Question

Let $y=f\left(x\right)$ be a non-negative function defined on $[0,2]$ satisfying the differential equation $y^3{y}^{\prime \prime }+1=0$. If $f^{\prime }\left(1\right)=0$ and $f\left(1\right)=1$, then

• A. the maximum value of $f\left(x\right)$ is $1.$
• B. the minimum value of $f\left(x\right)$ is $0.$
• C. the solution $y=f\left(x\right)$ of the given differential equation represents a semi-circle with centre $(1,0).$
• D. the area bounded by the curve $y=f\left(x\right)$ and $x-\text{axis}$ is $\frac{\pi }{4}\text{sq. units}.$

Answer 1

Answer: (A), (B), (C)

the solution $y=f\left(x\right)$ of the given differential equation represents a semi-circle with centre $(1,0).$

$y^3{y}^{\prime \prime }=-1$
$\Rightarrow y^{\prime }{y}^{\prime \prime }=-\frac{y^{\prime }}{y^3}$
Integrating both the sides,
$\frac{(y^{\prime }{)}^2}{2}=\frac{1}{2y^2}+C$
$f^{\prime }\left(1\right)=0,f\left(1\right)=1\Rightarrow C=-\frac{1}{2}$
$\Rightarrow \frac{(y^{\prime }{)}^2}{2}=\frac{1}{2y^2}-\frac{1}{2}$
$\Rightarrow y^{\prime }=\pm \frac{\sqrt{1-y^2}}{y}$
$\Rightarrow \frac{y}{\sqrt{1-y^2}}{y}^{\prime }=\pm 1$
Integrating both the sides,
$\int \frac{y}{\sqrt{1-y^2}}{y}^{\prime }dx=\pm \int dx$

Put $1-y^2=z^2\Rightarrow -2y{y}^{\prime }dx=2zdz$
$\Rightarrow -\int \frac{zdz}{z}=\pm x+c$
$\Rightarrow -z=\pm x+c$
$\Rightarrow -\sqrt{1-y^2}=\pm x+c$

$y\left(1\right)=1\Rightarrow c=\mp 1$
$\Rightarrow -\sqrt{1-y^2}=\pm x\mp 1$
$\Rightarrow -\sqrt{1-y^2}=x-1\text{or}-x+1$
Squaring both the sides, we get
$1-y^2=x^2-2x+1$
$\Rightarrow y^2=2x-x^2...\left(1\right)$
$\Rightarrow y=\sqrt{2x-x^2}=f\left(x\right)$

From eqn $\left(1\right)$,
$x^2+y^2-2x=0$
$\Rightarrow (x-1{)}^2+y^2=1,x\in [0,2]$


Clearly, $x=1$ is the point of maxima.
Maximum value of $f\left(x\right)=\sqrt{2-1}=1$

Minimum value of $f\left(x\right)=0$ at $x=0,2$

Area bounded by the curve $y=f\left(x\right)$ and $x-\text{axis}$ is :
$\frac{\pi \times 1^2}{2}=\frac{\pi }{2}\text{sq. units}.$