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If sum of maximum and minimum value of y = log2(x⁴ + x² + 1) - log2(x⁴ + 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(x⁴ + x² + 1) - log2(x⁴...
Given Function:
y = log2(x⁴ + x² + 1) - log2(x⁴ + x³ + 2x² + x + 1)
Finding Maximum and Minimum Values:
To find the maximum and minimum values of y, we need to find the critical points by setting the derivative of y with respect to x equal to 0 and solving for x.
After finding the critical points, we can determine the maximum and minimum values of y by plugging those critical points back into the original function.
Calculating Sum of Maximum and Minimum Values:
Once we have the maximum and minimum values of y, we can calculate their sum. Let the maximum value be M and the minimum value be m.
Therefore, the sum of the maximum and minimum values is M + m.
Expressing the Sum in Required Form:
We are required to express the sum in the form ((log2 m) - n), where m and 2 are coprime.
So, the sum can be written as log2(M) - n, where M is the maximum value and n is the minimum value.
Calculating (m + n):
Finally, we need to calculate (m + n) by comparing the obtained sum with the required form ((log2 m) - n) and identifying the values of m and n.
By following these steps, you can solve the given function and compute the required sum (m + n) in the specified form.
y = log2(x⁴ + x² + 1) - log2(x⁴ + x³ + 2x² + x + 1)
Finding Maximum and Minimum Values:
To find the maximum and minimum values of y, we need to find the critical points by setting the derivative of y with respect to x equal to 0 and solving for x.
After finding the critical points, we can determine the maximum and minimum values of y by plugging those critical points back into the original function.
Calculating Sum of Maximum and Minimum Values:
Once we have the maximum and minimum values of y, we can calculate their sum. Let the maximum value be M and the minimum value be m.
Therefore, the sum of the maximum and minimum values is M + m.
Expressing the Sum in Required Form:
We are required to express the sum in the form ((log2 m) - n), where m and 2 are coprime.
So, the sum can be written as log2(M) - n, where M is the maximum value and n is the minimum value.
Calculating (m + n):
Finally, we need to calculate (m + n) by comparing the obtained sum with the required form ((log2 m) - n) and identifying the values of m and n.
By following these steps, you can solve the given function and compute the required sum (m + n) in the specified form.
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If sum of maximum and minimum value of y = log2(x⁴ + x² + 1) - log2(x⁴ + 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(x⁴ + x² + 1) - log2(x⁴ + 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(x⁴ + x² + 1) - log2(x⁴ + 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(x⁴ + x² + 1) - log2(x⁴ + 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(x⁴ + x² + 1) - log2(x⁴ + 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(x⁴ + x² + 1) - log2(x⁴ + 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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