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