A) \[-6\]
B) \[-3\]
C) \[3\]
D) \[6\]
Correct Answer: A
Solution :
[a] \[f(x)={{x}^{4}}+b{{x}^{2}}+8x+1\] \[\Rightarrow \,\,\,\,\,f'(x)=4{{x}^{3}}+2bx+8\] \[\Rightarrow \,\,\,\,\,f'(x)=12{{x}^{2}}+2b\] For point of inflection, \[f''(x)=0\] \[\Rightarrow \,\,\,12{{x}^{2}}+2b=0\Rightarrow 6{{x}^{2}}+b=0\] \[\Rightarrow \,\,\,\,-6{{x}^{2}}=b\] ?.(1) For horizontal tangent, \[f'(x)=0\] \[\Rightarrow \,\,\,4{{x}^{3}}+2bx+8=0\] \[\Rightarrow \,\,\,4{{x}^{3}}+(-12{{x}^{2}})x+8=0\] \[[\therefore \,\,\,\,b=-6{{x}^{2}}]\] \[\Rightarrow \,\,\,\,\,{{x}^{3}}=1\] \[\Rightarrow \,\,\,\,\,x=1\] \[\therefore \,\,\,b=-6{{x}^{2}}=-6(1)=-6\]You need to login to perform this action.
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