Functions Question 480
Question: Let $ f(x)=\frac{x}{1+x^{2}} $ and $ g(x)=\frac{{e^{-x}}}{1+[x],} $ , where $ [x] $ is the greatest integer less than or equal to x, then
Options:
A) $ D(f+g)=R-[-2,0) $
B) $ D(f+g)=R-[-1,0) $
C) $ R(f)\cap R(g)=[ -2,\frac{1}{2} ] $
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
[d] $ D(f)=R;D(g)=R-[-1,0) $
$ \therefore D(f+g)=D(f)\cap D(g)=R\cap (R-[-1,0)=R\cap [-1,0) $ $ R(f)=[ -\frac{1}{2},\frac{1}{2} ];R(g)=R-{0} $
$ \therefore R(f)\cap R(g)=[ -\frac{1}{2},\frac{1}{2} ]\cap (R-{0}) $ $ =[ -\frac{1}{2},\frac{1}{2} ]-{0} $
BETA
