The function y = f xsatisfying the condition fx+1x=x3+1x3 is

fx+1x=x3+1x3put x+1x=yC.B.S⇒ x3+1x3+3.x.1xx+1x=y3⇒ x3+1x3+3y=y3⇒ x3+1x3=y3−3y∴ fy=y3−3y∴ fx=x3−3x

The function y = f xsatisfying the condition fx+1x=x3+1x3 is

fx+1x=x3+1x3put x+1x=yC.B.S⇒ x3+1x3+3.x.1xx+1x=y3⇒ x3+1x3+3y=y3⇒ x3+1x3=y3−3y∴ fy=y3−3y∴ fx=x3−3x

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$T h e f u n c t i o n y = f (x) s a t i s f y i n g t h e c o n d i t i o n f (x + \frac{1}{x}) = x^{3} + \frac{1}{x^{3}} i s$

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$f (x) = x^{2} - 2$

$f (x) = x^{3} - 3 x$

$f (x) = x^{2}$

$f (x) = x^{2} + 2$

answer is D.

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Detailed Solution

$\begin{array}{l} f (x + \frac{1}{x}) = x^{3} + \frac{1}{x^{3}} \\ p u t x + \frac{1}{x} = y \\ C . B . S \\ \Rightarrow x^{3} + \frac{1}{x^{3}} + 3. x . \frac{1}{x} (x + \frac{1}{x}) = y^{3} \\ \Rightarrow x^{3} + \frac{1}{x^{3}} + 3 y = y^{3} \\ \Rightarrow x^{3} + \frac{1}{x^{3}} = y^{3} - 3 y \\ ∴ f (y) = y^{3} - 3 y \\ ∴ f (x) = x^{3} - 3 x \end{array}$