(0) = 0, f(1) = 1 \) and \( f(2) = 2 \), then :

• A. \( f'(x) = 0 \) for some \( x \in [0, 2] \)
• B. \( f''(x)>0 \) for all \( x \in (0, 2) \)
• C. \( f''(x) = 0 \) for some \( x \in (0, 2) \)
• D. \( f''(x) = 0 \) for all \( x \in (0, 2) \)

Hint:

If \( n+1 \) points are collinear for a function that is \( n \)-times differentiable, then the \( n \)-th derivative must vanish at some point in the interval containing those points.
Here, \( (0,0), (1,1), (2,2) \) are collinear, so \( f''(x) = 0 \) at some point.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This question tests the application of the Mean Value Theorem (MVT) and Rolle's Theorem.
Step 2: Key Formula or Approach:
Apply Lagrange's Mean Value Theorem (LMVT) on sub-intervals \( [0, 1] \) and \( [1, 2] \).
Then apply Rolle's Theorem on the derivative function \( f'(x) \).
Step 3: Detailed Explanation:
On the interval \( [0, 1] \), by LMVT, there exists \( c_1 \in (0, 1) \) such that:
\[ f'(c_1) = \frac{f(1) - f(0)}{1 - 0} = \frac{1 - 0}{1} = 1 \]
On the interval \( [1, 2] \), by LMVT, there exists \( c_2 \in (1, 2) \) such that:
\[ f'(c_2) = \frac{f(2) - f(1)}{2 - 1} = \frac{2 - 1}{1} = 1 \]
Now, consider the function \( f'(x) \) on the interval \( [c_1, c_2] \).
Since \( f'(c_1) = f'(c_2) = 1 \), and \( f'(x) \) is differentiable (because \( f \) is twice differentiable), by Rolle's Theorem applied to \( f'(x) \), there exists some \( x \in (c_1, c_2) \subset (0, 2) \) such that the derivative of \( f'(x) \) is zero.
That is, \( f''(x) = 0 \) for some \( x \in (0, 2) \).
Step 4: Final Answer:
\( f''(x) = 0 \) for some \( x \in (0, 2) \).