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- 0D- 1

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

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Standard XII

Mathematics

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

The correct option is D ${0}$
$[x] = ϵ Z \Leftrightarrow s i n [x] π = t a n [x] π = 0$ ; Where Z is an integer.

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If f:R→R is defined by f(x)=sin[x]π+tan[x]π1+[x2], then the range of f(x) (where [x] denotes integral part of x)

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

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)

The correct option is D ${0}$
$[x] = ϵ Z \Leftrightarrow s i n [x] π = t a n [x] π = 0$ ; Where Z is an integer.