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Question Number 139190 by mathdanisur last updated on 23/Apr/21

a;b∈R , ((∣ax+b∣)/(1+x^2 )) ≤ 1 , ∀x∈R prove: ∣a∣≤2 ; ∣b∣≤1

$${a};{b}\in\mathbb{R}\:,\:\frac{\mid{ax}+{b}\mid}{\mathrm{1}+{x}^{\mathrm{2}} }\:\leqslant\:\mathrm{1}\:,\:\forall{x}\in\mathbb{R} \\ $$$${prove}:\:\:\mid{a}\mid\leqslant\mathrm{2}\:;\:\mid{b}\mid\leqslant\mathrm{1} \\ $$

Answered by mitica last updated on 24/Apr/21

x=0⇒∣b∣≤1 x=1⇒∣a+b∣≤2 x=−1⇒∣a−b∣≤2 ∣a∣=∣((2a)/2)∣=((∣a+b+a−b∣)/2)≤((∣a+b∣+∣a−b∣)/2)≤((2+2)/2)=2

$${x}=\mathrm{0}\Rightarrow\mid{b}\mid\leqslant\mathrm{1} \\ $$$${x}=\mathrm{1}\Rightarrow\mid{a}+{b}\mid\leqslant\mathrm{2} \\ $$$${x}=−\mathrm{1}\Rightarrow\mid{a}−{b}\mid\leqslant\mathrm{2} \\ $$$$\mid{a}\mid=\mid\frac{\mathrm{2}{a}}{\mathrm{2}}\mid=\frac{\mid{a}+{b}+{a}−{b}\mid}{\mathrm{2}}\leqslant\frac{\mid{a}+{b}\mid+\mid{a}−{b}\mid}{\mathrm{2}}\leqslant\frac{\mathrm{2}+\mathrm{2}}{\mathrm{2}}=\mathrm{2} \\ $$

Commented by mathdanisur last updated on 24/Apr/21

thank you sir, but ∣b∣≤1? please

$${thank}\:{you}\:{sir},\:{but}\:\mid{b}\mid\leqslant\mathrm{1}?\:{please} \\ $$

Commented by mitica last updated on 24/Apr/21

x=0⇒((∣a∙0+b∣)/(0^2 +1))≤1⇒((∣b∣)/1)≤1⇒∣b∣≤1

$${x}=\mathrm{0}\Rightarrow\frac{\mid{a}\centerdot\mathrm{0}+{b}\mid}{\mathrm{0}^{\mathrm{2}} +\mathrm{1}}\leqslant\mathrm{1}\Rightarrow\frac{\mid{b}\mid}{\mathrm{1}}\leqslant\mathrm{1}\Rightarrow\mid{b}\mid\leqslant\mathrm{1} \\ $$

Answered by mr W last updated on 24/Apr/21

Commented by mr W last updated on 24/Apr/21

let′s generally look at the function y=x^2 +bx+c there are three cases. case 1: y>0 for x∈R x^2 +bx+c=0 has no real root, i.e. Δ=b^2 −4c<0 case 2: y≥0 for x∈R x^2 +bx+c=0 has one double real root, i.e. Δ=b^2 −4c=0 case 3: x^2 +bx+c=0 has two real rootx, i.e. Δ=b^2 −4c>0 y>0 if x<x_1 or x>x_2 y<0 if x_1 <x<x_2 y=0 if x=x_1 or x_2 ⇒such that x^2 +bx+c≥0 for x∈R, Δ=b^2 −4c≤0

$${let}'{s}\:{generally}\:{look}\:{at}\:{the}\:{function} \\ $$$${y}={x}^{\mathrm{2}} +{bx}+{c} \\ $$$${there}\:{are}\:{three}\:{cases}. \\ $$$${case}\:\mathrm{1}:\:{y}>\mathrm{0}\:{for}\:{x}\in{R} \\ $$$${x}^{\mathrm{2}} +{bx}+{c}=\mathrm{0}\:{has}\:{no}\:{real}\:{root},\:{i}.{e}. \\ $$$$\Delta={b}^{\mathrm{2}} −\mathrm{4}{c}<\mathrm{0} \\ $$$${case}\:\mathrm{2}:\:{y}\geqslant\mathrm{0}\:{for}\:{x}\in{R} \\ $$$${x}^{\mathrm{2}} +{bx}+{c}=\mathrm{0}\:{has}\:{one}\:{double}\:{real}\:{root},\:{i}.{e}. \\ $$$$\Delta={b}^{\mathrm{2}} −\mathrm{4}{c}=\mathrm{0} \\ $$$${case}\:\mathrm{3}: \\ $$$${x}^{\mathrm{2}} +{bx}+{c}=\mathrm{0}\:{has}\:{two}\:{real}\:{rootx},\:{i}.{e}. \\ $$$$\Delta={b}^{\mathrm{2}} −\mathrm{4}{c}>\mathrm{0} \\ $$$${y}>\mathrm{0}\:{if}\:{x}<{x}_{\mathrm{1}} \:{or}\:{x}>{x}_{\mathrm{2}} \\ $$$${y}<\mathrm{0}\:{if}\:{x}_{\mathrm{1}} <{x}<{x}_{\mathrm{2}} \\ $$$${y}=\mathrm{0}\:{if}\:{x}={x}_{\mathrm{1}} \:{or}\:{x}_{\mathrm{2}} \\ $$$$ \\ $$$$\Rightarrow{such}\:{that}\:{x}^{\mathrm{2}} +{bx}+{c}\geqslant\mathrm{0}\:{for}\:{x}\in{R}, \\ $$$$\Delta={b}^{\mathrm{2}} −\mathrm{4}{c}\leqslant\mathrm{0} \\ $$

