A) \[\sqrt{3}{{\log }_{e}}\sqrt{3}\]
B) \[-\sqrt{3}{{\log }_{e}}\sqrt{3}\]
C) \[-\sqrt{3}{{\log }_{e}}3\]
D) \[-\sqrt{3}{{\log }_{e}}3\]
Correct Answer: A
Solution :
Given \[f(x)=\sin \left( \frac{2\times {{3}^{x}}}{1+{{9}^{x}}} \right)\] Let \[{{3}^{x}}=\tan (t)\] \[\Rightarrow f(x)=\arcsin \left( \frac{2\tan (t)}{1+{{\tan }^{2}}(t)} \right)\] As \[\sin (2t)=\frac{2\tan (t)}{1+{{\tan }^{2}}(t)}\] \[\Rightarrow f(x)=\arcsin (\sin (2t))\] \[\therefore f(x)=2t=2\arctan ({{3}^{x}})\] \[\to \frac{df}{dx}=\frac{2}{1+{{({{3}^{x}})}^{2}}}\times {{3}^{x}}.{{\log }_{e}}3\] \[\text{at x=}\frac{1}{2}\to \frac{df}{dx}=\frac{2}{1+{{\left( {{3}^{\frac{1}{2}}} \right)}^{2}}}\times {{3}^{\frac{1}{2}}}.{{\log }_{e}}3\] \[=\frac{1}{2}\times \sqrt{3}\times {{\log }_{e}}3\] \[=\sqrt{3}\times {{\log }_{e}}\sqrt{3}\] Hence correct option is \['\overset{.\,\,.}{\mathop{A}}\,'\].
You need to login to perform this action.
You will be redirected in
3 sec
