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AIEEE 2004

JEE Main 2004 Mathematics question on Sets Relations and Functions

If f(x)=sin⁡x+cos⁡xf(x) = \sin x + \cos x, g(x)=x2−1g(x) = x^2 - 1, then g{f(x)}g\{f(x)\} is invertible in the domain
  1. A.
    [0,π2]\left[0, \dfrac{\pi}{2}\right]
  2. B.
    [−π4,π4]\left[-\dfrac{\pi}{4}, \dfrac{\pi}{4}\right]
  3. C.
    [−π2,π2]\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]
  4. D.
    [0,π]\left[0, \pi\right]
Show answer and solution

Correct option: B

Solution

g{f(x)} = (sin x + cos x)² - 1 = 1 + 2sin x cos x - 1 = sin 2x.

For invertibility, sin 2x must be one-to-one (strictly monotonic). sin 2x is strictly increasing when 2x ∈ [-π/2, π/2], i.e., x ∈ [-π/4, π/4]. This is exactly option 2, and it's the standard interval containing 0 where sin 2x is monotonically increasing without repeating values.

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