About which point is this curve y = cot^-1 x…

About which point is this curve y = cot^-1 x symmetric?

Question 76

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Consider the curve $y = \cot^{-1} x$ , defined for every real $x$ , with its values on the principal branch $(0, \pi)$ .

About which point is this curve $y = \cot^{-1} x$ symmetric?

(0, 0)

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\left(\frac{\pi}{2}, 0\right)

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\left(0, \frac{\pi}{2}\right)

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(0, \pi)

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CorrectNot quiteAnswer C

Correct answer: C

$\cot^{-1}(-x) = \pi - \cot^{-1} x$ , so with $(x, y)$ the curve also passes through $(-x, \pi - y)$ .
The midpoint of those two points is $\left(0, \frac{\pi}{2}\right)$ for every $x$ : that is the centre of symmetry.

Worth noting

$(0, 0)$ is the centre for $y = \tan^{-1} x$ , where $\tan^{-1}(-x) = -\tan^{-1} x$ .

Ten questions from Inverse Trigonometric Functions, at your own pace.

Ten questions from Inverse Trigonometric Functions, at your own pace.