Let \[f(x)=\sin x+\cos x,\ g(x)={{x}^{2}}-1\]. Thus \[g(f(x))\] is invertible for \[x\in \] [IIT Screening 2004]

\[\left[ 0,\ \frac{\pi }{2} \right]\]

\[\left[ -\frac{\pi }{2},\ 0 \right]\]

\[\left[ -\frac{\pi }{2},\ \pi \right]\]

\[\left[ -\frac{\pi }{2},\ \frac{\pi }{4} \right]\]

Let \[f(x)=\sin x+\cos x,\ g(x)={{x}^{2}}-1\]. Thus \[g(f(x))\] is inv

1.	Let \[f(x)=\sin x+\cos x,\ g(x)={{x}^{2}}-1\]. Thus \[g(f(x))\] is invertible for \[x\in \] [IIT Screening 2004]
A.	\[\left[ -\frac{\pi }{2},\ 0 \right]\]
B.	\[\left[ -\frac{\pi }{2},\ \pi \right]\]
C.	\[\left[ -\frac{\pi }{2},\ \frac{\pi }{4} \right]\]
D.	\[\left[ 0,\ \frac{\pi }{2} \right]\]
Answer» D. \[\left[ 0,\ \frac{\pi }{2} \right]\]

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Let \[f(x)=\sin x+\cos x,\ g(x)={{x}^{2}}-1\]. Thus \[g(f(x))\] is invertible for \[x\in \] [IIT Screening 2004]

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