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

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(A) Given $f(x) = \sin x + \cos x$ and $g(x) = x^2 - 1$. $g(f(x)) = g(\sin x + \cos x) = (\sin x + \cos x)^2 - 1$. $= \sin^2 x + \cos^2 x + 2 \sin x \

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$-\frac{\pi}{4} \leq x \leq \frac{\pi}{4}$

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(A) Given $f(x) = \sin x + \cos x$ and $g(x) = x^2 - 1$. $g(f(x)) = g(\sin x + \cos x) = (\sin x + \cos x)^2 - 1$. $= \sin^2 x + \cos^2 x + 2 \sin x \cos x - 1$. $= 1 + \sin 2x - 1 = \sin 2x$. For a function to be invertible,it must be one-one and onto in the given domain. The function $h(x) = \sin 2x$ is one-one in the interval where the argument $2x$ lies within $[-\frac{\pi}{2}, \frac{\pi}{2}]$. Thus,$-\frac{\pi}{2} \leq 2x \leq \frac{\pi}{2}$. Dividing by $2$,we get $-\frac{\pi}{4} \leq x \leq \frac{\pi}{4}$. Therefore,the function is invertible in the interval $[-\frac{\pi}{4}, \frac{\pi}{4}]$.

$f(x) = \sin x + \cos x, g(x) = x^2 - 1$. Then $g(f(x))$ is invertible if:

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