Q133 / 200 · WBJEE 2013 · Mathematics

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

  • A
    $f(x) \leq 1$
  • B
    $1 < f(x) \leq 2$
  • C
    $2 < f(x) \leq 3$
  • Correct answer
    D
    $f(x)>3$

Solution

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

Hence, the correct answer is option: 'D'.

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