Question Number 58648 by rahul 19 last updated on 27/Apr/19
Commented by rahul 19 last updated on 27/Apr/19
$${Ans}:\:{b},{d}. \\ $$$${i}'{m}\:{getting}\:{that}\:{it}\:{cannot}\:{exceed}\:\mathrm{2}. \\ $$
Commented by maxmathsup by imad last updated on 27/Apr/19
![we have ∫_a ^b ∣f(x)∣∣g(x)∣dx ≤ (∫_a ^b f^2 (x)dx)^(1/2) .(∫_a ^b g^2 (x)dx) (cauchy shwarz) let take f(x) =(√(1+x)) and g(x) =(√(1+x^3 ))dx ⇒ ∫_0 ^1 (√(1+x))(√(1+x^3 ))dx ≤ (√(∫_0 ^1 (1+x)dx)). (√(∫_0 ^1 (1+x^3 )dx)) but ∫_0 ^1 (1+x)dx =[x+(x^2 /2)]_0 ^1 =(3/2) ∫_0 ^1 (1+x^3 )dx =[x+(x^4 /4)]_0 ^1 =(5/4) ⇒ I ≤(√(3/2)).(√(5/4)) ⇒ I ≤(√((15)/8)) so the answer is (d).](https://www.tinkutara.com/question/Q58695.png)
$${we}\:{have}\:\:\int_{{a}} ^{{b}} \mid{f}\left({x}\right)\mid\mid{g}\left({x}\right)\mid{dx}\:\leqslant\:\left(\int_{{a}} ^{{b}} \:{f}^{\mathrm{2}} \left({x}\right){dx}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:.\left(\int_{{a}} ^{{b}} \:{g}^{\mathrm{2}} \left({x}\right){dx}\right)\:\left({cauchy}\:{shwarz}\right) \\ $$$$\:{let}\:{take}\:{f}\left({x}\right)\:=\sqrt{\mathrm{1}+{x}}\:{and}\:{g}\left({x}\right)\:=\sqrt{\mathrm{1}+{x}^{\mathrm{3}} }{dx}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\sqrt{\mathrm{1}+{x}}\sqrt{\mathrm{1}+{x}^{\mathrm{3}} }{dx}\:\leqslant\:\sqrt{\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+{x}\right){dx}}.\:\sqrt{\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+{x}^{\mathrm{3}} \right){dx}} \\ $$$${but}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+{x}\right){dx}\:=\left[{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+{x}^{\mathrm{3}} \right){dx}\:=\left[{x}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}}\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\frac{\mathrm{5}}{\mathrm{4}}\:\:\Rightarrow\:{I}\:\leqslant\sqrt{\frac{\mathrm{3}}{\mathrm{2}}}.\sqrt{\frac{\mathrm{5}}{\mathrm{4}}}\:\:\Rightarrow\:{I}\:\leqslant\sqrt{\frac{\mathrm{15}}{\mathrm{8}}} \\ $$$${so}\:{the}\:{answer}\:{is}\:\left({d}\right). \\ $$
Commented by rahul 19 last updated on 28/Apr/19
$${thank}\:{U}\:{prof}.{Abdo} \\ $$
Commented by maxmathsup by imad last updated on 30/Apr/19
$${you}\:{are}\:{welcome}. \\ $$
Answered by tanmay last updated on 27/Apr/19
$$\frac{\mathrm{1}+{x}}{\mathrm{2}}\geqslant\left(\mathrm{1}×{x}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\frac{\mathrm{1}+{x}^{\mathrm{3}} }{\mathrm{2}}\geqslant\left(\mathrm{1}×{x}^{\mathrm{3}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{x}^{\mathrm{3}} \right)\geqslant\mathrm{4}\left({x}.