If `f ''(x)|le 1 AA x in R, and f (0) =0=f' (0),` then which of the following can not be true ?

Options

A. `f (-(1)/(2)) =1/6`
B. `f (2) =-4`
C. `f (-2) =3`
D. `f ((1)/(2)) =1/5`

Answer

To solve the problem, we need to analyze the given conditions and derive the implications step by step.

### Step 1: Understanding the Given Information
We are given that:
1. \( f''(x) \leq 1 \) for all \( x \in \mathbb{R} \)
2. \( f(0) = 0 \)
3. \( f'(0) = 0 \)

### Step 2: Analyzing the Second Derivative
Since \( f''(x) \leq 1 \), we can integrate this inequality to find bounds on the first derivative \( f'(x) \).

Integrating \( f''(x) \):
\[ f'(x) = \int f''(x) \, dx \leq \int 1 \, dx = x + C \]
Where \( C \) is a constant of integration.

### Step 3: Applying the Initial Condition for \( f'(0) \)
Using the condition \( f'(0) = 0 \):
\[ f'(0) = 0 + C \implies C = 0 \]
Thus, we have:
\[ f'(x) \leq x \]

### Step 4: Integrating the First Derivative
Now we integrate \( f'(x) \):
\[ f(x) = \int f'(x) \, dx \leq \int x \, dx = \frac{x^2}{2} + D \]
Where \( D \) is another constant of integration.

### Step 5: Applying the Initial Condition for \( f(0) \)
Using the condition \( f(0) = 0 \):
\[ f(0) = 0 + D \implies D = 0 \]
Thus, we have:
\[ f(x) \leq \frac{x^2}{2} \]

### Step 6: Finding the Lower Bound
Since \( f''(x) \leq 1 \), we can also consider that \( f''(x) \) could be negative. The most we can say is that \( f''(x) \geq -1 \) (if we assume it can take negative values). Integrating this gives:
\[ f'(x) \geq -x + C' \]
Using \( f'(0) = 0 \):
\[ 0 = C' \implies C' = 0 \]
Thus:
\[ f'(x) \geq -x \]

Integrating again:
\[ f(x) \geq -\frac{x^2}{2} + D' \]
Using \( f(0) = 0 \):
\[ 0 = D' \implies D' = 0 \]
Thus:
\[ f(x) \geq -\frac{x^2}{2} \]

### Step 7: Conclusion on the Bounds of \( f(x) \)
From the analysis, we conclude:
\[ -\frac{x^2}{2} \leq f(x) \leq \frac{x^2}{2} \]

### Step 8: Evaluating the Options
Now we need to evaluate the options provided to see which cannot be true based on the bounds we derived.

### Final Answer
The values of \( f(x) \) must lie between \( -\frac{x^2}{2} \) and \( \frac{x^2}{2} \). Any option that lies outside this range cannot be true.