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Question Number 102336 by 175mohamed last updated on 08/Jul/20

if  f(x)≤2l +1   and  ∫_1 ^3 f(x)dx≤l^2   find the value of l

$${if}\:\:{f}\left({x}\right)\leqslant\mathrm{2}{l}\:+\mathrm{1}\: \\ $$$${and}\:\:\int_{\mathrm{1}} ^{\mathrm{3}} {f}\left({x}\right){dx}\leqslant{l}^{\mathrm{2}} \\ $$$${find}\:{the}\:{value}\:{of}\:{l} \\ $$

Commented by mr W last updated on 08/Jul/20

no unique solution

$${no}\:{unique}\:{solution} \\ $$

Answered by PRITHWISH SEN 2 last updated on 08/Jul/20

let f_(max) =2l+1    at x=c  ∴ ∫_1 ^3 f(x)dx≤(3−1)f(c)=l^2   ⇒f(c)=(l^2 /2)  ∴(l^2 /2)=2l+1⇒l=2±(√6)   please check

$$\mathrm{let}\:\mathrm{f}_{\mathrm{max}} =\mathrm{2}{l}+\mathrm{1}\:\:\:\:\mathrm{at}\:\boldsymbol{\mathrm{x}}=\boldsymbol{\mathrm{c}} \\ $$$$\therefore\:\int_{\mathrm{1}} ^{\mathrm{3}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}\leqslant\left(\mathrm{3}−\mathrm{1}\right)\mathrm{f}\left(\mathrm{c}\right)={l}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{c}\right)=\frac{{l}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\therefore\frac{{l}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{2}{l}+\mathrm{1}\Rightarrow{l}=\mathrm{2}\pm\sqrt{\mathrm{6}}\:\:\:\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{check}} \\ $$

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