A) \[-\frac{1}{3}\]
B) \[-\frac{1}{14}\]
C) \[-4\]
D) \[-\frac{1}{2}\]
Correct Answer: B
Solution :
Given curve is \[y={{x}^{3}}+2x+6\] On differentiation w. r. t. x, we get \[\frac{dy}{dx}=3{{x}^{2}}+2\] Now, slope of normal \[=\frac{-1}{(dy/dx)}=\frac{-1}{3{{x}^{2}}+2}\] Since, the normal is parallel to the line \[x+14y+4=0\] \[\Rightarrow \] \[-\frac{1}{14}=-\frac{1}{3{{x}^{2}}+2}\,[{{m}_{1}}={{m}_{2}}]\] \[\Rightarrow \] \[3{{x}^{2}}+2=14\] \[\Rightarrow \] \[3{{x}^{2}}=12\,\,\,\Rightarrow \,\,\,\,\,{{x}^{2}}=4\] \[\therefore \] \[x=\pm 2\] \[\therefore \] Required slope of normal \[=\frac{-1}{3(4)+2}=\frac{-1}{14}\]
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