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)) 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 [π8,π8] \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 ϵ [π2,π2]x\ \epsilon\ [\frac{-\pi}{2},\frac{\pi}{2}]
  • x ϵ [π4,π4]x\ \epsilon\ [\frac{-\pi}{4},\frac{\pi}{4}]
  • x ϵ [0,π4]x\ \epsilon\ [0,\frac{\pi}{4}]
  • x ϵ [π8,π8]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 ϵ [π8,π8]x\ \epsilon\ [\frac{-\pi}{8},\frac{\pi}{8}].
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Approach Solution -2

The correct answer is: (D) x[π8,π8] 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)) g(f(x)) is invertible.

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

g(f(x))=(sin(2x)+cos(2x))21 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(2x)+cos(2x))2=sin2(2x)+2sin(2x)cos(2x)+cos2(2x) (sin(2x) + cos(2x))^2 = sin^2(2x) + 2sin(2x)cos(2x) + cos^2(2x)

Using the Pythagorean identity sin2(2x)+cos2(2x)=1 sin^2(2x) + cos^2(2x) = 1 , we get:

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

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

g(f(x))=1+2sin(2x)cos(2x)1=2sin(2x)cos(2x) 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) 2sin(2x)cos(2x) = sin(4x) , we get:

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

To make g(f(x))=sin(4x) 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) sin(4x) is one-to-one in the interval [π8,π8] \left[\frac{-\pi}{8}, \frac{\pi}{8}\right] , since 4x 4x will cover only one period of the sine wave in this interval. Therefore, g(f(x)) g(f(x)) is invertible in the domain x[π8,π8] x \in \left[\frac{-\pi}{8}, \frac{\pi}{8}\right] . Thus, the correct answer is (D) x[π8,π8] x \in \left[\frac{-\pi}{8}, \frac{\pi}{8}\right] .
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