MCQ
If ${\sin ^{ - 1}}x + {\cot ^{ - 1}}\left( {\frac{1}{2}} \right) = \frac{\pi }{2},$ then $x $ is
- A$0$
- ✓$\frac{1}{{\sqrt 5 }}$
- C$\frac{2}{{\sqrt 5 }}$
- D$\frac{{\sqrt 3 }}{2}$
Hence given equation can be written as
${\sin ^{ - 1}}x + {\cos ^{ - 1}}\frac{1}{{\sqrt 5 }} = \frac{\pi }{2}$
$ \Rightarrow $ $x = \frac{1}{{\sqrt 5 }}$.
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$g(x) = \left\{ \begin{array}{l}0,\,\,\,\,{\rm{when\,\,}}\,x\,{\rm{\,\,is\,\,}}\,{\rm{\,\,rational\,}}\\x,\,\,\,\,{\rm{\,\,when\,\,}}\,x\,{\rm{\,\,is\,\, irrational\,}}\end{array} \right.$ then $(f - g)$ is