Question
How many real solutions does the equation x7+ 14x5+ 16x3+ 30x – 560 = 0 have?
-
7
-
1
-
3
-
5
Solution
B.
1
x7+ 14x5+ 16x3+ 30x – 560 = 0
Let f(x) = x7+ 14x5+ 16x3+ 30x
⇒ f′(x) = 7x6+ 70x4+ 48x2+ 30 > 0 ∀ x.
∴ f (x) is an increasing function ∀ x.
