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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $f(x) = \sin 2x + \cos 2x$ and $g(x) = x^2 - 1$. First,calculate the composite function $g(f(x))$: $g(f(x)) = (\sin 2x + \cos 2x)^2 - 1$ $g(

Vedclass · Accepted Answer

$x \in \left[\frac{-\pi}{8}, \frac{\pi}{8}\right]$

Vedclass · Answer

(A) Given $f(x) = \sin 2x + \cos 2x$ and $g(x) = x^2 - 1$. First,calculate the composite function $g(f(x))$: $g(f(x)) = (\sin 2x + \cos 2x)^2 - 1$ $g(f(x)) = (\sin^2 2x + \cos^2 2x + 2 \sin 2x \cos 2x) - 1$ Since $\sin^2 	heta + \cos^2 	heta = 1$ and $2 \sin 	heta \cos 	heta = \sin 2	heta$: $g(f(x)) = (1 + \sin 4x) - 1 = \sin 4x$. $A$ function is invertible if it is one-to-one (injective) and onto (surjective) in the given domain. The function $y = \sin 	heta$ is invertible in the interval $	heta \in \left[\frac{-\pi}{2}, \frac{\pi}{2}ight]$. Here,$	heta = 4x$,so we set: $\frac{-\pi}{2} \le 4x \le \frac{\pi}{2}$ Dividing by $4$: $\frac{-\pi}{8} \le x \le \frac{\pi}{8}$. Thus,$g(f(x))$ is invertible in the domain $x \in \left[\frac{-\pi}{8}, \frac{\pi}{8}ight]$.

Let $f(x)=\sin 2x + \cos 2x$ and $g(x)=x^2-1$. Then $g(f(x))$ is invertible in the domain:

Similar Questions

Let $g(x)$ be the inverse of an invertible function $f(x)$ which is differentiable at $x = c$. Then $g'(f(c))$ equals:

Let $e^{f(x)} = \ln x$. If $g(x)$ is the inverse function of $f(x)$,then $g'(x)$ is equal to:

Let $f: A \rightarrow B$ and $g: B \rightarrow A$ be defined as $f(x)=x^2 \forall x \in A$ and $g(x)=x^{1/2} \forall x \in B$. $f(x)$ and $g(x)$ are inverse functions to each other when

Let $f(x) > 0$ for all $x$ and $f^{\prime}(x)$ exists for all $x$. If $f$ is the inverse function of $h$ and $h^{\prime}(x) = \frac{1}{1 + \log x}$,then $f^{\prime}(x)$ will be

If $y = f(x) = \frac{x + 2}{x - 1}$,then $x = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module