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if 2f(x^2)+3f(1/x^2)=x^2-1 for all x belongs to Real,then f(x^2) is??
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if 2f(x^2)+3f(1/x^2)=x^2-1 for all x belongs to Real,then f(x^2) is??
Given Equation:
2f(x^2) + 3f(1/x^2) = x^2 - 1

Explanation:
To find the value of f(x^2), we need to manipulate the given equation to isolate f(x^2).

Step 1 - Substitute x with 1/x:
When we substitute x with 1/x in the given equation, we get:
2f(1/x^2) + 3f(x^2) = 1/x^2 - 1

Step 2 - Add the two equations:
Adding the original equation and the equation obtained by substituting x with 1/x, we get:
2f(x^2) + 3f(1/x^2) + 2f(1/x^2) + 3f(x^2) = x^2 - 1 + 1/x^2 - 1
5(f(x^2) + f(1/x^2)) = x^2 + 1/x^2 - 2

Step 3 - Simplify:
(f(x^2) + f(1/x^2)) = (x^2 + 1/x^2 - 2)/5

Step 4 - Multiply by 2:
Multiplying both sides by 2, we get:
2(f(x^2) + f(1/x^2)) = 2(x^2 + 1/x^2 - 2)/5
2f(x^2) + 2f(1/x^2) = (2x^2 + 2/x^2 - 4)/5

Step 5 - Subtract the original equation:
Subtracting the original equation from the above equation, we get:
2f(x^2) + 2f(1/x^2) - 2f(x^2) - 3f(1/x^2) = (2x^2 + 2/x^2 - 4)/5 - (x^2 - 1)
-f(1/x^2) = (2x^2 + 2/x^2 - 4)/5 - x^2 + 1

Step 6 - Simplify further:
-f(1/x^2) = (2x^4 + 2 - 4x^2)/5x^2 - 5x^4 + 5x^2
-f(1/x^2) = (2x^4 - 4x^2 + 2 - 5x^4 + 5x^2)/5x^2
-f(1/x^2) = (-3x^4 + x^2 + 2)/5x^2

Conclusion:
Therefore, f(x^2) = (-3x^2 + 1 + 2x^4)/5x^2.
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if 2f(x^2)+3f(1/x^2)=x^2-1 for all x belongs to Real,then f(x^2) is??
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if 2f(x^2)+3f(1/x^2)=x^2-1 for all x belongs to Real,then f(x^2) is?? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about if 2f(x^2)+3f(1/x^2)=x^2-1 for all x belongs to Real,then f(x^2) is?? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for if 2f(x^2)+3f(1/x^2)=x^2-1 for all x belongs to Real,then f(x^2) is??.
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