If $f (x) = sin x + cos x$ and $g (x) = x^{2} - 1$ then $g {f (x)}$ is invertible in the domain

[0, \frac{π}{2}]

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[- \frac{π}{4}, \frac{π}{4}]

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[- \frac{π}{2}, \frac{π}{2}]

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[0, π]

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Solution

The correct option is B $[- \frac{π}{4}, \frac{π}{4}]$
Let $y = g {f (x)} = (sin x + cos x)^{2} - 1 = 2 sin x cos x = sin 2 x$

Hence $y$ is invertible in $[- \frac{π}{4}, \frac{π}{4}]$

Note: $sin x$ is invertible in $[- \frac{π}{2}, \frac{π}{2}]$

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f (x) = s i n x + c o s x, g (x) = x^{2} - 1

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Let $f (x) = s i n x + c o s x$ , $g (x) = x^{2} - 1$ . Thus $g (f (x))$ is invertible for $x ϵ$

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[0, \frac{π}{2}]

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