2x³ – 6x + 5 is an increasing function, if

The correct option is: x < -1 or x > 1.

To determine the interval in which the function is increasing, we first define the function:
$f(x)=2x^3-6x+5$

A function is strictly increasing on an interval where its first derivative is positive, i.e., $f^{\prime }(x)>0$.

Let us find the first derivative of $f(x)$ with respect to $x$ using the power rule of differentiation:
$f^{\prime }(x)=\frac{d}{dx}(2x^3-6x+5)$
$f^{\prime }(x)=6x^2-6$

For the function to be increasing, we set the derivative strictly greater than zero:
$6x^2-6>0$

Divide the entire inequality by 6:
$x^2-1>0$

We can factor the left-hand side as a difference of squares:
$(x-1)(x+1)>0$

The critical values where the derivative is zero are $x=-1$ and $x=1$. These points divide the real number line into three intervals:
1. $(-\infty ,-1)$ or $x<-1$
2. $(-1,1)$ or $-1<x<1$
3. $(1,\infty )$ or $x>1$

Let's analyze the sign of the product $(x-1)(x+1)$ in each region:
- If $x<-1$, both factors $(x-1)$ and $(x+1)$ are negative, so their product is positive ($>0$).
- If $-1<x<1$, $(x-1)$ is negative and $(x+1)$ is positive, so their product is negative ($<0$).
- If $x>1$, both factors are positive, so their product is positive ($>0$).

Therefore, the inequality holds true when:
$x<-1\text{or}x>1$