Question

If $f\left(x\right)=x(\frac{1}{x-1}+\frac{1}{x}+\frac{1}{x+1}),x>1$. then

• A. $f\left(x\right)\le 3$
• B. $1<f\left(x\right)\le 2$
• C. $2<f\left(x\right)\le 3$
• D. $f\left(x\right)>3$

Answer 1

Answer: (D)

$f\left(x\right)>3$

Given function is
$f\left(x\right)=x\{\frac{1}{x-1}+\frac{1}{x}+\frac{1}{x+1}\},x>1$
$=x\{\frac{2x}{x^2-1}+\frac{1}{x}\}=\frac{2x^2}{x^2-1}+1$
$=\{\frac{2}{(1-\frac{1}{x^2})}+1\}>3(\because x>1)$
$\therefore$ $f\left(x\right)>3$