Question

The number of distinct real roots of the equation x5(x3 – x2 – x + 1) + x (3x3 – 4x2 – 2x + 4) – 1 = 0 is ______ .

Answer 1

x8 - x7 - x6 + x5 + 3x4 - 4x3 - 2x2 + 4x - 1 = 0
⇒ x7(x - 1) - x5(x - 1) + 3x3(x - 1) - x (x2 - 1) + 2x (1 - x) + (x - 1) = 0
⇒ (x - 1) (x7 - x5 + 3x3 - x(x + 1) - 2x + 1) = 0
⇒ (x - 1) (x7 - x5 + 3x3 -x2 - 3x + 1) = 0
⇒ (x - 1) (x5 (x2 - 1) + 3x (x2 - 1) - 1 (x2 - 1)) = 0
⇒ (x - 1) (x2 - 1) (x5 + 3x - 1) = 0
∴ x = ± 1 are roots of above equation and x5 + 3x - 1 is a monotonic term hence vanishs at exactly one value of x other then 1 or - 1.
∴ 3 real roots.
So, the correct answer is 3.