Question
If $$f\left( x \right) = \sin x + \cos x,g\left( x \right) = {x^2} - 1,$$ then $$g\left( {f\left( x \right)} \right)$$ is invertible in the domain
A.
$$\left[ {0,\frac{\pi }{2}} \right]$$
B.
$$\left[ { - \frac{\pi }{4},\frac{\pi }{4}} \right]$$
C.
$$\left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]$$
D.
$$\left[ {0,\pi } \right]$$
Answer :
$$\left[ { - \frac{\pi }{4},\frac{\pi }{4}} \right]$$
Solution :
$$\eqalign{
& f\left( x \right) = \sin x + \cos x,g\left( x \right) = {x^2} - 1, \cr
& \Rightarrow g\left( {f\left( x \right)} \right) = {\left( {\sin x + \cos x} \right)^2} - 1 = \sin 2x \cr} $$
Clearly $$g\left( {f\left( x \right)} \right)$$ is invertible in $$ - \frac{\pi }{2} \leqslant 2x \leqslant \frac{\pi }{2}$$
$$\eqalign{
& \left[ {\because \sin \theta \,{\text{is}}\,{\text{invertible}}\,{\text{when}}\, - \frac{\pi }{2} \leqslant \theta \leqslant \frac{\pi }{2}} \right] \cr
& \Rightarrow - \frac{\pi }{4} \leqslant x \leqslant \frac{\pi }{4} \cr} $$