Q. Let $f(x) = 2^{2x - 1\ }$and $\phi(x) - 2^{x} + 2x\ log\ 2.$ If $f^{'}(x) > \phi^{'}(x),\$then
A) $0 < x < 1$
B) $0 \leq x < 1$
C) $x > 0$
D) $x \geq 0$
Explanation
(c)
Since, $f^{'}(x) > \phi^{'}(x)$
$\Rightarrow \ 2^{2x - 1}2\log{2 > - 2^{x}\log{2 + 2\log 2}}$
$\Rightarrow \ 2^{2x} > - 2^{x} + 2$
$\Rightarrow \ 2^{2x} + 2^{x} - 2 > 0$
$\Rightarrow \ \left( 2^{x} - 1 \right)\left( 2^{x} + 2 \right) >$ 0
$\Rightarrow \ 2^{x} - 1 > 0$ [$\because 2^{x} + 2 > 0\ for\ all\ x\rbrack$
$\Rightarrow \ 2^{x} > 1$
$\therefore\ \ x > 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.