Question

If a monotonic and continuous function \(y=f(x)\) has exactly one root in the interval \(x_1<x<x_2\), then:

• A. \(f(x_1)f(x_2) > 0\)
• B. \(f(x_1)f(x_2) = 0\)
• C. \(f(x_1)f(x_2) < 0\)
• D. \(f(x_1) - f(x_2) = 0\)

Hint:

For continuous monotonic functions, exactly one root in an interval implies a strict sign change at the endpoints. Always check \(f(x_1)\) and \(f(x_2)\).

Answer 1

The Correct Option is C

Step 1: Recall Intermediate Value Theorem.
A continuous function crossing zero between \(x_1\) and \(x_2\) must take opposite signs at the endpoints.

Step 2: Use monotonicity.
If \(f(x)\) is monotonic, then there can be at most one crossing. Given there is exactly one root, the sign at \(x_1\) and \(x_2\) must differ.

Step 3: Mathematical condition.
\[ f(x_1)\cdot f(x_2) < 0. \]

Step 4: Eliminate wrong options.
- (A) Same sign → no root. Contradiction.
- (B) Product zero → would mean root lies at endpoint, but problem states root in \((x_1, x_2)\).
- (D) Equality of values → contradicts monotonicity unless constant (which would not give one root).

Final Answer:
\[ \boxed{f(x_1)f(x_2) < 0} \]