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