Inverse Domain

Geometry Level 4

For this question, take the range of cot ⁡ − 1 x \cot^{-1} x as ( − π 2 , π 2 ] \left( - \dfrac{\pi}{2}, \dfrac{\pi}{2} \right] .

f ( x ) = 1 ln ⁡ ( cot ⁡ − 1 x ) \large f(x) = \dfrac{1}{\sqrt{\ln(\cot^{-1}x)}}

Find the domain of the function f ( x ) f(x) above.

( − ∞ , cot ⁡ 1 ) (-\infty,\cot1) ( 0 , cot ⁡ 1 ) (0, \cot1) [ 0 , cot ⁡ 1 ) [0,\cot1) [ cot ⁡ 1 , ∞ ] [\cot1,\infty] ( cot ⁡ 1 , ∞ ) (\cot1,\infty) R − cot ⁡ 1 \mathbb{R} - \cot1 ( − ∞ , 0 ) ∪ ( 0 , cot ⁡ 1 ) (-\infty,0) \cup (0,\cot1)

Akhil Bansal by Akhil Bansal , 22, India

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1 solution

1 Helpful 0 Interesting 0 Brilliant 0 Confused

how cot(1)? \

rakshith lokesh - 3 years, 3 months ago

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cot inverse x = 1 implies x = cot(1).

Venkata Karthik Bandaru - 3 years, 3 months ago

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