Let fx- x 1-x2 and gx- e-x 1-x- where -x- is the greater less than or equal to x- Then

The correct option is B dom(f+g)=R∼ ( 1,0) f(x)= x 1+x^2 ; xϵ R g(x)= e^ x 1+[x] 1+[x]≠ 0 [x]≠ 1 xϵ R ( 1,0) ...

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Byju's Answer

Standard XII

Mathematics

Definition of Functions

Let fx= x 1...

Question

Let $f (x) = \frac{x}{1 + x^{2}}$ and $g (x) + \frac{e^{- x}}{1 + [x]}$ where [x] is the greater less than or equal to x. Then

d o m (f + g) = R \sim [- 2, 0)

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d o m (f + g) = R \sim (- 1, 0)

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d o m (f + g) = [- 2, \frac{1}{2}]

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d o m (f + g) = [- \frac{1}{2}, \frac{1}{2}] \sim {0}

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Solution

The correct option is B $d o m (f + g) = R \sim (- 1, 0)$
$f (x) = \frac{x}{1 + x^{2}}; x ϵ R$

$g (x) = \frac{e^{- x}}{1 + [x]}$

$1 + [x] \neq 0$

$[x] \neq - 1$

$x ϵ R - (- 1, 0)$

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Let f(x)=x1+x2 and g(x)+e−x1+[x] where [x] is the greater less than or equal to x. Then

The correct option is B dom(f+g)=R∼(−1,0)f(x)=x1+x2;xϵRg(x)=e−x1+[x]1+[x]≠0[x]≠−1xϵR−(−1,0)

Let $f (x) = \frac{x}{1 + x^{2}}$ and $g (x) + \frac{e^{- x}}{1 + [x]}$ where [x] is the greater less than or equal to x. Then

The correct option is B $d o m (f + g) = R \sim (- 1, 0)$
$f (x) = \frac{x}{1 + x^{2}}; x ϵ R$

$g (x) = \frac{e^{- x}}{1 + [x]}$

$1 + [x] \neq 0$

$[x] \neq - 1$

$x ϵ R - (- 1, 0)$