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Question

If $$f(x)=a\log |x|+bx^2+x$$ has its extremum values 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 the above.
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Solution

The correct option is B $$a=2,b=-\dfrac 12$$
The given information in the question is:

$$f(x) = a\log \left | x \right | + bx^2 + x$$

An extremum is calculated from the derivative of the function about a point where the derivative is equal to 0.

So we calculate the derivative and equate it to 0.

$$\Rightarrow f'(x) = a \dfrac{\left | x \right |}{x}\times \dfrac {1}{\left | x \right |} +2bx + 1$$

$$\Rightarrow f'(x) =\dfrac{a}{x} +2bx +1$$

Now equating $$f'(x)$$ to zero.

$$ \Rightarrow f'(x) = 0$$

$$\Rightarrow \dfrac{a}{x} +2bx +1 = 0$$

Now it is given that the extremum is at $$x = -1$$ and$$x = 2$$

So substituting the value of extremum in the derivative equation which is its solution, we get

$$\Rightarrow -a -2b +1 = 0$$     .....(I)

$$\Rightarrow \dfrac{a}{2} +4b +1 = 0$$     .....(II)

Solving equations (I) and (II) we get,

$$\Rightarrow a + 2b =1$$     .....(III)

$$\Rightarrow a + 8b = -2$$     .....(IV)

Subtracting (Iv) from (III) we get,

$$\Rightarrow b = \dfrac{-1}{2}$$

Substituting the value of b In equation (III) we get,

$$\Rightarrow a = 2$$

For the function $$f(x)$$ to have an the extremum at the required point the value of $$a=2$$. and $$b = \dfrac{-1}{2}$$.     .....Answer[Option(B)]

Mathematics

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