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If sum of maximum and minimum value of y = log2(x² + x2 + 1) - log2(x² + x3 + 2x2 + x + 1) can be expressed in form ((log2 m) -n), where m and 2 are coprime then compute (m + n).?
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If sum of maximum and minimum value of y = log2(x² + x2 + 1) - log2(x²...
Analysis of the given function
- Let's analyze the given function y = log2(x² + x² + 1) - log2(x² + x³ + 2x² + x + 1).
- We can simplify the function to y = log2(2x² + 1) - log2(x² + x³ + 2x² + x + 1).
Finding the maximum and minimum values
- To find the maximum and minimum values of the function, we need to take the derivative of y with respect to x and set it equal to 0.
- After finding the critical points, we can determine the maximum and minimum values of y.
Calculating the sum of maximum and minimum values
- Once we have the maximum and minimum values of y, we can calculate their sum.
- This sum can be expressed in the form ((log2 m) - n), where m and n are coprime.
Finding the value of (m + n)
- Finally, we can determine the values of m and n from the expression ((log2 m) - n) and calculate their sum (m + n).
By following the steps outlined above, you can effectively analyze the given function, find the maximum and minimum values, calculate their sum in the desired form, and determine the value of (m + n) for the given function.
- Let's analyze the given function y = log2(x² + x² + 1) - log2(x² + x³ + 2x² + x + 1).
- We can simplify the function to y = log2(2x² + 1) - log2(x² + x³ + 2x² + x + 1).
Finding the maximum and minimum values
- To find the maximum and minimum values of the function, we need to take the derivative of y with respect to x and set it equal to 0.
- After finding the critical points, we can determine the maximum and minimum values of y.
Calculating the sum of maximum and minimum values
- Once we have the maximum and minimum values of y, we can calculate their sum.
- This sum can be expressed in the form ((log2 m) - n), where m and n are coprime.
Finding the value of (m + n)
- Finally, we can determine the values of m and n from the expression ((log2 m) - n) and calculate their sum (m + n).
By following the steps outlined above, you can effectively analyze the given function, find the maximum and minimum values, calculate their sum in the desired form, and determine the value of (m + n) for the given function.
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If sum of maximum and minimum value of y = log2(x² + x2 + 1) - log2(x² + x3 + 2x2 + x + 1) can be expressed in form ((log2 m) -n), where m and 2 are coprime then compute (m + n).? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about If sum of maximum and minimum value of y = log2(x² + x2 + 1) - log2(x² + x3 + 2x2 + x + 1) can be expressed in form ((log2 m) -n), where m and 2 are coprime then compute (m + n).? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If sum of maximum and minimum value of y = log2(x² + x2 + 1) - log2(x² + x3 + 2x2 + x + 1) can be expressed in form ((log2 m) -n), where m and 2 are coprime then compute (m + n).?.
If sum of maximum and minimum value of y = log2(x² + x2 + 1) - log2(x² + x3 + 2x2 + x + 1) can be expressed in form ((log2 m) -n), where m and 2 are coprime then compute (m + n).? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about If sum of maximum and minimum value of y = log2(x² + x2 + 1) - log2(x² + x3 + 2x2 + x + 1) can be expressed in form ((log2 m) -n), where m and 2 are coprime then compute (m + n).? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If sum of maximum and minimum value of y = log2(x² + x2 + 1) - log2(x² + x3 + 2x2 + x + 1) can be expressed in form ((log2 m) -n), where m and 2 are coprime then compute (m + n).?.
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