Question

Let a, b $\in R$ be such that the function f given by $f\left( x \right) = \ln \left| x \right| + b{x^2} + ax$ , $x \ne 0$ has extreme values at $x = - 1$ and $x = 2$ .
Statement $1$ : f has local maximum at $x = - 1$ and at $x = 2$
Statement $2$ : $a = \dfrac{1}{2}$ and $b = - \dfrac{1}{4}$
A.$\left( 1 \right)$ Statement $1$ is true, statement $2$ is true; statement $2$ is a correct explanation for statement $1$
B.$\left( 2 \right)$ Statement $1$ is true, statement $2$ is true; statement $2$ is not a correct explanation for statement $1$
C.$\left( 3 \right)$ Statement $1$ is true, statement $2$ is false
D.$\left( 4 \right)$ Statement $1$ is false, statement $2$ is true

Answer 1

Hint : First find the first derivative of the given function then by using the given values of x in the first derivative find the values of a and b. After that find the second derivative of the function and put these a and b values in the second derivative then check whether the value of the second derivative is negative or positive. By this you can check that f has local maxima or local minima at the given values of x.

Complete step-by-step answer :
An extreme or extreme value of a function is a point at which a maximum or minimum value of the function is obtained at some interval.
Here we will use the second derivative test according to which if the value of the second derivative is less than zero then it is a local maxima, if greater than zero then it is a local minimum and if equal to zero then the test fails.
So now let’s start the solving the question, as it is given to us that the function f is $f\left( x \right) = \ln \left| x \right| + b{x^2} + ax$ -------- (i)
On differentiating the equation (i) with respect to x we get
$f'\left( x \right) = \dfrac{1}{x} + 2bx + a$ ----------- (ii)
It is given to us that the extreme values of function are at $x = - 1$ and at $x = 2$ .Hence $f'\left( { - 1} \right) = 0 = f'\left( 2 \right)$
Therefore the value of $f'\left( x \right)$ at $x = - 1$ ,equation (ii) becomes
$\Rightarrow f'\left( { - 1} \right) = \dfrac{1}{{ - 1}} + 2b\left( { - 1} \right) + a$
As $f'\left( { - 1} \right) = 0$ so we can write it as
$\Rightarrow a - 2b = 1$ --------- (iii)
The value of $f'\left( x \right)$ at $x = 2$ ,equation (ii) becomes
$\Rightarrow f'\left( 2 \right) = \dfrac{1}{2} + 2b\left( 2 \right) + a$
As $f'\left( 2 \right) = 0$ so we can write it as
$\Rightarrow a + 4b = - \dfrac{1}{2}$ --------- (iv)
On solving equations (iii) and (iv) we get
$\Rightarrow - 6b = \dfrac{3}{2}$
On further solving we get
$\Rightarrow b = - \dfrac{1}{4}$
By putting this value of b in equation (iii) we get
$\Rightarrow a - 2\left( { - \dfrac{1}{4}} \right) = 1$
$\Rightarrow a + \dfrac{1}{2} = 1$
$\Rightarrow a = 1 - \dfrac{1}{2}$
$\Rightarrow a = \dfrac{{2 - 1}}{2}$
$\Rightarrow a = \dfrac{1}{2}$
Hence, we can say that statement $2$ is true.
Now we will find the second derivative of the function. So for this differentiate the equation (ii) with respect to x. On differentiating we get
Here the constant a is absent but b is there in the above expression. So, by putting the value of b here we get
On further solving we get
By taking L.C.M. we get
$\Rightarrow f''\left( x \right) = - \left( {\dfrac{{2 + {x^2}}}{{2{x^2}}}} \right)$
As the value of the second derivative is negative that is less than zero. So, we can say that f has local maxima at both $x = - 1$ and $x = 2$ .That statement $1$ is correct.
Hence, the correct option is $\left( 1 \right)$ Statement $1$ is true, statement $2$ is true; statement $2$ is a correct explanation for statement $1$
So, the correct answer is “Option A”.

Note : Remember that the second derivative test used to determine local extremum of a function under certain conditions. Keep in mind that if a function has a critical point for which f’(x) is zero and if the second derivative is negative then f has local maxima here and if it is positive then f has local minima here.