Question

Let the curve $y=f\left(x\right)$ passing through $(4,-2)$ satisfies the differential equation $y(x+y^3)dx=x(y^3-x)dy$ and let $y=g\left(x\right)=\underset{1/8}{\overset{{\sin }^{2}x}{\int }}{\sin }^{-1}\left(\sqrt{t}\right)dt+\underset{1/8}{\overset{{\cos }^{2}x}{\int }}{\cos }^{-1}\left(\sqrt{t}\right)dt,0\le x\le \frac{\pi }{2}$ be the second curve, then

• A. If the equation of curve $y=f\left(x\right)$ satisfies $a{y}^3+bx=0,$ then $a+b=3;\text{where}a\text{and}b$ are positive integers and co-prime.
• B. If the equation of curve $y=f\left(x\right)$ satisfies $a{y}^3+bx=0,$ then $a=b;\text{where}a\text{and}b$ are positive integers.
• C. Area of the region bounded by $y=f\left(x\right),y=g\left(x\right)$ and $x=0$ is $\frac{1}{8}{\left(\frac{3\pi }{8}\right)}^4$ sq. units.
• D. Area of the region bounded by $y=f\left(x\right),y=g\left(x\right)$ and $x=0$ is $\frac{1}{8}{\left(\frac{3\pi }{16}\right)}^4$ sq. units.

Answer 1

Answer: (A), (D)

The correct options are
A If the equation of curve $y=f\left(x\right)$ satisfies $a{y}^3+bx=0,$ then $a+b=3;\text{where}a\text{and}b$ are positive integers and co-prime.
D Area of the region bounded by $y=f\left(x\right),y=g\left(x\right)$ and $x=0$ is $\frac{1}{8}{\left(\frac{3\pi }{16}\right)}^4$ sq. units.

$y(x+y^3)dx=x(y^3-x)dy$
$\Rightarrow (y^{4}dx-x{y}^{3}dy)+(xydx+x^{2}dy)=0$
$\Rightarrow x^2{y}^3\left(\frac{ydx-xdy}{x^2}\right)+x(xdy+ydx)=0$
$\Rightarrow \frac{y}{x}.d\left(\frac{y}{x}\right)-\frac{d\left(xy\right)}{x^2{y}^2}=0$
$\Rightarrow \frac{1}{2}{\left(\frac{y}{x}\right)}^2+\frac{1}{xy}=C$

$y=f\left(x\right)$ passes through $(4,-2).$
$\Rightarrow C=0$
$\text{So,}{y}^3=-2x$
$\Rightarrow y=f\left(x\right)=-\sqrt[3]{32x}$

$g\left(x\right)=\underset{1/8}{\overset{{\sin }^{2}x}{\int }}{\sin }^{-1}\left(\sqrt{t}\right)dt+\underset{1/8}{\overset{{\cos }^{2}x}{\int }}{\cos }^{-1}\left(\sqrt{t}\right)dt$
$g^{\prime }\left(x\right)={\sin }^{-1}\left(\sin x\right)\times \sin 2x+{\cos }^{-1}\left(\cos x\right)\times (-\sin 2x)=0$
$\Rightarrow g^{\prime }\left(x\right)=0$
$\Rightarrow g\left(x\right)=C$
$\text{At}x=\frac{\pi }{4},g\left(x\right)=\frac{3\pi }{16}$
$\text{So,}g\left(x\right)=\frac{3\pi }{16}$


Required shaded area $=\underset{0}{\overset{3\pi /16}{\int }}\frac{y^3}{2}dy=\frac{1}{8}{\left(\frac{3\pi }{16}\right)}^4$