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Continuity Analysis of a Piecewise Floor Function
Let f: ℝ → ℝ be defined as f(x)=[e^x], x<0; ae^x+[x-1], 0 ≤ x<1; b+[sin (π x)], 1 ≤ x<2; [e^(-x)]-c, x ≥ 2 where a, b, c ∈ ℝ and [t] denotes greatest integer less than or equal to t. Then, which of the following statements is true?
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Hint 1 of 4
What are the simplified values of the function pieces on their respective open intervals?
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Step-by-step solutionView
Correct answer
By analyzing the left-hand and right-hand limits at the boundary points x=0,1,2, we find that making f(x) continuous at any two points forces a+b+c=1, so if f has exactly one discontinuity, a+b+c=1 must hold.
Option analysis
Why each option works or fails
A · There exists a,b,c ∈ ℝ such that f is continuous of ℝ.
Believing that with three free parameters (a,b,c) matching the three boundary points (x=0,1,2), a simultaneous solution always exists. Evaluate the continuity conditions explicitly: continuity requires a=1, ae−1=b−1, and b−1=−c, which gives b=e and c=1−e, but checking continuity inside the sub-intervals (or matching limits) reveals an inconsistency among all three points simultaneously.
B · If f is discontinuous at exactly one point, then a+b+c=1.
Making an algebraic sign error when solving the system of limit equations, leading to a+b+c=1 instead of a+b+c=1. Sum the derived parameters carefully: if f is made continuous at any two points, the sum a+b+c yields an expression involving Euler's number e, which cannot equal 1.
C · If f is discontinuous at exactly one point, then a+b+c ≠ 1.
This option correctly states that requiring exactly one discontinuity implies a+b+c=1. For f to have exactly one discontinuity, it must be continuous at two of the three transition points. Solving for any pair of continuous transitions leads to values of a,b,c whose sum contains terms in e and never equals 1.
D · f is discontinuous at atleast two points, for any values of a, b and c.
Overlooking that parameters a,b,c can be chosen to satisfy the continuity conditions at two of the transition points simultaneously. Check whether any two boundary conditions can be satisfied together; two conditions impose two linear constraints on the three free parameters a,b,c, which always has solutions.
Analyze continuity at x=0,1,2 and determine which option regarding the points of discontinuity is true.
03Approach
Evaluate left-hand limit, right-hand limit, and function value at each transition point (x=0,1,2) by determining the exact values of the greatest integer terms in small neighborhoods.
04Execute
At x=0:
- For x→0−, x<0⟹ex∈(0,1), so [ex]=0. Hence, f(0−)=0.
- For x→0+, x∈(0,1)⟹x−1∈(−1,0), so [x−1]=−1. Hence, f(0+)=ae0+(−1)=a−1.
- f(0)=ae0+[0−1]=a−1.
For continuity at x=0, we need f(0−)=f(0+)⟹0=a−1⟹a=1.
05Execute
At x=1:
- For x→1−, x∈(0,1)⟹[x−1]=−1. Thus, f(1−)=ae1−1=ae−1.
- For x→1+, x∈(1,2)⟹πx∈(π,2π)⟹sin(πx)∈(−1,0), so [sin(πx)]=−1. Thus, f(1+)=b−1.
- f(1)=b+[sin(π)]=b+0=b.
Since f(1+)=b−1=b=f(1), the right-hand limit f(1+) never equals f(1) for any real b. Therefore, f(x) is intrinsically discontinuous at x=1 regardless of a,b,c.
06Execute
At x=2:
- For x→2−, x∈(1,2)⟹πx∈(π,2π)⟹[sin(πx)]=−1. Thus, f(2−)=b−1.
- For x→2+, x>2⟹−x<−2⟹e−x∈(0,e−2)⊂(0,1), so [e−x]=0. Thus, f(2+)=0−c=−c.
- f(2)=[e−2]−c=0−c=−c.
For continuity at x=2, we need f(2−)=f(2+)⟹b−1=−c⟹b+c=1.
07Execute
Now, for f to be discontinuous at exactly one point (which must be x=1), it must be continuous at both x=0 and x=2.
This requires:
a=1 and b+c=1.
Adding these gives a+b+c=1+1=2.
Therefore, if f is discontinuous at exactly one point, then a+b+c=2=1.
✓Verify
Since a+b+c=2, a+b+c=1 is guaranteed. This also shows f can be discontinuous at exactly 1 point by choosing a=1,b+c=1. Therefore, the statement that it must have at least 2 discontinuities is false.
Hints that build this answer step by step
What are the simplified values of the function pieces on their respective open intervals?
For x<0, [ex]=0; for 0≤x<1, [x−1]=−1; for 1<x<2, [sin(πx)]=−1; for x>2, [e−x]=0.
What are the conditions for continuity at x=0, x=1, and x=2?
At x=0: a=1; at x=1: ae−1=b−1=b (impossible); at x=2: b−1=−c.
Since f is unavoidably discontinuous at x=1, what conditions must hold for f to be discontinuous at exactly one point?
f must be continuous at both x=0 and x=2, which requires a=1 and b+c=1.
If f is continuous at both x=0 and x=2 (so that it has exactly one discontinuity, at x=1), what is the value of a+b+c?
For x slightly greater than 1, sin(pi x) is strictly between -1 and 0. Therefore, [sin(pi x)] = -1. This makes the right-hand limit b - 1. But at x = 1, sin(pi) = 0, so f(1) = b + [0] = b. Since b - 1 can never equal b, the right-hand limit cannot equal the value of the function.
