Functions Question 588
Question: Let $ f(x)={{(x+1)}^{2}}-1,\ \ (x\ge -1) $ . Then the set $ S={x:f(x)={f^{-1}}(x)} $ is
[IIT 1995]
Options:
A) Empty
B) {0, -1}
C) {0, 1, -1}
D) $ { 0,\ -1,\ \frac{-3+i\sqrt{3}}{2},\ \frac{-3-i\sqrt{3}}{2} } $
Show Answer
Answer:
Correct Answer: D
Solution:
Let $ f(x)={{(x+1)}^{2}}-1,x\ge -1. $ Since $ f(x)={f^{-1}}(x) $
$ \therefore {{(x+1)}^{2}}-1=\sqrt{1+x}-1 $ $ ( \because {f^{-1}}(x)=\sqrt{1+x}-1 ) $
$ \Rightarrow {{(x+1)}^{4}}=1+x\Rightarrow (x+1)[{{(x+1)}^{3}}-1]=0 $
$ \Rightarrow x=-1 $ or $ {{(x+1)}^{3}}=1,\Rightarrow x+1=1,\omega ,{{\omega }^{2}} $
$ \Rightarrow x=0,-1,,\frac{-3+i\sqrt{3}}{2},\frac{-3-i\sqrt{3}}{2}. $
BETA
