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Let f(x) = [2x3 – 5]; then number of points in (1, 2) where the function is discontinuous are where [.] → G.I.F.
Correct answer is '13'. Can you explain this answer?
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Let f(x) = [2x3– 5]; then number of points in (1, 2) where the f...
Understanding the Function
The function given is f(x) = [2x^3 - 5], where [.] denotes the greatest integer function (G.I.F.). This function takes any real number input and returns the greatest integer less than or equal to that number.
Identifying Discontinuities
Discontinuities in the G.I.F. occur at points where the function jumps from one integer value to the next. For the function to be discontinuous, the expression inside the greatest integer function must equal an integer.
Finding Points of Discontinuity
1. Set the expression equal to an integer:
We need to solve for x where:
2x^3 - 5 = n (where n is any integer).
2. Rearranging:
This leads to:
2x^3 = n + 5
x^3 = (n + 5)/2
x = ((n + 5)/2)^(1/3)
3. Bounding n:
We are interested in values of x within the interval (1, 2). Thus, we evaluate:
- For x = 1: f(1) = 2(1)^3 - 5 = -3
- For x = 2: f(2) = 2(2)^3 - 5 = 9
Hence, n can take values from -3 to 9 (inclusive).
Counting Integer Values
- The integer values n range from -3 to 9, which gives us:
- -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
- This results in a total of 13 integer values.
Conclusion
Thus, there are 13 points in the interval (1, 2) where the function f(x) is discontinuous. Each integer value corresponds to a point where the G.I.F. function jumps, confirming the total count of discontinuities as 13.
The function given is f(x) = [2x^3 - 5], where [.] denotes the greatest integer function (G.I.F.). This function takes any real number input and returns the greatest integer less than or equal to that number.
Identifying Discontinuities
Discontinuities in the G.I.F. occur at points where the function jumps from one integer value to the next. For the function to be discontinuous, the expression inside the greatest integer function must equal an integer.
Finding Points of Discontinuity
1. Set the expression equal to an integer:
We need to solve for x where:
2x^3 - 5 = n (where n is any integer).
2. Rearranging:
This leads to:
2x^3 = n + 5
x^3 = (n + 5)/2
x = ((n + 5)/2)^(1/3)
3. Bounding n:
We are interested in values of x within the interval (1, 2). Thus, we evaluate:
- For x = 1: f(1) = 2(1)^3 - 5 = -3
- For x = 2: f(2) = 2(2)^3 - 5 = 9
Hence, n can take values from -3 to 9 (inclusive).
Counting Integer Values
- The integer values n range from -3 to 9, which gives us:
- -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
- This results in a total of 13 integer values.
Conclusion
Thus, there are 13 points in the interval (1, 2) where the function f(x) is discontinuous. Each integer value corresponds to a point where the G.I.F. function jumps, confirming the total count of discontinuities as 13.
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Let f(x) = [2x3– 5]; then number of points in (1, 2) where the function is discontinuous are where [.] → G.I.F.Correct answer is '13'. Can you explain this answer?
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Let f(x) = [2x3– 5]; then number of points in (1, 2) where the function is discontinuous are where [.] → G.I.F.Correct answer is '13'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Let f(x) = [2x3– 5]; then number of points in (1, 2) where the function is discontinuous are where [.] → G.I.F.Correct answer is '13'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let f(x) = [2x3– 5]; then number of points in (1, 2) where the function is discontinuous are where [.] → G.I.F.Correct answer is '13'. Can you explain this answer?.
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