Problem 10
Question
The function \(f(x)=[x]^{2}-\left[x^{2}\right]\) (where \([x]\) is the greatest integer less than or equal to \(x\) ), is discontinuous at (A) all integers (B) all integers except 0 and 1 (C) all integers except 0 (D) all integers except 1
Step-by-Step Solution
Verified Answer
(B) all integers except 0 and 1
1Step 1: Understand the Function Components
The function given is \( f(x) = [x]^2 - [x^2] \). Here, \([x]\) denotes the greatest integer less than or equal to \(x\), known as the floor function or greatest integer function.
2Step 2: Analyze Discontinuity Conditions
A function is generally discontinuous where the greatest integer function has a discontinuity, which occurs at integer points. We need to explore how the values of \([x]^2\) and \([x^2]\) change as \(x\) crosses an integer.
3Step 3: Evaluate at Integer Points for \([x]^2\)
When \(x\) is an integer, \(n\), \([n] = n\), so \([x]^2 = n^2\). For \(x=n+0^+\), \([x]=n\) and for \(x=n-0^+\), \([x]=n-1\). This means \([x]^2\) exhibits continuous behavior when moving through integer values.
4Step 4: Evaluate at Integer Points for \([x^2]\)
For \(x\) being an integer \(n\), \(x^2 = n^2\) gives \([x^2] = n^2\). Now for small \( \epsilon > 0 \), at \(x = n + \epsilon\), \(x^2 \approx n^2 + 2n\epsilon\), thus \([x^2]\) can change if \(2n\epsilon\) crosses the next integer, which can make it discontinuous at certain points.
5Step 5: Identify Continuity and Discontinuity
For non-negative integers like 0 and 1, \([x^2] = x^2\) does not cause a floor function discontinuity since its integer square itself is unaffected (\(n(n)\) does not increase). However, for integers \(n\geq2\), \([n^2 + 2n\epsilon]\) changes the integer part of the result abruptly, showing discontinuous jump in \([x^2]\).
6Step 6: Determine the Correct Solution
From previous analyses, \(f(x)\) is discontinuous at all integers for integer values \(n > 1\); however, it remains continuous at \(n = 0 \) and \(n = 1\). Therefore, the correct answer is (B) all integers except 0 and 1.
Key Concepts
Floor FunctionGreatest Integer FunctionDiscontinuity in Functions
Floor Function
The floor function, often denoted by \[x\], is a pivotal mathematical concept. It represents the greatest integer less than or equal to any given number. This function "rounds down" a real number to the nearest integer. For example:
- For \( x = 3.7 \), \[x\] = 3.
- For \( x = -2.3 \), \[x\] = -3.
Greatest Integer Function
Closely related to the floor function, the greatest integer function also returns the greatest integer less than or equal to a given number. It is, essentially, another way to refer to the floor function. Notably, the greatest integer function steps down whenever the argument crosses an integer boundary. For instance:
- \( [3] = 3 \) and \( [3-0^+] = 2 \).
Discontinuity in Functions
Discontinuity in functions is where a function jumps, breaks, or is not smooth. With respect to the floor or greatest integer function, discontinuities are typical at integer points. Since these functions can shift values dramatically when crossing an integer, the resulting function may not smoothly transition at these points. Specifically, for \( f(x) = [x]^2 - [x^2] \):
- At integer values, the behavior of \( [x]^2 \) remains stable, while \( [x^2] \) may not, due to changes in \( x^2 \) itself.
- As \( x = n + \epsilon \), even a minute \( \epsilon \) can cause \( [x^2] \) to vary if \( 2n\epsilon \), an increment over \( n^2 \), bypasses a whole number.
Other exercises in this chapter
Problem 8
Let \([x]\) denotes the greatest integer less than or equal to \(x\). If \(f(x)=[x \sin \pi x]\), then \(f(x)\) is (A) continuous at \(x=0\) (B) continuous in \
View solution Problem 9
The function \(f(x)=[x]^{2}-\left[x^{2}\right]\) (where \([x]\) is the greatest integer less than or equal to \(x\) ), is discontinuous at (A) all integers (B)
View solution Problem 11
Let \(f(x)\) be a function satisfying the condition \(f(-x)=f(x)\), for all real \(x\). If \(f^{\prime}(0)\) exists, then its value is (A) 0 (B) 1 (C) \(-1\) (D
View solution Problem 12
If \(f(x)=\left\\{\begin{array}{cc}\frac{x\left(3 e^{1 / x}+4\right)}{2-e^{1 / x}}, x \neq 0 \\ 0, x=0\end{array}\right.\), then \(f(x)\) is (A) continuous as w
View solution