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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 2.

$${Ans}:\:{b},{d}. \\ $$$${i}'{m}\:{getting}\:{that}\:{it}\:{cannot}\:{exceed}\:\mathrm{2}. \\ $$

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).

$${we}\:{have}\:\:\int_{{a}} ^{{b}} \mid{f}\left({x}\right)\mid\mid{g}\left({x}\right)\mid{dx}\:\leqslant\:\left(\int_{{a}} ^{{b}} \:{f}^{\mathrm{2}} \left({x}\right){dx}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:.\left(\int_{{a}} ^{{b}} \:{g}^{\mathrm{2}} \left({x}\right){dx}\right)\:\left({cauchy}\:{shwarz}\right) \\ $$$$\:{let}\:{take}\:{f}\left({x}\right)\:=\sqrt{\mathrm{1}+{x}}\:{and}\:{g}\left({x}\right)\:=\sqrt{\mathrm{1}+{x}^{\mathrm{3}} }{dx}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\sqrt{\mathrm{1}+{x}}\sqrt{\mathrm{1}+{x}^{\mathrm{3}} }{dx}\:\leqslant\:\sqrt{\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+{x}\right){dx}}.\:\sqrt{\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+{x}^{\mathrm{3}} \right){dx}} \\ $$$${but}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+{x}\right){dx}\:=\left[{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+{x}^{\mathrm{3}} \right){dx}\:=\left[{x}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}}\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\frac{\mathrm{5}}{\mathrm{4}}\:\:\Rightarrow\:{I}\:\leqslant\sqrt{\frac{\mathrm{3}}{\mathrm{2}}}.\sqrt{\frac{\mathrm{5}}{\mathrm{4}}}\:\:\Rightarrow\:{I}\:\leqslant\sqrt{\frac{\mathrm{15}}{\mathrm{8}}} \\ $$$${so}\:{the}\:{answer}\:{is}\:\left({d}\right). \\ $$

Commented by rahul 19 last updated on 28/Apr/19

thank U prof.Abdo

$${thank}\:{U}\:{prof}.{Abdo} \\ $$

Commented by maxmathsup by imad last updated on 30/Apr/19

you are welcome.

$${you}\:{are}\:{welcome}. \\ $$

Answered by tanmay last updated on 27/Apr/19

((1+x)/2)≥(1×x)^(1/2)   ((1+x^3 )/2)≥(1×x^3 )^(1/2)   (1+x)(1+x^3 )≥4(x.x^3 )^(1/2)   (1+x)(1+x^3 )≥4x^2   (√((1+x)(1+x^3 ))) ≥2x  ((1+x+1+x^3 )/2)≥{(1+x)(1+x^3 )}^(1/2)   ((2+x+x^3 )/2)≥(√((1+x)(1+x^3 )))   so  ∫_0 ^1 ((2+x+x^3 )/2)dx≥∫_0 ^1 (√((1+x)(1+x^3 ))) dx≥∫_0 ^1 2xdx  (1/2)∣2x+(x^2 /2)+(x^4 /4)∣_0 ^1 ≥I≥2×∣(x^2 /2)∣_0 ^1   (1/2)×((11)/4)≥I≥1  ((11)/8)≥I≥1  ((11)/8)=1.375   and (√((15)/8)) =(√(1.875)) =1.369≈1.37

