A) \[\frac{x}{\sqrt{1+3{{x}^{2}}}}\]
B) \[\frac{x}{\sqrt{1+{{x}^{2}}}}\]
C) \[\frac{x}{\sqrt{1+2{{x}^{2}}}}\]
D) None of these
Correct Answer: A
Solution :
\[f(x)=\frac{x}{\sqrt{1+{{x}^{2}}}}\] \[\therefore \] \[(fof)x=f(f(x))=f\left( \frac{x}{\sqrt{1+{{x}^{2}}}} \right)\] \[=\frac{\frac{x}{\sqrt{1+{{x}^{2}}}}}{\sqrt{1+{{\left( \frac{x}{\sqrt{1+{{x}^{2}}}} \right)}^{2}}}}\] \[=\frac{x}{\sqrt{1+{{x}^{2}}}\sqrt{1+\frac{{{x}^{2}}}{1+{{x}^{2}}}}}\] \[=\frac{x\sqrt{1+{{x}^{2}}}}{\sqrt{1+{{x}^{2}}}\sqrt{1+{{x}^{2}}+{{x}^{2}}}}\] \[=\frac{x}{\sqrt{1+2{{x}^{2}}}}\] \[(fofof)x=f(fof)x\] \[=f\left( \frac{x}{\sqrt{1+2{{x}^{2}}}} \right)\] \[=\frac{\frac{x}{\sqrt{1+2{{x}^{2}}}}}{\sqrt{1+{{\left( \frac{x}{\sqrt{1+2{{x}^{2}}}} \right)}^{2}}}}\] \[=\frac{x\sqrt{1+2{{x}^{2}}}}{\sqrt{1+2{{x}^{2}}}\sqrt{1+2{{x}^{2}}+{{x}^{2}}}}\] \[=\frac{x}{\sqrt{1+3{{x}^{2}}}}\]
You need to login to perform this action.
You will be redirected in
3 sec
