Problem 98
Question
The range of the function \(y=\sin ^{-1}\left[x^{2}+\frac{1}{2}\right]+\cos ^{-1}\left[x^{2}-\frac{1}{2}\right]\), where \([\cdot]\) denotes the integral part, is (A) \((0, \pi)\) (B) \([0, \pi]\) (C) \(\\{\pi\\}\) (D) \(\\{0, \pi\\}\)
Step-by-Step Solution
Verified Answer
The range of the function is \(\{\pi\}\) (Option C).
1Step 1: Understand the Domain of Each Component
For the function to be defined, \( \sin^{-1} \left( [x^2 + \frac{1}{2} ]\right) \) must be between \(-1\) and \(1\) and the same applies for \( \cos^{-1} \left( [x^2 - \frac{1}{2} ] \right) \).
2Step 2: Simplify the Function Components
Let's consider simplifying the components:1. \([x^2 + \frac{1}{2}]\) takes integer values, and it must be \(-1 \leq [x^2 + \frac{1}{2}] \leq 1\) 2. \([x^2 - \frac{1}{2}]\) takes integer values, and it must be \(-1 \leq [x^2 - \frac{1}{2}] \leq 1\)Thus, both values must be integers between \(-1\) and \(1\).
3Step 3: Assess the Possible Integer Combinations
Through comprehensive domain analysis, it can be determined that they must satisfy the condition:1. \([x^2 + \frac{1}{2}] = 1, [x^2 - \frac{1}{2}] = 0\)since integral parts can only lead to these constrained combinations.
4Step 4: Calculate \( \sin^{-1} \) and \( \cos^{-1} \) Values
With identified values, calculate \( \sin^{-1}(1) + \cos^{-1}(0) = \frac{\pi}{2} + \frac{\pi}{2}\).This simplifies to \( \pi \).
5Step 5: Verify the Result
Check whether these values respect the original constraints.
The combination satisfies all required domain constraints for both inverse operations ensuring the expression and hence the range.
Key Concepts
Domain and Range of FunctionsInteger Part FunctionComposite Functions Analysis
Domain and Range of Functions
When studying functions, understanding the concepts of domain and range is crucial. The domain refers to all possible input values a function can accept, while the range represents all possible output values the function can produce.
In the exercise, the given function involves inverse trigonometric functions:
In our specific problem, ensuring that \([x^2 + \frac{1}{2}]\) and \([x^2 - \frac{1}{2}]\) lie within this interval is key to defining the domain. The domain constraints are essential because they ensure that each expression inside the functions remains valid, ultimately influencing the outcome of the composite function's range.
In the exercise, the given function involves inverse trigonometric functions:
- \( \sin^{-1}\left([x^2 + \frac{1}{2}]\right) \)
- \( \cos^{-1}\left([x^2 - \frac{1}{2}]\right) \)
In our specific problem, ensuring that \([x^2 + \frac{1}{2}]\) and \([x^2 - \frac{1}{2}]\) lie within this interval is key to defining the domain. The domain constraints are essential because they ensure that each expression inside the functions remains valid, ultimately influencing the outcome of the composite function's range.
Integer Part Function
The integer part function, denoted usually by square brackets \([\cdot]\), plays a critical role in understanding the problem given. This mathematical operation outputs the greatest integer less than or equal to a given number.
Consider \([x^2 + \frac{1}{2}]\) and \([x^2 - \frac{1}{2}]\) from the exercise. These expressions must be evaluated to integers that lie between -1 and 1 for the inverse trigonometric functions to be defined properly. The integer part function effectively truncates any fractional part of these expressions, converting them to a workable form for the inverse trigonometric calculations.
This operation is crucial because it limits the possible values \(x^2\) can take, thus narrowing down which integers the function components can resolve to. The simplification through integer part evaluation is a common technique in such problems as it reduces complexity, making it easier to determine valid values that satisfy all function conditions.
Consider \([x^2 + \frac{1}{2}]\) and \([x^2 - \frac{1}{2}]\) from the exercise. These expressions must be evaluated to integers that lie between -1 and 1 for the inverse trigonometric functions to be defined properly. The integer part function effectively truncates any fractional part of these expressions, converting them to a workable form for the inverse trigonometric calculations.
This operation is crucial because it limits the possible values \(x^2\) can take, thus narrowing down which integers the function components can resolve to. The simplification through integer part evaluation is a common technique in such problems as it reduces complexity, making it easier to determine valid values that satisfy all function conditions.
Composite Functions Analysis
Composite functions involve applying one function to the results of another. Here, our composite function is built from the integer part followed by the inverse trigonometric functions, specifically
Exploring the possible integer combinations leading to function definitions shows that only one possible valid combination (\([x^2 + \frac{1}{2}] = 1\) and \([x^2 - \frac{1}{2}] = 0\)) satisfies both components and hence the original function. This arrangement results in outputs for
- \( \sin^{-1} \left([x^2 + \frac{1}{2} ]\right)\)
- \( \cos^{-1} \left( [x^2 - \frac{1}{2} ] \right)\)
Exploring the possible integer combinations leading to function definitions shows that only one possible valid combination (\([x^2 + \frac{1}{2}] = 1\) and \([x^2 - \frac{1}{2}] = 0\)) satisfies both components and hence the original function. This arrangement results in outputs for
- \( \sin^{-1}(1) = \frac{\pi}{2} \)
- \( \cos^{-1}(0) = \frac{\pi}{2} \)
Other exercises in this chapter
Problem 95
The domain of the function \(f(x)=\ln \left\\{\operatorname{sgn}\left(9-x^{2}\right)\right\\}+\sqrt{[x]^{3}-4[x]}\), where \([\cdot]\) denotes integral part, is
View solution Problem 97
The range of the function \(y=\left[x^{2}\right]-[x]^{2}, x \in[0,2]\) where [-] denotes the integral part, is (A) \(\\{0\\}\) (B) \(\\{0,1\\}\) (C) \(\\{1,2\\}
View solution Problem 100
Let \(f: N \rightarrow N\), where \(f(x)=x+(-1)^{x-1}\). Then, (A) \(f^{-1}(x)=x+(-1)^{x-1}\) (B) \(f^{-1}(x)=x\) (C) \(f^{-1}(x)=x-(-1)^{x-1}\) (D) None of the
View solution Problem 102
Domain of definition of the function \(f(x)=\sqrt{\sin x}+\sin ^{-1}\left(\frac{2|x|}{1+x^{2}}\right)\) (A) \((2 n \pi,(2 n+1) \pi), n \in I\) (B) \([2 n \pi,(2
View solution