If $f (x) = s i n x + c o s x, g (x) = x^{2} - 1$ , 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 $[- \frac{π}{4}, \frac{π}{4}]$
By definition of composition of function,
g(f (x)) $= (s i n x + c o s x)^{2} - 1$ , is invertible (i.e. bijective)
$\Rightarrow$ g{f(x)} = sin 2x is bijective.
We know, sin x is bijective, only when $x ϵ [- \frac{π}{2}, \frac{π}{2}]$
Thus, g{f(x)} is bijective, if - $\frac{π}{2} \leq 2 x \leq \frac{2 x}{2}$
$\Rightarrow - \frac{π}{4} \leq x \leq \frac{π}{4}$

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