f(x)=x2+1x2+2f(x) = \frac{x^2+1}{x^2+2}f(x)=x2+2x2+1
For x=-1 => f(x)=1+11+2=23f(x) = \frac{1+1}{1+2}= \frac{2}{3}f(x)=1+21+1=32
For x=1 => f(x)=1+11+2=23f(x) = \frac{1+1}{1+2}= \frac{2}{3}f(x)=1+21+1=32
=> Not one-one function
=> Not a bijective (no inverse exists)
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