Question

Let $y=f(x)$ satisfies the differential equation $\left(1+x^2\right)\frac{dy}{dx}+2xy=2x$ and $f(0)=2$. Then the number of integers in the range of $f(x)$ is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (A)

Given, $\frac{dy}{dx}+\left(\frac{2x}{1+x^2}\right)y=\left(\frac{2x}{1+x^2}\right)$ (Linear differential equation)
$\therefore$ I.F. $=e^{\ln \left(1+x^2\right)}=1+x^2.$
So, general solution is

$y\cdot \left(1+x^2\right)=\int \left(\frac{2x}{1+x^2}\right)\cdot \left(1+x^2\right)dx+C$
$\Rightarrow y\left(1+x^2\right)=x^2+C$
$\text{ As }y(0)=2\Rightarrow 2=0+c$
$\therefore y=f(x)=\left(\frac{x^2+2}{x^2+1}\right)=\left(1+\frac{1}{x^2+1}\right)$
Range of $f(x)=(1,2]$.