Question

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

• A. $x\ \epsilon\ [\frac{-\pi}{2},\frac{\pi}{2}]$
• B. $x\ \epsilon\ [\frac{-\pi}{4},\frac{\pi}{4}]$
• C. $x\ \epsilon\ [0,\frac{\pi}{4}]$
• D. $x\ \epsilon\ [\frac{-\pi}{8},\frac{\pi}{8}]$

Hint:

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.

Answer 1

The Correct Option is D

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

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