Question

If positive numbers a^–1, b^–1, c^–1 are in A.P., then product of roots equation x² – kx + 2b¹⁰¹ – a¹⁰¹ – c¹⁰¹ = 0 (k Î R), has

• A. +ve sign
• B. –ve sign
• C. 0
• D. None of these

Answer 1

Answer: (B)

–ve sign

Answer 2

Given a^–1, b^–1, c^–1 are in A.P. ̃ a, b, c are in H.P.

̃ $\frac{a^{101} + c^{101}}{2}$> ($\sqrt{ac}$)¹⁰¹ > b¹⁰¹ ( Q $\sqrt{ac}$> b)

̃ 2b¹⁰¹ –a¹⁰¹ – c¹⁰¹ < 0

Let f(x) = x² –kx + 2b¹⁰¹ –a¹⁰¹ – c¹⁰¹

f(–¥) = (¥) > 0

f(0) < 0; f(¥) > 0

Hence equation f(x) = 0 has one root in (–¥, 0) and other in (0, ¥)