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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?

$$\mathrm{The}\:\mathrm{range}\:\mathrm{of}\:{f}\left({x}\right)\:=\:\frac{\left\{{x}\right\}^{\mathrm{2}} \:−\:\left\{{x}\right\}\:+\:\mathrm{1}}{\left\{{x}\right\}^{\mathrm{2}} \:+\:\left\{{x}\right\}\:+\:\mathrm{1}}; \\ $$$$\left(\mathrm{where}\:\left\{\centerdot\right\}\:\mathrm{denotes}\:\mathrm{fractional}\:\mathrm{function}\right) \\ $$$$\mathrm{is}? \\ $$

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)

$$\:{f}\left({x}\right)\:=\:\frac{\left\{{x}\right\}^{\mathrm{2}} \:−\:\left\{{x}\right\}\:+\:\mathrm{1}}{\left\{{x}\right\}^{\mathrm{2}} \:+\:\left\{{x}\right\}\:+\:\mathrm{1}} \\ $$$$\mathrm{let}\:\mathrm{g}=\left\{\mathrm{x}\right\} \\ $$$$−\mathrm{1}<\mathrm{g}<\mathrm{1} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{g}^{\mathrm{2}} −\mathrm{g}+\mathrm{1}}{\mathrm{g}^{\mathrm{2}} +\mathrm{g}+\mathrm{1}}=\mathrm{1}−\frac{\mathrm{2g}}{\mathrm{g}^{\mathrm{2}} +\mathrm{g}+\mathrm{1}} \\ $$$$\mathrm{g}\rightarrow+\mathrm{1}\:\mathrm{f}\left(\mathrm{x}\right)\rightarrow\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{g}=\mathrm{0},\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{1} \\ $$$$\mathrm{g}\rightarrow−\mathrm{1}\:\mathrm{f}\left(\mathrm{x}\right)\rightarrow\mathrm{3} \\ $$$$ \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\in\left(\frac{\mathrm{1}}{\mathrm{3}},\mathrm{3}\right) \\ $$

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 .

$$\mathrm{0}.\mathrm{3}\:\:\:\:{i}\:{believe} \\ $$$$\left[{x}\right]\:{is}\:{the}\:{greatest}\:{integer}\:{that} \\ $$$${the}\:{value}\:{of}\:{x}\:{has}\:{crossed}\:{going} \\ $$$${along}\:{the}\:{numberline}\:{in}\:{the}\: \\ $$$${positive}\:{x}\:{direction}. \\ $$$$\:\:\left\{{x}\right\}\:{is}\:{the}\:{value}\:{that}\:{need}\:{be} \\ $$$${added}\:{to}\:\left[{x}\right]\:{to}\:{make}\:{it}\:{reach}\:\boldsymbol{{x}}\:. \\ $$

Commented by mrW1 last updated on 07/Jun/17

{−2.7}=−0.7

$$\left\{−\mathrm{2}.\mathrm{7}\right\}=−\mathrm{0}.\mathrm{7} \\ $$

Commented by mrW1 last updated on 07/Jun/17

(one) definition of fractional part of a number see diagram:

$$\left(\mathrm{one}\right)\:\mathrm{definition}\:\mathrm{of}\:\mathrm{fractional}\:\mathrm{part}\:\mathrm{of}\:\mathrm{a}\:\mathrm{number} \\ $$$$\mathrm{see}\:\mathrm{diagram}: \\ $$

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

$$\mathrm{i}\:\mathrm{have}\:\mathrm{corrected}.\:\mathrm{it}\:\mathrm{is}\:\mathrm{y}\in\left(\frac{\mathrm{1}}{\mathrm{3}},\mathrm{3}\right) \\ $$$$\mathrm{you}\:\mathrm{can}\:\mathrm{check}\:\mathrm{with}\:\mathrm{x}=−\mathrm{1}.\mathrm{5}: \\ $$$$\left\{\mathrm{x}\right\}=−\mathrm{0}.\mathrm{5} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\left(−\mathrm{0}.\mathrm{5}\right)^{\mathrm{2}} −\left(−\mathrm{0}.\mathrm{5}\right)+\mathrm{1}}{\left(−\mathrm{0}.\mathrm{5}\right)^{\mathrm{2}} +\left(−\mathrm{0}.\mathrm{5}\right)+\mathrm{1}}=\frac{\mathrm{1}.\mathrm{75}}{\mathrm{0}.\mathrm{75}}=\mathrm{2}.\mathrm{33}>\mathrm{1} \\ $$

