QJEEM12Macfdf: JEE Main 2012 Mathematics - Class 12 JEEM12Macfdf Solution: Let be such that the function given by has extreme values at…

Question

Verified Solution • 2025–26 Session moderate

Let

a,b \in R

be such that the function

f

given by

f\left( x \right) = In\left| x \right| + b{x^2} + ax,\,x \ne 0

has extreme values at

x=-1

and

x=2

Statement-1 : $f$ has local maximum at $x=-1$ and at $x=2$ .

Statement-2 : $a = {1 \over 2}$ and $b = {-1 \over 4}$

A Statement - 1 is false, Statement - 2 is true.
B Statement - 1 is true , Statement - 2 is true; Statement - 2 is a correct explanation for Statement - 1.
C Statement - 1 is true, Statement - 2 is true; Statement - 2 is not a correct explanation for Statement - 1.
D Statement - 1 is true, Statement - 2 is false.

Accepted Answer

Given,

f\left( x \right) = \ln \left| x \right| + b{x^2} + ax

\therefore

f'\left( x \right) = {1 \over x} + 2bx + a

At

x=-1,

f'\left( x \right) = - 1 - 2b + a = 0

\Rightarrow a - 2b = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)

At

x=2,

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

\Rightarrow a + 4b = - {1 \over 2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)

On solving

(i)

and

(ii)

we get

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

Thus,

f'\left( x \right) = {1 \over x} - {x \over 2} + {1 \over 2} = {{2 - {x^2} + x} \over {2x}}

= {{ - {x^2} + x + 2} \over {2x}} = {{ - \left( {{x^2} - x - 2} \right)} \over {2x}} = {{ - \left( {x + 1} \right)\left( {x - 2} \right)} \over {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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