Question Number 59846 by bhanukumarb2@gmail.com last updated on 15/May/19
Commented by MJS last updated on 15/May/19
$$\mathrm{do}\:\mathrm{you}\:\mathrm{have}\:\mathrm{an}\:\mathrm{answer}? \\ $$$$\mathrm{it}\:\mathrm{seems}\:\mathrm{to}\:\mathrm{be}\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Commented by bhanukumarb2@gmail.com last updated on 16/May/19
$${thanks}\:{sir} \\ $$
Commented by bhanukumarb2@gmail.com last updated on 16/May/19
$${right} \\ $$
Commented by bhanukumarb2@gmail.com last updated on 16/May/19
$${ur} \\ $$
Commented by bhanukumarb2@gmail.com last updated on 16/May/19
$${ur}\:{approch} \\ $$
Commented by MJS last updated on 16/May/19
![f(x)= { ((1; 0≤x≤a)),((−c; a≤x≤1)) :} I_1 =∫_0 ^a 1dx+∫_a ^1 −cdx=0 ⇒ c=(a/(1−a)) I_2 =∫_0 ^a 1^3 dx+∫_a ^1 ((a/(1−a)))^3 dx=((a(1−2a))/((a−1)^2 )) (d/da)[((a(1−2a))/((a−1)^2 ))]=0 ((3a−1)/((a−1)^3 ))=0 ⇒ a=(1/3) ⇒ c=(1/2) ⇒ I_2 =(1/4) the idea is, make f(x)=1 in the longest possible interval and keep it constant and <1 within the rest, because 1^3 =1 and c^3 <c](https://www.tinkutara.com/question/Q59949.png)
$${f}\left({x}\right)=\begin{cases}{\mathrm{1};\:\mathrm{0}\leqslant{x}\leqslant{a}}\\{−{c};\:{a}\leqslant{x}\leqslant\mathrm{1}}\end{cases} \\ $$$${I}_{\mathrm{1}} =\underset{\mathrm{0}} {\overset{{a}} {\int}}\mathrm{1}{dx}+\underset{{a}} {\overset{\mathrm{1}} {\int}}−{cdx}=\mathrm{0}\:\Rightarrow\:{c}=\frac{{a}}{\mathrm{1}−{a}} \\ $$$${I}_{\mathrm{2}} =\underset{\mathrm{0}} {\overset{{a}} {\int}}\mathrm{1}^{\mathrm{3}} {dx}+\underset{{a}} {\overset{\mathrm{1}} {\int}}\left(\frac{{a}}{\mathrm{1}−{a}}\right)^{\mathrm{3}} {dx}=\frac{{a}\left(\mathrm{1}−\mathrm{2}{a}\right)}{\left({a}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\frac{{d}}{{da}}\left[\frac{{a}\left(\mathrm{1}−\mathrm{2}{a}\right)}{\left({a}−\mathrm{1}\right)^{\mathrm{2}} }\right]=\mathrm{0} \\ $$$$\frac{\mathrm{3}{a}−\mathrm{1}}{\left({a}−\mathrm{1}\right)^{\mathrm{3}} }=\mathrm{0}\:\Rightarrow\:{a}=\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\:{c}=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:{I}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{the}\:\mathrm{idea}\:\mathrm{is},\:\mathrm{make}\:{f}\left({x}\right)=\mathrm{1}\:\mathrm{in}\:\mathrm{the}\:\mathrm{longest} \\ $$$$\mathrm{possible}\:\mathrm{interval}\:\mathrm{and}\:\mathrm{keep}\:\mathrm{it}\:\mathrm{constant}\:\mathrm{and} \\ $$$$<\mathrm{1}\:\mathrm{within}\:\mathrm{the}\:\mathrm{rest},\:\mathrm{because}\:\mathrm{1}^{\mathrm{3}} =\mathrm{1}\:\mathrm{and}\:{c}^{\mathrm{3}} <{c} \\ $$
Commented by bhanukumarb2@gmail.com last updated on 16/May/19
$${how}\:{u}\:{get}\:{such}\:{thought}.{how}\:{i}\:{can}\:{think}\: \\ $$$${like}\:{this} \\ $$
Commented by bhanukumarb2@gmail.com last updated on 16/May/19
$${c}\:{should}\:{be}\:{like}\:{c}={a}/…{i}\:{hve}\:{seen}\:{this}\:{approch} \\ $$$${in}\:{mse}\:{to}..{same}\:{doubt}\:{that}\:{time}.. \\ $$$$ \\ $$