Commented by Tinkutara last updated on 07/Jun/17

But answer is ((1/3), 1]

$$\mathrm{But}\:\mathrm{answer}\:\mathrm{is}\:\left(\frac{\mathrm{1}}{\mathrm{3}},\:\mathrm{1}\right] \\ $$

Commented by Tinkutara last updated on 07/Jun/17

Sir, if g = {x}, then 0 ≤ g < 1. Why −1 < g < 1 ?

$$\mathrm{Sir},\:\mathrm{if}\:{g}\:=\:\left\{{x}\right\},\:\mathrm{then}\:\mathrm{0}\:\leqslant\:{g}\:<\:\mathrm{1}. \\ $$$$\mathrm{Why}\:−\mathrm{1}\:<\:{g}\:<\:\mathrm{1}\:? \\ $$

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

$$\mathrm{that}\:\mathrm{is}\:\mathrm{only}\:\mathrm{for}\:\mathrm{x}\geqslant\mathrm{0} \\ $$$$\mathrm{for}\:\mathrm{x}\leqslant\mathrm{0},\:−\mathrm{1}<\left\{\mathrm{x}\right\}\leqslant\mathrm{0} \\ $$$$ \\ $$$$\left\{\mathrm{2}.\mathrm{5}\right\}=\mathrm{0}.\mathrm{5} \\ $$$$\left\{−\mathrm{2}.\mathrm{5}\right\}=−\mathrm{0}.\mathrm{5} \\ $$

Commented by Tinkutara last updated on 07/Jun/17

What will be {−2.7}?

$$\mathrm{What}\:\mathrm{will}\:\mathrm{be}\:\left\{−\mathrm{2}.\mathrm{7}\right\}? \\ $$

Commented by ajfour last updated on 07/Jun/17

0≤{x}<1 , i have read this.

$$\mathrm{0}\leqslant\left\{{x}\right\}<\mathrm{1}\:,\:{i}\:{have}\:{read}\:{this}. \\ $$

Commented by mrW1 last updated on 07/Jun/17

there are 2 kinds of definition for x<0 which are used in mathematic world.

$$\mathrm{there}\:\mathrm{are}\:\mathrm{2}\:\mathrm{kinds}\:\mathrm{of}\:\mathrm{definition}\:\mathrm{for}\:\mathrm{x}<\mathrm{0} \\ $$$$\mathrm{which}\:\mathrm{are}\:\mathrm{used}\:\mathrm{in}\:\mathrm{mathematic}\:\mathrm{world}. \\ $$

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.

$$\mathrm{Fraction}\:\mathrm{part}\:\mathrm{in}\:\mathrm{Indian}\:\mathrm{textbooks} \\ $$$$\mathrm{is}\:\mathrm{defined}\:\mathrm{as}\:\left\{{x}\right\}={x}−\lfloor{x}\rfloor \\ $$$$\mathrm{where}\:\lfloor\centerdot\rfloor\:\mathrm{is}\:\mathrm{floor}\:\mathrm{function}. \\ $$

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

$$\mathrm{We}\:\mathrm{have}\:\mathrm{always} \\ $$$$\left\{{x}\right\}\:=\:{x}\:−\:\left[{x}\right] \\ $$$${e}.{g}.\:{x}\:\in\:{Z}\:\left(\mathrm{say}\:\mathrm{5}\right),\:\left\{\mathrm{5}\right\}\:=\:\mathrm{5}\:−\:\left[\mathrm{5}\right]\:=\:\mathrm{0} \\ $$$${x}\:\in\:{R}^{+} \:\left(\mathrm{say}\:\mathrm{4}.\mathrm{8}\right),\:\left\{\mathrm{4}.\mathrm{8}\right\}\:=\:\mathrm{4}.\mathrm{8}\:−\:\left[\mathrm{4}.\mathrm{8}\right] \\ $$$$=\:\mathrm{4}.\mathrm{8}\:−\:\mathrm{4}\:=\:\mathrm{0}.\mathrm{8} \\ $$$${x}\:\in\:{R}^{−} \:\left(\mathrm{say}\:−\mathrm{3}.\mathrm{78}\right), \\ $$$$\left\{−\mathrm{3}.\mathrm{78}\right\}\:=\:−\mathrm{3}.\mathrm{78}\:−\:\left[−\mathrm{3}.\mathrm{78}\right] \\ $$$$=\:−\mathrm{3}.\mathrm{78}\:+\:\mathrm{4}\:=\:\mathrm{0}.\mathrm{22} \\ $$$$\mathrm{Hence}\:\left\{{x}\right\}\:\mathrm{is}\:\mathrm{always}\:\mathrm{positive}\:\mathrm{and} \\ $$$$\mathrm{0}\:\leqslant\:\left\{{x}\right\}\:<\:\mathrm{1}. \\ $$$$\mathrm{In}\:\mathrm{this}\:\mathrm{way}\:\left\{−\mathrm{0}.\mathrm{5}\right\}\:=\:\mathrm{0}.\mathrm{5}\:\mathrm{and}\:\mathrm{thus} \\ $$$${f}\left(−\mathrm{0}.\mathrm{5}\right)\:=\:\frac{\mathrm{3}}{\mathrm{7}} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\:<\:{f}\left(−\mathrm{0}.\mathrm{5}\right)\:<\:\mathrm{1} \\ $$

