Search any question & find its solution
Question:
Answered & Verified by Expert
If $f(x)=x\left(\frac{1}{x-1}+\frac{1}{x}+\frac{1}{x+1}\right), x>1$. Then
Options:
Solution:
2934 Upvotes
Verified Answer
The correct answer is:
$f(x)>3$
Given function is
$$
\begin{aligned}
f(x) &=x\left\{\frac{1}{x-1}+\frac{1}{x}+\frac{1}{x+1}\right\}, x>1 \\
&=x\left\{\frac{2 x}{x^{2}-1}+\frac{1}{x}\right\}=\frac{2 x^{2}}{x^{2}-1}+1 \\
&=\left\{\frac{2}{\left(1-\frac{1}{x^{2}}\right)}+1\right\}>3
\end{aligned}
$$
$\therefore \quad f(x)>3$
$$
\begin{aligned}
f(x) &=x\left\{\frac{1}{x-1}+\frac{1}{x}+\frac{1}{x+1}\right\}, x>1 \\
&=x\left\{\frac{2 x}{x^{2}-1}+\frac{1}{x}\right\}=\frac{2 x^{2}}{x^{2}-1}+1 \\
&=\left\{\frac{2}{\left(1-\frac{1}{x^{2}}\right)}+1\right\}>3
\end{aligned}
$$
$\therefore \quad f(x)>3$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.