Let f(x) = sin 2x + cos 2x and g(x)=x 2 -1, then g(f(x)) is invertible in the domain

$x\ \epsilon\ [\frac{-\pi}{2},\frac{\pi}{2}]$

$x\ \epsilon\ [\frac{-\pi}{4},\frac{\pi}{4}]$

$x\ \epsilon\ [0,\frac{\pi}{4}]$

Let f(x) = sin 2x + cos 2x and g(x)=x 2 -1, then g(f(x)) is invertible in the domain

$x\ \epsilon\ [\frac{-\pi}{2},\frac{\pi}{2}]$

$x\ \epsilon\ [\frac{-\pi}{4},\frac{\pi}{4}]$

$x\ \epsilon\ [0,\frac{\pi}{4}]$

Question:

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

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To make a function like $ g(f(x)) $ invertible, check if it is one-to-one. For trigonometric functions, like sine, you can use the property that the function is one-to-one in specific intervals where it does not repeat. In this case, the interval $ \left[\frac{-\pi}{8}, \frac{\pi}{8}\right] $ ensures that the sine function is one-to-one and thus invertible.

Updated On: Mar 29, 2025

$x\ \epsilon\ [\frac{-\pi}{2},\frac{\pi}{2}]$
$x\ \epsilon\ [\frac{-\pi}{4},\frac{\pi}{4}]$
$x\ \epsilon\ [0,\frac{\pi}{4}]$
$x\ \epsilon\ [\frac{-\pi}{8},\frac{\pi}{8}]$

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The Correct Option is D

Approach Solution - 1

The correct answer is (D) : $x\ \epsilon\ [\frac{-\pi}{8},\frac{\pi}{8}]$.

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Approach Solution -2

The correct answer is: (D) $ x \in \left[\frac{-\pi}{8}, \frac{\pi}{8}\right] $.

We are given the functions:

f(x) = sin(2x) + cos(2x)
g(x) = x² - 1

We need to find the domain where $ g(f(x)) $ is invertible.

First, we compute the composite function $ g(f(x)) $:

$ g(f(x)) = (sin(2x) + cos(2x))^2 - 1 $

To simplify this, let's first expand $ (sin(2x) + cos(2x))^2 $:

$ (sin(2x) + cos(2x))^2 = sin^2(2x) + 2sin(2x)cos(2x) + cos^2(2x) $

Using the Pythagorean identity $ sin^2(2x) + cos^2(2x) = 1 $, we get:

$ (sin(2x) + cos(2x))^2 = 1 + 2sin(2x)cos(2x) $

Thus, $ g(f(x)) $ becomes:

$ g(f(x)) = 1 + 2sin(2x)cos(2x) - 1 = 2sin(2x)cos(2x) $

Using the double angle identity for sine, $ 2sin(2x)cos(2x) = sin(4x) $, we get:

$ g(f(x)) = sin(4x) $

To make $ g(f(x)) = sin(4x) $ invertible, the function must be one-to-one. A sine function is one-to-one in the interval where it doesn't repeat. The sine function $ sin(4x) $ is one-to-one in the interval $ \left[\frac{-\pi}{8}, \frac{\pi}{8}\right] $, since $ 4x $ will cover only one period of the sine wave in this interval. Therefore, $ g(f(x)) $ is invertible in the domain $ x \in \left[\frac{-\pi}{8}, \frac{\pi}{8}\right] $. Thus, the correct answer is (D) $ x \in \left[\frac{-\pi}{8}, \frac{\pi}{8}\right] $.

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Top Questions on composite of functions

Let $f:\R \rightarrow \R$ and $g:\R \rightarrow\R$ be functions defined by
$f(x)=\left\{ \begin{array}{ll} x|x|\sin(\frac{1}{x}), & x\ne0 \\ 0, & x = 0, \end{array} \right.\text{and} \ g(x)=\left\{ \begin{array}{ll} 1-2x, & 0\leq x\leq \frac{1}{2}, \\ 0, & \text{otherwise.} \end{array} \right.$
Let $a,b,c,d \in \R$. Define the function $h:\R\rightarrow\R$ by
$h(x)=af(x)+b(g(x)+g(\frac{1}{2}-x))+c(x-g(x))+d\ g(x), x\in\R$.
Match each entry in List-I to the correct entry in List-II.

List - I		List - II
(P)	If a = 0, b = 1, c = 0 and d = 0, then	(1)	h is one-one.
(Q)	If a = 1, b = 0, c = 0 and d = 0, then	(2)	h is onto.
(R)	If a = 0, b = 0, c = 1 and d = 0, then	(3)	h is differentiable on $\R$
(S)	If a = 0, b = 0, c = 0 and d = 1, then	(4)	the range of h is [0, 1].
		(5)	the range of h is {0, 1}.

The correct option is