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  If f : R ---->R defined by f(x) = x2 + 1, then values of f -1 (17) and  f -1(-3) are

A.

{Ø}, (4, - 4)

B.

{3,-3},{Ø}

C.

{Ø},{3,-3}

D.

{4,-4},Ø

Answer is "D". But i dint understand how

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firstly,

if f(x) is defined on  R ---->R, how can function be invertible since it is not onto because f(x) can never take negative values and codomain consist of all real numbers...

secondly,

if still you want to calculate f-1 using formula than option D is correct,

f(x)=y=x2 + 1, 

=>f-1= +/-sqrt(x-1)

now subsituting, x=17, f-1= +/-4  and

x=-3, there will be imaginary value wich is not in domain so f-1

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