Functions Question 468
Let $ g(x)=1+x^2-
[x] $ and $ f(x)= \begin{cases} & -1,\ x<0 \\ & 0,\ \ x=0,\ \\ & \text{1,}\ \ \ x>0 \\ \end{cases} . $ then for all $ x,\ f(g(x)) $ is equal to [IIT Screening 2001; UPSEAT 2001]
Options:
x
1
C) $ f(x) $
D) $ g(x) $
Show Answer
Answer:
Correct Answer: B
Solution:
Here $ g(x)=1+n-n=1,,x=n\in Z $ $ 1+n+k-n=1+k $ , $ x=n+k $ (where $ n\in Z,,0<k<1 $ ) Now $ f(g(x))= \begin{cases} & -1,,g(x)<0 \\ & ,0,g(x)=0 \\ & ,1,,g(x)>0 \\ \end{cases} . $ Clearly, $ g(x)>0 $ for all x. So, $ f(g(x))=1 $ for all x.
BETA
