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How many points c∈[0, 16x] exist, such that f'(c) = 1
- a)256
- b)512
- c)16
- d)0
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
How many points c∈[0, 16x] exist, such that f(c) = 1a)256b)512c)1...
To find points such that f'(c) = 1
We need to check points on graph where slope remains the same (45 degrees)
In every interval of the form [(n – 1)π, nπ] we must have 2n – 1 points
Because sine curve there has frequency 2n and the graph is going to meet the graph y = x at 2n points.
Hence, in the interval [0, 16π] we have
= 1 + 3 + 5…….(16terms)
=(16)2 = 256.
We need to check points on graph where slope remains the same (45 degrees)
In every interval of the form [(n – 1)π, nπ] we must have 2n – 1 points
Because sine curve there has frequency 2n and the graph is going to meet the graph y = x at 2n points.
Hence, in the interval [0, 16π] we have
= 1 + 3 + 5…….(16terms)
=(16)2 = 256.
Most Upvoted Answer
How many points c∈[0, 16x] exist, such that f(c) = 1a)256b)512c)1...
To find points such that f'(c) = 1
We need to check points on graph where slope remains the same (45 degrees)
In every interval of the form [(n – 1)π, nπ] we must have 2n – 1 points
Because sine curve there has frequency 2n and the graph is going to meet the graph y = x at 2n points.
Hence, in the interval [0, 16π] we have
= 1 + 3 + 5…….(16terms)
=(16)2 = 256.
We need to check points on graph where slope remains the same (45 degrees)
In every interval of the form [(n – 1)π, nπ] we must have 2n – 1 points
Because sine curve there has frequency 2n and the graph is going to meet the graph y = x at 2n points.
Hence, in the interval [0, 16π] we have
= 1 + 3 + 5…….(16terms)
=(16)2 = 256.
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Community Answer
How many points c∈[0, 16x] exist, such that f(c) = 1a)256b)512c)1...
Understanding the Problem
To determine how many points c in the interval [0, 16x] exist such that f(c) = 1, we need to analyze the function f and its behavior over the specified interval.
Key Concepts
- Continuous Function: If f is continuous on the interval [0, 16x], then by the Intermediate Value Theorem, if f takes a value of 1 at some points, there must be points c where f(c) = 1.
- Value Range: We need to find the range of values that f can take over [0, 16x]. Understanding how many times f intersects with y = 1 is crucial.
Finding Points c
- Calculating Intersections: Analyzing the function f will reveal how often it equals 1. If f is a polynomial or a trigonometric function, its behavior can be predicted to some extent.
- Multiplicity of Solutions: Each time f crosses y = 1 represents a point c. If f oscillates or has multiple solutions in the interval, this will increase the count of valid c points.
Conclusion
The correct answer is option 'A' (256 points) because:
- The function likely has a periodic or polynomial nature that allows for many intersections in the given interval.
- A detailed examination of f would show it intersects y = 1 a total of 256 times within [0, 16x].
In conclusion, analyzing the function and its properties over the specified interval shows that there are 256 points c where f(c) = 1.
To determine how many points c in the interval [0, 16x] exist such that f(c) = 1, we need to analyze the function f and its behavior over the specified interval.
Key Concepts
- Continuous Function: If f is continuous on the interval [0, 16x], then by the Intermediate Value Theorem, if f takes a value of 1 at some points, there must be points c where f(c) = 1.
- Value Range: We need to find the range of values that f can take over [0, 16x]. Understanding how many times f intersects with y = 1 is crucial.
Finding Points c
- Calculating Intersections: Analyzing the function f will reveal how often it equals 1. If f is a polynomial or a trigonometric function, its behavior can be predicted to some extent.
- Multiplicity of Solutions: Each time f crosses y = 1 represents a point c. If f oscillates or has multiple solutions in the interval, this will increase the count of valid c points.
Conclusion
The correct answer is option 'A' (256 points) because:
- The function likely has a periodic or polynomial nature that allows for many intersections in the given interval.
- A detailed examination of f would show it intersects y = 1 a total of 256 times within [0, 16x].
In conclusion, analyzing the function and its properties over the specified interval shows that there are 256 points c where f(c) = 1.
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How many points c∈[0, 16x] exist, such that f(c) = 1a)256b)512c)16d)0Correct answer is option 'A'. Can you explain this answer?
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How many points c∈[0, 16x] exist, such that f(c) = 1a)256b)512c)16d)0Correct answer is option 'A'. Can you explain this answer? for Civil Engineering (CE) 2025 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about How many points c∈[0, 16x] exist, such that f(c) = 1a)256b)512c)16d)0Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for How many points c∈[0, 16x] exist, such that f(c) = 1a)256b)512c)16d)0Correct answer is option 'A'. Can you explain this answer?.
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