Let `f(x) = min [1/2 - 3x^2/4, 5x^2/4]`, for `0 <= x <= 1` then maximum value of f(x) is

Options

A. 0
B. `5/64`
C. `5/4`
D. `5/16`

Answer

To find the maximum value of the function \( f(x) = \min\left(\frac{1}{2} - \frac{3x^2}{4}, \frac{5x^2}{4}\right) \) for \( 0 \leq x \leq 1 \), we will follow these steps:

### Step 1: Set the two expressions equal to each other
To find the point where the minimum function changes, we need to set the two expressions equal to each other:
\[ \frac{1}{2} - \frac{3x^2}{4} = \frac{5x^2}{4} \]

### Step 2: Solve for \( x \)
Rearranging the equation:
\[ \frac{1}{2} = \frac{5x^2}{4} + \frac{3x^2}{4} \]
\[ \frac{1}{2} = \frac{8x^2}{4} \]
\[ \frac{1}{2} = 2x^2 \]
Now, multiply both sides by \( \frac{1}{2} \):
\[ x^2 = \frac{1}{4} \]
Taking the square root gives:
\[ x = \frac{1}{2} \quad (\text{since } x \geq 0) \]

### Step 3: Evaluate \( f(x) \) at \( x = \frac{1}{2} \)
Now we will substitute \( x = \frac{1}{2} \) into either of the original expressions to find the maximum value:
Using the second expression:
\[ f\left(\frac{1}{2}\right) = \frac{5\left(\frac{1}{2}\right)^2}{4} \]
Calculating this gives:
\[ = \frac{5 \cdot \frac{1}{4}}{4} = \frac{5}{16} \]

### Step 4: Check the endpoints
We also need to check the values of \( f(x) \) at the endpoints \( x = 0 \) and \( x = 1 \):
- For \( x = 0 \):
\[ f(0) = \min\left(\frac{1}{2} - 0, 0\right) = \min\left(\frac{1}{2}, 0\right) = 0 \]
- For \( x = 1 \):
\[ f(1) = \min\left(\frac{1}{2} - \frac{3}{4}, \frac{5}{4}\right) = \min\left(-\frac{1}{4}, \frac{5}{4}\right) = -\frac{1}{4} \]

### Step 5: Conclusion
The maximum value of \( f(x) \) occurs at \( x = \frac{1}{2} \):
\[ \text{Maximum value of } f(x) = \frac{5}{16} \]

### Final Answer
The maximum value of \( f(x) \) is \( \frac{5}{16} \).

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