The function $f(x)=\cot ^{-1} x+x$ is increasing in the interval.
The function $f(x)=\cot ^{-1} x+x$ is increasing in the interval.
- $(-\infty, \infty)$
- $(0,3)$
- $(1, \infty)$
- $(-1, \infty)$
Solution
$\begin{aligned}
& \mathrm{f}(\mathrm{x})=\cot ^{-1} \mathrm{x}+\mathrm{x} \\
& \therefore \mathrm{f}^{\prime}(\mathrm{x})=\frac{-1}{1+\mathrm{x}^2}+1=\frac{-1+1+\mathrm{x}^2}{1+\mathrm{x}^2}=\frac{\mathrm{x}^2}{1+\mathrm{x}^2} \\
& \text { Here } \mathrm{x}^2 \geq 0 \Rightarrow \frac{\mathrm{x}^2}{1+\mathrm{x}^2} \geq 0
\end{aligned}$
Hence $\mathrm{f}(\mathrm{x})$ is always increasing.
Asked in: MHT CET 2021 (21 Sep Shift 1)
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