{x}^{\mathrm{3}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{x}^{\mathrm{3}} \right)\geqslant\mathrm{4}{x}^{\mathrm{2}} \\ $$$$\sqrt{\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{x}^{\mathrm{3}} \right)}\:\geqslant\mathrm{2}{x} \\ $$$$\frac{\mathrm{1}+{x}+\mathrm{1}+{x}^{\mathrm{3}} }{\mathrm{2}}\geqslant\left\{\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{x}^{\mathrm{3}} \right)\right\}^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\frac{\mathrm{2}+{x}+{x}^{\mathrm{3}} }{\mathrm{2}}\geqslant\sqrt{\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{x}^{\mathrm{3}} \right)}\: \\ $$$${so} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}+{x}+{x}^{\mathrm{3}} }{\mathrm{2}}{dx}\geqslant\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{x}^{\mathrm{3}} \right)}\:{dx}\geqslant\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{2}{xdx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mid\mathrm{2}{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}}\mid_{\mathrm{0}} ^{\mathrm{1}} \geqslant{I}\geqslant\mathrm{2}×\mid\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\mid_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{11}}{\mathrm{4}}\geqslant{I}\geqslant\mathrm{1} \\ $$$$\frac{\mathrm{11}}{\mathrm{8}}\geqslant{I}\geqslant\mathrm{1} \\ $$$$\frac{\mathrm{11}}{\mathrm{8}}=\mathrm{1}.\mathrm{375}\:\:\:{and}\:\sqrt{\frac{\mathrm{15}}{\mathrm{8}}}\:=\sqrt{\mathrm{1}.\mathrm{875}}\:=\mathrm{1}.\mathrm{369}\approx\mathrm{1}.\mathrm{37} \\ $$
Commented by tanmay last updated on 28/Apr/19
Commented by rahul 19 last updated on 28/Apr/19
$${Sir},\:{if}\:{I}\:{do}\:{the}\:{same}\:{procedure}\:{as}\:{in}\:\:{Q}.\:\mathrm{58626} \\ $$$${then}\:{i}\:{will}\:{get}: \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{1}\:\Rightarrow{m}=\mathrm{1} \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{2}\Rightarrow{M}=\mathrm{2} \\ $$$$\Rightarrow\Rightarrow\:\mathrm{1}<{I}<\mathrm{2}. \\ $$$${What}'{s}\:{wrong}\:{in}\:{this}\:{method}? \\ $$
Commented by tanmay last updated on 28/Apr/19
$${i}\:{am}\:{co}\:{relating}\:{with}\:{mean}\:{value}\:{theorem}\:{of}\: \\ $$$${intregal} \\ $$
Commented by tanmay last updated on 28/Apr/19
Commented by tanmay last updated on 28/Apr/19
Commented by rahul 19 last updated on 29/Apr/19
$${still}\:{confused}… \\ $$$${why}\:{can}\:{we}\:{apply}\:{in}\:{Q}.\mathrm{58626}\:{then}\:? \\ $$$$\left({in}\:{interval}\:\mathrm{1}\:{to}\:\mathrm{3}\:{given}\:{function}\:{does}\right. \\ $$$$\left.{not}\:{have}\:{min}/{max}.\:{value}.\right) \\ $$
Answered by tanmay last updated on 27/Apr/19
Commented by tanmay last updated on 27/Apr/19
$${in}\:{books}\:{answer}\:{is}\:{above}\:{along}\:{with}\:{formula} \\ $$$${however}\:\frac{\mathrm{11}}{\mathrm{8}}=\mathrm{1}.\mathrm{375}\:\leftarrow{what}\:{i}\:{calculated} \\ $$$$\sqrt{\frac{\mathrm{15}}{\mathrm{8}}}\:=\sqrt{\mathrm{1}.\mathrm{875}}\:=\mathrm{1}.\mathrm{369}\leftarrow{what}\:{given}\:{in}\:{book} \\ $$
Commented by rahul 19 last updated on 28/Apr/19
$${thank}\:{u}\:{sir}. \\ $$