Problem 43
Question
The function $$ f(x)=\sin ^{-1}\left(x-x^{2}\right)+\sqrt{1-\frac{1}{|x|}}+\frac{1}{\left[x^{2}-1\right]} $$ is defined in the interval (where [-] is the greatest integer) (A) \(x \in\left(\sqrt{2}, \frac{1+\sqrt{5}}{2}\right)\) (B) \(x \in\left(1, \frac{1+\sqrt{5}}{2}\right)\) (C) \(x \in\left[\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right]\) (D) \(x \in\left(-\sqrt{2}, \frac{1+\sqrt{5}}{2}\right)\)
Step-by-Step Solution
Verified Answer
The function is defined in the interval (B): \(x \in \left(1, \frac{1+\sqrt{5}}{2}\right)\).
1Step 1: Analyze the Inverse Sine Function
For the component \( \sin^{-1}(x - x^2) \) to be defined, the expression inside the inverse sine, \( x - x^2 \), must be within the range of \([-1, 1]\). We need to find the values of \( x \) that satisfy this inequality.
2Step 2: Simplify the Quadratic Inequality
Set up the inequality as \( -1 \leq x - x^2 \leq 1 \). Simplify:1. \( x - x^2 \geq -1 \) simplifies to \( (x - \frac{1}{2})^2 \leq \frac{5}{4} \).2. \(-x^2 + x - 1 \leq 0\) solves to \( (x - 1)^2 \geq 0 \), so \( x eq 1 \).
3Step 3: Handle the Square Root Expression
For \( \sqrt{1 - \frac{1}{|x|}} \) to be defined, \( |x| > 1 \). Thus, the interval for \( x \) is \( x < -1 \) or \( x > 1 \).
4Step 4: Analyze the Greatest Integer Function
The term \( \frac{1}{[x^2 - 1]} \) is defined only if \( [x^2 - 1] eq 0 \), meaning \( x eq \pm 1 \). This makes the function undefined at these points as well.
5Step 5: Combine Conditions
From step 2, we have valid \( x \) values for \( x \in \left(-\sqrt{2}, \frac{1+\sqrt{5}}{2}\right) \); step 3 and step 4 further suggest \( x > 1 \) and \( x eq \pm 1 \). Thus, the compatible interval for all parts is \( 1 < x < \frac{1+\sqrt{5}}{2} \), which matches option (B).
Key Concepts
Quadratic InequalitiesDomain of FunctionsGreatest Integer FunctionInterval Notation
Quadratic Inequalities
Quadratic inequalities involve expressions in which a quadratic function is set to be greater than or less than a reference value. The simplest form of a quadratic inequality is like the one in the exercise: \( -1 \leq x - x^2 \leq 1 \). Solving quadratic inequalities typically requires finding the roots of the equation and analyzing the intervals between the roots where the inequality holds true.
When we express \( x - x^2 \) within \([-1, 1]\), we first manipulate the inequality \( x - x^2 \geq -1 \) into a standard form by shifting terms. Completing the square can sometimes be helpful for this analysis. By rewriting the inequality, it becomes \((x - \frac{1}{2})^2 \leq \frac{5}{4}\).
This reveals the interval where the condition holds. The solution to such inequalities often results in two primary conditions that can be expressed in interval or set notations for simplicity and clarity.
When we express \( x - x^2 \) within \([-1, 1]\), we first manipulate the inequality \( x - x^2 \geq -1 \) into a standard form by shifting terms. Completing the square can sometimes be helpful for this analysis. By rewriting the inequality, it becomes \((x - \frac{1}{2})^2 \leq \frac{5}{4}\).
This reveals the interval where the condition holds. The solution to such inequalities often results in two primary conditions that can be expressed in interval or set notations for simplicity and clarity.
Domain of Functions
The domain of a function is the complete set of possible values of the independent variable that will give us a real number as an output. In other words, it's the set of 'input' values within which the function operates without errors, such as undefined operations like division by zero or taking the square root of a negative number in real-number arithmetic.
