Q81 / 150 · MHT CET 2021 (21 Sep Shift 1) · Mathematics

The function $f(x)=\cot ^{-1} x+x$ is increasing in the interval.

  • Correct answer
    A
    $(-\infty, \infty)$
  • B
    $(0,3)$
  • C
    $(1, \infty)$
  • D
    $(-1, \infty)$

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.

Hence, the correct answer is option: 'A'.

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