Let f(x) = sin 2x + cos 2x and g(x) = x2- 1, then g (f(x)) will be invertible for the domain

Question

Let f(x) = sin 2x + cos 2x and g(x) = x2- 1, then g (f(x)) will be invertible for the domain (A) x ∈ [-(π/2), (π/2)] (B) x ∈ [-(π/8), (π/8)] (C) x ∈ [-(π/4), (π/4)] (D) x ∈ [0, (π/4))

Tardigrade Admin · Accepted Answer

g(f(x)) = g(sin 2x + cos 2x) = (sin 2x + cos 2x)2 - 1 = sin 4x Now, sin 4x will be invertible for - (π/2) le 4x le (π/2) or -(π/8) le x le (π/8)

Q. Let $f (x) = s in 2 x + cos 2 x$ and $g (x) = x^{2} - 1$ , then $g (f (x))$ will be invertible for the domain

$x \in [- \frac{π}{2}, \frac{π}{2}]$

$x \in [- \frac{π}{8}, \frac{π}{8}]$

$x \in [- \frac{π}{4}, \frac{π}{4}]$

$x \in [0, \frac{π}{4})$

$g (f (x)) = g (s in 2 x + cos 2 x)$
$= (s in 2 x + cos 2 x)^{2} - 1 = s in 4 x$
Now, $s in 4 x$ will be invertible for
$- \frac{π}{2} \leq 4 x \leq \frac{π}{2}$
or $- \frac{π}{8} \leq x \leq \frac{π}{8}$

Q. Let f(x)=sin2x+cos2x and g(x)=x2−1, then g(f(x)) will be invertible for the domain

x∈[−2π​,2π​]

x∈[−8π​,8π​]

x∈[−4π​,4π​]

x∈[0,4π​)