If f(x) is defined on [0,1], then the domain of `f(3x^(2))` , is

Options

A. `[0,1//sqrt(3)]`
B. `[-1//sqrt(3),1//sqrt(3)]`
C. `[-sqrt(3),sqrt(3)]`
D. none of these

Answer

To find the domain of \( f(3x^2) \) given that \( f(x) \) is defined on the interval \([0, 1]\), we need to determine the values of \( x \) for which \( 3x^2 \) lies within the interval \([0, 1]\).

### Step-by-Step Solution:

1. **Identify the condition for \( f(3x^2) \)**:
Since \( f(x) \) is defined on \([0, 1]\), we need \( 3x^2 \) to be within this interval:
\[ 0 \leq 3x^2 \leq 1 \]

2. **Break down the inequality**:
We can break this down into two parts:
- \( 3x^2 \geq 0 \)
- \( 3x^2 \leq 1 \)

3. **Solve the first part**:
The inequality \( 3x^2 \geq 0 \) is always true for all \( x \) in the real numbers, since \( x^2 \) is non-negative.

4. **Solve the second part**:
For the inequality \( 3x^2 \leq 1 \):
\[ x^2 \leq \frac{1}{3} \]
Taking the square root of both sides, we get:
\[ -\sqrt{\frac{1}{3}} \leq x \leq \sqrt{\frac{1}{3}} \]

5. **Express the interval**:
Since \( x^2 \) is non-negative, we only consider the non-negative part:
\[ 0 \leq x \leq \sqrt{\frac{1}{3}} \]

6. **Final domain**:
Therefore, the domain of \( f(3x^2) \) is:
\[ x \in \left[0, \frac{1}{\sqrt{3}}\right] \]

### Conclusion:
The domain of \( f(3x^2) \) is \( \left[0, \frac{1}{\sqrt{3}}\right] \).

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