If f - R - R is defined by f x -sin - x - -tan - x - -1- x 2- then the range of f - where - x - denotes integral part of x A- -1-1-B- -1-1C- 1D- 0

The correct option is D 0 [x]=n ϵ Z ⇔ sin[x]π=tan[x] π =0 ...

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Idea of a Set

if f : R → R ...

Question

$i f f : R \to R i s d e f i n e d b y f (x) = \frac{s i n [x] π + t a n [x] π}{1 + [x^{2}]}, t h e n t h e r a n g e o f f = (w h e r e [x] d e n o t e s i n t e g r a l p a r t o f (x)$

[-1,1]

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Solution

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Similar questions

f : R \to R

f (x) = \frac{s i n [x] π + t a n [x] π}{1 + [x^{2}]}

f : R \to R

f (x) = \frac{s i n [x] π + t a n [x] π}{1 + [x^{2}]}

f (x) = \frac{sin ([x] π)}{x^{2} + x + 1}

[x]

x

f (x)

f : R \to R

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} α + \frac{sin [x]}{x} & i f x > 0 2 & i f x = 0 β + [\frac{sin x - x}{x^{3}}] & i f x < 0 \end{matrix}

[x]

x = 0

β - α =

f : R \to R

f (x) = \frac{x}{1 + x^{2}}, x \in R .

f

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