Q51 / 120 · VITEEE 2018 · Mathematics
The interval in which the function $f(x)=\frac{4 x^{2}+1}{x}$ is decreasing is :
- Correct answerA$\left(-\frac{1}{2}, \frac{1}{2}\right)$
- B$\left[-\frac{1}{2}, \frac{1}{2}\right]$
- C$(-1,1)$
- D$[-1,1]$
Solution
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 $\left(-\frac{1}{2}, \frac{1}{2}\right)$
Hence, the correct answer is option: 'A'.