A) \[-1+\sqrt{x+1}\]
B) \[-1-\sqrt{x+1}\]
C) Does not exist because f is not one-one
D) Does not exist because f is not onto
Correct Answer: A
Solution :
| [a] Let \[x,y\in R\] such that \[x\ge -1,y\ge -1\] |
| Then, \[f(x)=f(y)\] |
| \[\Rightarrow {{(x+1)}^{2}}-1={{(y+1)}^{2}}-1\] |
| \[\Rightarrow {{x}^{2}}+2x={{y}^{2}}+2y\] |
| \[\Rightarrow {{x}^{2}}-{{y}^{2}}=-2(x-y)\] |
| \[\Rightarrow (x-y)(x+y+2)=0\] |
| \[\Rightarrow x-y=0orx+y+2=0\] |
| \[\Rightarrow x=y\,\,or\,\,x=y=-1\] |
| \[\therefore \] f is one-one. |
| Also, f is onto as for all \[y\ge -1,\] there exists |
| \[x=-1+\sqrt{y+1}\ge -1\] Such that \[f(x)=y\] |
| \[\therefore \] f is invertible. |
| Let \[f(x)=y\Rightarrow {{(x+1)}^{2}}-1=y\] |
| \[\Rightarrow x=-1\pm \sqrt{y+1}\] |
| But \[x\ge -1\] |
| \[\therefore x=-1+\sqrt{y+1}\] |
| \[\Rightarrow {{f}^{-1}}(y)=-1+\sqrt{y+1}\] |
| Hence, \[{{f}^{-1}}(x)=-1+\sqrt{x+1}\] |
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