Question

2f(x²) + 3f(1/x²) = x² - 1 for all x belongs to R-{0} . find f(x²) and f(x⁴) ?

Answer 1

2f( x²) + 3f( 1/x²) = x² -1 ----------(1)

put x = 1/x

2f(1/x²) + 3f(x²) = 1/x² -1 --------(2)

solve both equation ,

4f( x²) - 9 f(x²) = 2(x² -1)- 3 ( 1/x² -1)

- 5 f( x²) = 2x² -2 -3/x² +3 =1 + 2x² -3/x²

f( x²) = { 2x⁴ + x² -3 }/(-5x²)

f( x²) = { 3 - x² -2x⁴ }/5x² ( answer)

now,

we can also write this in f( x ) form
e.g f( x ) = ( 3 - x -2x²)/5x

now , put x = x⁴

so,

f( x⁴) = ( 3 -x⁴ - 2x^8 )/5x⁴ ( answer )

Answer 2

Answer:

$f( {x}^{4} ) = (1 - {x}^{4} )(2 {x}^{4} + 3)\div5 {x}^{4}$