Functions Question 190
Let $ f(x)=\sin x+\cos x,\ g(x)=x^{2}-1 $ . Thus $ g(f(x)) $ is invertible for $ x\in \mathbb{R} $
[IIT Screening 2004]
Options:
A) $ [ -\frac{\pi }{2},\ 0 ] $
B) $ [ -\frac{\pi }{2},\ \pi ] $
C) $ [ -\frac{\pi }{2},\ \frac{\pi }{4} ] $
D) $ [ 0,\ \frac{\pi }{2} ] $
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Answer:
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 [ -\frac{\pi }{2},\frac{\pi }{2} ] $ 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} $ .
BETA
