Question

The function $f\left(x\right)=\frac{2x^2-1}{x^4},x>0$, decreases in the interval

• A. $[1,\infty )$
• B. $\left(1,\infty \right)$
• C. $\left(0,\infty \right)$
• D. $[0,\infty )$

Answer 1

The correct option is B $\left(1,\infty \right)$

Given that, $f\left(x\right)=\frac{2x^2-1}{x^4},x>0\Rightarrow f\left(x\right)=\frac{2}{x^2}-\frac{1}{x^4},x>0\Rightarrow f\text{'}\left(x\right)=-\frac{4}{x^3}+\frac{4}{x^5}=\frac{4\left(1-x^2\right)}{x^5}f\text{'}\left(x\right)=0\Rightarrow \frac{4\left(1-x^2\right)}{x^5}=0\Rightarrow \left(1-x^2\right)=0\Rightarrow x=1\mathrm{or}-1.\mathrm{and}{x}^5=0\Rightarrow x=0\therefore x=1,0\mathrm{and}-1\mathrm{are}\mathrm{the}\mathrm{critical}\mathrm{points}.$
This can be represented on number line as

From the figure, f '(x) < 0 for $x\in \left(1,\infty \right)$ where f(x) decreases.