The non-negative values of b for which the function \frac{16x^{3}}{3}-4bx^{2}+x has neither <strong>MAXIMUM</strong> nor <strong>MINIMUM</strong> in the range x \gt 0 is
- A. 0 \lt b \lt 1
- B. 1 \lt b \lt 2
- C. b \gt 2
- D. 0 \leq b \lt 1 ✓
Correct Answer: D. 0 \leq b \lt 1
Explanation
f'(x) = 16x^2 - 8bx + 1. For f(x) to have no extremum, f'(x) must not change sign. Since 16 \gt 0, f'(x) \geq 0 for all x, requiring discriminant D \leq 0. D = 64b^2 - 64 \leq 0 \implies b^2 \leq 1 \implies -1 \leq b \leq 1. Given b \geq 0, we have 0 \leq b \leq 1. The closest strict matching option is 0 \leq b \lt 1.
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