F(X)=|cos^​-1(sgn x)| / cos^-1(sgn x) then which of the following is (are) not correct?
​​a) range of f(x) contains no integer
​b)graph of f(x) is symmetric about y-axis
c)the equation f(x)=0 has two distinct real solutions
​d) inverse of f(x) is not defined

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{F}\left(\mathrm{x}\right)=\frac{\left|{\cos }^{-1}\left(\mathrm{sgnx}\right)\right|}{{\cos }^{-1}\left(\mathrm{sgnx}\right)} \\ \mathrm{Range} \mathrm{of} \mathrm{sgn} \left(\mathrm{x}\right) \mathrm{is} \left\{-1,0,1\right\}, \mathrm{for} \mathrm{x}<0, \mathrm{x}=0 \mathrm{and} \mathrm{x}>0 \mathrm{respectively} \\ \mathrm{And} \mathrm{as} {\cos }^{-1}\left(1\right)=0 \mathrm{and} \mathrm{denominator} \mathrm{cannot} \mathrm{be} 0, \mathrm{hence} \\ \mathrm{F}\left(\mathrm{x}\right) \mathrm{is} \mathrm{not} \mathrm{defined} \mathrm{when} \mathrm{x}>0 \\ \mathrm{F}\left(0\right)= \frac{\left|{\cos }^{-1}\left(0\right)\right|}{{\cos }^{-1}\left(0\right)}=\frac{\left|\mathrm{\pi }/2\right|}{\mathrm{\pi }/2}=1 \\ \mathrm{For} \mathrm{x}<0 \\ \mathrm{F}\left(\mathrm{x}\right)=\frac{\left|\mathrm{\pi }\right|}{\mathrm{\pi }}=1 \\ \mathrm{Hence} \mathrm{range} \mathrm{of} \mathrm{f}\left(\mathrm{x}\right) \mathrm{is} 1 \mathrm{for} \mathrm{x}\le 0 \mathrm{and} \mathrm{it} \mathrm{is} \mathrm{undefined} \mathrm{for} \mathrm{x}>0 \\ \mathrm{As} \mathrm{F}\left(\mathrm{x}\right)=1 \mathrm{for} \mathrm{all} \mathrm{x}\le 0, \mathrm{hence} \mathrm{it} \mathrm{is} \mathrm{many}-\mathrm{one}. \\ \Rightarrow {\mathrm{F}}^{-1}\left(\mathrm{x}\right) \mathrm{does} \mathrm{not} \mathrm{exist}...\mathrm{Option}\left(\mathrm{D}\right) \end{gathered}$$

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