Lf f-R- R and g-R- R defined by fx-x- and gx-x-3- for x- R- - denotes greatest integer function- then gfx-8-5-x-8-5-

The correct option is C { 3, 2} We have, f(x) =|x| and G(x)=[x 3]=[x] 3 Since, #160; 85 lt;x lt;85⇒ 0≤ |x| ≤85⇒ 0≤ f(x) ≤85 Now cas ...

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Standard XII

Mathematics

Greatest Integer Function

lf f:R→ R a...

Question

lf $f : R \to R$ and $g : R \to R$ defined by $f (x) = | x |$ and $g (x) = [x - 3]$ for $x \in R$ .
$[.]$ denotes greatest integer function, then ${g (f (x)) : \frac{- 8}{5} < x < \frac{8}{5}} =$

{0, 1}

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{- 1, - 2}

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{- 3, - 2}

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{2, 3}

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Solution

The correct option is C ${- 3, - 2}$
We have, $f (x) = | x |$ and $G (x) = [x - 3] = [x] - 3$
Since, $- \frac{8}{5} < x < \frac{8}{5} \Rightarrow 0 \leq | x | \leq \frac{8}{5} \Rightarrow 0 \leq f (x) \leq \frac{8}{5}$
Now case 1. If $0 \leq x < 1 \Rightarrow [x] = 0$
$\Rightarrow g (f (x)) = 0 - 3 = - 3$
case 2. If $1 \leq x < \frac{8}{5} \Rightarrow [x] = 1$
$\Rightarrow g (f (x)) = 1 - 3 = - 2$
Hence ${g (f (x)) : - \frac{8}{5} < x < \frac{8}{5}} = {- 3, - 2}$

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lf f:R→R and g:R→R defined byf(x)=|x| and g(x)=[x−3] for x∈R.[.] denotes greatest integer function, then{g(f(x)):−85<x<85}=

The correct option is C {−3,−2}We have, f(x)=|x| and G(x)=[x−3]=[x]−3Since, −85<x<85⇒0≤|x|≤85⇒0≤f(x)≤85Now case 1. If 0≤x<1⇒[x]=0⇒g(f(x))=0−3=−3case 2. If 1≤x<85⇒[x]=1⇒g(f(x))=1−3=−2Hence {g(f(x)):−85<x<85}={−3,−2}

lf $f : R \to R$ and $g : R \to R$ defined by $f (x) = | x |$ and $g (x) = [x - 3]$ for $x \in R$ .
$[.]$ denotes greatest integer function, then ${g (f (x)) : \frac{- 8}{5} < x < \frac{8}{5}} =$