Question

Let $a,b\in R$ be such that the function $f$ is 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 $2:a=\frac{1}{2}$ and $b=\frac{-1}{4}.$

• 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.

Answer 1

The correct option is B Statement $1$ is true, statement $2$ is true; statement $2$ is a correct explanation for statement $1.$

We have,
$f\left(x\right)=\ln \left|x\right|+b{x}^2+ax$
$\therefore f^{\prime }\left(x\right)=\frac{1}{x}+2bx+a$

Now at $x=-1,f^{\prime }\left(x\right)=0$
$\Rightarrow -1-2b+a=0$
$\Rightarrow a-2b=1\cdots \left(1\right)$

Also at $x=2,f^{\prime }\left(x\right)=0$
$\Rightarrow \frac{1}{2}+4b+a=0$
$\Rightarrow a+4b=\frac{-1}{2}\cdots \left(2\right)$

Solving $\left(1\right)$ and $\left(2\right)$ we get,
$b=\frac{-1}{4}$ and $a=\frac{1}{2}$

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

Now, $f^{\prime \prime }(-1)=-\frac{3}{2}<0$
and $f^{\prime \prime }\left(2\right)=-\frac{3}{4}<0$

Hence, $f$ has local maximum at $x=-1$ and at $x=2.$

$\Rightarrow$ Statement $1$ is true, statement $2$ is true; statement $2$ is a correct explanation for statement $1.$