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What is the least positive integer satisfying the condition: $$x^{2} -4x > \cot^{-1} (x)$$

Graphically entering the value $y=x^{2} -4x$ and $y= \cot^{-1}(x)$, I concluded that the answer is $x=5$ as it satisfies the condition of a positive integer, but I would like to solve this question using a mathematical equation if possible.

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2 Answers 2

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Let $$ f(x)=x^2-4x-\cot^{-1}x. $$ Note that $$ f'(x)=2x-4+\frac{1}{x^2+1}>0 $$ implies $f(x)$ is increasing for $x\ge 2$. Clearly when $x=1,2,3,4$, the inequality does not hold since the LHS is non-positive. When $x=5$, the inequality holds. Thus $x=5$ is the smallest integer satisfying the inequality.

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  • $\begingroup$ xpaul's answer might well be better than mine; it depends on the intent of the questioner. If the question is for a pre-calculus class, he probably intended that you strive for the insight in my answer. If the question is for a calculus class, xpaul's answer is probably preferable. $\endgroup$ May 3, 2018 at 18:24
  • $\begingroup$ Can this be found using the concept tan x>x>sin x $\endgroup$ May 3, 2018 at 18:27
  • $\begingroup$ @SamarImamZaidi Re my previous comment, I suspect that the intended answer is either xpaul's answer or mine. $\endgroup$ May 3, 2018 at 18:28
  • $\begingroup$ can we use the concept $cot^{-1 } x>\frac{1}{x}$ $\endgroup$ May 3, 2018 at 18:29
  • $\begingroup$ @SamarImamZaidi, I doubt. $\endgroup$
    – xpaul
    May 5, 2018 at 20:47
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$x \in \mathbb{Z^+} \Rightarrow cot^{-1} x > 0.\;\;$ 5 is the smallest positive integer $x$ such that $x^2 - 4x > 0.$

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  • $\begingroup$ I know that the condition of x is x>4 and x<0 but still can it be derived mathematically if possible $\endgroup$ May 3, 2018 at 18:19

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