Let f x - x -1- x 2- and g x - e - x -1- x - where - x - is the greatest integer less than or equal to x- Then-A- R f - R g -2- 1-2-B- None of the aboveC- Df-g-R-1-0D- Df-g-R-2-0

The correct option is C D(f + g) = R – [ 1, 0)D (f) = R ; D (g) = R [ 1,0) ∴ D (f+g) = D(f) ∩ D (g) = R cap (R [ 1,0)) R ∩ [ 1,0) R(f)= [ 1/2,1/2 ] ; R (g) ...

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Standard XII

Mathematics

Higher Order Derivatives

Let f x = x /...

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,

D(f + g) = R – [- 2, 0 )

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D(f + g) = R – [ -1, 0)

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R (f) \cap R (g) = [- 2, \frac{1}{2}]

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None of the above

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Solution

The correct option is B D(f + g) = R – [ -1, 0)
D (f) = R ; D (g) = R - [-1,0)
$∴$ D (f+g) = D(f) $\cap$ D (g) = R $c a p$ (R-[-1,0))
R $\cap$ [-1,0)
$R (f) = [- \frac{1}{2}, \frac{1}{2}]$ ; R (g) = R - {0}
$∴ R (f) \cap R (g) = [- \frac{1}{2}, \frac{1}{2}] \cap (R - {0})$
$= [- \frac{1}{2}, \frac{1}{2}] - {0}$

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

The correct option is B D(f + g) = R – [ -1, 0)D (f) = R ; D (g) = R - [-1,0) ∴ D (f+g) = D(f) ∩ D (g) = R cap (R-[-1,0)) R ∩ [-1,0) R(f)=[−12,12] ; R (g) = R - {0} ∴ R(f)∩R(g)=[−12,12]∩(R−{0}) =[−12,12]−{0}

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,