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 \[f(x)\,=(1,\,\,2]\].You need to login to perform this action.
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