f(x) = 2x + cot–1x + log ( 1 + x2 - x)

fx=2x+cot-1x+log1+x2-x fIx=2-1/1+x2+1/1+x2-x1/1+x222x-1 =2-1/1+x2+x-1+x2/1+x2-x*1/1+x2 =2-1/1+x2-1/1+x2 x∈R⇒fIx>0fx is increasing for -∞,∞

f(x) = 2x + cot–1x + log ( 1 + x2 - x)

fx=2x+cot-1x+log1+x2-x fIx=2-1/1+x2+1/1+x2-x1/1+x222x-1 =2-1/1+x2+x-1+x2/1+x2-x*1/1+x2 =2-1/1+x2-1/1+x2 x∈R⇒fIx>0fx is increasing for -∞,∞

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$f (x) = 2 x + c o t^{- 1} x + \log (\sqrt{1 + x^{2}} - x)$

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Decreases on (– $\infty$ , 0)

Neither increases nor decreases on (- $\infty$ , $\infty$ )

Decreases on (0, $\infty$ )

Increases on (– $\infty$ , $\infty$ )

answer is D.

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Detailed Solution

$f (x) = 2 x + \cot^{- 1} x + \log (\sqrt{1 + x^{2}} - x) f^{I} (x) = 2 - 1 / 1 + x^{2} + 1 / \sqrt{1 + x^{2}} - x (1 / \sqrt[2]{1 + x^{2}} (2 x) - 1) = 2 - 1 / 1 + x^{2} + x - \sqrt{1 + x^{2}} / \sqrt{1 + x^{2}} - x * 1 / \sqrt{1 + x^{2}} = 2 - 1 / 1 + x^{2} - 1 / \sqrt{1 + x^{2}} x \in R \Rightarrow f^{I} (x) > 0$