A) 4, 1
B) 4, 0
C) 3, 2
D) None of these
Correct Answer: D
Solution :
Let \[y={{x}^{2}}+1\] \[\Rightarrow \] \[{{x}^{2}}=y-1\] \[\Rightarrow \] \[x=\pm \sqrt{y-1}\] \[\Rightarrow \] \[{{f}^{-1}}(y)=\pm \sqrt{y-1}\] \[\Rightarrow \] \[{{f}^{-1}}(x)=\pm \sqrt{x-1}\] \[\therefore \] \[{{f}^{-1}}(17)=\pm \sqrt{17-1}=\pm 4\] and \[{{f}^{-1}}(-3)=\pm \sqrt{-3-1}\] \[=\pm \sqrt{-4},\] which is not possible. Alternate Method : Since\[f(x)\]is an even function, so its inverse cannot be defined.
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