Problem 42
Question
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 \(\\{x \in R: g[f(x)] \leq f\) \([g(x)]\\}=\) (A) \(Z \cup(-\infty, 0)\) (B) \((-\infty, 0)\) (C) \(Z\) (D) \(R\)
Step-by-Step Solution
Verified Answer
The solution set is (A) \(Z \cup (-\infty, 0)\).
1Step 1: Understand the Functions
We have two functions: \(f(x) = |x|\), which is the absolute value function, and \(g(x) = [x]\), which is the floor function, mapping any real number \(x\) to the greatest integer less than or equal to \(x\).
2Step 2: Evaluate g(f(x))
Since \(f(x) = |x|\), the expression \(g(f(x))\) becomes \(g(|x|)\). This is equal to \([|x|]\), which is the greatest integer less than or equal to \(|x|\).
3Step 3: Evaluate f(g(x))
Using \(g(x) = [x]\), the expression \(f(g(x))\) becomes \(f([x])\), which results in \(|[x]|\). This is the absolute value of the greatest integer less than or equal to \(x\).
4Step 4: Compare g(f(x)) with f(g(x))
The inequality is \(g(f(x)) \leq f(g(x))\), which translates to \([|x|] \leq |[x]|\). We check the possible range of \(x\) for which this holds.
5Step 5: Analyze Outcomes for x ≥ 0
If \(x ≥ 0\), both \(|x|\) and \([x]\) are non-negative. Hence, \([x] = x\) if \(x\) is an integer, and \([x] \leq [x]\) generally holds, satisfying the inequality regardless of the decimal part.
6Step 6: Analyze Outcomes for x < 0
If \(x < 0\), \(|x| = -x\), a positive value, and \( [x] \leq 0 \). Thus, \([|x|] = |-x| - 1\), not satisfying \([-x|] ≯ ||[x]||\) for any non-zero \(x\), because \(|[x]| = |-[x]| \ge 0\) emphasizing the need for \(x < 0\).
7Step 7: Determining the Answer Set
The valid solution set needs to satisfy \([|x|] \leq ||[x]|| \), which holds true primarily when \(x\) is negative or \(x\) is an integer. Thus, \({x \in R : g(f(x)) \leq f(g(x))}\) encompasses integers and negative reals.
Key Concepts
Absolute Value FunctionFloor FunctionInequality Solutions
Absolute Value Function
The absolute value function, denoted as \( f(x) = |x| \), is a function that transforms a real number \( x \) into its non-negative counterpart. Essentially, absolute value tells us how far a number is from zero on the number line, without considering the direction. For instance, both \(|3|\) and \(|-3|\) equal 3 because both are 3 units away from zero.
Key properties of the absolute value function include:
Understanding this function helps in analyzing inequalities, as it helps in estimating only positive outputs, irrespective of whether the input \( x \) is positive or negative.
Key properties of the absolute value function include:
- \( |x| \geq 0 \) for all \( x \in R \)
- \( |x| = 0 \) if and only if \( x = 0 \)
- \( |xy| = |x||y| \) for any real numbers \( x \) and \( y \)
- \( |x+y| \leq |x| + |y| \) (triangle inequality)
Understanding this function helps in analyzing inequalities, as it helps in estimating only positive outputs, irrespective of whether the input \( x \) is positive or negative.
Floor Function
The floor function, denoted \( g(x) = [x] \), maps a real number \( x \) to the greatest integer less than or equal to \( x \). This function can be seen as 'rounding down' any real number to an integer.
Let's consider some examples:
The emphasis on integer results makes the floor function crucial for evaluating the comparison between \( g(f(x)) \) and \( f(g(x)) \). This can determine whether integer values or specific inequalities are satisfied.
Let's consider some examples:
- \([3.7] = 3\)
- \([ -1.2 ] = -2\)
- \([5] = 5\) (as the number is already an integer)
The emphasis on integer results makes the floor function crucial for evaluating the comparison between \( g(f(x)) \) and \( f(g(x)) \). This can determine whether integer values or specific inequalities are satisfied.
Inequality Solutions
Solving inequalities involves finding the range of values within which the inequalities hold true. In the exercise, we're comparing \( g(f(x)) \) and \( f(g(x)) \) to establish \([|x|] \leq |[x]|\).
To solve this inequality, consider the following:
To solve this inequality, consider the following:
- When \( x \geq 0 \), \( |x| = x \), and thus \([|x|] = [x]\) which maintains equality.
- For \( x < 0 \), the function \( |x| = -x \), and the inequality \([|x|] \leq |[x]|\) means we require \([-|x|] \leq |[x]|\).
- When \( x \) is a negative integer, \([|x|] = |-x| \), and since \( x \) already rounds down, the requirement holds well for any \( x < 0 \), or when \( x \) is an integer.
Other exercises in this chapter
Problem 39
The domain of definition of $$ f(x)=\sqrt{\frac{\log _{0.3}|x-2|}{|x|}} $$ (A) \([1,2) \cup(2,3]\) (B) \([1,3]\) (C) \(\mathbb{R}-(1,3]\) (D) None of these
View solution 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 43
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 greate
View solution Problem 47
Let \(f: \mathbb{R} \rightarrow A=\left\\{y: 0 \leq y
View solution