Question

If $f\left(x\right)=a\log |x|+b{x}^2+x$ has its extremum values at $x=-1$ and $x=2$, then

• A. $a=2,b=-1$
• B. $a=2,b=-\frac{1}{2}$
• C. $a=-2,b=\frac{1}{2}$
• D. None of the above.

Answer 1

The correct option is B $a=2,b=-\frac{1}{2}$

The given information in the question is:

$f\left(x\right)=a\log \left|x\right|+b{x}^2+x$

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.

$\Rightarrow f^{\prime }\left(x\right)=a\frac{\left|x\right|}{x}\times \frac{1}{\left|x\right|}+2bx+1$

$\Rightarrow f^{\prime }\left(x\right)=\frac{a}{x}+2bx+1$

Now equating $f^{\prime }\left(x\right)$ to zero.

$\Rightarrow f^{\prime }\left(x\right)=0$

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

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)

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

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

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

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

Subtracting (Iv) from (III) we get,

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

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

$\Rightarrow a=2$

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