Commented by mr W last updated on 24/Apr/21

((∣ax+b∣)/(1+x^2 ))≤1 for x∈R ⇒ −1≤((ax+b)/(1+x^2 ))≤1 ⇒ −1−x^2 ≤ax+b≤1+x^2 1) ax+b≤1+x^2 x^2 −ax+1−b≥0 Δ=a^2 +4(b−1)≤0 ...(i) 4(b−1)≤0 ⇒b≤1 2) −1−x^2 ≤ax+b x^2 +ax+b+1≥0 Δ=a^2 −4(b+1)≤0 ...(ii) 4(b+1)≥a^2 ≥0 ⇒b≥−1 ⇒−1≤b≤1 ⇒∣b∣≤1 (i)+(ii): 2a^2 −4−4≤0 a^2 ≤4 ⇒∣a∣≤2

$$\frac{\mid{ax}+{b}\mid}{\mathrm{1}+{x}^{\mathrm{2}} }\leqslant\mathrm{1}\:{for}\:{x}\in{R} \\ $$$$\Rightarrow\:−\mathrm{1}\leqslant\frac{{ax}+{b}}{\mathrm{1}+{x}^{\mathrm{2}} }\leqslant\mathrm{1} \\ $$$$\Rightarrow\:−\mathrm{1}−{x}^{\mathrm{2}} \leqslant{ax}+{b}\leqslant\mathrm{1}+{x}^{\mathrm{2}} \\ $$$$ \\ $$$$\left.\mathrm{1}\right)\:{ax}+{b}\leqslant\mathrm{1}+{x}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} −{ax}+\mathrm{1}−{b}\geqslant\mathrm{0} \\ $$$$\Delta={a}^{\mathrm{2}} +\mathrm{4}\left({b}−\mathrm{1}\right)\leqslant\mathrm{0}\:\:\:...\left({i}\right) \\ $$$$\mathrm{4}\left({b}−\mathrm{1}\right)\leqslant\mathrm{0} \\ $$$$\Rightarrow{b}\leqslant\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\:−\mathrm{1}−{x}^{\mathrm{2}} \leqslant{ax}+{b} \\ $$$${x}^{\mathrm{2}} +{ax}+{b}+\mathrm{1}\geqslant\mathrm{0}\:\:\: \\ $$$$\Delta={a}^{\mathrm{2}} −\mathrm{4}\left({b}+\mathrm{1}\right)\leqslant\mathrm{0}\:\:\:...\left({ii}\right) \\ $$$$\mathrm{4}\left({b}+\mathrm{1}\right)\geqslant{a}^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$$\Rightarrow{b}\geqslant−\mathrm{1} \\ $$$$\Rightarrow−\mathrm{1}\leqslant{b}\leqslant\mathrm{1} \\ $$$$\Rightarrow\mid{b}\mid\leqslant\mathrm{1} \\ $$$$\left({i}\right)+\left({ii}\right): \\ $$$$\mathrm{2}{a}^{\mathrm{2}} −\mathrm{4}−\mathrm{4}\leqslant\mathrm{0} \\ $$$${a}^{\mathrm{2}} \leqslant\mathrm{4} \\ $$$$\Rightarrow\mid{a}\mid\leqslant\mathrm{2} \\ $$

Commented by mathdanisur last updated on 24/Apr/21

Perfect Sir thanks.. How can this be Sir pliase.. if: a;b∈R, ∣ax^3 +bx∣≤1, ∀∣x∣≤1 prov: ∣bx^3 +ax∣≤1, ∣3ax^2 +b∣≤9, ∀∣x∣≤1

$${Perfect}\:{Sir}\:{thanks}.. \\ $$$${How}\:{can}\:{this}\:{be}\:{Sir}\:{pliase}.. \\ $$$${if}:\:{a};{b}\in\mathbb{R},\:\mid{ax}^{\mathrm{3}} +{bx}\mid\leqslant\mathrm{1},\:\forall\mid{x}\mid\leqslant\mathrm{1} \\ $$$${prov}:\:\mid{bx}^{\mathrm{3}} +{ax}\mid\leqslant\mathrm{1},\:\mid\mathrm{3}{ax}^{\mathrm{2}} +{b}\mid\leqslant\mathrm{9},\:\forall\mid{x}\mid\leqslant\mathrm{1} \\ $$

Commented by mr W last updated on 24/Apr/21

please open a new thread for this new question!

$${please}\:{open}\:{a}\:{new}\:{thread}\:{for}\:{this}\: \\ $$$${new}\:{question}! \\ $$

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