Question

If a₁ < a₂ < a₃ < a₄ < a₅ < a₆ then the equation (x – a₁) (x – a₃) (x – a₅) + 2(x – a₂) (x – a₄) (x – a₆) = 0 has – 

(A) Four real roots 

(B) One real root 

(C) One real root in each interval (a₁, a₂), (a_3, a₄) and (a_5, a₆) 

(D) None of these 

Answer 1

Correct option (C) One real root in each interval (a_1, a₂), (a₃, a₄) and (a_5, a₆)

Explanation: 

∴ At least one real root lies in (a₁, a₂) Similarly, at least one real roots lies in each interval (a_3, a₄) and (a_5, a₆₎ But f(x) is cubic, therefore there are only three roots. 

 Hence the equation f(x) = 0 has one real roots in each interval (a₁,a₂) (a₃,a₄) and (a_5,a₆) 

Comments

Do you have another method