A) \[\frac{2}{{{c}^{2}}}\]
B) \[\frac{-2}{{{c}^{2}}}\]
C) \[\frac{2}{c}\]
D) \[\frac{-2}{c}\]
Correct Answer: A
Solution :
\[\sqrt{x+y}+\sqrt{y-x}=c\] ?(i) \[\Rightarrow \]\[\frac{(y+x)-(y-x)}{\sqrt{x+y}-\sqrt{y-x}}=c\] \[\Rightarrow \]\[\sqrt{y+x}-\sqrt{y-x}=\frac{2x}{c}\] ?(2) By adding (1) and (2), we get \[2\sqrt{y+x}=c+\frac{2x}{c}\] \[\Rightarrow \]\[4(y+x)={{c}^{2}}+\frac{4{{x}^{2}}}{{{c}^{2}}}+4x\] \[\therefore \] \[4\frac{dy}{dx}=\frac{8x}{{{c}^{2}}}\] \[\therefore \] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{2}{{{c}^{2}}}\]
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