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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). answer is said to be 5 how though?
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If sum of maximum and minimum value of y = log2(x⁴ + x² + 1) - log2(x⁴...
- **Finding the Maximum and Minimum Value of y**
Given function: y = log2(x⁴ + x² + 1) - log2(x⁴ + x³ + 2x² + x + 1)
- **Step 1: Finding the Critical Points**
To find the critical points, we need to calculate the derivative of y with respect to x and set it equal to 0.
- **Step 2: Classifying the Critical Points**
After finding the critical points, we need to classify them as either maximum or minimum by using the second derivative test.
- **Step 3: Calculating the Maximum and Minimum Value of y**
Once we have identified the critical points as maximum or minimum, we can substitute these values back into the original function to find the corresponding maximum and minimum values of y.
- **Expressing the Sum in the Required Form**
Now that we have the maximum and minimum values of y, we can calculate their sum and express it in the form ((log2 m) - n).
- **Computing (m + n)**
Finally, add the values of m and n to get the result, which should be equal to 5 as given in the question.
Given function: y = log2(x⁴ + x² + 1) - log2(x⁴ + x³ + 2x² + x + 1)
- **Step 1: Finding the Critical Points**
To find the critical points, we need to calculate the derivative of y with respect to x and set it equal to 0.
- **Step 2: Classifying the Critical Points**
After finding the critical points, we need to classify them as either maximum or minimum by using the second derivative test.
- **Step 3: Calculating the Maximum and Minimum Value of y**
Once we have identified the critical points as maximum or minimum, we can substitute these values back into the original function to find the corresponding maximum and minimum values of y.
- **Expressing the Sum in the Required Form**
Now that we have the maximum and minimum values of y, we can calculate their sum and express it in the form ((log2 m) - n).
- **Computing (m + n)**
Finally, add the values of m and n to get the result, which should be equal to 5 as given in the question.
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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). answer is said to be 5 how though?
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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). answer is said to be 5 how though? for JEE 2024 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). answer is said to be 5 how though? covers all topics & solutions for JEE 2024 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). answer is said to be 5 how though?.
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). answer is said to be 5 how though? for JEE 2024 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). answer is said to be 5 how though? covers all topics & solutions for JEE 2024 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). answer is said to be 5 how though?.
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