Question

Let $f\left(x\right)=x^3+\frac{1}{x^3},x\ne 0$. If the intervals in which f(x) increases are $(-\infty ,a]$ and $[b,\infty )$ then min(b - a) is equal to

• A. 0
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is B

2

Here $f^{\prime }\left(x\right)=3x^2-\frac{3}{x^4}\ge 0\Rightarrow x^6-1\ge 0\Rightarrow x\in (-\infty ,-1]\cup [1,\infty )$

$\therefore$ $min(b-a)=min\left(b\right)-max\left(a\right)=1-(-1)=2$