Functions Question 480
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)=\mathbb{R}-[-2,0) $
B) $ D(f+g)=\mathbb{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(f)=R;D(g)=R\setminus[-1,0) $ $ \therefore D(f+g)=D(f)\cap D(g)=\mathbb{R}\cap (\mathbb{R}\setminus [-1,0])=\mathbb{R}\cap [-1,0) $ $ R(f)=[ -\frac{1}{2},\frac{1}{2} ];R(g)=\mathbb{R}\setminus {0} $ $ \therefore R(f)\cap R(g)=[ -\frac{1}{2},\frac{1}{2} ]\cap (R-{0}) $ $ =[ -\frac{1}{2},\frac{1}{2} ]-{0} $
BETA
