A) \[\frac{3}{1+9{{x}^{3}}}\]
B) \[\frac{9}{1+9{{x}^{3}}}\]
C) \[\frac{3x\sqrt{x}}{1-9{{x}^{3}}}\]
D) \[\frac{3x}{1-9{{x}^{3}}}\]
Correct Answer: B
Solution :
Let \[y={{\tan }^{-1}}\left( \frac{6x\sqrt{x}}{1-9{{x}^{3}}} \right)\]where \[x\in \left( 0,\frac{1}{4} \right)\] \[={{\tan }^{-1}}\left( \frac{2.(3{{x}^{3/2}})}{1-(3{{x}^{3/2}})} \right)=2{{\tan }^{-1}}(3{{x}^{3/2}})\] As \[3{{x}^{3/2}}\in \left( 0,\frac{3}{8} \right)\] \[\therefore \] \[g(x)=\frac{9}{1+9{{x}^{3}}}\]
You need to login to perform this action.
You will be redirected in
3 sec
