Question
Let $$f\left( x \right) = {\cos ^{ - 1}}\left( {\frac{{{x^2}}}{{1 + {x^2}}}} \right).$$ The range of $$f$$ is :
A.
$$\left[ {0,\,\frac{\pi }{2}} \right]$$
B.
$$\left[ { - \frac{\pi }{2},\,\frac{\pi }{2}} \right]$$
C.
$$\left[ { - \frac{\pi }{2},\,0} \right]$$
D.
none of these
Answer :
none of these
Solution :
$$\left| {\frac{{{x^2}}}{{1 + {x^2}}}} \right| \leqslant 1.$$ This is true for all $$x\, \in \,R.$$ So, the domain $$=R$$
$$\eqalign{
& {\text{Now }}\frac{{{x^2}}}{{1 + {x^2}}} = 1 - \frac{1}{{1 + {x^2}}} \cr
& \therefore \,\,0 \leqslant \frac{{{x^2}}}{{1 + {x^2}}} < 1 \cr
& \therefore \,\,{\cos ^{ - 1}}0 \geqslant {\cos ^{ - 1}}\frac{{{x^2}}}{{1 + {x^2}}} > {\cos ^{ - 1}}1 \cr
& \Rightarrow \frac{\pi }{2} \geqslant {\cos ^{ - 1}}\frac{{{x^2}}}{{1 + {x^2}}} > 0 \cr
& \therefore {\text{The range }} = \left( {0,\,\frac{\pi }{2}} \right] \cr} $$