A) 360
B) -360
C) \[\frac{1}{360}\]
D) \[-\frac{1}{360}\]
Correct Answer: D
Solution :
\[\frac{dy}{dx}=f'(x)=12(4{{x}^{3}}-x)\] Slope of normal \[=-\frac{1}{12x(4{{x}^{2}}-1)}\] At point (2, 3), slope \[=-\frac{1}{12\times 2(16-1)}=-\frac{1}{12\times 15}=-\frac{1}{360}\]
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