Q. How many real solutions does the equation $x^{7} + 14x^{5} + 16x^{3} + 30x - 560 = 0$ have?
A) 5
B) 7
C) 1
D) 3
Explanation
(c)
Let $f(x) = x^{7} + 14x^{5} + 16x^{3} + 30x - 560$
$f^{'}(x) = 7x^{6} + 70x^{4} + 48x^{2} + 30 > 0,\ \forall x \in R$
$\therefore f(x)$ is increasing
$\therefore\ \ \ f(x) = 0$ has only one solution
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.