$$\frac{\mathrm{1}+{x}}{\mathrm{2}}\geqslant\left(\mathrm{1}×{x}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\frac{\mathrm{1}+{x}^{\mathrm{3}} }{\mathrm{2}}\geqslant\left(\mathrm{1}×{x}^{\mathrm{3}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{x}^{\mathrm{3}} \right)\geqslant\mathrm{4}\left({x}.{x}^{\mathrm{3}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{x}^{\mathrm{3}} \right)\geqslant\mathrm{4}{x}^{\mathrm{2}} \\ $$$$\sqrt{\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{x}^{\mathrm{3}} \right)}\:\geqslant\mathrm{2}{x} \\ $$$$\frac{\mathrm{1}+{x}+\mathrm{1}+{x}^{\mathrm{3}} }{\mathrm{2}}\geqslant\left\{\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{x}^{\mathrm{3}} \right)\right\}^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\frac{\mathrm{2}+{x}+{x}^{\mathrm{3}} }{\mathrm{2}}\geqslant\sqrt{\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{x}^{\mathrm{3}} \right)}\: \\ $$$${so} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}+{x}+{x}^{\mathrm{3}} }{\mathrm{2}}{dx}\geqslant\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{x}^{\mathrm{3}} \right)}\:{dx}\geqslant\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{2}{xdx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mid\mathrm{2}{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}}\mid_{\mathrm{0}} ^{\mathrm{1}} \geqslant{I}\geqslant\mathrm{2}×\mid\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\mid_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{11}}{\mathrm{4}}\geqslant{I}\geqslant\mathrm{1} \\ $$$$\frac{\mathrm{11}}{\mathrm{8}}\geqslant{I}\geqslant\mathrm{1} \\ $$$$\frac{\mathrm{11}}{\mathrm{8}}=\mathrm{1}.\mathrm{375}\:\:\:{and}\:\sqrt{\frac{\mathrm{15}}{\mathrm{8}}}\:=\sqrt{\mathrm{1}.\mathrm{875}}\:=\mathrm{1}.\mathrm{369}\approx\mathrm{1}.\mathrm{37} \\ $$

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. 58626  then i will get:  f(0)=1 ⇒m=1  f(1)=2⇒M=2  ⇒⇒ 1<I<2.  What′s wrong in this method?

$${Sir},\:{if}\:{I}\:{do}\:{the}\:{same}\:{procedure}\:{as}\:{in}\:\:{Q}.\:\mathrm{58626} \\ $$$${then}\:{i}\:{will}\:{get}: \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{1}\:\Rightarrow{m}=\mathrm{1} \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{2}\Rightarrow{M}=\mathrm{2} \\ $$$$\Rightarrow\Rightarrow\:\mathrm{1}<{I}<\mathrm{2}. \\ $$$${What}'{s}\:{wrong}\:{in}\:{this}\:{method}? \\ $$

Commented by tanmay last updated on 28/Apr/19

i am co relating with mean value theorem of   intregal

$${i}\:{am}\:{co}\:{relating}\:{with}\:{mean}\:{value}\:{theorem}\:{of}\: \\ $$$${intregal} \\ $$

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.58626 then ?  (in interval 1 to 3 given function does  not have min/max. value.)

$${still}\:{confused}... \\ $$$${why}\:{can}\:{we}\:{apply}\:{in}\:{Q}.\mathrm{58626}\:{then}\:? \\ $$$$\left({in}\:{interval}\:\mathrm{1}\:{to}\:\mathrm{3}\:{given}\:{function}\:{does}\right. \\ $$$$\left.{not}\:{have}\:{min}/{max}.\:{value}.\right) \\ $$

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 ((11)/8)=1.375 ←what i calculated  (√((15)/8)) =(√(1.875)) =1.369←what given in book

$${in}\:{books}\:{answer}\:{is}\:{above}\:{along}\:{with}\:{formula} \\ $$$${however}\:\frac{\mathrm{11}}{\mathrm{8}}=\mathrm{1}.\mathrm{375}\:\leftarrow{what}\:{i}\:{calculated} \\ $$$$\sqrt{\frac{\mathrm{15}}{\mathrm{8}}}\:=\sqrt{\mathrm{1}.\mathrm{875}}\:=\mathrm{1}.\mathrm{369}\leftarrow{what}\:{given}\:{in}\:{book} \\ $$

Commented by rahul 19 last updated on 28/Apr/19

thank u sir.

$${thank}\:{u}\:{sir}. \\ $$

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