A) \[x=-1\]and \[x=2,\]
B) \[a=2,\,b=-\frac{1}{2}\]
C) \[a=2,=-\frac{1}{2}\]
D) none of the above
Correct Answer: B
Solution :
\[y=a\log x+b{{x}^{2}}+x\]has extremum at \[x=-1\]and\[x=2\] \[\therefore \]\[\frac{dy}{dx}=0\]at \[x=-1\]and \[x=2\] \[\Rightarrow \]\[\frac{a}{x}+2bx+1=0,\]at \[x=-1\]and \[x=2\] \[\therefore \] \[-a-2b+1=0\]and \[\frac{a}{2}+4b+1=0\] \[\Rightarrow \]\[a=2\]and \[b=-\frac{1}{2}\]
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