How many real solutions does the equation x7 + 14x5 + 16x3 + 30x − 560 = 0 have?
(A) 7
(B) 1
(C) 3
(D) 5
Answer is (B) 1
x7 + 14x5 + 16x3 + 30z - 560 = 0
Let f(x) = x7 + 14x5 + 16x3 + 30x. Then
f'(x) = 7x6 + 70x4 + 48x2 + 30 > 0 ∀ x
Therefore, f(x) is a strictly increasing function for all x.
So, it can have at the most one solution.
