Let [math: f] be a continuous function defined on the domain [math: [0,2]][0,2]. If [math: f(0)=1] and [math: f(2)=3], then the equation [math: f(x)=0] has no solution.

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Here we examine the statement about the continuous function f on [0,2] with f(0)=1 and f(2)=3.
Option 1: True. The claim that f(x) = 0 has no solution would require that the function never crosses the horizontal axis on the entire interval. However, continuity alone with endpoint values of the same sign does not guarantee the absence of a root. A function could dip below zero somewhere in (0,2) and then return to a positive value at x=2, still remaining continuous. The intermediate value property tells us about values between f(0) and f(2) specifically and guarantees a root if f(0) and f(2) have opposite signs or if either endpoint is exactly zero, but it does not rule out internal crossing when both endpoints are positive. Consider a hypothetical continuous curve that starts at 1, dips to a negative value, then comes back to 3 at x=2; such a function would satisfy the given endpoint conditions yet have a point where f(x)=0. Therefore, the claim that there is no solution is not guaranteed by the given information.

Option 2: False. The counterexample argument above illustrates that a continuous function on [0,2] can cross zero in the interior even if f(0)=1 and f(2)=3. The key concept is that continuity ensures values between endpoints are attained, but it does not restrict the graph from crossing zero in between when both endpoints are positive. The IVT tells us about existence of a root specifically when there is a sign change between the endpoints, which is not the case here. Hence, asserting that there is no solution to f(x)=0 is incorrect in general under the given hypotheses.

Option 3: There is nothing in the statement that implies f(x)=0 must occur only if signs differ at endpoints; this misstates the scope of the intermediate value property. The IVT guarantees a root if f(a) and f(b) have opposite signs or if one is zero, but it does not imply that roots cannot exist when both endpoint values are positive. This makes the assertion in the problem unnecessarily restrictive.

Option 4: The claim does not consider the possibility of multiple internal extremums. Even with a single peak or trough, continuity permits crossing zero multiple times as long as the function remains continuous and endpoints have the specified positive values. Therefore, the zero-set is not precluded by the endpoint data alone.

_MATH1013_1ABCD_2025 Subsection 3.3 (closed on 11 Oct)

Let [math: f] be a continuous function defined on the domain [math: [0,2]][0,2]. If [math: f(0)=1] and [math: f(2)=3], then the equation [math: f(x)=0] has no solution.

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