A) \[\left[ -\frac{\pi }{2},\ 0 \right]\]
B) \[\left[ -\frac{\pi }{2},\ \pi \right]\]
C) \[\left[ -\frac{\pi }{2},\ \frac{\pi }{4} \right]\]
D) \[\left[ 0,\ \frac{\pi }{2} \right]\]
Correct Answer: C
Solution :
By definition of composition of function, \[g(f(x))={{(\sin x+\cos x)}^{2}}-1\] Þ \[g(f(x))=\sin 2x\] We know sin x is bijective only, when \[x\in \left[ -\frac{\pi }{2},\frac{\pi }{2} \right]\] Thus \[g(x)\] is bijective if \[-\frac{\pi }{2}\le 2x\le \frac{\pi }{2}\Rightarrow \frac{-\pi }{4}\le x\le \frac{\pi }{4}\].
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