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AIEEE 2012 Solved Paper

Section: Mathematics

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Question : 71 of 89

Marks: +1, -0

Let

a, b \in R

be such that the function

f

f(x)=\ln|x|+b x^2+a x, x \neq 0

has extreme values at

x=-1

x=2

f

has local maximum at

x=-1

x=2

a=\frac{1}{2}

b=\frac{-1}{4}

Statement $-1$ is false, Statement $-2$ is true.
Statement $-1$ is true, Statement $-2$ is true; Statement - 2 is a correct explanation for Statement - 1.
Statement - 1 is true, Statement - 2 is true; Statement - 2 is not a correct explanation for Statement $-1$ .
Statement $-1$ is true, Statement $-2$ is false.

Solution:

Given,

f(x)=\ln |x|+b x^2+a x

\therefore f'(x)=\frac{1}{x}+2b x+a

At

x=-1, f'(x)=-1-2b+a=0

\Rightarrow a-2b=1

At

x=2, f'(x)=\frac{1}{2}+4b+a=0

\Rightarrow a+4b=-\frac{1}{2}

On solving

(i)

and

(ii)

we get

a=\frac{1}{2}, b=-\frac{1}{4}

Thus,

f'(x)=\frac{1}{x}-\frac{x}{2}+\frac{1}{2}=\frac{2-x^2+x}{2x}

=\frac{-x^2+x+2}{2x}=\frac{- (x^2-x-2)}{2x}=\frac{-(x+1)(x-2)}{2x}

So maximum at

x=-1,2

Hence both the statements are true and statement 2 is a correct explanation for 1.

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