I was solving this question :$I = \\int_0^1xf(x)\\,dx = \\frac{1}{6}J = \\int_0^1 (f(x))^2\\,dx = \\frac{1}{12}f\\left( \\frac{1}{2} \\right) = ?$Here$f(x)$iscontinuous.So,tosolvethisItriedtogetanintegralthatcontainedboththegivenintegrals.So,Iassumeaparameter$t$and:$\\int_0^1 (f(x) - tx)^2\\,dx =0 \\int_0^1(f(x))^2,dx -2t\\int_0^1xf(x)\\,dx +t^2\\int_0^1x^2\\,dx =0$Puttinginthevaluesandsolving,Igot:$t = \\frac{1}{2}$.Sothen,$(f(x) - 0.5x)^2$isalwayspositive,soinorderfortheintegralIassumedtoevaluateto$0$,thefunctionhadtobe$0$.$(f(x)-0.5x)^2 =0f(x) = 0.5xf(0.5) = 0.25$Thiswascorrectaccordingtotheanswerkey,butmydoubtisthat:Therecanbeanotherfunction$g(x) \\neq tx$whichalsosatisfiesthegivenconditions,and$g(0.5) \\neq 0.25$.So,howcanweprovethateither$tx$istheonlyfunctionsatisfyingthegivenconditions,orthatforeverypossible$g(x)$,$g(0.5)$ will have to be 0.25 ?

The Dragonborn , 4 Years ago

Grade 12th pass