If y=f(x)satisfy the conditionfx+1x=x2+1x2 (x≠0)then f(x)is

Given,fx+1x=x2+1x2x≠0⇒fx+1x=x+1x2-2put x+1x=t⇒f(t)=t2-2 ⇒f(x)=x2-2 ∴x∈[2,∞)

If y=f(x)satisfy the conditionfx+1x=x2+1x2 (x≠0)then f(x)is

Given,fx+1x=x2+1x2x≠0⇒fx+1x=x+1x2-2put x+1x=t⇒f(t)=t2-2 ⇒f(x)=x2-2 ∴x∈[2,∞)

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If y=f(x)satisfy the condition $f (x + \frac{1}{x}) = x^{2} + \frac{1}{x^{2}}$ (x $\neq$ 0)then f(x)is

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x²-2,x $\in$ R-{0}

-x²-2

x²-2,x $\in$ $|x|$ $\in$ $[2, \infty)$

-x²+2

answer is D.

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Detailed Solution

Given,

$f (x + \frac{1}{x}) = x^{2} + \frac{1}{x^{2}}$ x $\neq$ 0

$\Rightarrow f (x + \frac{1}{x}) = {(x + \frac{1}{x})}^{2} - 2$

put $x + \frac{1}{x}$ =t

$\Rightarrow f (t) = t^{2} - 2 \Rightarrow f (x) = x^{2} - 2 ∴ |x| \in [2, \infty)$

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If y=f(x)satisfy the condition $f (x + \frac{1}{x}) = x^{2} + \frac{1}{x^{2}}$ (x $\neq$ 0)then f(x)is