Question

Let $y=f\left(x\right)$ be a real-valued differentiable function on the set of all real numbers $R$ such that $f\left(1\right)=1$. If $f\left(x\right)$ satisfies $x{f}^{\prime }\left(x\right)=x^2+f\left(x\right)-2$, then

• A. $f\left(x\right)$ is even function
• B. $f\left(x\right)$ is odd function
• C. minimum value of $f\left(x\right)$ is $0$
• D. $y=f\left(x\right)$ represent a parabola with focus $(1,\frac{5}{4})$

Answer 1

The correct option is D $y=f\left(x\right)$ represent a parabola with focus $(1,\frac{5}{4})$

$x{f}^{\prime }\left(x\right)=x^2+f\left(x\right)-2$
Let $y=f\left(x\right)$
$\Rightarrow \frac{dy}{dx}=f^{\prime }\left(x\right)$
Now, $\frac{dy}{dx}+\left(\frac{-1}{x}\right)y=x-\frac{2}{x}$
which is a linear differential equation.
$I.F.=\exp \left(\int \frac{-1}{x}dx\right)=e^{-\ln x}=\frac{1}{x}$

The general solution is
$y\left(\frac{1}{x}\right)=\int (x-\frac{2}{x})\frac{1}{x}dx$
$\therefore \frac{y}{x}=x+\frac{2}{x}+C$
As $y\left(1\right)=1\Rightarrow C=-2$
$\therefore \frac{y}{x}=x+\frac{2}{x}-2$
$\Rightarrow f\left(x\right)=x^2-2x+2=(x-1{)}^2+1$
Clearly, $f\left(x\right)$ is neither even nor odd.
Minimum value of $f\left(x\right)$ is $1.$

$y=(x-1{)}^2+1$
$\Rightarrow (x-1{)}^2=4\cdot \frac{1}{4}(y-1)$
$(a=\frac{1}{4},h=1,k=1)$
$\therefore y=f\left(x\right)$ represent a parabola with focus $(1,\frac{5}{4}).$