A) \[\frac{x}{y}\]
B) \[\frac{y}{{{x}^{2}}}\]
C) \[\frac{y}{x}\]
D) \[0\]
Correct Answer: D
Solution :
We have\[{{\sin }^{-1}}\left( \frac{{{x}^{2}}-{{y}^{2}}}{{{x}^{2}}+{{y}^{2}}} \right)=\log a\] \[\Rightarrow \]\[\frac{{{x}^{2}}-{{y}^{2}}}{{{x}^{2}}+{{y}^{2}}}=\sin (\log a)\] \[\Rightarrow \]\[\frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta }=\sin (\log a)\] (on putting\[y=x\tan \theta \]) \[\Rightarrow \]\[\cos 2\theta =\sin (\log a)\] \[\Rightarrow \]\[2\theta ={{\cos }^{-1}}\,(\sin \log a)\] \[\Rightarrow \,\,\theta =\frac{1}{2}{{\cos }^{-1}}\,\{\sin (\operatorname{loga})\}\] \[{{\tan }^{-1}}\left( \frac{y}{x} \right)=\frac{1}{2}{{\cos }^{-1}}\{\sin \log a\}\] \[\Rightarrow \]\[\frac{1}{1+\frac{{{y}^{2}}}{{{x}^{2}}}}\cdot \frac{x\frac{dy}{dx}-y}{{{x}^{2}}}=0\] \[\Rightarrow \]\[x\frac{dy}{dx}-y=0\] \[\Rightarrow \]\[\frac{dy}{dx}=\frac{y}{x}\] ? (i) \[\Rightarrow \]\[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{-y}{{{x}^{2}}}+\frac{1}{x}\cdot \frac{dy}{dx}=\frac{-y}{{{x}^{2}}}+\frac{1}{x}\left( \frac{y}{x} \right);\] {from (i)} \[\Rightarrow \]\[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{-y}{{{x}^{2}}}+\frac{y}{{{x}^{2}}}=0\]
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