Q50 / 100 · BITSAT 2023 (Memory Based Paper 1) · 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 $f(x)=\frac{4 x^2+1}{x}$ Thus $f^{\prime}(x)=4-\frac{1}{x^2}$
$\mathrm{f}(\mathrm{x})$ 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 $f(x)$ is decreasing, is $\left(-\frac{1}{2}, \frac{1}{2}\right)$.
$\mathrm{f}(\mathrm{x})$ 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 $f(x)$ is decreasing, is $\left(-\frac{1}{2}, \frac{1}{2}\right)$.
Hence, the correct answer is option: 'A'.