Question

How many real solutions does the equation x⁷ + 14x⁵ + 16x³ + 30x – 560 = 0 have?
(a) 5 

(b) 7

(c) 1 

(d) 3

Answer 1

(c) : Let f (x) = x⁷ + 14x⁵ + 16x³ + 30x – 560
∴ f '(x) = 7x⁶ + 70x⁴ + 48x² + 30
=> f '(x) > 0 ∀ x ∈ R
i.e. f(x) is an strictly increasing function.
so it can have at the most one solution. It can be shown that it has exactly one solution.