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⌋

$$\mathrm{Wikipedia}\:\mathrm{page}\:\mathrm{on}\:\mathrm{this}\:\mathrm{has}\:\mathrm{three} \\ $$$$\mathrm{different}\:\mathrm{definitions}\:\mathrm{for}\:\mathrm{fractional} \\ $$$$\mathrm{part}. \\ $$$${frac}\left({x}\right)={x}−\lfloor{x}\rfloor \\ $$$${frac}\left\{{x}\right\}=\begin{cases}{{x}−\lfloor{x}\rfloor}&{{x}\geqslant\mathrm{0}}\\{{x}−\lceil{x}\rceil}&{{x}<\mathrm{0}}\end{cases} \\ $$$${fract}\left({x}\right)={x}−\lfloor\mid{x}\mid\rfloor \\ $$$$\lfloor\centerdot\rfloor=\mathrm{floor}\:\mathrm{function} \\ $$$$\lceil\centerdot\rceil=\mathrm{ceiling}\:\mathrm{function} \\ $$$${frac}\left(−\mathrm{1}.\mathrm{2}\right)\:\mathrm{can}\:\mathrm{take}\:.\mathrm{8},−.\mathrm{2}\:\mathrm{or}\:.\mathrm{2} \\ $$$$\mathrm{depending}\:\mathrm{upon}\:\mathrm{the}\:\mathrm{definition}\:\mathrm{u}\:\mathrm{use}. \\ $$$$\mathrm{JEE}\:\mathrm{exams}\:\mathrm{will}\:\mathrm{use}\:\left\{{x}\right\}={x}−\lfloor{x}\rfloor \\ $$

Commented by mrW1 last updated on 07/Jun/17

thank you mr prakash jain for this explanation.

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{mr}\:\mathrm{prakash}\:\mathrm{jain}\:\mathrm{for}\:\mathrm{this} \\ $$$$\mathrm{explanation}. \\ $$

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]

$$\mathrm{Question}\:\mathrm{done}! \\ $$$${f}\left({x}\right)\:=\:\mathrm{1}\:−\:\frac{\mathrm{2}\left\{{x}\right\}}{\left(\left\{{x}\right\}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\:\frac{\mathrm{3}}{\mathrm{4}}} \\ $$$$\mathrm{Now}\:\mathrm{assuming}\:\mathrm{0}\:\leqslant\:\left\{{x}\right\}\:<\:\mathrm{1}, \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{1}\:\leqslant\:\left(\left\{{x}\right\}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\:\frac{\mathrm{3}}{\mathrm{4}}\:<\:\mathrm{3} \\ $$$$\therefore\:\mathrm{When}\:\left\{{x}\right\}\:=\:\mathrm{0},\:{f}\left({x}\right)\:=\:\mathrm{1} \\ $$$$\mathrm{And}\:\mathrm{as}\:\left\{{x}\right\}\:\rightarrow\:\mathrm{1},\:{f}\left({x}\right)\:\rightarrow\:\mathrm{1}\:−\:\frac{\mathrm{2}}{\mathrm{3}}\:=\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\therefore\:{f}\left({x}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{decreasing}\:\mathrm{function}. \\ $$$$\mathrm{R}_{{f}\left({x}\right)} \:=\:\left(\frac{\mathrm{1}}{\mathrm{3}},\:\mathrm{1}\right] \\ $$

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