If-for-nonzero-x-2f-x-2-3f-1-x-2-x-2-1-then-f-x-2-

Question Number 96222 by bobhans last updated on 30/May/20

If for nonzero x; 2 f (x^{2}) + 3 f (\frac{1}{x^{2}}) = x^{2} - 1

t h e n f (x^{2}) = ?

Commented by bobhans last updated on 31/May/20

we have 2 f (x^{2}) + 3 f (\frac{1}{x^{2}}) = x^{2} - 1

\Rightarrow 4 f (x^{2}) + 6 f (\frac{1}{x^{2}}) = 2 x^{2} - 2 \dots (i)

a l s o 2 f (\frac{1}{x^{2}}) + 3 f (x^{2}) = \frac{1}{x^{2}} - 1 \dots (i i)

3 \times (i i) - (i) \Rightarrow 5 f (x^{2}) = \frac{3 - 3 x^{2}}{x^{2}} - (2 x^{2} - 2)

f (x^{2}) = \frac{3 - 3 x^{2} - 2 x^{4} + 2 x^{2}}{5 x^{2}} = \frac{3 - x^{2} - 2 x^{4}}{5 x^{2}}

Commented by PRITHWISH SEN 2 last updated on 31/May/20

put x = \frac{1}{x}

2 f (\frac{1}{x^{2}}) + 3 f (x^{2}) = \frac{1 - x^{2}}{x^{2}} \dots \dots . (ii)

(i) \times 2 - (ii) \times 3 given equ . is (i)

f (x^{2}) = \frac{3 - x^{2} - {2 x}^{4}}{{5 x}^{2}}

Answered by mathmax by abdo last updated on 01/Jun/20

we have 2 f (x^{2}) + 3 f (\frac{1}{x^{2}}) = x^{2} - 1 \Rightarrow 2 f (\frac{1}{x^{2}}) + 3 f (x^{2}) = \frac{1}{x^{2}} - 1 (change x by \frac{1}{x}) \Rightarrow

we get the system {\begin{cases} 2 f (x^{2}) + 3 f (\frac{1}{x^{2}}) = x^{2} - 1 \\ 3 f (x^{2}) + 2 f (\frac{1}{x^{2}}) = \frac{1 - x^{2}}{x^{2}} \end{cases}

Δ_{s} = | \begin{matrix} 2 3 \\ 3 2 \end{matrix} | = 4 - 9 = - 5

Δ_{f (x^{2})} = | \begin{matrix} x^{2} - 1 3 \\ \frac{1 - x^{2}}{x^{2}} 2 \end{matrix} | = {2 x}^{2} - 2 - \frac{3 (1 - x^{2})}{x^{2}} = \frac{{2 x}^{4} - {2 x}^{2} - 3 + {3 x}^{2}}{x^{2}} = \frac{{2 x}^{4} + x^{2} - 3}{x^{2}}

\Rightarrow f (x^{2}) = \frac{{2 x}^{4} + x^{3} - 3}{- {5 x}^{2}} = \frac{- {2 x}^{4} - x^{3} + 3}{{5 x}^{2}}

Commented by mathmax by abdo last updated on 01/Jun/20

error of typo f (x^{2}) = \frac{{2 x}^{4} + x^{2} - 3}{- {5 x}^{2}} = \frac{- {2 x}^{4} - x^{2} + 3}{{5 x}^{2}}