First slide

Functions (XII)

Question

If $f {(x)}^{2} f (\frac{1 - x}{1 + x}) = x^{3}, x \neq - 1, 1$ and $f (x) \neq 0,$ then ${f (- 2)}$ (the fractional part of f(-2) is equal to

Moderate

A

2 /3
B

1/ 3
C

1/ 2
D

0

Solution

Replacing x by $\frac{1 - x}{1 + x}$ in the equation
$(f (x))^{2} f (\frac{1 - x}{1 + x}) = x^{3}$ …(i)
we have
${(f (\frac{1 - x}{1 + x}))}^{2} f (x) = {(\frac{1 - x}{1 + x})}^{3}$ …(ii)
solving (i) and (ii), we have
$(\frac{x^{3}}{(f {(x))}^{2}}) f (x) = {(\frac{1 - x}{1 + x})}^{3}$
$\Rightarrow \frac{(f {(x))}^{3}}{x^{6}} = {(\frac{1 - x}{1 + x})}^{3}$
$\Rightarrow f (x) = x^{2} (\frac{1 + x}{1 - x})$
$f (- 2) = 4 (\frac{- 1}{3}) = - \frac{4}{3}$
${f (- 2)} = - \frac{4}{3} - [- \frac{4}{3}] = - \frac{4}{3} + 2 = \frac{2}{3}$

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