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Question Number 161461 by ZiYangLee last updated on 18/Dec/21

Let the region bounded by the curve   y=x(1−x) and the x-axis be R.  The line y=mx divides R into two parts,  find the value of (1−m)^3 .

$$\mathrm{Let}\:\mathrm{the}\:\mathrm{region}\:\mathrm{bounded}\:\mathrm{by}\:\mathrm{the}\:\mathrm{curve}\: \\ $$$${y}={x}\left(\mathrm{1}−{x}\right)\:\mathrm{and}\:\mathrm{the}\:{x}-\mathrm{axis}\:\mathrm{be}\:\mathrm{R}. \\ $$$$\mathrm{The}\:\mathrm{line}\:{y}={mx}\:\mathrm{divides}\:\mathrm{R}\:\mathrm{into}\:\mathrm{two}\:\mathrm{parts}, \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\left(\mathrm{1}−{m}\right)^{\mathrm{3}} . \\ $$

Commented by cortano last updated on 18/Dec/21

Total area = (2/3)×1(1/4)= (1/6)  Area I=((Δ(√Δ))/(6a^2 ))   where x^2 +(m−1)x=0   Δ = (m−1)^2 ; (√Δ) =∣(m−1)∣   since m<1    ⇒area I = (1/2) total area  ⇒(((1−m)^3 )/6) = (1/(12))  ⇒(1−m)^3  =(1/2)

$${Total}\:{area}\:=\:\frac{\mathrm{2}}{\mathrm{3}}×\mathrm{1}\frac{\mathrm{1}}{\mathrm{4}}=\:\frac{\mathrm{1}}{\mathrm{6}} \\ $$$${Area}\:{I}=\frac{\Delta\sqrt{\Delta}}{\mathrm{6}{a}^{\mathrm{2}} } \\ $$$$\:{where}\:{x}^{\mathrm{2}} +\left({m}−\mathrm{1}\right){x}=\mathrm{0} \\ $$$$\:\Delta\:=\:\left({m}−\mathrm{1}\right)^{\mathrm{2}} ;\:\sqrt{\Delta}\:=\mid\left({m}−\mathrm{1}\right)\mid \\ $$$$\:{since}\:{m}<\mathrm{1} \\ $$$$\:\:\Rightarrow{area}\:{I}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:{total}\:{area} \\ $$$$\Rightarrow\frac{\left(\mathrm{1}−{m}\right)^{\mathrm{3}} }{\mathrm{6}}\:=\:\frac{\mathrm{1}}{\mathrm{12}} \\ $$$$\Rightarrow\left(\mathrm{1}−{m}\right)^{\mathrm{3}} \:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented by som(math1967) last updated on 18/Dec/21

Is the line y=mx  divides R into  two equal parts  sir ?

$${Is}\:{the}\:{line}\:{y}={mx}\:\:{divides}\:{R}\:{into}\:\:{two}\:{equal}\:{parts} \\ $$$${sir}\:? \\ $$

Commented by som(math1967) last updated on 18/Dec/21

x−x^2 =0  x(1−x)=0 ∴x=0,1  area of R     ∫_0 ^1 (x−x^2 )dx=(1/2)−(1/3)=(1/6) sq unit  again mx=x−x^2   x=0,(1−m)  ∴ ∫_0 ^(1−m) (x−x^2 )dx −∫_0 ^(1−m) mxdx=(1/(12))    (((1−m)^2 )/2) −(((1−m)^3 )/3) −((m(1−m)^2 )/2)=(1/(12))    (1−m)^2 ((1/2) −((1−m)/3) −(m/2))=(1/(12))  (1−m)^2 (((3−2+2m−3m)/6))=(1/(12))  (((1−m)^3 )/6)=(1/(12))  ∴(1−m)^3 =(1/2)

$${x}−{x}^{\mathrm{2}} =\mathrm{0} \\ $$$${x}\left(\mathrm{1}−{x}\right)=\mathrm{0}\:\therefore{x}=\mathrm{0},\mathrm{1} \\ $$$${area}\:{of}\:{R}\: \\ $$$$\:\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\left({x}−{x}^{\mathrm{2}} \right){dx}=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{6}}\:{sq}\:{unit} \\ $$$${again}\:{mx}={x}−{x}^{\mathrm{2}} \\ $$$${x}=\mathrm{0},\left(\mathrm{1}−{m}\right) \\ $$$$\therefore\:\underset{\mathrm{0}} {\overset{\mathrm{1}−{m}} {\int}}\left({x}−{x}^{\mathrm{2}} \right){dx}\:−\underset{\mathrm{0}} {\overset{\mathrm{1}−{m}} {\int}}{mxdx}=\frac{\mathrm{1}}{\mathrm{12}} \\ $$$$\:\:\frac{\left(\mathrm{1}−{m}\right)^{\mathrm{2}} }{\mathrm{2}}\:−\frac{\left(\mathrm{1}−{m}\right)^{\mathrm{3}} }{\mathrm{3}}\:−\frac{{m}\left(\mathrm{1}−{m}\right)^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{12}} \\ $$$$\:\:\left(\mathrm{1}−{m}\right)^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\:−\frac{\mathrm{1}−{m}}{\mathrm{3}}\:−\frac{{m}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{12}} \\ $$$$\left(\mathrm{1}−{m}\right)^{\mathrm{2}} \left(\frac{\mathrm{3}−\mathrm{2}+\mathrm{2}{m}−\mathrm{3}{m}}{\mathrm{6}}\right)=\frac{\mathrm{1}}{\mathrm{12}} \\ $$$$\frac{\left(\mathrm{1}−{m}\right)^{\mathrm{3}} }{\mathrm{6}}=\frac{\mathrm{1}}{\mathrm{12}} \\ $$$$\therefore\left(\mathrm{1}−{m}\right)^{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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