A) \[1\]
B) \[0\]
C) \[1/2\]
D) \[2\]
Correct Answer: A
Solution :
[a] \[y={{e}^{{{\sin }^{-1}}x}}+{{e}^{-{{\cos }^{-1}}x}}={{e}^{{{\sin }^{-1}}x}}+{{e}^{\frac{\pi }{2}+{{\sin }^{-1}}x}}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,y={{e}^{{{\sin }^{-1}}x}}(1+{{e}^{-\pi /2}})\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,y'=\frac{{{e}^{{{\sin }^{-1}}x}}}{\sqrt{1-{{x}^{2}}}}\,\,(1+{{e}^{-\pi /2}})\Rightarrow y'=\frac{y}{\sqrt{1-{{x}^{2}}}}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,(1-{{x}^{2}})\,{{(y')}^{2}}={{y}^{2}}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,2y'y''(1-{{x}^{2}})+{{(y')}^{2}}(-2x)=2yy'\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,y''(1-{{x}^{2}})+y'(-x)=y\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\left| \frac{({{x}^{2}}-1)\,y''+y}{y'x} \right|=1\]
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