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Let `f : R to R` be a function such that `f(0)=1` and for any `x,y in R, f(xy+1)=f(x)f(y)-f(y)-x+2.` Then `f` isA. one-one and ontoB. one-one but not ontoC. many one but ontoD. many one and into |
Answer» Correct Answer - A `f(xy+1)=f(yx+1)` `f(x)f(y)-f(y)-x+2=f(y)f(x)-f(x)-y+2` `f(x) -f(y)=x-y` Putting `y=0` `f(x)-1=x-0` `f(x) = x+1` |
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