question_answer

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

A)  1                                            

B)  2  

C)  3                                            

D)  4

Correct Answer: A

Solution :

Graph of \[f(x)\,=\frac{{{x}^{2}}+2}{{{x}^{2}}+1}\]             Given, \[\frac{dy}{dx}\,+\left( \frac{2x}{1+{{x}^{2}}} \right)\,y=\left( \frac{2x}{1+{{x}^{2}}} \right)\]                         (Linear differential equation)             \[\therefore \,\,\,I.F.\,\,={{e}^{\ell n\,\,(1+{{x}^{2}})}}\,=1+{{x}^{2}}\]             So, general solution is             \[y.(1+{{x}^{2}})=\int_{{}}^{{}}{\left( \frac{2x}{1+{{x}^{2}}} \right)\,\left( 1+{{x}^{2}} \right)dx+C}\]             \[\Rightarrow \,\,y(1+{{x}^{2}})\,={{x}^{2}}+C\]             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 \[f(x)\,=(1,\,\,2]\].