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f(x)=cos⁡[ex–sin⁡(x)]f(x)=\cos[e^x–\sin(x)]

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Full solution

Q. f(x)=cos⁡[ex–sin⁡(x)]f(x)=\cos[e^x–\sin(x)]
  1. Understand Cosine Function Behavior: We need to understand the behavior of the cosine function to determine the range of f(x)f(x). The cosine function, cos⁡(θ)\cos(\theta), has a range of [−1,1][-1, 1] for any real number θ\theta.
  2. Consider Argument of f(x)f(x): Now, let's consider the argument of the cosine function in f(x)f(x), which is ex–sin⁡(x)e^x – \sin(x). The exponential function exe^x is always positive for all real xx, and sin⁡(x)\sin(x) oscillates between −1-1 and 11.
  3. Positivity of ex−sin⁡(x)e^x - \sin(x): Since exe^x is always greater than sin⁡(x)\sin(x), the expression ex–sin⁡(x)e^x – \sin(x) will always be positive.\newlineHowever, the exact value of ex–sin⁡(x)e^x – \sin(x) can vary greatly depending on xx.
  4. Cosine Range Constraint: Regardless of the value of ex−sin⁡(x)e^x - \sin(x), the cosine of this expression will always fall within the range of the cosine function itself, which is [−1,1][-1, 1].
  5. Final Range of f(x)f(x): Therefore, the range of f(x)=cos⁡[ex–sin⁡(x)]f(x) = \cos[e^x – \sin(x)] is the same as the range of the cosine function.\newlineThe range of f(x)f(x) is [−1,1][-1, 1].

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