Find range for f(X)=sin^-1[x²+1/2]+cos^-1 [x²-1/2] ,where [ ] denotes greatest integer function.

To find the range of the function \( f(X) = \sin^{-1}[x^2 + \frac{1}{2}] + \cos^{-1}[x^2 - \frac{1}{2}] \), where \([ ]\) denotes the greatest integer function, we need to analyze the components of the function carefully. The greatest integer function, also known as the floor function, rounds down to the nearest integer. This means we will first determine the values of \( x^2 + \frac{1}{2} \) and \( x^2 - \frac{1}{2} \) and how they interact with the floor function.

Understanding the Components

Let's break down the two components of the function:

• For \( [x^2 + \frac{1}{2}] \): The expression \( x^2 \) is always non-negative, so \( x^2 + \frac{1}{2} \) will always be at least \( \frac{1}{2} \). The floor function will take integer values starting from 0. For example:
• If \( 0 \leq x^2 < \frac{1}{2} \), then \( [x^2 + \frac{1}{2}] = 0 \).
• If \( \frac{1}{2} \leq x^2 < \frac{3}{2} \), then \( [x^2 + \frac{1}{2}] = 1 \).
• If \( \frac{3}{2} \leq x^2 < \frac{5}{2} \), then \( [x^2 + \frac{1}{2}] = 2 \).

• For \( [x^2 - \frac{1}{2}] \): This expression can take negative values as well. For instance:
• If \( 0 \leq x^2 < \frac{1}{2} \), then \( [x^2 - \frac{1}{2}] = -1 \).
• If \( \frac{1}{2} \leq x^2 < \frac{3}{2} \), then \( [x^2 - \frac{1}{2}] = 0 \).
• If \( \frac{3}{2} \leq x^2 < \frac{5}{2} \), then \( [x^2 - \frac{1}{2}] = 1 \).

Combining the Components

Now, let's combine these results to find the possible values of \( f(X) \):

• When \( 0 \leq x^2 < \frac{1}{2} \):
  • \( f(X) = \sin^{-1}(0) + \cos^{-1}(-1) = 0 + \pi = \pi \)
• When \( \frac{1}{2} \leq x^2 < \frac{3}{2} \):
  • \( f(X) = \sin^{-1}(1) + \cos^{-1}(0) = \frac{\pi}{2} + \frac{\pi}{2} = \pi \)
• When \( \frac{3}{2} \leq x^2 < \frac{5}{2} \):
  • \( f(X) = \sin^{-1}(2) + \cos^{-1}(1) \) is not valid since \( \sin^{-1}(2) \) is undefined.

Finalizing the Range

From our analysis, we see that the only valid output for \( f(X) \) occurs in the first two intervals, both yielding \( \pi \). Therefore, the range of the function \( f(X) \) is simply:

This means that regardless of the input \( x \), the output of the function will always be \( \pi \) as long as \( x^2 \) remains within the specified bounds. Thus, the function is constant over the defined intervals.