Q. Let$f(x) = 1 + 2x^{2} + 2^{2}x^{4}\ + ... + 2^{10}x^{20}$. Then, $f(x)$ has
A) More than one minimum
B) Exactly one minimum
C) at least one maximum
D) None of the above
Explanation
(a)
$f^{'}(x) = 4x + 16x^{3} + ... + 2^{10}.20.x^{19}$
$= 4x(1 + 4x^{2} + ... + {5.2}^{10}x^{18})$ …(i)
$\Rightarrow \ \ f^{''}(x) = 4 + 48x^{2} + \ldots + 2^{10}380.x^{18} > 0$
$\therefore f(x)$ is minimum at $x = 0.$ [from Eq. (i)]
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.