A) \[\frac{3}{2g(e)}\]
B) \[\frac{1}{2g(e)}\]
C) \[\frac{2}{3g(e)}\]
D) \[\frac{1}{3g(e)}\]
Correct Answer: A
Solution :
\[I.F.={{e}^{\int{\frac{g'(x)}{g(x)}dx}}}={{e}^{In\,g(x)}}=g(x).\] \[\Rightarrow \,y.g(x)=\frac{1}{2}\int{\frac{2g(x)g'(x)}{1+{{g}^{2}}(x)}}dx=\frac{1}{2}\ell n(1+{{g}^{2}}(x))+C\] \[C=1-\frac{\ell n\,\,2}{2}\] \[\tan ({{90}^{o}}\,-{{60}^{o}})\,=\tan {{30}^{o}}=\frac{1}{\sqrt{3}}\]You need to login to perform this action.
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