$f (x) = {tan}^{- 1} x - x$ is decreasing in

(- \infty, \infty)

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(0, \infty) o n l y

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(- 1, 1) o n l y

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(0, 1)

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Solution

The correct option is B $(- \infty, \infty)$
$f (x) = {tan}^{- 1} x - x ⟹ f^{'} (x) = \frac{1}{1 + x^{2}} - 1 = \frac{- x^{2}}{1 + x^{2}}$
Since $f^{'} (x) \leq 0$ for all $x ϵ R$
So, $f (x)$ is decreasing in $R$ or $(- \infty, \infty)$

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