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Question Number 15093 by Tinkutara last updated on 07/Jun/17
The range of f(x) = (({x}^2 − {x} + 1)/({x}^2 + {x} + 1)); (where {∙} denotes fractional function) is?
Therangeoff(x)={x}2−{x}+1{x}2+{x}+1;(where{⋅}denotesfractionalfunction)is?
Answered by mrW1 last updated on 07/Jun/17
f(x) = (({x}^2 − {x} + 1)/({x}^2 + {x} + 1)) let g={x} −1<g<1 f(x)=((g^2 −g+1)/(g^2 +g+1))=1−((2g)/(g^2 +g+1)) g→+1 f(x)→1−(2/3)=(1/3) g=0, f(x)=1 g→−1 f(x)→3 ⇒f(x)∈((1/3),3)
f(x)={x}2−{x}+1{x}2+{x}+1letg={x}−1<g<1f(x)=g2−g+1g2+g+1=1−2gg2+g+1g→+1f(x)→1−23=13g=0,f(x)=1g→−1f(x)→3⇒f(x)∈(13,3)
Commented by ajfour last updated on 07/Jun/17
0.3 i believe [x] is the greatest integer that the value of x has crossed going along the numberline in the positive x direction. {x} is the value that need be added to [x] to make it reach x .
0.3ibelieve[x]isthegreatestintegerthatthevalueofxhascrossedgoingalongthenumberlineinthepositivexdirection.{x}isthevaluethatneedbeaddedto[x]tomakeitreachx.
Commented by mrW1 last updated on 07/Jun/17
{−2.7}=−0.7
{−2.7}=−0.7
Commented by mrW1 last updated on 07/Jun/17
(one) definition of fractional part of a number see diagram:
(one)definitionoffractionalpartofanumberseediagram:
Commented by mrW1 last updated on 07/Jun/17
Commented by mrW1 last updated on 07/Jun/17
i have corrected. it is y∈((1/3),3) you can check with x=−1.5: {x}=−0.5 f(x)=(((−0.5)^2 −(−0.5)+1)/((−0.5)^2 +(−0.5)+1))=((1.75)/(0.75))=2.33>1
ihavecorrected.itisy∈(13,3)youcancheckwithx=−1.5:{x}=−0.5f(x)=(−0.5)2−(−0.5)+1(−0.5)2+(−0.5)+1=1.750.75=2.33>1
Commented by Tinkutara last updated on 07/Jun/17
But answer is ((1/3), 1]
Butansweris(13,1]
Commented by Tinkutara last updated on 07/Jun/17
Sir, if g = {x}, then 0 ≤ g < 1. Why −1 < g < 1 ?
Sir,ifg={x},then0⩽g<1.Why−1<g<1?
Commented by mrW1 last updated on 07/Jun/17
that is only for x≥0 for x≤0, −1<{x}≤0 {2.5}=0.5 {−2.5}=−0.5
thatisonlyforx⩾0forx⩽0,−1<{x}⩽0{2.5}=0.5{−2.5}=−0.5
Commented by Tinkutara last updated on 07/Jun/17
What will be {−2.7}?
Whatwillbe{−2.7}?
Commented by ajfour last updated on 07/Jun/17
0≤{x}<1 , i have read this.
0⩽{x}<1,ihavereadthis.
Commented by mrW1 last updated on 07/Jun/17
there are 2 kinds of definition for x<0 which are used in mathematic world.
thereare2kindsofdefinitionforx<0whichareusedinmathematicworld.
Commented by prakash jain last updated on 07/Jun/17
Fraction part in Indian textbooks is defined as {x}=x−⌊x⌋ where ⌊∙⌋ is floor function.
FractionpartinIndiantextbooksisdefinedas{x}=x−⌊x⌋where⌊⋅⌋isfloorfunction.
Commented by Tinkutara last updated on 07/Jun/17
We have always {x} = x − [x] e.g. x ∈ Z (say 5), {5} = 5 − [5] = 0 x ∈ R^+ (say 4.8), {4.8} = 4.8 − [4.8] = 4.8 − 4 = 0.8 x ∈ R^− (say −3.78), {−3.78} = −3.78 − [−3.78] = −3.78 + 4 = 0.22 Hence {x} is always positive and 0 ≤ {x} < 1. In this way {−0.5} = 0.5 and thus f(−0.5) = (3/7) (1/3) < f(−0.5) < 1
Wehavealways{x}=x−[x]e.g.x∈Z(say5),{5}=5−[5]=0x∈R+(say4.8),{4.8}=4.8−[4.8]=4.8−4=0.8x∈R−(say−3.78),{−3.78}=−3.78−[−3.78]=−3.78+4=0.22Hence{x}isalwayspositiveand0⩽{x}<1.Inthisway{−0.5}=0.5andthusf(−0.5)=3713<f(−0.5)<1
Commented by prakash jain last updated on 07/Jun/17
Wikipedia page on this has three different definitions for fractional part. frac(x)=x−⌊x⌋ frac{x}= { ((x−⌊x⌋),(x≥0)),((x−⌈x⌉),(x<0)) :} fract(x)=x−⌊∣x∣⌋ ⌊∙⌋=floor function ⌈∙⌉=ceiling function frac(−1.2) can take .8,−.2 or .2 depending upon the definition u use. JEE exams will use {x}=x−⌊x⌋
Wikipediapageonthishasthreedifferentdefinitionsforfractionalpart.frac(x)=x−⌊x⌋frac{x}={x−⌊x⌋x⩾0x−⌈x⌉x<0fract(x)=x−⌊∣x∣⌋⌊⋅⌋=floorfunction⌈⋅⌉=ceilingfunctionfrac(−1.2)cantake.8,−.2or.2dependinguponthedefinitionuuse.JEEexamswilluse{x}=x−⌊x⌋
Commented by mrW1 last updated on 07/Jun/17
thank you mr prakash jain for this explanation.
thankyoumrprakashjainforthisexplanation.
Answered by Tinkutara last updated on 08/Jun/17
Question done! f(x) = 1 − ((2{x})/(({x} + (1/2))^2 + (3/4))) Now assuming 0 ≤ {x} < 1, we get 1 ≤ ({x} + (1/2))^2 + (3/4) < 3 ∴ When {x} = 0, f(x) = 1 And as {x} → 1, f(x) → 1 − (2/3) = (1/3) ∴ f(x) is a decreasing function. R_(f(x)) = ((1/3), 1]
Questiondone!f(x)=1−2{x}({x}+12)2+34Nowassuming0⩽{x}<1,weget1⩽({x}+12)2+34<3∴When{x}=0,f(x)=1Andas{x}→1,f(x)→1−23=13∴f(x)isadecreasingfunction.Rf(x)=(13,1]

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