Functions Question 467
Question: Let $ f:R\to R $ be given by $ f(x)={{(x+1)}^{2}}-1,x\ge -1. $ Then, $ {f^{-1}}(x), $ is
Options:
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
Show Answer
Answer:
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=yorx=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} $
BETA
