Menu Close

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-

Tinku Tara 0 Comments
Pin




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

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *