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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

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Correct option is A

Replacing x by 1-x1+x in the equation (f(x))2f1−x1+x=x3 …(i)we have f1−x1+x2f(x)=1−x1+x3 …(ii)solving (i) and (ii), we have x3(f(x))2 f(x) = 1-x1+x3⇒ (f(x))3x6=1-x1+x3⇒ f(x) = x21+x1-x f(-2)=4-13=-43{f(-2)} = -43--43 =-43+2=23

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If f(x)2 f1-x1+x= x3 , x≠ -1 , 1 and f(x)≠0 , then {f(-2)} (the fractional part of f(-2) is equal to

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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