Let f(x) = sin x + cos x and g(x) = x^2 - 1. | vedclass

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(C) Given $f(x) = \sin x + \cos x$ and $g(x) = x^2 - 1$. First,we find the composition $g(f(x))$: $g(f(x)) = (\sin x + \cos x)^2 - 1$ $= (\sin^2 x + \

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$[ - \frac{\pi }{4}, \frac{\pi }{4} ]$

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(C) Given $f(x) = \sin x + \cos x$ and $g(x) = x^2 - 1$. First,we find the composition $g(f(x))$: $g(f(x)) = (\sin x + \cos x)^2 - 1$ $= (\sin^2 x + \cos^2 x + 2 \sin x \cos x) - 1$ $= (1 + \sin 2x) - 1 = \sin 2x$. For a function to be invertible,it must be bijective (one-to-one and onto). The function $h(x) = \sin 2x$ is invertible in the interval where the argument $2x$ lies within the principal branch of the inverse sine function,i.e.,$[ -\frac{\pi}{2}, \frac{\pi}{2} ]$. Thus,we require: $-\frac{\pi}{2} \le 2x \le \frac{\pi}{2}$ Dividing by $2$,we get: $-\frac{\pi}{4} \le x \le \frac{\pi}{4}$. Therefore,$g(f(x))$ is invertible for $x \in [ -\frac{\pi}{4}, \frac{\pi}{4} ]$.

Let $f(x) = \sin x + \cos x$ and $g(x) = x^2 - 1$. Then $g(f(x))$ is invertible for $x \in $

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