In this exercise, determining the domain involves understanding each component separately. For example, the inverse sine function \( \sin^{-1}(x - x^2) \) requires its input to be within \([-1, 1]\). Similarly, the square root function \( \sqrt{1 - \frac{1}{|x|}} \) is only defined when \( |x| > 1 \). Lastly, we examine the term with the greatest integer function where the expression inside must not lead to division by zero. Balancing these conditions identifies the valid range of \( x \) values.
In this exercise, determining the domain involves understanding each component separately. For example, the inverse sine function \( \sin^{-1}(x - x^2) \) requires its input to be within \([-1, 1]\). Similarly, the square root function \( \sqrt{1 - \frac{1}{|x|}} \) is only defined when \( |x| > 1 \). Lastly, we examine the term with the greatest integer function where the expression inside must not lead to division by zero. Balancing these conditions identifies the valid range of \( x \) values.
Greatest Integer Function
The greatest integer function, denoted by \([x]\), represents the largest integer less than or equal to \( x \). It is also known as the floor function. In this context, the term \( \frac{1}{[x^2 - 1]} \) is defined only when \([x^2 - 1] eq 0\).
This means that \( x^2 - 1 \) must not be an integer multiple of zero. If \( x^2 - 1 = 0 \), it would imply division by zero, causing undefined behavior. Therefore, avoiding values \( x = \pm 1 \) is necessary in maintaining the function’s well-defined condition. This aspect is critical when defining the function's domain, ensuring all parts of the function are operable.
This means that \( x^2 - 1 \) must not be an integer multiple of zero. If \( x^2 - 1 = 0 \), it would imply division by zero, causing undefined behavior. Therefore, avoiding values \( x = \pm 1 \) is necessary in maintaining the function’s well-defined condition. This aspect is critical when defining the function's domain, ensuring all parts of the function are operable.
Interval Notation
Interval notation is a mathematical notation used to represent a set of numbers along a number line. It is concise and clarifies the intervals over which a particular function is defined or an inequality holds.
Using parentheses \(( )\) in interval notation indicates that the endpoints are not included in the interval, while square brackets \([ ]\) denote inclusion. For example, in this exercise, the interval \((1, \frac{1+\sqrt{5}}{2})\) indicates that values between 1 and \(\frac{1+\sqrt{5}}{2}\) are included, but the endpoints are not.
Interval notation simplifies the expression of solution sets for inequalities and domains, making it easier for students to grasp the range of possible values at a glance. This tool is crucial in mathematics, enhancing the clarity and brevity of expressing domains, ranges, and solutions.
Using parentheses \(( )\) in interval notation indicates that the endpoints are not included in the interval, while square brackets \([ ]\) denote inclusion. For example, in this exercise, the interval \((1, \frac{1+\sqrt{5}}{2})\) indicates that values between 1 and \(\frac{1+\sqrt{5}}{2}\) are included, but the endpoints are not.
Interval notation simplifies the expression of solution sets for inequalities and domains, making it easier for students to grasp the range of possible values at a glance. This tool is crucial in mathematics, enhancing the clarity and brevity of expressing domains, ranges, and solutions.
Other exercises in this chapter
Problem 40
Let \(f: R \rightarrow R\) be a function defined by, \(f(x)=\) \(-\frac{|x|^{3}+|x|}{1+x^{2}}\), then the graph of \(f(x)\) lies in which quadrant \((s) ?\) (A)
View solution Problem 42
If \(f: R \rightarrow R\) and \(g: R \rightarrow R\) are given by \(f(x)=|x|\) and \(g(x)=[x]\) for each \(x \in R\), then \(\\{x \in R: g[f(x)] \leq f\) \([g(x
View solution Problem 47
Let \(f: \mathbb{R} \rightarrow A=\left\\{y: 0 \leq y
View solution Problem 48
Suppose \(f:[2,2] \rightarrow R\) is defined by, $$ f(x)= \begin{cases}-1 & \text { for }-2 \leq x \leq 0 \\ x-1 & \text { for } 0 \leq x \leq 2\end{cases} $$ t
View solution