Domain of the function f(x) = frac{1}{sqrt{ln | vedclass

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(D) For the function $f(x) = \frac{1}{\sqrt{\ln(\cot^{-1}x)}}$ to be defined,the expression inside the square root must be strictly positive: $\ln(\co

Vedclass · Accepted Answer

$( -\infty, \cot 1 )$

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(D) For the function $f(x) = \frac{1}{\sqrt{\ln(\cot^{-1}x)}}$ to be defined,the expression inside the square root must be strictly positive: $\ln(\cot^{-1}x) > 0$ Since the base of the logarithm is $e > 1$,we have: $\cot^{-1}x > e^0$ $\cot^{-1}x > 1$ Since the function $y = \cot^{-1}x$ is a strictly decreasing function,the inequality reverses when we apply the cotangent function: $x Thus,the domain of the function is $( -\infty, \cot 1 )$.

Domain of the function $f(x) = \frac{1}{\sqrt{\ln(\cot^{-1}x)}}$ is

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