Question

Let f(x) = $\left( 1 - \frac{\sqrt{21 - 4b - b^{2}}}{b + 1} \right)$ x³ + px – 100. Then f(x) is strictly increasing in R for –

• A. b Ī (2, 3]
• B. b Ī [2, 3)
• C. b Ī (1, 3)
• D. b Ī [1, 3]

Answer 1

Answer: (A)

b Ī (2, 3]

Answer 2

f ¢ (x) > 0 for x Ī R

so 1 – $\frac{\sqrt{21 - 4b - b^{2}}}{b + 1}$ > 0

So 2b² + 6b – 1 < 0 and – b² – 4b + 21 ³ 0