Question:
For any continuous and differentiable function \( f(x) \) with first derivative \( f'(x) \neq 0 \) for all values of \( x \), which one of the following is ALWAYS TRUE for \( a \neq b \)?
For any continuous and differentiable function \( f(x) \) with first derivative \( f'(x) \neq 0 \) for all values of \( x \), which one of the following is ALWAYS TRUE for \( a \neq b \)?
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If \( f'(x) \neq 0 \) for all \( x \), the function is one-to-one, meaning \( f(a) \neq f(b) \) for \( a \neq b \).
Updated On: Apr 14, 2025
- \( f(a)>f(b) \)
- \( f(a) = f(b) \)
- \( f(a) \neq f(b) \)
- \( f(a)<f(b) \)
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The Correct Option is C
Solution and Explanation
Concept:
If the first derivative \( f'(x) \) is never zero for any value of \( x \), then the function is either strictly increasing or strictly decreasing throughout its domain.
Case 1: Strictly Increasing
If \( f'(x) > 0 \), then for any two values \( a < b \), we have: \[ f(a) < f(b) \] This implies \( f(a) \neq f(b) \).
Case 2: Strictly Decreasing
If \( f'(x) < 0 \), then for any two values \( a < b \), we have: \[ f(a) > f(b) \] Again, \( f(a) \neq f(b) \).
Conclusion:
In either case, the function is injective (one-to-one), meaning it never takes the same value at two different points.
✅ Correct Answer: \( f(a) \neq f(b) \)
If the first derivative \( f'(x) \) is never zero for any value of \( x \), then the function is either strictly increasing or strictly decreasing throughout its domain.
Case 1: Strictly Increasing
If \( f'(x) > 0 \), then for any two values \( a < b \), we have: \[ f(a) < f(b) \] This implies \( f(a) \neq f(b) \).
Case 2: Strictly Decreasing
If \( f'(x) < 0 \), then for any two values \( a < b \), we have: \[ f(a) > f(b) \] Again, \( f(a) \neq f(b) \).
Conclusion:
In either case, the function is injective (one-to-one), meaning it never takes the same value at two different points.
✅ Correct Answer: \( f(a) \neq f(b) \)
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