If $f (x) = sin x + cos x$ , $g (x) = x^{2} - 1$ 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 $[- \frac{π}{2}, \frac{π}{2}]$
$f (x) = sin x + cos x, g (x) = x^{2} - 1 T h e n g [f (x)] g [f (x)] = g (sin x + cos x) = {(sin x + cos x)}^{2} - 1 g [f (x)] = {sin}^{2} x + {cos}^{2} x + 2 sin x . cos x - 1 = 1 + 2 sin x . cos x - 1 [∵ {sin}^{2} x + {cos}^{2} = 1] g (f (x)) = sin 2 x A n s . [∵ 2 sin x . cos x = sin 2 x] L e t g (f (x)) = y y = sin 2 x [∵ D o m a i n o f sin x i s \frac{- π}{2}, \frac{π}{2}] ∴ D o m a i n o f sin 2 x = [\frac{- π}{2}, \frac{π}{2}] A n s .$

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Similar questions

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

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

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

g (x) = x^{2} - 1

\in

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[0, \frac{π}{2}]

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