Q. $2x^{3} - 6x + 5$ is an increasing function, if
A) $0 < x < 1$
B) $- 1 < x < 1$
C) $0 < - 1\ or\ x > 1$
D) $- 1 < x < - \frac{1}{2}$
Explanation
(c)
Let $f(x) = 2x^{3} - 6x + 5$
On differentiating w.r.t. $x$, we get
$f^{'}(x) = 6x^{2} - 6$
Since, it is increasing function
$\Rightarrow \ 6x^{2} - 6 > 0$
$\Rightarrow \ \ (x - 1)(x + 1) > 0$
$\Rightarrow \ \ x > 1\ \ \ or\ \ \ x < - 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.