A) \[y=\sin \frac{1}{x}+\cos \frac{1}{x}\]
B) \[y=\frac{x+1}{x\sin \,(1/x)}\]
C) \[y=\sin \frac{1}{x}-\cos \frac{1}{x}\]
D) \[y=\frac{x+1}{x\cos \,(1/x)}\]
Correct Answer: C
Solution :
\[\because \] \[\frac{dy}{dx}-\frac{\tan \,(1/x)}{{{x}^{2}}}\] \[y=-\frac{\sec \,(1/x)}{{{x}^{2}}}\] ...(1) \[\because \] \[I.F.={{e}^{-\int{\frac{\,\,\tan \,\,(1/x)}{{{x}^{2}}}\,\,dx}}}={{e}^{\ln \,\,\sec \,(1/x)}}=\sec \,(1/x)\] \[\therefore \] \[y\,\sec \,(1/x)=-\int{\frac{{{\sec }^{2}}\,(1/x)}{{{x}^{2}}}dx}\] \[\Rightarrow \] \[y\sec \,(1/x)=\tan \,(1/x)+c\] ?(2) If \[x\to \infty \,\,;\,\,y\to -1\] \[\Rightarrow \] \[(-1)\,\,(1)=0+c\]\[\Rightarrow \]\[c=-1\]put in (2) \[y=\sin \,(1/x)-cos\,(1/x)\]
You need to login to perform this action.
You will be redirected in
3 sec
