Suppose that  f  is continuous on [0, 6] and that the only solutions of the equation  f (x) = 3 are x = 1 and x = 5. If  f (4) = 5, then which of the following statements must be true?

(i)  f (2)  >  3
(ii)  f (0)  <  3
(iii)  f (6)  <  3

We are given that \( f \) is continuous on the interval \([0, 6]\) and that the only solutions to the equation \( f(x) = 3 \) occur at \( x = 1 \) and \( x = 5 \). This implies that between \( x = 1 \) and \( x = 5 \), the function \( f(x) \) does not equal 3 anywhere else. Additionally, we know that \( f(4) = 5 \), which is greater than 3.

We will now analyze each of the three statements:

### Statement (i): \( f(2) > 3 \)
- The function is continuous, and the only points where \( f(x) = 3 \) are \( x = 1 \) and \( x = 5 \).
- From the Intermediate Value Theorem (IVT), since \( f(1) = 3 \) and \( f(4) = 5 \), \( f(x) \) must be greater than 3 between \( x = 1 \) and \( x = 4 \). Hence, \( f(2) > 3 \).

So, **statement (i) is true**.

### Statement (ii): \( f(0) < 3 \)
- Since \( f(x) = 3 \) only at \( x = 1 \), and \( f(x) \) must be either greater than or less than 3 outside the point \( x = 1 \).
- If \( f(0) \geq 3 \), the continuity of \( f \) would force the function to reach 3 before \( x = 1 \), violating the condition that \( f(x) = 3 \) only at \( x = 1 \).
- Therefore, \( f(0) \) must be less than 3 to avoid crossing 3 anywhere between \( x = 0 \) and \( x = 1 \).

So, **statement (ii) is true**.

### Statement (iii): \( f(6) < 3 \)
- Similarly, \( f(x) = 3 \) only at \( x = 5 \), and for \( f(x) \) to stay above or at 3 after \( x = 5 \), it would have to cross 3 again after \( x = 5 \), which would contradict the given condition.
- Therefore, \( f(6) \) must be less than 3 to ensure that the function does not cross the value 3 after \( x = 5 \).

So, **statement (iii) is true**.

### Conclusion:
All three statements **must be true**. Therefore, the correct answer is that statements (i), (ii), and (iii) are all true.

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Do you want more details or have any questions? Here are some additional questions to explore:

1. Can the Intermediate Value Theorem be applied to any function that is not continuous?
2. What happens if \( f(2) = 3 \)? Would this affect the results?
3. How would the situation change if \( f(x) = 3 \) had more solutions on \([0, 6]\)?
4. Could there be another point where \( f(x) = 5 \) apart from \( x = 4 \)?
5. What would happen if \( f(x) \) was discontinuous at any point in the interval?

**Tip:** The Intermediate Value Theorem is a powerful tool when analyzing continuous functions, as it guarantees that a function takes on all values between any two points on its graph.

Explore the analysis of a continuous function on the interval [0,6], where the only solutions of f(x) = 3 occur at x = 1 and x = 5. We apply the Intermediate Value Theorem to determine which statements about f(2), f(0), and f(6) are true. This problem highlights key concepts in continuity and function behavior, suitable for Grades 10-12.

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