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

The correct option is D { 0 }[x]=ϵ Z⇔ sin[x]π=tan[x]π=0 ; Where Z is an integer. ...

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Using Graph to Find Range of a Function

If f : R → R ...

Question

If $f : R \to R$ is defined by $f (x) = \frac{s i n [x] π + t a n [x] π}{1 + [x^{2}]}$ , then the range of f(x) (where [x] denotes integral part of x)

[- 1, 1]

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{- 1, 1}

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{1}

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{0}

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

f : R \to R

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

f

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