Search any question & find its solution
Question:
Answered & Verified by Expert
The interval in which the function $f(x)=\frac{4 x^{2}+1}{x}$ is decreasing is :
Options:
Solution:
1801 Upvotes
Verified Answer
The correct answer is:
$\left(-\frac{1}{2}, \frac{1}{2}\right)$
Given $\mathrm{f}(\mathrm{x})=\frac{4 \mathrm{x}^{2}+1}{\mathrm{x}}$ Thus $\mathrm{f}^{\prime}(\mathrm{x})=4-\frac{1}{\mathrm{x}^{2}}$ $\mathrm{f}(\mathrm{x})$ will be decreasing if $\mathrm{f}^{\prime}(\mathrm{x}) < 0$
Thus $4-\frac{1}{x^{2}} < 0$
$$
\Rightarrow \frac{1}{x^{2}}>4 \Rightarrow \frac{-1}{2} < x < \frac{1}{2}
$$
Thus interval in which $\mathrm{f}(\mathrm{x})$ is decreasing, is $\left(-\frac{1}{2}, \frac{1}{2}\right)$
Thus $4-\frac{1}{x^{2}} < 0$
$$
\Rightarrow \frac{1}{x^{2}}>4 \Rightarrow \frac{-1}{2} < x < \frac{1}{2}
$$
Thus interval in which $\mathrm{f}(\mathrm{x})$ is decreasing, is $\left(-\frac{1}{2}, \frac{1}{2}\right)$
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.