Functions

To find the domain of the function , we need to consider the conditions for both the numerator and the denominator.

• Numerator ( ): The domain of the inverse cosine function is . So, .
• Denominator ( ): The denominator cannot be zero. is the greatest integer function (floor function). We must find where . This occurs for all such that . So we must exclude the interval from our domain.

Combining both conditions: We start with the domain from the numerator: . We must exclude the interval where the denominator is zero: . So, the domain is . This can be written as the union of two parts: Part 1: The interval from -1 up to (but not including) 0, which is . Part 2: The single point at the end of the original interval, which is . Therefore, the domain is .

We are given the piecewise function: Now, let's check each of the given statements to identify the FALSE one.

Option (a): For , we have , so . Thus, this statement is correct.

Option (b): For , , so .

Since for , we get: Thus, this statement is also correct.

Option (c): For , and .

So, For , . Hence, this statement is false.

Option (d): For , and . Also, . So, Thus, this statement is correct.

Conclusion: The FALSE statement is option (c). Thus, the correct answer is .

The function involves the composition of several functions. Let's break it down:
- is a real-valued function defined for all . The domain of is the set of real numbers where is defined.

- represents the sine function, which is defined for all real numbers, so it does not introduce any restrictions on the domain.

- represents the signum function, which is also defined for all real numbers and gives the sign of its argument.

Thus, the domain of is ultimately determined by the domain of , since both and are defined everywhere. The functions involved do not introduce any new domain restrictions. Therefore, the domain of is the same as the domain of , and option (a) is incorrect.
Now, let's analyze each option:

• (a) The domain of h(x) is different from the domain of g(x) at the same point.
• (b) h(x) is discontinuous at g(x)=0.
• (c) The domain of h(x) is the same as the domain of g(x).
• (d) The domain of continuity of h(x) equals the domain of continuity of g(x)−{x∈R∣g(x)=0}.​

Thus, the correct answer is (a).