A) \[a=2,b=-1\]
B) \[a=2,b=\frac{-1}{2}\]
C) \[a=-2,b=\frac{1}{2}\]
D) None of these
Correct Answer: B
Solution :
We have, \[f(x)=a\log |x|+b{{x}^{2}}+x\] \[\therefore \] \[f(x)=\frac{a}{x}+2bx+1\] Since,\[f(x)\]attains its extremum value at \[x=-1,2\]. \[\therefore \] \[f(-1)\frac{-a}{1}-2b+1=0\] \[\Rightarrow \] \[f(-1)=-a-2b+1=0\] and \[f(2)=\frac{a}{2}+4b+1=0\] \[\Rightarrow \] \[a=2,b=-\frac{1}{2}\]You need to login to perform this action.
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