A) \[(-2,2)\]
B) \[(-1,1)\]
C) (0, 2)
D) None of these
Correct Answer: A
Solution :
Given, \[{{x}^{3}}=12y\] \[\Rightarrow \] \[3{{x}^{2}}\frac{dx}{dy}=12\] \[\Rightarrow \] \[\frac{dx}{dy}=\frac{4}{{{x}^{2}}}\] But it is given \[\left| \frac{dy}{dx} \right|>1\] \[\therefore \] \[\frac{4}{{{x}^{2}}}>1\] \[\Rightarrow \] \[{{x}^{2}}-4<0\] \[\Rightarrow \] \[-2<x<2\]
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