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) | ||
\[y=\sin \,\,(1/x)-cos\,\,(1/x)\] | ||
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