Question

# If $$f(x)=a\log { \left| x \right| } +b{ x }^{ 2 }+x$$ has is extremum value at $$x=-1$$ and $$x=2$$, then

A
a=2,b=1
B
a=2,b=12
C
a=2,b=12
D
None of these

Solution

## The correct option is B $$a=2,b=-\cfrac { 1 }{ 2 }$$We have $$f(x)=a\log { \left| x \right| } +b{ x }^{ 2 }+x$$On differentiating w.r.t $$x$$ we get$$f'(x)=\cfrac { a }{ x } +2bx+1\quad$$Since, $$f(x)$$ attains its extremum value at $$x=-1,2$$$$\therefore f'(-1)=-\cfrac { a }{ 1 } -2b+1$$$$\Rightarrow -a-2b+1=0...(i)\quad$$and $$\cfrac { a }{ 2 } +4b+1=0...(ii)\quad$$From Eqs(i) and (ii)$$\Rightarrow a=2,b=-\cfrac { 1 }{ 2 }$$Mathematics

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